# Deriving the Heat capacity from Fermi-Dirac statistics

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?

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 $$E_F$$.
• @RishabhJain For an analytical expression you would first get rid of the $\sqrt{E}$ term of the free-electron DoS and assume that temperature does not affect the chemical potential $\mu = E_F \gg kT$. This must have been done for determining the effective electron mass from the linear term in $c_v$. – Pieter Sep 21 '19 at 17:51