Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 33C (Statistical Physics)

The Bose-Einstein distribution gives $n_i=[\exp ((E_i-\mu )/(k_{B}T))-1{]}^{-1}$, with chemical potential below the lowest energy when its discrete occupation is finite. In the continuum approximation,If the zero-energy ground state is macroscopically occupied, its discrete occupation $N_0$ must be added to the first integral; it contributes no energy to the second. This distinction is essential when applying the formulas to Bose-Einstein condensation.

The photon number in thermal blackbody radiation is not fixed, so $\mu =0$. With $x=E/(k_{B}T)$,The integral is a temperature-independent finite constant (its explicit value follows from the gamma-zeta integral). Differentiating at fixed volume gives

For fixed particle number, Bose-Einstein condensation means that a macroscopic fraction occupies the single-particle ground state. As $\mu \uparrow 0$, the excited-state population reaches its maximumAt the transition this equals $N$, soBelow $T_c$, the chemical potential is approximately zero in the large-$N$ continuum description. Additional particles occupy the ground state, givingFor finite $N$, the actual chemical potential remains slightly negative; the saturation statement is the macroscopic approximation used here.

Ordinary equilibrium blackbody photons do not undergo this fixed-number condensation. Their number adjusts through absorption and emission, while their chemical potential stays zero; cooling reduces the equilibrium photon number rather than forcing an excess into the ground state. This answer concerns the thermal blackbody model, not specially constrained systems where an effective photon number is maintained.