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

The dispersion relation in the free electron model is,

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

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

\begin{equation} 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. \end{equation}

Assuming a single relaxation time,

\begin{equation} 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. \end{equation}

The differential volume is,

\begin{equation} d^3k =k^2\sin\theta dk d\theta d\varphi \qquad k_z = k\cos\theta \end{equation} \begin{equation} 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. \end{equation}

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

\begin{equation} 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. \end{equation}

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

\begin{equation} 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. \end{equation}

The derivative of the Fermi function is,

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

where

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

Differentiating to find $dk$

\begin{equation} dk= \frac{m^*k_BT}{\hbar^2k}dx \end{equation}

The thermal conductivity can be written as,

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

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 $\frac{\pi^2}{3}$.

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

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.