Why entropy in ideal BEC is independent of density I am confused about one statement from the book Bose-Einstein Condensation in Dilute Gases.
In the chapter discussing superfluid, the auther claims:
The pressure and the entropy density $\frac{S}{V}$ of ideal Bose-Einstein condensed gas depend on temperature but not on density.
Here, ideal means no interaction, homogeneous external field or no external field, and infinite volume. 
This statement is perhaps easily obtained from the quantum partition function.
For classical particles, assuming the number of micro states for one particle is $a$, then for $n$ non-interacting particles, the number of microstates would be $a^n$. Therefore the entropy is dependent on the total particle number. Given that the volume is fixed, the entropy is dependent on density.
For the BEC case, it's different in the fact that symmetry and indistinguishability must be considered. And these two effects will effectively reduce the number of allowed microstates. 
However, I cannot obtain the above statement directly without writing down the partition function and hence no clearer picture or intuition. Could anyone help to better understand this?
 A: For an homogeneous ideal gas in 3D below $T_c$ we get 
\begin{align}
p &= \zeta(5/2) \left(
\frac{m}{2\pi \hbar^2}\right)^{3/2} \;
(k_B T)^{5/2}\propto T^{5/2}
\\
\frac{S}{N} &= k_B \frac{5}{2} 
\frac{\zeta(5/2)}{\zeta(3/2)}
\left(\frac{T}{T_c}\right)^{3/2} 
\propto \left(\frac{T}{T_c}\right)^{3/2} 
\propto \left(\frac{T}{n^{2/3}}\right)^{3/2} 
= \frac{T^{3/2} }{n} = \frac{T^{3/2} }{N} V 
\end{align}
where $\zeta(\alpha) = \sum_{j=1}^{\infty} j^{-\alpha}$ is the Riemann zeta function. Hence,  below $T_c$ the entropy does not depend on the total number of particles. The reason for this is that only particles in the excited states contribute to $S$.  Consequently, the number of particles in the macroscopically occupied state is irrelevant.
Why do the particles in the ground state do not contribute to the entropy?
The energy of the ground state is arbitrary, and we usually choose $E_0=0$ -- the energy in the ground state is zero. Now, the specific heat is given by $C = \frac{\partial E}{\partial T}$ which yields
$$
C
\propto \frac{E}{T} 
\propto T^\alpha
$$
where we used $E \propto T^{\alpha +1}$. Since we also know $C = T \frac{\partial S}{\partial T}$ we obtain
$$
S = \int \frac{C}{T} dT 
\propto T^\alpha
\propto \frac{E}{T} 
$$
Therefore, a particle in the ground state does not contribute to the entropy. 
