## Free electron model: Electrical contribution to the thermal conductivity

The dispersion relation in the free electron model is,

$$E(\vec{k})= \frac{\hbar^2 k^2}{2m^*}.$$

Here $m^*$ is the effective mass. For an isotropic system, the electrical contribution to the thermal conductivity is,

$$K=\frac{1}{4\pi^3\hbar^2T}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu}\left( \left( E(\vec{k}) - \mu \right) \nabla_{\vec{k}}E(\vec{k})\cdot\hat{z}\right)^2 d^3k.$$

Assuming a single relaxation time,

$$K=\frac{\tau\hbar^2 k_B^2T}{4\pi^3 m^{*2}}\int \frac{\partial f_0}{\partial \mu}\left( \frac{E(\vec{k}) - \mu}{k_BT} \right)^2 k_z^2 d^3k.$$

The differential volume is,

$$d^3k =k^2\sin\theta dk d\theta d\varphi \qquad k_z = k\cos\theta$$ $$K=\frac{\tau\hbar^2 k_B^2T}{4\pi^3 m^{*2}}\int \frac{\partial f_0}{\partial \mu}\left( \frac{E(\vec{k}) - \mu}{k_BT} \right)^2 k^4\cos^2\theta\sin\theta dk d\theta d\varphi.$$

The integral over $\varphi$ contributes a factor of $2\pi$.

$$K=\frac{\tau\hbar^2 k_B^2T}{2\pi^2 m^{*2}}\int \frac{\partial f_0}{\partial \mu}\left( \frac{E(\vec{k}) - \mu}{k_BT} \right)^2 k^4\cos^2\theta\sin\theta dk d\theta.$$

The integral over $\theta$ contributes a factor of $2/3$.

$$K=\frac{\tau\hbar^2 k_B^2T}{3\pi^2 m^{*2}}\int \frac{\partial f_0}{\partial \mu}\left( \frac{E(\vec{k}) - \mu}{k_BT} \right)^2 k^4 dk.$$

The derivative of the Fermi function is,

$$\frac{\partial f_0}{\partial \mu}= \frac{\exp(x)}{k_BT\left(\exp(x)+1\right)^2}$$

where

$$x= \frac{E-\mu}{k_BT}=\frac{\frac{\hbar^2k^2}{2m^*}-\mu}{k_BT}.$$

Differentiating to find $dk$

$$dk= \frac{m^*k_BT}{\hbar^2k}dx$$

The thermal conductivity can be written as,

$$K=\frac{\tau k_B^2T}{3\pi^2m^*}\int \frac{x^2\exp(x)}{\left(\exp(x)+1\right)^2} k^3 dx.$$

At low temperatures, the derivative of the Fermi function is sharply peaked around the Fermi energy so $k$ is approximately $k_F$ and it can be pulled out the the integral. The remaining integral over $x$ evaluates to $\frac{\pi^2}{3}$.

$$K=\frac{\tau k_B^2Tk_F^3}{9m^*}.$$

$$n =\frac{k_F^3}{3\pi^2}.$$
$$K=\frac{\pi^2\tau n k_B^2T}{3m^*}.$$
The electron contribution to the thermal conductivity is often linear at low temperatures but decreases at higher temperatures due to the increase of electron-phonon scattering. This increase simply comes from there being more phonons at high temperatures and a consequence is that the relaxation time $\tau$ gets shorter at higher temperatures.