Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

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)


\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)


\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)


\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)


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


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: