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I have designed this simple thought experiment that seems to contradict 2nd law of thermodynamics. Could you please find a mistake in my reasoning? 

Fixed box with reflective (white) walls 
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|_______ | __________|                    
         ^
         This part is free to move horizontally and is black on the left and white on the right side

In the right half of the box we don't have any photons since box is completely white inside and the moving part is white on its right side. The left part however is filled with bouncing photons as they are emitted by the left side of the movable wall which is black.

According to my reasoning the moving part should move to the right because of the radiation pressure. Photons will lose energy due to doppler effect.

Since every part of the device is at the same temperature movement of the wall inside would contradict 2nd rule of thermodynamics.

Whole device is in the vacuum box which in turn is placed on a table in a laboratory.

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  • $\begingroup$ I think you have misunderstood the idea of a "blackbody". An ideal blackbody (which is approximated best by black objects) emits photons with a blackbody spectrum. However, that doesn't mean that white or red or blue objects don't emit any photons at all! White objects above absolute zero absolutely do emit photons, just with a different, possibly more complicated, spectrum. If a white object and a black object are in thermal equilibrium, they will also emit the same average power in photons. $\endgroup$ – Brionius May 4 '15 at 22:17
  • $\begingroup$ @Brionius Therefore why Stefan-Boltzmann’s law states that bodies with emissivity equal to 0 don't emit any energy? $\endgroup$ – user1354439 May 4 '15 at 22:20
  • $\begingroup$ That's forbidden by the third law. :-) $\endgroup$ – CuriousOne May 4 '15 at 22:22
  • $\begingroup$ @user1354439 I should have said "emit or reflect the same average power" $\endgroup$ – Brionius May 4 '15 at 23:48
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There is also, a gas of photons in the right side.

It was trapped there when you assembled your box, and since you are assuming a perfectly zero emissivity, these photons must be perfectly reflected from all surfaces. That means they are blue shifted if the wall moves toward the right and red-shifted if the wall moves toward the left.

Result:

  • If you assembled the apparatus at the test temperature then it was and remains in equilibrium without motion.

  • If you assembled it at a different temperature then your experiment is equivalent to heating or cooling one side while the other is adiabatic. This case includes all attempts to exclude the photon gas from the right-hand side.

    What happens is exactly what you expect: as the photon gas on the left warms (cools) the wall moves to the right (left) causing the gas on the right to warm (cool) until equilibrium is re-established.

Thermodynamics for the win.

To compute the equilibrium position for the wall, you'll need several different equations of state: (a) that for the gas on the right of a fixed number of always-reflected photons and (b) for the gas on the left of photons in thermal contact with a blackbody (the wall) whose number will vary and one for the wall (well, at least you'll need the heat capacity).

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  • $\begingroup$ That makes sense. Shame on me, I did not think about it :) Thank you for this clear and exhaustive explanation $\endgroup$ – user1354439 May 4 '15 at 23:24
  • $\begingroup$ @user1354439 I'm not sure that it is entirely obvious. I had to think for a while before I realized that there had to be a trapped photon gas, because for more ordinary emissivities the original photons would be quickly absorbed. $\endgroup$ – dmckee♦ May 5 '15 at 0:37

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