Why is the chemical potential of a photon gas inside a black body exactly zero?
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After several years, I still don't see the reason clearly. In the texts I have read (lecture notes, Wikipedia, Balian vol. 1), they say something about how the photon number is not a conserved quantity, but neither is the amount of one component in a chemical reaction or of one phase in a phase transition and still in these systems, we find a nonzero chemical potential. What happens wrong if I compute the grandcanonical ensemble of ultrarelativistic bosonic ideal gas particles with a generic chemical potential? Alternatively, how can I see that the corresponding derivative of free energy with respect to N is exactly zero if I use the canonical ensemble? I can work around this piece of knowledge using statistical treatment of quantum optics (taking the black body as an ensemble of independent modes, each behaving as an ideal, one-dimensional, quantum mechanical harmonic oscillator) but that's not what they do in the above texts. What's the full form of the semi-classical argument?
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Answer:
There is a discussion of this in the link (which is an article from Am. J. Phys.). Find p428 (about half-way through the document). It might help.
☮ Vašek at Yahoo! Answers Visit the source
Other answers
Vasek, if after "several years" you're unable to see the reason clearly, might it be overoptimistic to find the answer here in the space of a few days? This is a fun problem, but I will definitely need time to review this subject. Edit: I don't deserve the 6 TUs I got on this, since I have not had found that block of time I need to understand this question. This isn't your average school problem. Once I get an adequate understanding of the matter, I will send my opinion to you via email. Thanks for posting this question, Vasek, it's something I want to and will have to think about.
Scythian1950
I think the point is that the free energy is completely independent of the photon number so the derivative (which is the chemical potential) is zero. You could have the same free energy with any number of photons, by just changing the energy of each photon. If the photon energy were fixed this would not be the case, but we are talking about a blackbody distribution so there are photons of all energies.
pegminer
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