# Why standing waves in Rayleigh-Jeans derivation?

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?

• 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. Mar 31 at 0:52
• I cropped your question to address a single question Mar 31 at 15:11
• Okay, that works, thank you Mar 31 at 22:23

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 $$kT/2$$.
• 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. Mar 31 at 23:57
• 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. Apr 2 at 16:44