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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.

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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.

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All the details and numerical factors can be found in books like Pethick & Smith, and Pitaevskii & Stringari.

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