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Recently I've been reading a lot about blackbody radiation, Rayleigh-Jeans law, Planck's law and the UV catastrophe.

In deriving the Rayleigh-Jeans and Planck's laws, we are examining a perfectly reflecting cavity filled with radiating ideal gas. The gas and the radiation are in thermal equilibrium. In Rayleigh-Jeans law, it is assumed that as an ideal gas has the average energy of $1/2kT$ per degree of freedom, the radiation has the same average energy per mode.

This is somewhat understandable to me, as the radiation originates from the charged particles having average energy of $1/2kT$, it would be reasonable to assume the radiation has the same average energy. But how does an ideal gas radiate in the first place?

My understanding of the ideal gas model is that it is a collection of particles moving with constant speeds and are non-interacting except during collisions, which are fully elastic. But to radiate, a particle has to accelerate. Only situation I can think of where electrons in ideal gas accelerate are during collisions (they collide and change directions), but the collisions are elastic and there is no change in kinetic energies of the particles. So if the collisions are elastic, where does the energy to produce radiation come from?

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  • $\begingroup$ "but the collisions are elastic and there is no change in kinetic energies of the particles" there is a dp/dt with the collision that turns to electromagnetic radiation. The elastic hypothesis fails with real atoms that have spill over fields from the orbitals . $\endgroup$
    – anna v
    Nov 11 '18 at 12:57
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In deriving the Rayleigh-Jeans and Planck's laws, we are examining a perfectly reflecting cavity filled with radiating ideal gas.

The gas does not have to be ideal. Ideal gas is too simple a model to provide description of interaction with the EM radiation.

Real gas interacts with EM radiation, because its molecules consist of charged particles. The collisions of the molecules are not elastic.

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  • $\begingroup$ But still we are using results that describe ideal gases, such as $1/2kT$ for energy. Isn't this a result from Maxwell-Boltzmann distribution, which specifically describes ideal gases? $\endgroup$
    – S. Rotos
    Nov 11 '18 at 14:54
  • $\begingroup$ @S.Rotos there is continuity in the regions of validity between classical and quantum mechanical frames.. we still use the 1/r potential for the hydrogen atom, which was derived from macroscopic observations. The ideal gas is a useful approximation as long as one knows the limits of accuracy where it can be used. $\endgroup$
    – anna v
    Nov 11 '18 at 15:47
  • $\begingroup$ @S.Rotos, the result "1/2kT per quadratic degree of freedom" is valid (in classical theory) for any gas with short-range interactions, the gas does not have to be ideal. $\endgroup$ Nov 11 '18 at 16:53

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