# 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.

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.