# Peeter Joot's (OLD) Blog.

• ## Archives

 papasu on PHY450H1S. Relativistic Electr… papasu on Energy term of the Lorentz for… lidiodu on PHY450H1S. Relativistic Electr… lidiodu on PHY450H1S. Relativistic Electr… lidiodu on bivector form of Stokes t…

• 307,992

## Low temperature Fermi gas chemical potential

Posted by peeterjoot on April 24, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

## Question: Low temperature Fermi gas chemical potential

[1] section 8.1 equation (33) provides an implicit function for $\mu \equiv k_{\mathrm{B}} T \ln z$

\begin{aligned}n = \frac{4 \pi g}{3} \left( \frac{2m}{h^2} \right)^{3/2}\mu^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{ (k_{\mathrm{B}} T)^2 }{ \mu^2 } + \cdots \right),\end{aligned} \hspace{\stretch{1}}(1.0.1)

or

\begin{aligned}E_{\mathrm{F}}^{3/2} = \mu^{3/2} \left( 1 + \frac{\pi^2}{8} \frac{ (k_{\mathrm{B}} T)^2 }{ \mu^2 } + \cdots \right).\end{aligned} \hspace{\stretch{1}}(1.0.2)

In class, we assumed that $\mu$ was quadratic in $k_{\mathrm{B}} T$ as a mechanism to invert this non-linear equation. Without making this quadratic assumption find the lowest order, non-constant approximation for $\mu(T)$.

To determine an approximate inversion, let’s start by multiplying eq. 1.0.2 by $\mu^{1/2}/E_{\mathrm{F}}^2$ to non-dimensionalize things

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}} \right)^{1/2} = \left( \frac{\mu}{E_{\mathrm{F}}} \right)^2 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.3)

or

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}} \right)^{1/2} =\frac{1}{{ 1 - \left( \frac{\mu}{E_{\mathrm{F}}} \right)^{3/2} }}\frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2.\end{aligned} \hspace{\stretch{1}}(1.0.4)

If we are looking for an approximation in the neighborhood of $\mu = E_{\mathrm{F}}$, then the LHS factor is approximately one, whereas the fractional difference term is large (with a corresponding requirement for $k_{\mathrm{B}} T/E_{\mathrm{F}}$ to be small. We must then have

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}} \right)^{3/2} \approx 1 - \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.4)

or

\begin{aligned}\mu\approx E_{\mathrm{F}}\left(1 - \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right)^{2/3}\approx E_{\mathrm{F}}\left(1 - \frac{2}{3} \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right).\end{aligned} \hspace{\stretch{1}}(1.0.4)

This gives us the desired result

\begin{aligned}\boxed{\mu \approx E_{\mathrm{F}}\left(1 - \frac{\pi^2}{12} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right).}\end{aligned} \hspace{\stretch{1}}(1.0.7)

# References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.