[Physics] Could the chemical potential of a Bose gas be zero

Question

Could the chemical potential of a Bose gas be zero ?

If it was the case, we will have an infinite number of particles in the ground state ! No ?

But I've heard that for $T < T_c$, $\mu = 0$, so I don't understand…

Answer 1

Throughout, let's assume that the ground state energy of the system under consideration is zero.

Chay Paterson has addressed your question in the case of a gas of bosons in which the number of particles is not conserved, but from the wording of your question, it seems that you're concerned about the case in which the total number of particles is fixed.

For a system of bosons with a fixed number $N$ of particles, the answer to your question is

No. The chemical potential is nonzero for all $T>0$.

You point out, however, that it is often said that below the critical temperature $T_c$, the chemical potential is zero, so what's going on? The resolution is essentially that

the chemical potential very well-approximated by zero for almost all temperatures below the critical temperature, but it is never exactly zero.

As you cool the system down from above the critical temperature, the number $N_e$ of particles in excited states get's lower and lower. On the other hand, the number of particles in the excited states at any given temperature is bounded above, and this bound decreases as a function of temperature like $T^{3/2}$ (for non-relativistic systems in three dimensions). At a particular sufficiently low temperature (the critical temperature), this upper bound is lower than the total number of particles in the system and particles are forced into the ground state. But at this point, the chemical potential does not drop exactly to zero. It does, however, get very small very quickly as the temperature decreases and the number of particles in the ground state increases.

In fact, If we look at the number of particles in the ground state as a function of temperature $$ N_0 = \frac{1}{e^{-\mu(T)/kT}-1} $$ which gives $$ \mu(T) = -kT\ln\left(1+\frac{1}{N_0}\right) $$ then we see that the chemical potential is always strictly less than zero because the argument of the log is always strictly greater than $1$. But as you dial the temperature down below the critical temperature, and as the number of particles in the ground state increases towards $N$, the total number of particles in the system, which is presumably very large, and the chemical potential decreases since the argument of the log approaches $1$.