NB: I understand that the model presented here is the "failing" classical model. I'm just trying to understand the formal reasoning of the model.

In The Feynman Lectures on Physics Vol. I Ch. 41-2 Thermal equilibrium of radiation Feynman begins with the model of a rarefied gas confined to a perfectly mirrored box, in thermal equilibrium with the ambient electromagnetic radiation in the box. He assumes the gas atoms are classical oscillators of natural frequency $\omega_{0},$ and, thereby establishes Equation 41-12


The derivation begins with the assumption that the atoms have one resonant frequency $\omega_{0}.$ But he tells us

Then we substitute the formula (41.6) [$\gamma=\frac{\omega_{0}}{Q}=\frac{2}{3}\frac{r_{0}\omega_{0}^{2}}{c},$] for gamma (do not worry about writing $\omega_{0};$ since it is true of any $\omega_{0}$, we may just call it $\omega$) and the formula for $I\left[\omega\right]$ then comes out


He goes on to say:

First, let us notice a remarkable feature of that expression. The charge of the oscillator, the mass of the oscillator, all properties specific to the oscillator, cancel out, because once we have reached equilibrium with one oscillator, we must be at equilibrium with any other oscillator of a different mass, or we will be in trouble.

I'm not sure how to comprehend that. My best guess is that it means, given a gas of atoms resonant at $\omega_{0}$ in thermal equilibrium with the ambient radiation of a perfectly reflective cavity, if another kind of atom with natural frequency $\omega_{1}$ were mixed in at the same temperature and pressure, there wold be no change in the ambient radiation profile. That is, the original gas "wouldn't know the difference".

We can therefore assume the radiation profile is independent of the kinds of oscillators present. That is, the intensity at frequency $\omega$ and temperature $T$ is the same, whether or not there are atomic oscillators resonant at $\omega$.

Is that a good way to think about this?

PS: In the past when reading these lectures, I attempted to "understand" the topics well enough to justify turning the page. This time I am attempting to summarize everything. It is far more difficult than I had expected.


I believe that in this model the black body is assumed to contain oscillators of all possible natural frequencies $\omega$ (since the black body is by definition able to absorb all frequencies). Each of these oscillators contribute to the ambient electromagnetic radiation present in the box when the whole system has reached thermal equilibrium.

Now we consider only oscillators with natural frequency $\omega_0$. At equilibrium we need that these emit exactly the same amount of radiation as they absorb from the ambient radiation (if not they would lose energy and the whole system would cool). From Feynman's analysis we see that in order for this to happen the intensity of the ambient radiation with frequency $\omega_0$ must go like $\omega_0^2$ (the intensities of other frequencies are not important since only those near $\omega_0$ are absorbed). But we can apply the same argument for all other oscillators of all other possible frequencies, deducing that for any frequency ω the intensity of ambient radiation of that frequency goes like $\omega^2$.

  • $\begingroup$ My reading of Feynman is that the classical black body radiation profile is the same whether there is one species of oscillator with natural frequency $\omega_{0}$; two types of oscillators with resonance at $\omega_{0}$ and $\omega_{1}$, respectively; $n$ types, each with its own resonant frequency, or a continuous distribution. Also, there is no specification of the relative number density of each type of oscillator. So, no matter what the mix of oscillators, the classically predicted distribution will be given by the same formula: $I\left[\omega\right]=\frac{\omega^{2}kT}{\pi^{2}c^{2}}.$ $\endgroup$ – Steven Thomas Hatton May 13 '18 at 19:38
  • $\begingroup$ No, we should assume oscillators of all frequencies are present. If we only had one species of oscillator with freq. ω0 we could only deduce from Feynman's argument that in the ambient radiation I[ω0] goes like ω0^2. The argument does not allow us to say anything about the intensity of general ω. Only if we assume oscillators of all frequencies are present can we replace ω0 by ω. You should be able to see that the relative number density of each type of oscillator does not matter in the argument (only one oscillator at any particular frequency would do!) $\endgroup$ – Guest May 14 '18 at 14:12
  • $\begingroup$ I am familiar with the model of black body radiation for a continuous distribution of oscillators. A metal with free "conduction" electrons and soot are two examples. Attend to the second quote included in the original post: "...once we have reached equilibrium with one oscillator, we must be at equilibrium with any other oscillator..." Applied to a single species of oscillator, with a single fundamental frequency, (as I understand him) Feynman asserts that the profile is indistinguishable from that of a continuous distribution of fundamental frequencies. $\endgroup$ – Steven Thomas Hatton May 15 '18 at 16:18
  • $\begingroup$ Feynman's quoted statement is correct. He means if we think one oscillator in the gas has reached equilibrium all others present must have too. So the formula for the ambient radiation better not depend on the mass of the oscillator since then two oscillators of the same resonant frequency w0 but different masses could never reach equilibrium because they would both need different intensities of the ambient radiation at w0 depending on their mass. $\endgroup$ – Guest May 16 '18 at 8:49
  • $\begingroup$ Your second statement is wrong. If we had a gas containing oscillators of a single natural frequency then clearly we would have only one frequency of radiation present in the ambient radiation - the resonant frequency of the oscillator. $\endgroup$ – Guest May 16 '18 at 8:52

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