I've tried to derive the pressure, $P$, for a weakly degenerate gas of fermions (and analogously for bosons). The strange thing is that the expression I calculated is correct except for the sign of a term. I checked the calculations and i don't see any error, so the error might be conceptual. Can you explain me why it is conceptually wrong to calculate the pressure as follow?

For weakly degenerate gases we can write, $\Omega=\Omega^{MB}(1 \mp \frac {e^{\beta \mu}} {4 \sqrt 2})$, where the upper sign is for fermions, and the lower one for bosons. MB stands for Maxwell Boltzmann. This equation is derived in my book, so i'm pretty confident that is correct in this context.

Now, $\Omega=-PV$, and since for a system described by Maxwell Boltzmann grand potential we have $PV=K_BNT$, we can write $\Omega^{MB}=-K_bNT$. Substituting in the first one: $\Omega=-K_bNT(1\mp \frac {e^{\beta \mu}} {4 \sqrt 2})$.

Then, $\mu=K_bT ln(N\Lambda^3/V)$, so, substituting in the previous one we get $\Omega=-K_bNT(1 \mp N \Lambda^3/V4 \sqrt2)$.

Finally, since $\Omega=-PV$ we have $P=\frac {K_bNT}{V}(1 \mp \frac {N \Lambda^3}{V4 \sqrt 2})$.

Notice that the correct expression should be $P=\frac {K_bNT}{V}(1 \pm \frac {N \Lambda^3}{V4 \sqrt 2})$


1 Answer 1


Your $\mu$ expression is correct in the classical limit only. There is a correction term which is the same order as the correction you have kept. That is you should substitute for $\Omega^{MB}(T,V,\mu) = -\frac{VT}{\Lambda^3} e^{\beta \mu}$ to get $\Omega = -\frac{VT}{\Lambda^3} e^{\beta\mu}\left [1 \mp 2^{-5/2} e^{\beta \mu} \right ]$.

Then calculate $N$ from the negative $\mu$ derivative. Solving for $e^{\beta \mu}$ as a function of $n$ gives $e^{\beta u} = \Lambda^3 n \left [ 1 \pm 2^{-3/2} n\Lambda^3 +...\right ]$. Substituting into $P(T,\mu)$ and keeping all terms at the same order will give the standard result.

  • $\begingroup$ Nope, because the equation from which i started is already a second order expansion. Introducing a correction term for $\mu$ is equivalent to introduce higher order terms. Indeed my expression for $P$ has just a wrong sign, but all the terms are already present $\endgroup$
    – SimoBartz
    Sep 30, 2022 at 21:48
  • $\begingroup$ As I pointed out in my answer, your original equation for $\Omega$ is correct. Your mistake is your equation for $\mu$. It is not correct at this order. If you worked through my answer, you will see you neglect a term that is twice the term you keep, but with opposite sign. Adding this gives the standard text book result. $\endgroup$
    – user200143
    Oct 1, 2022 at 18:37

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