chemical potential vs pressure and temperature; difficulty with Fermi gasesby jfizzix Tags: chemical, difficulty, fermi, gases, potential, pressure, temperature 

#19
May1812, 10:32 AM

Sci Advisor
P: 3,375

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... 



#20
May2112, 05:42 AM

Sci Advisor
P: 3,375

[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. 



#21
May2112, 07:18 AM

P: 219

could you show how you derived this? James 



#22
May2112, 08:10 AM

Sci Advisor
P: 3,375

[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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