# Chemical potential of Bose-Einstein condensation

This is a problem in An Introduction to Thermal Physics.

(a)(b) is easy to work out. In (c), I use the formula $N_{excited}=(\frac{T}{T_c})^{3/2}N, \ (T<T_c)$ to calculate how many atoms are there in the ground state. But I am completely lost about the difference between chemical potential and ground-state energy, because the book always assumes $\mu \approx 0$ when $T<T_c$. I don't think the answer should simply be $\epsilon_0$ but don't know to how to calculate.

I know this question seems to be silly but please provide some hints about it.

1.The ground-state energy of your system is here $\epsilon_\textbf{k}=0$ correponding to the ground-state $|\textbf{k=0}\rangle$, which is macroscopically occupied in a BEC (i.e. $N\sim N_0$).
2.The chemical potential $\mu$ is the increment in energy that you give to your condensate by adding one particle. It is generally defined using free energy $F$ : $$\mu=\left(\frac{\partial F}{\partial N}\right)_{V,\;T}$$
If you want to evaluate $\mu$, considere the ground-state occupation : $$N_0=\frac{1}{e^{-\beta\mu}-1}$$
If you want $N_0$ to be the biggest possible (macroscopic occupation), you must fulfill the condition $e^{-\beta\mu}\rightarrow 1$, in other words $\mu\rightarrow 0$.
For larger resolution on $\mu$, one can do a Taylor expansion on the expression of $N_0$ : $$N_0\underset{\mu\rightarrow 0}{\sim}-\frac{1}{\beta\mu}$$ Then, $$\mu\simeq -\frac{1}{N_0\beta}=-\frac{k_B T}{N_0}=-\frac{k_B T}{N-N_{exc}}$$