4
$\begingroup$

If have been thought, that Bose Einstein condensation occurs of the ground-state is occupied macroscopically, so $n_0\in \mathcal{O}(N)$ when performing the thermodynamic limit.

So naively, this seems odd as of course, if I put in more particles without increasing system size this will always be fulfilled (when keeping the temperature constant). The problem with my imagination is now, that of course when performing the thermodynamic limit, I have to change the systems volume while I put in new particles and thus "create" more low energy states.

How do I see, that above the critical temperature these effects cancel, so that due to the additional states $n_0$ does not scale linearly with $N$?

$\endgroup$
1
$\begingroup$

The short answer is that you have to work with particle densities, namely, $$n=N/V,$$ where $N$ is the number of particles, and $V$ is the volume of your system.

The long answer is as follows.

Working in the thermodynamic ensemble with a fixed chemical potential $\mu$, and knowing that the Bose-Einstein distribution gives the average number density of non-condensed particles $$n_{nc} = \frac{1}{V} \sum_{\boldsymbol k\neq 0}\frac{1}{\exp(\beta [\hbar^2 \boldsymbol k^2/2m - \mu])-1},$$ where $\beta = 1/k_B T$ is the inverse temperature, $k_B$ is the Boltzmann coefficient, $m$ is the mass of the particles, and $\hbar\boldsymbol k$ is the momentum. Fixing the temperature $T$ and the number of excited particles $n_{nc}$ allows one to find the chemical potential for the situation at hand. Note that $\mu = - \infty$ gives a very small particle density, which increases with increasing $\mu$.

However, requiring a low enough temperature in combination with a high enough density, gives a problem. Namely, satisfying the equation means that the chemical potential has to be positive. This means that some low-momentum mode is occupied negatively, as $$\frac{1}{\exp(-\beta\mu)-1}\simeq \frac{1}{-\beta\mu},$$ for small $\beta\mu$. We thus conclude that positive $\mu$'s are not allowed.

This paradox is resolved by realizing that the correct formula for the total number of particles is actually $$n = n_0 + n_{nc}.$$ Therefore, if after fixing the total number of particles $n$ and the temperature $T$ one sees that $n_{nc}$ with $\mu=0$ is still less than $n$, one has to conclude that the rest of the particles reside in the condensate, and hence $n_0\neq 0$.

All this is standard textbook knowledge. It is elaborated on, for example, in the book Bose–Einstein Condensation in Dilute Gases by Pethick and Smith.

$\endgroup$

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.