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.
Best Answer
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.