In the real world, the ultraviolet catastrophe doesn't happen because the quantization of photons modifies the classical behavior of light at frequencies comparable to and higher than the temperature. But classical electromagnetism is a mathematically self-consistent theory, so we could imagine a world where $\hbar = 0$ and electromagnetism remains classical to arbitrarily high frequencies. How would the ultraviolet catastrophe work in such a world? Classically, all frequencies are equally populated at all temperatures and each has energy $\frac{1}{2} k_B T$, apparently leading to infinite energy radiation at any temperature, which doesn't seem compatible with conservation of energy. What would happen if you put a theoretical fully classical system in contact with a thermal bath (which seems like a physically reasonable set-up)?

My guess is that since this system has an infinite number of quadratic degrees of freedom, the usual canonical-ensemble derivation of thermal equilibrium breaks down. I think that if you couple a perfectly conducing cavity to any thermal bath, no matter how large, then the cavity will absorb an unboundedly large amount of energy from the bath. In the usual derivation, we assume that the bath has so much more energy than the system that its energy density is independent of the state of the system, but in this case that assumption will eventually be violated. The cavity's energy will become comparable to the bath's, so to find its equilibrium state, we will need to consider the details of the bath and treat the combined cavity-bath system in the microcanonical ensemble. So the Boltzmann distribution and the equipartition theorem will no longer apply, and the ultraviolet catastrophe will be avoided.

  • $\begingroup$ I think a fully classical system is at least problematic: and in particular I think you want to distinguish between mathematically self-consistent and physically plausible. GR for instance is mathematically at least reasonable yet clearly has catastrophic problems with physical quantities diverging (curvature singularities): so just because the theory is OK mathematically does not mean it is OK physically (in all cases: GR clearly works very well in many cases!). $\endgroup$
    – user107153
    Mar 29, 2016 at 8:17
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    $\begingroup$ I fail to see the justification for the upvotes. It's quite clear that the so-called "classical " radiation equation leads to an impossible universe. Why bother? $\endgroup$ Mar 29, 2016 at 13:03
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    $\begingroup$ @CarlWitthoft: In actuality the term "classical" is way too vague to distinguish what is going wrong with the Rayleigh-Jeans formula. The classical theory of EM is used both with statistical mechanics and a particular model of matter and many things can go wrong in the two latter ingredients. For starters, with any real material, all frequencies higher than the plasmon frequency of the conducting (black body) cavity can actually escape from it and won't equilibrate so that the equilibrated sum should go only up to the plasmon frequency more or less. $\endgroup$
    – gatsu
    Mar 29, 2016 at 19:49
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    $\begingroup$ > "in the real world" better to say "in quantum theory" here - this can be explained in more ways. For example, there may not be a way to achieve equilibrium for high enough frequencies, like when metal can't stop gamma rays. $\endgroup$ May 2, 2019 at 10:55
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    $\begingroup$ > "classical electromagnetism is a mathematically self-consistent theory" this is somewhat problematic claim. Electromagnetism is a broad term and includes things such as EM energy being given by Poynting formulae and matter being made of point charges. But those two are inconsistent. So something has to be altered or removed to get a self-consistent theory. $\endgroup$ May 2, 2019 at 10:57

2 Answers 2


I think this is an interesting question. If one tries to couple the EM field to matter somewhat realistically and classically, he needs a model for matter. One possible model would be that of an assembly of uncorrelated dipoles. If we assume these dipoles to be point dipoles then the energy they emit is related to the wavelength as $\varepsilon_{\lambda} \sim \lambda^{-4}$. There is no discussion that this is correct as this corresponds to the Rayleigh scattering result and we more or less witness its effect everyday when we go outside.

So, it could be that the "ultraviolet catastrophe" attributed to the Rayleigh-Jeans formula, which incidentally has also an EM energy density that goes as $\sim \lambda^{-4}$, assumes somewhere that the scattering material objects are point-like.

At the very least for consistency reason, at equilibrium, the energy density emitted by an assembly of dipoles has to agree with that calculated independently for the EM field and the two above results have to agree.

Now, the point is that at high frequency, the wavelength becomes of the order of the size of the scatterers and we could get for instance something close to Mie scattering instead of Rayleigh scattering. The effect it would have is that the energy fraction emitted by a scatterer becomes roughly independent of the wavelength in this regime and so something will probably happen and prevent a fully diverging energy density at short wavelength.

So in effect, one would have to cut off the spectrum somewhere and make it at least saturate at some value.

Note that this is something ubiquitous in statistical mechanics (on a quite different note the partition function of a single hydrogen atom diverges and one needs often to put by hand a cutoff corresponding to an atom radius big enough to create an ambiguity between a pair of H atoms and an H2 molecule) and it is probable that people are often too eager to claim, a posteriori, that a new theory was definitely needed while something could possibly have been done with a better model.

This of course does not mean that quantum mechanics is not required in the end but it means that the dramatic failure of the Rayleigh-Jeans formula is a problem owing to all the assumptions leading to the said formula, not just one.


What would happen if you put a theoretical fully classical system in contact with a thermal bath (which seems like a physically reasonable set-up)?

I assume by the system you mean perfectly reflecting cavity so it has infinite number of modes and can theoreticaly contain unlimited amount of EM energy, in high frequency modes. And by thermal bath, I assume you mean some system in thermodynamic equilibrium but with finite energy.

For example, we can put a glowing piece of graphite into this perfect reflecting cavity.

What would happen depends on whether there is mechanism that can carry energy from low modes into arbitrarily high modes. If so, the cavity would transform the radiation of the graphite into gamma and higher frequency radiation. For a system whose energy is quadratic function of state variables (the simplest model of continuum field such as elastic medium or EM field), it is much more probable that energy is in the high frequency modes than for it to stay in the low frequency ones, because density of states increases with frequency. No equilibrium could be achieved, this would be a run away process.

If there is no such mechanism of transfer, or if it works only up to some frequency (most likely in reality), then perhaps some equilibrium could be achieved. Depending on the size of the bath, the assumption about it not being susceptible to change could be valid or not.


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