Why can't lenses transfer from colder body to a hotter one?

Recently I came up with an idea of a system that apparently is able to transfer heat from a colder body to a hotter one. Obviously, this violates the 2nd rule of thermodynamics, so I would be grateful if you could point out a flaw in my understanding :)

Below is a diagram of a system. There are 2 perfectly black balls (initially both at 100K), inside 2 perfectly reflective spheres, the whole system is in a vacuum. However, parts of the spheres were removed and replaced with lenses which focus electromagnetic radiation emitted by 2 black balls.

Lenses are chosen so that they focus all the incoming radiation from one ball on another ball. No heat escapes the system since both spheres are perfectly reflective. Now, the left ball transfers a big fraction of its own radiation to the right ball, but receives little in return from the right ball. Therefore, we can deduce that there should be a heat transfer from left ball to the right one. We can deduce the equilibrium temperature by equating heat transfers:

Where F1 is a fraction of left sphere occupied by the lens; T1 is the temperature of left ball. And F2 is a fraction of right sphere occupied by the lens; T2 is the temperature of right ball.

Equilibrium temperature does not depend on initial conditions, so if we start with both balls at 100K the equilibrium should be like 95K on left and 105K on right sphere. Also by choosing the right lens to be very small (F2 goes to 0) we can extract almost all the heat from the left ball so say 1K and 199K in equilibrium, that would be very useful!

Could you please point out a flaw in my reasoning?

• Part of your mistake is that each element of the surface of the spheres, acting as black-bodies, will radiate in every direction, not just radially outward from the center of the spheres. – Sean E. Lake Oct 20 '16 at 21:43
• If the ball is small compared to the size of the sphere then won't this be negligible? – user1354439 Oct 20 '16 at 21:50

The flaw in your argument is that you've assumed that all the radiation is emitted perpendicular to the blackbodies. This is incorrect; it actually comes out in every direction.

All the perpendicular rays you drew indeed make it to the other blackbody, but not all rays in general do. Since the lenses magnify the left sphere, the angular tolerance is less for the sphere on the left. So you get more area contributing, but less contribution from each patch of area. The total radiation flux between the two spheres is equal in both directions, and neither heats up the other.

The general optical principle here is known as the conservation of etendue. Etendue is roughly "area times angle", and measures the amount of radiation flux received. There's an explicit proof at the link above that etendue is conserved in refractions, which proves it for your case, since lenses just refract twice. On a deeper level, it comes from time reversal invariance: every path forward is balanced by a path back. For another example of this reasoning, see this question.

The first thing I noticed which you overlooked is that because they are enclosed in perfectly reflecting spheres with a part cut out for the lens, you're going to be leaking radiation in directions that do not correspond to the ray diagram you've drawn. The bigger the lens area, the more radiation is going to be lost from the sphere into the vacuum. The radiation reflecting from the bottom left of the left sphere for example will not be "seen" by the lens on the right sphere.

This alone means there is no equilibrium temperature as it's not a closed system and energy will be lost to the vacuum.

• Thanks, but wouldn't it be possible to construct the lenses so that there is no radiation leak? Or maybe some kind of light fibres leading from one sphere to another? – user1354439 Oct 20 '16 at 21:49
• I think you have been misled by your own diagram. Ray tracing has to be exact and this is pretty crude. Even if you connected the two lenses with a tube with perfect reflection on the inside this still wouldn't work. Take a look here: livephysics.com/problems-and-answers/optics/… and think about how much of the radiation from the larger lens the smaller lens actually "sees". Then think about how much radiation is leaking from the smaller lens in the other direction. You should find that what you've suggested is impossible. – Doug Oct 20 '16 at 21:57
• Hmm, such a shame :( Would save us from global warming... – user1354439 Oct 20 '16 at 22:13
• I don't think this answers the core question. So what if it leaks radiation? Just place the entire system in a thermal bath of temperature $T$. Then all radiation leaking out is balanced with radiation going in, but it still looks like the radiation from left to right isn't balanced by the radiation from right to left. – knzhou Oct 21 '16 at 0:23
• This makes the assumption that it's already in equilibrium with radiation in each sphere. You also didn't consider my first comment. – Doug Oct 21 '16 at 0:35

Both perfectly reflecting spheres actually become themselves blackbodies at $T=100K$ connected with different window areas that emit and receive radiation only to and from the other blackbody. As no radiation and heat can escape, both reflecting spheres together with their enclosed perfectly black balls become one single blackbody at $T=100K$. Therefore no net transport of radiation occurs from one black ball to the other and no temperature difference appears in accordance with the second law of thermodynamics.

Clausius statement : it is not possible to transfer heat from low to high by automatic machine without external energy.

If the perimeter of both the sphere is equal then T2>T1. So your conclusion is correct. But the problem is that the radiation from left and right are not equal since F1>F2. But initially they are emitting the same amount of radiation. One body is is discharging more heat than other which is violating the principle of calorimetry. This radiation difference is acting as the external energy which is making it possible to transfer heat from low to high.