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I never understood how the equipartition theorem was applied electromagnetic waves inside the metallic blackbody. As hyperphysics puts it (http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/mod7.html)

The classical view treats all electromagnetic modes of the cavity as equally likely because you can add an infinitesmal amount of energy to any mode

But the existence of the modes is conditioned to the existence of certain movements by the particle, so I don't see it as a "degree of freedom". For example, for a high frequency mode to exist there should be electrons jiggling with a certain frequency. I expected some sort of statistic on electrons speed and acceleration for a given temperature, like from the Maxwell-Boltzmann distribution, to derive the expected electromagnetic radiation spectrum for the blackbody.

1) Why are the electromagnetic modes are considered "degrees of freedom" if their existence is conditioned to the motion of the electrons? If they actually aren't, please explain it

2) Does an alternative (non cavity + modes) approach exist? I was thinking fluctuations of charge density on the surface of a solid metal sphere due to temperature.

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1) Why are the electromagnetic modes considered "degrees of freedom" if their existence is conditioned to the motion of the electrons?

You assume that EM waves are determined by the motion of the electrons and have no freedom of their own. Although that is a reasonable picture of things (for example, if fields are assumed purely retarded, which is very natural), it is not necessary.

The common derivations of thermal radiation spectrum do not assume that.

On the contrary, they assume that EM field is a thing that can exist in vacuum independently of charged particles.

This is possible because the Maxwell equations are only conditions that fields have to satisfy. They alone do not determine the fields. In order to determine the fields, some additional boundary conditions need to be assumed.

The common derivations of thermal radiation spectrum are based on the equations for vacuum

$$ \nabla \cdot \mathbf E = 0 $$

$$ \nabla \cdot \mathbf B = 0 $$

$$ \nabla \times \mathbf E = -\frac{1}{c}\frac{\partial \mathbf B}{\partial t} $$

$$ \nabla \times \mathbf B = \frac{1}{c}\frac{\partial \mathbf E}{\partial t} $$

and boundary conditions appropriate for closed ideal metal shell.

This system then admits infinite number of solutions $\mathbf E_H, \mathbf B_H$ . Every solution can be written as discrete sum of standing waves - "modes" and these coefficients are considered as coordinates determining the configuration of the field, hence the term "degrees of freedom".

The motion of the charged matter does not enter the derivation explicitly.

So the answer to 1) is "in the derivation of thermal EM spectrum, the existence of EM waves is not considered to be conditioned by the motion of the charged matter".

Now I agree with you that this approach is far from satisfactory. The boundary condition of ideal metallic shell is unrealistic for high frequencies. In practice, such boundary condition is an easy but only an approximately correct way to account for the motion of electrons in the metal.

2) Does an alternative (non cavity + modes) approach exist? I was thinking fluctuations of charge density on the surface of a solid metal sphere due to temperature.

The closest thing I know is later Planck's derivation, where he assumes the EM field is emitted by the matter oscillators in steps $\hbar \omega$. You can read about it in his great book

M. Planck, The theory of heat radiation, P. BLAKISTON'S SON & Co. 1914

https://archive.org/details/theheatradiation00planrich

There are other kinds of derivations, for example Timothy Boyer's derivations based on the zero-point radiation:

http://prola.aps.org/abstract/PR/v182/i5/p1374_1

You can find other related papers by Timothy Boyer on arxiv.

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  • $\begingroup$ Doesn't it matter that the boundary conditions imply moving charges? $\endgroup$ – galmeida Oct 12 '19 at 11:08
  • $\begingroup$ Not for the derivation of the Rayleigh-Jeans formula (which, if assumed valid for all frequencies, implies infinite EM energy). The boundary condition of perfect reflection isn't actually necessary; as long as energy loss from cavity is slow and radiation is in equilibrium, what happens at the boundary of the region doesn't influence the result, except perhaps at points very close to the surface. One can do the R-J derivation for imaginary cube without walls, in vacuum, with the same result. $\endgroup$ – Ján Lalinský Oct 12 '19 at 16:42
  • $\begingroup$ The reason the perfectly reflecting cavity is often mentioned is for the reason that originally, people measuring radiation properties of thermal radiation worked with radiation from "oven", a metallic cavity, to prevent energy losses by radiation, except for the thin beam measured. Extending this concept in theory, a perfectly reflecting cavity was introduced as a concept. Radiation energy cannot escape from such hypothetical cavity. But it is not necessary for the R-J derivation. $\endgroup$ – Ján Lalinský Oct 12 '19 at 16:43
  • $\begingroup$ "One can do the R-J derivation for imaginary cube without walls, in vacuum, with the same result." > Where can I find this? $\endgroup$ – galmeida Nov 4 '19 at 22:58
  • $\begingroup$ Just repeat the standard derivation without any boundary condition on the field (use Fourier series with both sines and cosines, so the field at the boundary isn't fixed). Express Poynting energy in terms of Fourier coefficients. Then apply the equipartition theorem. The result is the same whether the box is reflecting EM waves or imaginary region of space. $\endgroup$ – Ján Lalinský Nov 6 '19 at 12:17

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