chemical potential vs pressure and temperature; difficulty with Fermi gases

So.. in 1D,

The fermi energy $E_{f}$ can be expressed in terms of the Fermi pressure $P_{f}$ using the expression for the internal energy $U$ at absolute zero. If we differentiate the internal energy $U$ with respect to the length of the 1D box $L$ and add a negative sign, we have an expression for the pressure.

$P = -(\frac{\partial U}{\partial L})_{\sigma,N}$

In 1D at absolute zero...

$U = \frac{1}{3} N E_{f}$ where $E_{f} = \frac{\hbar^{2}}{8 m}(\frac{N \pi}{L})^{2}$

From these expressions, we can express the Fermi energy in terms of the pressure at absolute zero.

$E_{f} = (\frac{3 \pi}{2})^{\frac{2}{3}}(\frac{\hbar^{2}}{8 m})^{\frac{1}{3}} P^{\frac{2}{3}} = C_{1} P^{\frac{2}{3}}$

Then the expression for the chemical potential becomes

$\mu ≈ C_{1} P^{\frac{2}{3}} + \frac{\pi^{2}}{12} \frac{\tau^{2}}{C_{1} P^{\frac{2}{3}}}$

With this as our form for the chemical potential, we see that it does decrease with temperature at constant pressure and particle number.

I'm pretty sure I've got it right this time, but feel free to correct me. Does this make sense?

-James

 Recognitions: Science Advisor 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...
 Recognitions: Science Advisor $E_F \approx C_1 P^{2/3}(1-\frac{\pi^2 k^2T^2}{9 (C_1P^{2/3})^2})$ Sufficient to render the dependence of mu on T negative.

 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

Recognitions:
 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.