Perpetuum Mobile - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:07:12Z http://mathoverflow.net/feeds/question/10876 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10876/perpetuum-mobile Perpetuum Mobile Anton Petrunin 2010-01-06T03:41:55Z 2011-11-13T16:21:22Z <p>In 2 hours after posting this, I realized that preserving Liouville measure solves the problem completely. Sorry for disturbing...</p> <p><hr /></p> <p><strong>Construction of perpetuum mobile:</strong> Consider room with mirror walls formed by two arcs of two ellipses with common foci and two segments of on bisecting perpendicular for the focuses as on the picture:</p> <p><img src="http://lh6.ggpht.com/%5Fugz98iOeD5o/S0QBmM5dRYI/AAAAAAAABGQ/iKo8SKli8nU/s144/Perpetuum.Mobile.png" alt="Perpetuum Mobile" /></p> <p>Place one-point-bodies with the same temperature in each focus --- they radiate and</p> <ul> <li><p>All rays from blue focus come to the red one.</p></li> <li><p>Big portion of rays from red focus goes back to it-self (that are all rays which reflect in bisector) while the rest goes to the blue focus. </p></li> </ul> <p>Thus red focus getting hotter than blue one; i.e., we have a <a href="http://en.wikipedia.org/wiki/Perpetual%5Fmotion" rel="nofollow">perpetual motion machine of the second kind</a>...</p> <p><strong>Questions:</strong> </p> <ul> <li><p>Why exactly it does not work? My guess is: if instead of one-point-bodies we have bodies with real size (no matter how small) it will no longer work, but I'm too lazy to do calculations, and I also it should be a good explanation (with no calculations).</p></li> <li><p>For those of you who think it is not math, here is math formulation: Assume instead of one-point-bodies we have very small bodies of arbitrary shape. Then physics tells us it should not longer work. BUT I do not see mathematical proof of it...</p></li> </ul> <p><strong>Comment:</strong></p> <ul> <li>I know this construction from Vladimir Troitsky. A similar (but not as elegant) construction appears in comment on the Brain Teaser in the September 1972 issue of Physics Education (page 414). (maybe earlier?) --- thanks to Scott Carnahan for the ref. On page 446, there is a "solution", it only says that <em>"it would not work because of the finite sizes of [the bodies]"</em>. Next year (June 1973, p.292) a letter with a better explanation was pubilshed in the same journal, this "better explanation" roughly says that it does not work by a "well established law".</li> </ul> http://mathoverflow.net/questions/10876/perpetuum-mobile/10882#10882 Answer by S. Carnahan for Perpetuum Mobile S. Carnahan 2010-01-06T04:36:53Z 2010-01-06T08:04:35Z <p>To answer your first two questions, <a href="http://scholar.google.com/scholar?q=%22perpetuum%20mobile%22%20%22ellipsoidal%20mirror%22" rel="nofollow">yes, it is a known construction</a> (physics education journal, 1973).</p> <p>If the mirrors are made of ordinary matter and the photons have any sort of wavelike properties, then the whole setup (including the mirrors) will approach thermal equilibrium. If you're asking about comparing an output integral to an input integral without such physical considerations, then you get an imbalance. I'm a little rusty on this, but I've been told that thermodynamics as a theory cannot be derived from physical principles without some quantum mechanical input. In particular, conventional notions of entropy seem to require some kind of finiteness for microstates.</p> <p><b>Edit:</b> <a href="http://answers.yahoo.com/question/index?qid=20081016084520AAe3WEC" rel="nofollow">This page</a> has a discussion of a very similar problem (but with spherical instead of flat pieces). <strike>The answer claims that if your body has small nonzero size, but idealized non-physical mirrors and photons, you still get a violation of the second law (contra your suggestion).</strike> <b>More Edit:</b> The previous sentence arose from a misreading - as Yuri Bakhtin's comments explain, nonzero size yields a homogeneous photon gas. I imagine the homogeneity is what you mean with the reference to Liouville measure.</p> http://mathoverflow.net/questions/10876/perpetuum-mobile/10883#10883 Answer by Yuri Bakhtin for Perpetuum Mobile Yuri Bakhtin 2010-01-06T04:42:01Z 2010-01-06T04:42:01Z <p>I guess, in the equilibrium some portion of the energy will be scattered along the rays.</p> <p>UPD. As I see, this agrees with the answer by Scott.</p> http://mathoverflow.net/questions/10876/perpetuum-mobile/10893#10893 Answer by JosÃ© Figueroa-O'Farrill for Perpetuum Mobile JosÃ© Figueroa-O'Farrill 2010-01-06T06:36:38Z 2010-01-06T06:36:38Z <p>This is not answer, but a comment. It just didn't fit.</p> <p>Am I the only one to whom the classification of perpetuum mobili into three kinds reminds them of this passage in Stanislaw Lem's Cyberiad?</p> <blockquote> <p>Everyone knows that dragons don't exist. But while this simplistic formulation may satisfy the layman, it does not suffice for the scientific mind. The School of Higher Neantical Nillity is in fact wholly unconcerned with what does exist. Indeed, the banality of existence has been so amply demonstrated, there is no need for us to discuss it any further here. The brilliant Cerebron, attacking the problem analytically, discovered three distinct kinds of dragon: the mythical, the chimerical, and the purely hypothetical. They were all, one might say, nonexistent, but each non-existed in an entirely different way. </p> </blockquote> http://mathoverflow.net/questions/10876/perpetuum-mobile/12807#12807 Answer by Mark Levi for Perpetuum Mobile Mark Levi 2010-01-24T01:53:40Z 2010-01-24T14:37:46Z <p>The impossibility of this perpetuum mobile is (I think) a consequence of the fact that the phase flow preserves the volume $dx\ dy\ d\theta$, AT LEAST IF THE PROBLEM IS MATHEMATICALLY FORMULATED in terms of classical billiards (I can imagine different formulations). Here is one possible mathematical formulation of the physical statements, avoiding the need to talk of the temperature. I apologize to experts for some of what follows. I claim that the phase volume of the billiard is a reasonable interpretation of the energy of the corresponding set of "photons". More precisely, our photons can be thought of as particles moving with unit speed and undergoing elastic collisions with obstacles. The phase space is three dimensional with coordinates $x,y,\theta$ where $\theta$ is the angle of the velocity with a fixed direction in the plane. To justify the above claim, let us start with a huge number of "photons" equidistributed in the phase space: the density is nearly constant. By volume preservation the density will remain constant for all time. Thus the volume really counts the number of photons, and thus the energy, since each "photon" carries one unit of energy. </p> <p>Now it is easy to see that the combined energy of "photons" hitting an obstacle $c$ (e.g., a disk) equals the combined energy of the ones that bounce off (during the same time interval). Indeed, $c$ is the projection onto the $xy$-plane of the cylinder $c\times S^1$ in the phase space. Half of the surface of this cylinder is entered by the fluid; another half is exited. By volume preservation, the entering flux equals the exiting flux. That is, as many photons hit the obstacle, as bounce off. Thus the temperature of the obstacle does not change. </p> http://mathoverflow.net/questions/10876/perpetuum-mobile/31950#31950 Answer by paulmurray for Perpetuum Mobile paulmurray 2010-07-15T03:35:28Z 2010-07-15T03:35:28Z <p><a href="http://scienceblogs.com/builtonfacts/2010/05/solar_heat_thermodynamics_pt_2.php" rel="nofollow">There's a great post relating to this on science blogs.</a> Define the "radiance" of a source of illumination as the amount of light coming from it to the observer per unit of solid arc. In any passive optical system, the radiance is conserved. When you use a curved mirror, for instance, to focus the sun's rays on an observer, or a solid object to partly shade an obsever, what the observer sees is that the sun appears to take up more or less of the sky - the surface temperature of the sun appears the same.</p> <p>The problem with the mirror setup is that you are proposing sources of light that are point sources - you are trying to divide by zero, to have a light source with infinite radiance. The moment you do this, all bets are off. </p> <p>For any non-point source of light, the arrangement will not work. From the point of view of many of the points on the surface of the blue light source, the red light source will not occupy the whole of its field of view - the image of the red source in the larger ellipse is too small to completely obscure the image of the blue source behind it.</p> http://mathoverflow.net/questions/10876/perpetuum-mobile/80825#80825 Answer by Rod Schmidt for Perpetuum Mobile Rod Schmidt 2011-11-13T16:21:22Z 2011-11-13T16:21:22Z <p>I think the finite size is a red herring; with ideal sources/sinks and particles it still doesn't "work".</p> <p>Incorrect approach: 100% of rays from from blue focus go to the red one; 80% of rays from red focus go to the blue one; 100% > 80%; therefore imbalance. Incorrect because we must look at absolute numbers, not percentages, when saying the flow from blue to red must equal the flow from red to blue.</p> <p>Incorrect approach: Initally both foci emit 100 particles, and during a time period epsilon, 100 particles go from blue to red, but only 80 go from red to blue. True so far, but Incorrect approach because the paths are not all the same length so the travel times are not all the same. When all paths are equally full of particles, the numbers arriving at each focus will be the same.</p> <p>Correct approach: Short path takes T1 time units to travel; long path takes T2. At equilibrium, during a time period epsilon, 100 particles will leave blue, of which the fraction that hit the tighter-curved mirror will arrive at red after T1, and the fraction that hit the longer-curved mirror will arrive at red after T2. During the same time, 100 particles will leave red; a fraction of those will go to blue in time T1, and a fraction will go back to red in time T2. I'm too lazy to work it all out right now, but that's the correct approach.</p>