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.



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.