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Here is a machine which seems to violate the second law of thermodynamics: 2 confocal ellipsoids and an annulus

  • $A$ and $B$ are point black bodies of the same temperature (initially).
  • everything is rotationally symmetric around the axis $AB$
  • $e$ and $f$ are ellipsoids with foci $A$ and $B$, made of a reflective material
  • $CD$ and $EF$ are sections of a reflective annulus
  • there is no air

The stable state of the machine is $A$ having higher temperature than $B$ because

  • The heat radiated by $B$ is all absorbed by $A$ (via paths $B\rightarrow K\rightarrow A$ and $B\rightarrow J\rightarrow A$).
  • The heat radiated by $A$ is either absorbed by $B$ (via paths $A\rightarrow K\rightarrow B$ and $A\rightarrow J\rightarrow B$) OR by $A$ (via paths $A\rightarrow G\rightarrow I\rightarrow A$)

This seems to violate the 2nd law.

So, where is the hole here?

PS. While point bodies and perfect mirrors do not exits, note that we have quite a lot of margin here: a huge left ellipsoid and a tiny right ellipsoid will lead to almost 50% of all radiation from $A$ reflecting back to $A$. So, "small" bodies and 90%-efficient mirrors should be fine.

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    $\begingroup$ Why downvote? Please explain! $\endgroup$ – sds Oct 6 '14 at 16:22
  • $\begingroup$ Perfect mirrors and point blackbodies are not physical. However, to your point, there may be a more "fundamental" reason. Someone smarter than I will have to address that. $\endgroup$ – garyp Oct 6 '14 at 16:38
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    $\begingroup$ Could you explicitly explain why this would contradict the second law? No word games please, calculate the entropies that you claim are violating the law - i.e. where does entropy decrease here? (Note that, though you have found a stable state, it is not granted that any initial configuration will actually reach that state) $\endgroup$ – ACuriousMind Oct 6 '14 at 16:59
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    $\begingroup$ Voting to close. This is yet another variant of the ellipsoid paradox in thermodynamics. The resolution is simple: You are assuming point particles. Non-point sources will bathe the entire structure in light. See, for example Yoder & Adkins, "Resolution of the ellipsoid paradox in thermodynamics." American Journal of Physics 79.8 (2011): 811-818.. $\endgroup$ – David Hammen Oct 6 '14 at 19:40
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    $\begingroup$ @David Hammen - asking a question about how a setup avoids violating some law is not the same as saying confidently that it does violate that law, I assumed the question was sincere and not a rhetorical way of saying "look, I've violated the 2nd law!" Physics textbooks often present "puzzler" questions like these for the purpose of aiding understanding. $\endgroup$ – Hypnosifl Oct 8 '14 at 22:37
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Ingenious. A and B are small, but they cannot be points.The image of B is magnified at A. Therefore if A and B are the same size, some of the light from B will miss A.

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  • $\begingroup$ Why "The image of B is magnified at A."? $\endgroup$ – sds Oct 6 '14 at 16:58
  • $\begingroup$ Can you actually do the calculation with a finite radius $R$ for the black bodies and check that the energy transfer balances out? What about the limit $R\to 0$? $\endgroup$ – Steven Mathey Oct 6 '14 at 17:10
  • $\begingroup$ @sds - that's based on mirror optics--concave parabolic mirrors create magnified virtual images of objects at any position closer than the center of curvature of the mirror (which is at twice the focal length, so your A and B would in this range), see here for a photo of what's actually seen in one, and for a ray diagram, see the bottom left illustration at the top of p. 332 in the pdf here (the object is the blue arrow, the virtual image is the orange arrow) $\endgroup$ – Hypnosifl Oct 6 '14 at 17:26
  • $\begingroup$ @StevenMathey In the limit that $R\rightarrow 0$ Stefan-Boltzmann Law has the emitted power tend to 0. Since this is a universal law for all black bodies, there is no parameter you can tune to keep the finite answer in the limit of zero surface area. $\endgroup$ – By Symmetry Oct 6 '14 at 17:33
  • $\begingroup$ @StevenMathey - I think a detailed calculation would probably be very laborious, since you need to take into account not only the images of A and B in each parabolic section and the flat section, but also the hall-of-mirrors effect where some of the light from the image of B from the mirror near it misses hitting A and is instead reflected in the mirror near A, creating another image whose size is different from that of the image of B created when the light from B takes a direct path to the mirror near A. I would bet that akrasia's answer is the basic key to the solution though. $\endgroup$ – Hypnosifl Oct 6 '14 at 17:56
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Position yourself on the surface of body A and shoot a ray in any direction. After a certain number of reflections from the mirror surfaces (possibly none) it will hit either body A or body B. Now expand the ray into a very narrow cone, such that all the rays in the cone finish on the same body. Looking into that cone you will see a radiant intensity characteristic of the temperature of that body. This is true regardless of any magnification effect (a consequence of the reciprocity theorem for view factors). And since the same is true for any initial ray the radiant intensity will be uniform in all directions, and characteristic of the temperature of both bodies. So there will be no net heat transfer.

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  • $\begingroup$ I don't quite follow this, can you explain where in the argument you made use of the assumption that the entire system has reached equilibrium? If the two bodies start out at different temperatures, both emitting blackbody radiation, it can't be true at the beginning that "there will be no net heat transfer", so either the argument is flawed or there is some step in the argument that makes use of an assumption which is valid if we assume the entire system is already at equilibrium, but not valid otherwise. $\endgroup$ – Hypnosifl Oct 7 '14 at 21:50

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