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Chemical potential vs pressure and temperature; difficulty with Fermi gases

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DrDu
#19
May18-12, 10:32 AM
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P: 3,593
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...
DrDu
#20
May21-12, 05:42 AM
Sci Advisor
P: 3,593
[itex] E_F \approx C_1 P^{2/3}(1-\frac{\pi^2 k^2T^2}{9 (C_1P^{2/3})^2})[/itex]
Sufficient to render the dependence of mu on T negative.
jfizzix
#21
May21-12, 07:18 AM
P: 221
Quote Quote by DrDu View Post
[itex] E_F \approx C_1 P^{2/3}(1-\frac{\pi^2 k^2T^2}{9 (C_1P^{2/3})^2})[/itex]
Sufficient to render the dependence of mu on T negative.

could you show how you derived this?

-James
DrDu
#22
May21-12, 08:10 AM
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P: 3,593
Quote Quote by jfizzix View Post
could you show how you derived this?

-James

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


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