In Wikipedia's article on Fermi Gases, they have the following equation for the chemical potential:

$$\mu = E_0 + E_F \left[ 1- \frac{\pi ^2}{12} \left(\frac{kT}{E_F}\right) ^2 - \frac{\pi^4}{80} \left(\frac{kT}{E_F}\right)^4 + \cdots \right]$$ where $E_0$ is the potential energy per particle, $k$ is the Boltzmann constant and $T$ is the temperature.

I don't understand how they get the third term, particularly the 1/80 factor. I've often seen this equation expressed just to the tau^2 term, and I understand how to get the 1/12 factor from a binomial expansion of $$\left( 1 + \frac{\pi^{2}}{8} \left(\frac{\tau}{E}\right)^{2}\right)^{-2/3} $$

However, I've tried continuing the binomial expansion and cannot figure out why the factor is 1/80.

  • $\begingroup$ Sorry about the erroneous edit - our bad. (+1 btw :) $\endgroup$
    – user10851
    Jul 1, 2013 at 6:37
  • $\begingroup$ No worries; I should've written it in Latek to start. Unfortunately I had somewhere I needed to be. $\endgroup$
    – MattS
    Jul 1, 2013 at 6:40

1 Answer 1


I not sure how you obtained the last expression. The standard Sommerfeld expansion (for details, see e.g. Ashcroft & Mermin) gives a slightly different result, which is $$ E_{F} \approx \mu\left[1+\frac{\pi^{2}}{8}\left(\frac{k_{B}T}{\mu}\right)^{2}\,\right]^{2/3} \approx \mu\left[1+\frac{\pi^{2}}{12}\left(\frac{k_{B}T}{\mu}\right)^{2}\right] $$ to leading non-trivial order in $k_{B}T/\mu$, i.e., $O\big((k_{B}T/\mu)^{2}\big)$. (I have set $E_{0}=0$ in the expression from Wikipedia.)

We can invert this relation by substituting $\mu = E_{F}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + \cdots \right]$ into the above. That is,

$$ E_{F} = E_{F}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + \cdots \right]\left\{1+\frac{\pi^{2}}{12}\left(\frac{k_{B}T}{E_{F}}\right)^{2}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + \cdots\right]^{-2}\right\}. $$

Comparing the zeroth order terms in $k_{B}T/E_{F}$ on both sides of the above equation, we simply obtain $E_{F}=E_{F}$. Comparing the second order terms, we have $0 = \frac{\pi^{2}}{12} + c_{2}$. Hence

$$ \mu = E_{F}\left[1-\frac{\pi^{2}}{12}\left(\frac{k_{B}T}{E_{F}}\right)^{2}\right] $$ up to $O\Big((k_{B}T/E_{F})^{2}\Big)$.

To determine the next-order correction to $\mu$, you should include one higher order term in the Sommerfeld expansion, which gives

$$ E_{F} \approx \mu\left[1+\frac{\pi^{2}}{8}\left(\frac{k_{B}T}{\mu}\right)^{2}+\frac{7\pi^{4}}{640}\left(\frac{k_{B}T}{\mu}\right)^{4}\,\right]^{2/3}. $$

Expanding this up to $O\big((k_{B}T/\mu)^{4}\big)$, then substituting $\mu = E_{F}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + c_{4} (k_{B}T/E_{F})^{4}+\cdots\right]$ (where we have already determined $c_{2}$ before) into it, and then matching the coefficients of both sides up to $O\Big((k_{B}T/E_{F})^{4}\Big)$ will lead to the desired result.

  • $\begingroup$ Where does $\mu = E_{F}\left[1 + c_{2} (k_{B}T/E_{F})^{2}\right]$ come from? $\endgroup$
    – MattS
    Jul 1, 2013 at 9:02
  • $\begingroup$ I'm expanding $\mu/E_{F}$ in powers of a small quantity $k_{B}T/E_{F}$, and we know that it is equal to one at zeroth order because $E_{F}=\mu$ at $T=0$. $\endgroup$
    – higgsss
    Jul 1, 2013 at 9:11
  • $\begingroup$ I'm sorry, I'm still not quite following. I'm trying to repeat your analysis and get to $\mu = E_{F} \left[1 - \frac{\pi^{2}}{12}\left(\frac{k_{B}T}{E_{F}}\right)^{2}\right]$, but I'm not getting it. $\endgroup$
    – MattS
    Jul 1, 2013 at 9:13
  • $\begingroup$ Substitute $\mu = E_{F}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + \cdots \right]$ into $E_{F} = \mu \left[1 + \frac{\pi^{2}}{12} \left(\frac{k_{B}T}{\mu}\right)^{2} \right]$ and match the coefficient of both sides up to $O\Big((k_{B}T/E_{F})^{2}\Big)$. $\endgroup$
    – higgsss
    Jul 1, 2013 at 9:23
  • $\begingroup$ I tried that, and I didn't achieve anything. What is $O\big((k_{B}T/\mu)^{2}\big)$? $\endgroup$
    – MattS
    Jul 1, 2013 at 9:27

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