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A central part of the Rayleigh-Jeans law is that the number of allowed states follows $dN \sim \nu ^{2} d\nu$ because the allowed frequencies have to be standing waves. Then by equipartition we get that the energy density in a given frequency band increases without bound for increasing $\nu$, which is obviously unphysical.

Regarding a non-cavity solid object emitting thermally, by contrast, I'm picturing a lattice of atoms and electrons oscillating and/or colliding in such a complicated way as to be random. Perhaps there are conduction electrons, perhaps not. Now, I don't believe the phonon motion contributes much to EM radiation (?); instead, I think, electron oscillations (classically; in QM this would be transitions) dominate. Such electron motion would not be characterized by standing waves or frequencies which depending on the dimensions of the solid, surely?

It seems reasonable that at a given temperature, random "dipole" (or whatever) oscillations in a solid object would classically follow something like a Boltzmann distribution with a maximum at a finite frequency. Intuitively, this could lead to an emission with the qualitative shape of the experimental blackbody curve. Is there a classical theory of random thermal oscillations in a lattice which is relevant here?

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  • $\begingroup$ There seem to be two different questions here, why many objects are similar enough to a blackbody and why standing waves in Rayleigh-Jeans calculation. Please edit to keep only one question per post. $\endgroup$
    – Mauricio
    Commented Mar 31, 2023 at 0:52
  • $\begingroup$ I cropped your question to address a single question $\endgroup$
    – Mauricio
    Commented Mar 31, 2023 at 15:11
  • $\begingroup$ Okay, that works, thank you $\endgroup$
    – Dennis
    Commented Mar 31, 2023 at 22:23

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Standing waves are not a necessary assumption in the Rayleigh-Jeans derivation. When the derivation works with radiation inside a cavity with walls impermeable to radiation, so the waves can't go through the walls, the radiation inside can be expressed as superposition of standing waves.

But the same kind of derivation can be made for imaginary region in space with imagined walls that do not influence the EM field in any way. The field can be expressed as Fourier sum in finite cuboid region of space, over travelling waves - this is sometimes called "periodic boundary conditions" (because Fourier sum produces function that repeats outside the main region). This is just mathematics, it does not impose any condition on the field inside. The rest of the derivation is otherwise the same, and leads to the same result.

Thus in the Rayleigh-Jeans derivation, standing waves is an unimportant detail; the important assumptions are 1) energy is given by the Poynting formulae; 2) EM field is in thermodynamic equilibrium; 3) equipartition is valid for EM field in thermodynamic equilibrium: every quadratic Fourier term in expression of EM energy contributes with average energy $kT/2$.

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  • $\begingroup$ Hi Jan, thank you for the response! I still don't get it. Why do we need to impose any boundary conditions outside the body? It's just atoms radiating. If the atoms have random frequencies in something like Boltzmann distribution, I'd expect the radiation to be emitted according to this distribution. There is no ultraviolet catastrophe, because the temperature restricts the number of atoms with very high frequency. You mention the EM field is in thermodynamic equilibrium. With what? The waves are moving out, away from the black body. Still confused. Thanks for help. $\endgroup$
    – Dennis
    Commented Mar 31, 2023 at 22:23
  • $\begingroup$ Forget atoms. The Rayleigh-Jeans derivation is about distribution of energy of EM radiation among frequencies, EM radiation is its own system, in thermodynamic equilibrium at some temperature $T$. It does not matter with what it is in equilibrium, this can be with absorbing walls of the cavity, or with a body inside cavity made of perfectly reflecting walls. Atoms won't produce UV catastrophe; UV catastrophe is a property of continuous systems with infinite number of degrees of freedom. $\endgroup$ Commented Mar 31, 2023 at 23:57
  • $\begingroup$ We do not have to impose any boundary conditions. However, the premise of the R-J derivation is that EM radiation in some region of space is in state of thermodynamic equilibrium at some temperature. This won't happen unless the radiation is trapped somehow, e.g. in cavity with material walls. Otherwise the EM radiation would expand, EM energy would dilute and it could not be in state of thermodynamic equilibrium. $\endgroup$ Commented Mar 31, 2023 at 23:59
  • $\begingroup$ Okay, hmm. I'm interested in normal objects,not cavities. Solids in thermal equilibrium absorb energy in various ways and radiate EM out; people seem to refer to this as blackbody radiation (e.g. a lightbulb filament). I hear what you're saying about the cavity being necessary for RJ; but as you explain, this model does not seem not relevant to unbound radiation from atoms in a real solid. I guess I misunderstood the significance of the RJ theory. Is there a classical theory of frequency of thermal radiation relevant to normal objects? Thank you again. $\endgroup$
    – Dennis
    Commented Apr 2, 2023 at 15:34
  • $\begingroup$ Normal objects emit thermal radiation too, but it is not a black body radiation; it can have lower or equal intensity to blackbody radiation at any frequency. If total emission intensity is higher than blackbody radiation, then the difference is considered radiation due to fluorescence or other process, not thermal emission. Difference of thermal emission of real solid bodies as compared to a blackbody is described by emissivity $\epsilon(\omega,T)$. Microscopic theory of such thermal emission probably exists in papers / specialized books on solid state theory, but I don't know much about it. $\endgroup$ Commented Apr 2, 2023 at 16:44

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