Why can we treat normal modes of black-body radiation as thermally independent?

Question

In my professor's lecture notes about black-body radiation, a black body modeled by an opaque hollow box with a small hole is considered.

After explaining why this is a sound model for a black-body, the notes go on to calculate the energy distribution of a normal mode of the radiation as follows:

The amount of energy in the normal mode $\omega$ is $E=n\hbar\omega$. Hence the probability of having $n$ photons existing in the normal mode $\omega$ is: $$P(n)=\frac{e^{-n\hbar\omega\beta}}{Z}$$ where the partition function is: $$Z=\sum_{n=0}^{\infty}e^{-n\hbar\omega\beta}$$

This seems to suggest that the entire normal mode, which I understand to mean "All photons of frequency $\omega$", can be treated as an isolated system in equilibrium with a reservoir at temperature $K_BT=\frac{1}{\beta}$.

How is this possible? If the system as a whole, that is the black-body, is at thermal equilibrium with the reservoir, why does that imply that if we only consider photons emitted by the body at a certain definite frequency then they are a at equilibrium with the reservoir by themselves, irrespective of all other radiation (and energy within the body)?

NOTE: I was uncertain whether this falls more under the "thermodynamics" or "quantum-mechanics" tag. Please feel free to correct my tagging if it is wrong.

Answer 1

Tag "thermodynamics" is appropriate for your question.

Let me start from a direct answer to your question. It is puzzling to realize that the equilibrium of the whole system, which implies a statistical distribution of properties within the system, corresponds to apparently separate equilibria for each degree of freedom, in the cases of separable Hamiltonians. This is a situation not peculiar of black-body radiation. Also a perfect gas is made of 3N independent degrees of freedom and one could wonder how can it be that all the independent degrees of freedom are able to "cooperate" to build up the Maxwell-Boltzmann's distribution of velocities? In a system of non-interacting atoms it could be possible to make all of them starting with the same velocity, i.e. very far from M&B.

Similarly, in the case of black-body radiation one could think a cavity where all the normal modes of the electromagnetic (EM) field correspond to radiation in just one direction and at a just one frequency. What about the Planck's distribution?

The answer is in the often overlooked condition of "thermal equilibrium".

Atoms of an ideal gas, as well as normal modes of the EM field of the cavity, do not interact among them. However, the equilibrium with the container, i.e. interaction of the degrees of freedom of the system with the boundary walls, is enough to allow them to reach equilibrium as a whole. It could be seen as a kind of boundary-mediated interaction which is not explicitly present in the system Hamiltonian, but it is there to ensure the possibility of thermal equilibrium. In typical applications of the ensemble theory, the equilibration process is not described explicitly, but its consequences are encoded in the formulae for the relevant equilibrium partition functions.

As side remark, I would take separate the description in term of EM normal modes and that in term of photons. They are not the same thing, although they are strongly related. Basically, photons are the particle-like interpretation of the quantized normal modes.

Answer 2

Let me summarize previous answer without mention of scary Hamiltonian. Basically The independent oscillation modes are not really completely independent and isolated otherwise nothing would ever change, but get fed at least in a small way from walls and whatever. The details of the leaking mechanism of energy from one mode to another or with walls is irrelevant to the final statistical distribution. At equilibrium each degree of freedom gets its 1/2 kT . Classically. Quantum effects modify things sometimes but that’s a separate issue.