Background - This post is closely related to this question. Planck considered that the electromagnetic waves inside a cavity are standing waves due to oscillating charges on the wall of the cavity. At equilibrium, the frequency of oscillation of the oscillating charge is equal to the frequency of the electromagnetic wave produced by it (i.e., at equilibrium, the energy of the oscillating charge is equal to the energy of the produced electromagnetic wave). Planck proposed that the energy of the oscillating charge is an integral multiple of $h\nu$ i.e. $\epsilon_n=nh\nu$ where $n=0,1,2,3...$ According to this, the radiation energy density per unit volume is

$$u(\nu)d\nu=\frac{8\pi \nu^2}{c^3}\left(\frac{h\nu}{e^{\frac{h\nu}{k_BT}}-1}\right)d\nu$$

However, we know that quantum harmonic oscillators have energy $\epsilon_n=(n+\frac{1}{2})h\nu$!

Question: Ultraviolet catastrophe from ground-state energy? If we consider $\epsilon_n=(n+\frac{1}{2})h\nu$, then the average energy per vibrational mode becomes

$$\left<E\right>=\frac{\sum_0^\infty(n+\frac{1}{2})h\nu \ e^{-\frac{(n+\frac{1}{2})h\nu}{k_BT}}}{\sum_0^\infty e^{-\frac{(n+\frac{1}{2})h\nu}{k_BT}}}=\frac{h\nu}{2}\left(1+\frac{2}{e^{\frac{h\nu}{k_BT}}-1}\right) \, .$$

From this, the radiation energy per volume becomes $$u(\nu)d\nu=\frac{4\pi h\nu^3}{c^3}\left(1+\frac{2}{e^{\frac{h\nu}{k_BT}}-1}\right)d\nu \, .$$

If we plot these $u(\nu)$ vs $\nu$, we get this plot

Energy Distribution

which shows that energy distribution for $\epsilon_n=(n+\frac{1}{2})h\nu$ results in ultraviolet catastrophe! Does this mean that there are no oscillating charges on the wall of the cavity supposed to be quantum harmonic oscillators?


1 Answer 1


At Equilibrium, the Frequency of oscillation of the oscillating charge is equal to the Frequency of the Electromagnetic Wave produced by it


i.e. at Equilibrium, Energy of the Oscillating Charge is equal to the Energy of the produced Electromagnetic Wave.

No. There is no such thing as "energy of the produced wave by a single wall oscillator" here. Poynting energy can be ascribed to either total volume of the cavity, or to a Fourier mode in that volume (one term of an infinite sum that makes total Poynting energy of that volume). But one cannot take just wave produced by a single wall oscillator and assign it energy.

What is more correct to say (a reasonable assumption), is that every EM radiation oscillator has the same expected average energy as the wall oscillator of the same frequency, provided there is such a wall oscillator at all, and provided it is in equilibrium with the field oscillator.

This plot clearly shows that, Energy Distribution for $\epsilon_n=(n+\frac{1}{2})h\nu$ results in Ultraviolet Catastrophe i.e. it is impossible!

On the contrary, it is the correct result of the adopted assumptions (that energy is given by the Hamiltonian with zero point terms). Yes, energy spectrum, in both classical and quantum theory, suffers from a problem that can be called UV catastrophe, but for different reasons.

The quantum variant is less bad, in the sense that the divergence is due to "bad terms" (in the expression of field energy) which are independent of temperature, so we can redefine energy expression to exclude those terms. In classical UV catastrophe, the divergence is due to bad multiplicative factor depending on temperature, we can't just remove some terms from the energy expression to get rid of the infinity problem there.

Using $\hbar\omega (n+\frac{1}{2})$ for oscillator energy instead of $n$ is more correct for single mechanical oscillator (and for the material oscillators in the wall), as there should be positive oscillator energy (of both kinds) even in the ground state, due to uncertainty relations. But for field oscillators, this leads to a problem with infinite energy in arbitrarily small volume, because there is infinity of field oscillators. So maybe $n\hbar \omega$ instead of $\hbar\omega (n+\frac{1}{2})$ is the better definition for field energy.

In quantum theory, there are two interesting viewpoints on this problem. First, it is not clear how that "bad term" would make any difference to predicted intensity of blackbody radiation. Zero-point energy is the same everywhere, but is not associated with any sort of direction or transport of energy or dependence on temperature. It may be present at the detector without registering in usual measurements of radiation intensity (via thermal or photoelectric effects). Of course, infinite total energy is an objectionable feature of the spectral distribution of energy in and of it self, that asks for some explanation/resolution. But the objectionable thing is EM energy expression that for any small volume gives infinity, not the fact the calculated spectral function differs from the measured blackbody radiation intensity by some finite term.

Second, since we still have the problem with infinite energy, we should fix the expression defining energy. This needs not be the same as the Hamiltonian we found first, because there is infinity of equivalent Hamiltonians, but we should choose only one expression as the definition of field energy. One standard procedure of canonical quantization of free EM field leads to harmonic-oscillator-like Hamiltonian with zero point terms. But we can simply scratch them off in the definition of the Hamiltonian, and then we have equivalent Hamiltonian without those terms, which predicts most (all?) things just the same, and the problem of infinite field energy due to zero point terms evaporates.

There is no single best solution to this issue of infinite energy that would be accepted in physics. There are other ways to resolve this. For example, in classical theory UV catastrophe won't happen, if the assumption of the derivation (*) is true only for some low enough frequencies $\omega < \omega_c$ but breaks down for higher ones. After all, there is not enough material oscillators in the walls to physically shield the cavity from radiation of extremely high frequencies, and there really can't be equilibrium at high enough frequencies. In quantum theory, maybe even the assumption we're fine building on top of Poynting's formula breaks down at high frequencies; that formula predicts big field energy due to big value of square of electric field, but maybe there is very little energy associated with that if such strong field is due to single charged particle. Especially in the gamma ray regime where the field oscillations are due to uncorrelated chaotic motions of elementary particles, rather than correlated motion of macroscopic charge distributions.

  • $\begingroup$ What do you mean by "moment of Electric Field"?, The extra $\frac{1}{2}h\nu$ term? $\endgroup$
    – Lusypher
    Apr 17, 2022 at 5:06
  • $\begingroup$ I mean the idea that EM energy is given by integral of square of total electric field. $\endgroup$ Apr 17, 2022 at 14:11

Not the answer you're looking for? Browse other questions tagged or ask your own question.