The absolutely crucial difference between (i) a gas of 4 He atoms or, eg., massive mesons and (ii) a gas of photons is that, while all of these are bosons, the photons are massless (they have no rest mass, and their dispersion relation is ω = ck (with energy E = hω/ 2π). This leads to a really important result-we know from thermodynamics that or that or that ie., no matter which variables we hold constant, µ measures the energy to add a particle But a key fact about the photon gas is that in equilibrium with, eg., a set of atoms, the photon number is arbitrary; one can change the number of photons without changing the total energy. Thus at equilibrium (where one of these thermodynamic potential is minimized), we have µ = 0, and N is completely undetermined. This is true even if for a set of photons inside a very large box: the box itself is made of matter, & so photons can be created or destroyed at the walls. To decouple photons from matter we need a perfect vacuum (ie., intergalactic space). The photon number is then conserved-it can only change via photon-photon interactions, which require creation of a very high-energy e + e-pair, with exponentially small probability (of order exp[-2mc 2 /kT]). Only then we can have non-zero µ
CITATION STYLE
Maciel, W. J. (2016). The Photon Gas. In Introduction to Stellar Structure (pp. 57–66). Springer International Publishing. https://doi.org/10.1007/978-3-319-16142-6_4
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