# A Problem with Bose-Einstein Condensation

1. Jan 6, 2017

### ShayanJ

In section 7.1 of his statistical mechanics, Pathria derives the formula $N_e=V\frac{(2\pi m k T)^{\frac 3 2}}{h^3}g_{\frac 3 2}(z)$ where $\displaystyle g_{\frac 3 2}(z)=\sum_{l=1}^\infty \frac{z^l}{l^{\frac 3 2}}$ and $z=e^{\frac \mu {kT}}$. This formula gives the number of particles that are not in the ground state w.r.t. the temperature.
The maximum of $g_{\frac 3 2}(z)$ happens at $z=1$ and is equal to $\zeta(\frac 3 2)$. So whenever $z=1$, $N_e$ reaches its maximum and any other particle has to go to the ground state and $z=1$ happens at any $T<T_c$.
My problem is with $T>T_c$. I can calculate $N_e$ for any temperature which gives me the capacity of the excited states at the given temperature. Now I put $N>N_e$ particles in the energy levels and so the excited states become full and the rest of the particles have to go to the ground state and I get Bose-Einstein condensation again, this time for $T>T_c$ which can't be right because we're supposed to have condensation only for $T<T_c$.
What's wrong here?
Thanks

2. Jan 7, 2017

### Fightfish

The critical temperature is dependent on the particle density, so if you keep the volume constant and add more particles, $T_{c}$ is going to change as well.

3. Jan 7, 2017

### ShayanJ

Yeah, good point. Thanks.