Recently I came across (or well, derived in a lecture) the isothermal compressibility for an ideal boson gas. This was done in the context of statistical physics, using the quantum version of the grand canonical ensemble. There, one finds that in the high temperature, low density limit ($\lambda << \frac{N}{V}$ and $e^{\beta \mu} << 1$) the isothermal compressibility $\kappa_T$ is given by

$\kappa_T = \frac{V}{Nk_BT}\frac{1}{1-\frac{\lambda^3N}{2^{3/2}V}}$

up to a first order quantum correction. Here, $\lambda$ is the thermal de Broglie wavelength.

So for high temperatures, this is approximately just the isothermal compressibility of a classical ideal gas, as $\lambda \propto T^{-1/2}$. But, especially for lower temperatures but also for higher temperatures, it is bigger than what you would find clasically. I'm trying to figure out why this is the case. The naive explanation I can come up with is that a part of the boson gas will already be a bose-einstein condensate, where the compressibility is infinite. So when averaging over the entire ensamble, you end up with a slightly higher compressibility than clasically. The reason I'm not sure is because the way I understand it is that the condensation doesn't happen until $T_c$, the critical temperature. But I suppose it does make some sense that somewhere in the ensamble there is a part that is at this temperature.

Another, probably wrong, explanation I can think of is coming from the other side, by looking at $\kappa_T$ for the ideal fermion gas. Here you find that the compressibility is actually lower than in the classical case, which is easy to understand from the Pauli exclusion principle. Could it then be that the classical case is sort of an average between bosons and fermions? I don't like that explanation personally, but I'm not sure about the validity of the first one either.

I left out a large part of the derivation (actually, all) of $\kappa_T$ of course, so if you feel as if more is required to analyse the situation I can add that in.


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