I am trying to understand 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 not certain 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) 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, the 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?

  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.

  • $\begingroup$ I think the most surprising part here is 1. Indeed we treat the electromagnetic field as if it was a gas of particles interacting with a bath at temperature $T$ (this is the usual canonical situation). It’s actually not clear how this equilibrium should be reached from first principles but it works! $\endgroup$ – lcv Dec 21 '18 at 8:01
  • $\begingroup$ To add to part 1. Whenever we say that a system is at temperature $T$ we mean that it is at equilibrium with a large bath of temperature $T$. Then you can of course say that the system itself is at temperature $T$,I.e., the photons in this case. $\endgroup$ – lcv Dec 21 '18 at 8:04
  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.

  • $\begingroup$ 1) You are right. Thank you. 2) You are wrong. The link you provided is for a positive mass particle. A photon is massless, thus does not obey the Schrodinger equation. 3) You are wrong. Look at Equation (1). It has explicitly $L$ the length of the box in it, and the form of the square root of the sum is implicitly obtained by the boundary being rectangular. The equation above Equation (3) is an approximation when $n$ is large and when $n$ assumes the square root sum form which is constrained by the geometry. $\endgroup$ – Hans Jul 1 '17 at 21:29
  • $\begingroup$ 2) Ok the problem is not exactly the same, but can be done in a similar way remembering that the momentum of a photon is $p=h\nu/c$. 3) What you usually mean by distribution is "particle number distribution", which is explicitly independent of $L$... but it is true it may depend of it via the energy per particle $\epsilon$. Of course the energy function will have a different form if you consider a different geometry of the system, e.g. photons in cylinder, sphere, but the thermodynamic limit, $N\rightarrow\infty$, $L\rightarrow\infty$ will always get rid of any dependency on the geometry $\endgroup$ – Saavestro Jul 1 '17 at 21:35
  • $\begingroup$ 2) What is the "similar way"? Please do not give me a hand waving argument, which I can manufacture myself easily. What is the equation, as it is not Schrodinger? It must be QFT then? 3) I understand it is for $N\rightarrow\infty$. The $L$ is divided away in the unit energy form. However, it is under the assumption that the cube is not changing as $N\rightarrow\infty$ and it depends on the square root of sum form of $n$ in the article. A different geometry may not give the same result. Will you be able to prove the same thing for, say, a fractal geometry? $\endgroup$ – Hans Jul 1 '17 at 21:58
  • $\begingroup$ 2) I think you are right, the rigorous way to do it is QFT 3) If you calculate the number of particles $N$ you would see that the energy has dependency $N/L$ which is constant in the thermodynamic limit. I don't know what is a fractal geometry, but the point is that you have photons inside some cavity with an arbitrary shape, and such shape doesn't really accounts when summing states in the thermodynamic limit. I recommend reading the notes of D. Tong about quantum gases damtp.cam.ac.uk/user/tong/statphys/three.pdf $\endgroup$ – Saavestro Jul 1 '17 at 22:21
  • $\begingroup$ 3) The note you link to has the exact same expression for $n$ and is premised on it being a cube --- even a rectangular box will make the expression different, not to mention a sphere which requires spherical harmonics or a cylinder which requires Bessel functions. By what do you conclude "photons inside some cavity with an arbitrary shape, and such shape doesn't really accounts when summing states in the thermodynamic limit"? That is a pretty big statement. Again, I have no use for hand waving arguments which come a dime a dozen, well actually come for free. Please derive it rigorously. $\endgroup$ – Hans Jul 1 '17 at 22:41

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