# Chemical potential vs pressure and temperature; difficulty with Fermi gases

 Sci Advisor P: 3,595 No, you need the expansion of U or P in T in the same order as that for mu. Come on, it's not so difficult...
 Sci Advisor P: 3,595 $E_F \approx C_1 P^{2/3}(1-\frac{\pi^2 k^2T^2}{9 (C_1P^{2/3})^2})$ Sufficient to render the dependence of mu on T negative.
P: 221
 Quote by DrDu $E_F \approx C_1 P^{2/3}(1-\frac{\pi^2 k^2T^2}{9 (C_1P^{2/3})^2})$ Sufficient to render the dependence of mu on T negative.

could you show how you derived this?

-James
P: 3,595
 Quote by jfizzix could you show how you derived this? -James

$Lp=-\Omega=\frac{2}{\beta}\int_0^\infty d\omega D(\omega) \ln (1 +\exp(-\beta(\omega-\mu))$
Here, $\Omega$ is the grand potential, D is the density of states $L\frac{\sqrt{2m}}{\pi \hbar}\omega^{-1/2}=AL \omega^{-1/2}$ with $A=\alpha_1^{-1/2}/2$.
Partial integration:
$p=-\frac{2A}{\beta} \int_0^\infty d\omega \left(\int_0^\omega d\omega' \omega'^{-1/2} \right) \frac{-\beta}{\exp(\beta(\omega-\mu))+1}$
Sommerfeld expansion...
$p\approx A\left[ \frac{3}{2}E_F^{3/2}+\frac{\pi^2}{6\beta^2}E_F^{-1/2}\right]$
Solve for E_F at same level of accuracy.

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