Deriving the Heat capacity from Fermi-Dirac statistics

Question

I was watching the lectures on Solid state physics by Steve Simon (Oxford). He was explaining how to find Heat capacity of metal due to electrons from Fermi-Dirac statistics. You can write the total number of electrons as $$N =g(E_f) \int_0^\infty \frac{E^\frac{1}{2}}{1+e^{\beta(E-u)}}dE.$$

Here, $g(E_f)$ is the density of states. Now, from this formula, if you know $N$ and temperature, you should be able to figure out $u$ (chemical potential). Then, you can find average energy by averaging the integral above with $E$ multiplied. From there, you can get specific heat by differentiating with respect to temperature. He didn't do it this way as it involves a lot of algebra according to him. I tried solving the integral but couldn't do it. Can anyone solve it or provide some reference?

Answer 1

There is the handwaving argument that the width of the edge is about $kT$ and that increasing the temperature will cause electrons to occupy states that are something like $kT$ higher in energy.

So the electron energy increases with $T^2$. The electronic heat capacity $c_v$ is the derivative, the cause of the linear term in the low-temperature specific heat of metals.

If you want to try calculate it is probably necessary to assume a constant DoS and a value of $kT$ much smaller than $E_F$.