I Questions about Jean-Rayleigh's derivation of Ultraviolet Catastrophe

[LightPhoton]

TL;DR

Questions about Jean-Rayleigh's derivation of Ultraviolet Catastrophe related to black body, EM waves in cavity and use of equipartition theorem

I am following this video and; Eisberg and Resnick's Book for this derivation, for I cannot find other sources that go as in-depth as they do.


$$\Large\text{Question 1)} $$

Jean's cube, or the metallic cube, is assumed to be a perfect absorber. On this fact alone, authors state


Now assume that the walls of the cavity are uniformly heated to temperature T. Then the walls will emit thermal radiation which will fill the cavity. The small fraction of this radiation incident from the inside upon the hole will pass through the hole. Thus the hole will act as an emitter of thermal radiation. Since the hole must have the properties of the surface of a blackbody, the radiation emitted by the hole must have a blackbody spectrum; but since the hole is merely sampling the thermal radiation present inside the cavity, it is clear that the radiation in the cavity must also have a blackbody spectrum. In fact, it will have a blackbody spectrum characteristic of the temperature T on the walls, since this is the only temperature defined for the system.


Now, I am not sure how having the absorption properties of a black body implies that it must also have emission properties of the black body as well. The closest answer to this was in the comments of this question's answer, which is basically "that's what experiments tell us."



$$\Large\text{Question 2)} $$


In deriving the spectrum of EM waves inside the cube, we assume that the waves can be broken up into three independent components. This seems logical, but given the complexity of Maxwell's equations, I have a hard time buying this. In the video @19:42 the diagram helps in clearing this up and is pretty satisfactory, but a mathematical proof would be better.



$$\Large\text{Question 3)} $$



Why did it seem reasonable at the time to use the Equipartition theorem when it clearly only adds to the kinetic energy of the system? There is no sensible way of talking about the kinetic energy of EM waves, even if it has a quadratic form $(\epsilon_0E^2/2)$

 

[Demystifier]

Now, I am not sure how having the absorption properties of a black body implies that it must also have emission properties of the black body as well. The closest answer to this was in the comments of this question's answer, which is basically "that's what experiments tell us."

We assume thermal equilibrium. In equilibrium, absorption is in equilibrium with emission.

In deriving the spectrum of EM waves inside the cube, we assume that the waves can be broken up into three independent components. This seems logical, but given the complexity of Maxwell's equations, I have a hard time buying this. In the video @19:42 the diagram helps in clearing this up and is pretty satisfactory, but a mathematical proof would be better.

Why do you say three? EM waves have two polarizations. There are 3 directions in space so naively one might expect 3 polarizations, but the longitudinal polarization (the one in the direction of motion) is missing due to the fact EM field is massless. I don't know a simple explanation of this, but if you are ready for a complicated one it can be found in many places.

Why did it seem reasonable at the time to use the Equipartition theorem when it clearly only adds to the kinetic energy of the system? There is no sensible way of talking about the kinetic energy of EM waves, even if it has a quadratic form $(\epsilon_0E^2/2)$

The equipartition theorem is valid whenever the Hamiltonian is quadratic in canonical coordinates and momenta, i.e. something of the form $H \sim ap^2+bx^2$, where $a$ and $b$ are constants. The EM Hamiltonian is exactly of this form, $H \sim aE^2+bB^2$, where $E$ is proportional to canonical momentum and $B$ to canonical coordinates. To see this, one has to write the EM theory in terms of gauge potentials, which are the fundamental canonical variables.