Conservation of entendu and thermodynamics for a black body embedded inside a high index material (xkcd what if)

Question

I was reading this xkcd what if, which lead to reading about entendue, and I was wondering what happens if we put a black body inside a lens. If it has a higher refractive index, then, since entendue has a factor of n^2 in there, power can be more concentrated inside the medium, and should be able to heat the surface surface more.

Here's an extreme example: consider a large black body next to a small black body suspended in the centre of a large and very high refractive index sphere

The diagram shows that incoming rays will be refracted close to perpendicular, and so will all hit the centre black body, concentrating power density from the edge of the lens to the centre.

In addition, the majority of outgoing rays will not be perpendicular to the surface, so the majority will be reflected back due to total internal reflection.

So the question is this: since the 2nd law states heat can't flow from cold to hot, how does the small black body reach equilibrium at the same temperature as the large one, considering the asymmetry in their radiation which would seem to disproportionally heat the small one?

Answer 1

The answer is that the black body inside the high index material simply emits more radiation. Planck's law (which governs black body radiance) has a factor of 1/c^2 in it, which isn't c in a vacuum as I and some other wikipedia editors assumed, but it's c in the medium that contains the black body.

Since c=c_0/n, the total emitted radiance has a factor of n^2 in there that cancels out the n^2 factor in entendue, and all is well, they reach equilibrium at the same temperature.

But what if there's a small air gap adding a refractive boundary?

This reverses the situation. Not all of the rays hitting the refractive sphere will enter the small black body. Since the angle to the inner surface is less than the outer surface, a lot of the incoming radiation will reflect from the inner surface and won't be absorbed. And all outgoing radiation will exit the sphere, none will be reflected back, like this terrible paint diagram shows