The derivation of the Planck distribution

Question

I am trying to understand the derivation of the Planck distribution and black body radiation. In the Wikipedia derivation of the Planck distribution, the photons confined within a cubic box, are emitting from and absorbed by, and are in equilibrium with the wall of the cube. I understand the calculation presented. However, I am uncertain about the following points.

1. Is the temperature here that of the photons alone, of the matter of the wall alone or the ensemble of the photon and the matter? Most likely it is the last case. How is the temperature defined and the Boltzmann distribution derived with the photons under consideration? It is not mentioned at all in the Wikipedia derivation.

2. I suppose Equation (1) in the aforementioned derivation for photon gas in a box, i.e. $$E_{n_1,n_2,n_3}\left(r\right)=\left(r+\frac{1}{2}\right)\frac{hc}{2L}\sqrt{n_1^2 + n_2^2 + n_3^2} \tag1$$ comes from solving a wave equation with zero boundary condition. I suppose this wave equation comes from the quantum field theory, describing the photons. Is this correct? In classic electrodynamics, Maxwell's equation has a zero boundary condition if the wall is a perfect conductor with zero electric or magnetic field in the interior of the wall so as to perfectly reflect the electromagnetic wave. Are we to impose the same condition here with the purpose to confine the energy of the photo inside of the box?

Edit: The box indeed has perfectly conducting wall whereby the parallel component of the electric field at the boundary indeed vanishes.

3. Apparently the size and geometry of the box affect the final distribution. I suppose if we construct an object with many small walled cavities with fractal-like geometry, we will get a different power distribution. Is this correct?

Edit: It turns out point 3. is a complicated question. The leading term of the eigenvalue distribution is proportional to volume, with some caveat on the geometric roughness of the boundary, according to Weyl's law. The proof concerning the geometric roughness of the boundary is complicated.

Answer 1

1. In order to obtain the probability distribution, statistical mechanics considers that the system has reached thermal equilibrium (see statistical ensemble). Hence the photons inside the box (subsystem) and the walls (reservoir) must have the same temperature. Photons do not behave as a Boltzmann distribution, they behave as a Bose-Einstein distribution with zero chemical potential.

2. I don't know QFT, but eq. (1) of Wikipedia can be obtained from QM's problem of a particle in a cubic box with homogeneous boundary conditions.

3. The particle distribution function is independent of the box size. The spectral energy density also is independent of the box characteristics.

Answer 2

In my answer to the physics stack question Discrete Cavity Radiation I answer your questions 2 and 3. The summary is:

2. In classical EM a cavity with infinite conductivity walls only supports modes of certain frequencies because $E_{Transverse}=0$ and $B_{Perpendicular}=0$ at infinite conductivity walls. Real walls have some resistivity and thus can be penetrated by standing waves that don't exactly fit in the box. These non-resonant modes have very low Q because their energy is quickly dissipated by the resistivity of the walls. Thus any non-resonant frequency wave train emitted into the box (say by black body oscillators in the walls) will exist as a standing wave for a few oscillations (low Q).

3. The Planck spectrum in a box of all one temperature is universal. There is no modification of it due to wall emissivity. There are no peaks in the spectrum due to the resonant modes of the box. The universality of this spectrum was argued by Kirchoff in ~1860 using the second law of thermodynamics, though the exact form of the Planck function was not discovered till 50 years later. Yes, this seems weird since the density of resonant states in the cavity was used in the derivation of the Planck distribution!

If you doubt this, consider the Johnson noise from a resistor. The noise from the resistor can be considered as the resistor being an antenna picking up the Planck radiaion from whatever box it is in. Now imagine making the box very small about the resistor (eg: a standard 50 ohm terminator). For frequencies below the lowest resonant frequency of this box, we would have eliminated the low frequency Johnson noise if the Planck radiation was only in the box's resonant peaks. Experimentally, Johnson noise is white (flat with frequency in the Rayleigh-Jeans region of the Planck spectrum).