Bose condensate in 4d Could a boson gas condensate in a hypervolume $V$ in 4D? How can I find its critical temperature and the heat capacity?
In the books it just said volume $V$, it does not specify the dimension.
My professor asked this and I have no ideia how to start this question. Please help me. It is a statistical mechanics subject at the university.
 A: Yes.
Th key thing here is that for non-interacting bosons the mean occupancy of each (single-particle) state $j$ is given by:
$$ f(E_j) = \frac{1}{e^{(E_j - \mu)/kT}-1} .$$
Now you see that the ground state $E_0$, the occupancy is infinity. This is because for the $E_0$ state the chemical potential $\mu$ also needs to be zero, in order to guarantee $f$ to still be positive. Physically, the chemical potential is defined as $\partial U/\partial N$, i.e. the energy added when you add one particle to the system. But if you add it to the $E=0$ state, then the extra energy is 0...
Bose-Einstein condensation begins when you saturate the excited states and start macroscopically occupying the ground state, which has infinite occupancy.
Below $T_c$, $f(E_0)$ starts blowing up so it does not make sense using the above distribution anymore, since the atoms start amassing into the ground state. 
So $T_c$ is extracted from when your total $N$ is equal to the number of atoms in the excited states, $N_{ex} = \int_0 ^{\infty}dE \, g(E) \, f(E)$ where $g(E)$ is the density of states, i.e. the number of states in a given interval $[E, E+dE]$. The sum should have been over the states, but I changed it to the energy $E$ just by introducing this density of states term.  
The density of states $g(E)$ scales with the number of dimensions $d$. For a free $d$ dimensional system it goes as $g(E) \propto E^{d/2  -1}$, while for $d$ dimensional harmonic potential it scales as $g(E) \propto E^{d-1}$.
In general you can write:
$$ g(E) \propto E^{\alpha  -1},$$
with $\alpha$ being the number of degrees of freedom in the system divided by 2. For free particles in $d$ dimensions, $\alpha = d/2$, and for a $d$ dimensional harmonic potential, the degrees of freedom are $2d$ ($d$ translations and $d$ oscillations) so $\alpha = d$. All agree with the above.
The integral above can be rewritten as:
$$N_{ex} = \int_0 ^{\infty}dE \, g(E) \, f(E) \propto  (k T_c)^{\alpha} \int_0 ^{\infty} dx\frac{x^{\alpha-1}}{e^x - 1} $$
where I defined $x$ as $E/k T_c$.
The intregral
$$ \int_0 ^{\infty} dx \frac{x^{\alpha-1}}{e^x - 1} = \Gamma(\alpha) \zeta(\alpha),  \qquad \alpha > 1$$
with $\Gamma$ being the gamma function, $\zeta$ being the Riemann zeta function.
Which gives you:
$$ k T_c \propto \frac{1}{[\Gamma(\alpha) \zeta(\alpha)]^{1/\alpha}}. $$
To have a BEC transiton, you want $T_c \neq 0$, i.e. a non-trivial solution.
In free space, $d=2,3,4$ have $\alpha = 1, 3/2, 2$:
$$
\begin{array}{ccc}
\alpha & \Gamma(\alpha) & \zeta(\alpha) \\
\hline
1/2 & \text{integral does not converge}      \\
1 & 1    & \infty      \\
3/2 & \sqrt{\pi}/2  & 2.612      \\
2 & 1    & \pi^2/6      \\
\dots & \dots & \dots
\end{array} $$
So in a free system with $d = 1,2$ the only solution is $T_c = 0$, but for $d>2$, $T_c$ is finite. 
--
All the details and numerical factors can be found in books like Pethick & Smith, and Pitaevskii & Stringari.
