Planck, in his treatment to solve blackbody radiation, considered that the electromagnetic waves inside the cavity of the blackbody 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, energy of the oscillating charge is equal to the energy of the produced electromagnetic wave.

He proposed that, the energy of the oscillating charge is integral multiple of $h\nu$ i.e. $\epsilon_n=nh\nu$ where $n=0,1,2,3,4,...$

According to this, radiation energy density per unit volume is $$u(\nu)d\nu=\frac{8\pi \nu^2}{c^3}\left(\frac{h\nu}{e^{\frac{h\nu}{k_BT}}-1}\right)d\nu$$ but we know that quantum harmonic oscillators have energy $\epsilon_n=(n+\frac{1}{2})h\nu$** where n=0,1,2,3,4,...

If we consider this $\epsilon_n=(n+\frac{1}{2})h\nu$, 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 i.e. it is impossible!

Then, are not the oscillating charges on the wall of the cavity supposed to be quantum harmonic oscillators?

  • 1
    $\begingroup$ Please note that in English, only the first word of a sentence, proper nouns, and the word "I" are capitalized. We do not capitalize other nouns. $\endgroup$
    – DanielSank
    Feb 9 at 1:20

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.(*)

Using $n+\frac{1}{2}$ for oscillator energy instead of $n$ is more correct, as there should be positive oscillator energy (of both kinds) even in the ground state, due to uncertainty relations.

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. Yes, Poynting energy spectrum, in both classical and quantum theory, suffers from UV catastrophe.

The quantum variant is better in the sense that the divergence is due to "bad term" that is the same everywhere in space, and independent of position and temperature. It is not clear how that "bad term" would make any difference to predicted radiation intensity. Zero-point radiation is the same everywhere, it may be present at detector without registering in usual measurement of radiation intensity (thermal or photoelectric effects of some rays of radiation).

Of course, infinite total energy is an objectionable feature of the spectral distribution of energy that asks for some explanation. But the objectionable feature is EM energy in any small volume being infinite, not measured radiation intensity differing from calculated energy spectral function by some constant term.

There is no single best solution to this issue of infinite energy that would be accepted in physics. There are many possibilities to resolve this. For example, the assumption of the derivation (*) may be true for some low enough frequencies $\omega < \omega_c$ but then break 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 such high frequencies. Or another way would be, maybe our formula for EM energy (Poynting's formula) breaks down at high frequencies; it predicts great zero point energy due to great value of second moment of electric field, but maybe there is very little energy associated with that second moment when the oscillations are very quick.

  • $\begingroup$ What do you mean by "moment of Electric Field"?, The extra $\frac{1}{2}h\nu$ term? $\endgroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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