## Thermoelectric current

A temperature gradient can cause a current to flow along a wire. The electrons move from the hot side to the cold side. Both charge and energy are transported in this case. The general expression for the electric current density is,

$$\vec{j}_{\text{elec}}= \frac{e}{4\pi^3\hbar^2}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\nabla_{\vec{k}}E(\vec{k})\left( \nabla_{\vec{k}}E(\vec{k})\cdot\left(\nabla_{\vec{r}}\tilde{\mu}+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T \right) \right) d^3k.$$

One end of the wire is grounded an the other is attached to an ammeter which is then also grounded. There is no voltage drop across a perfect ammeter so the gradient of the electrochemical potential is zero. The thermoelectric current produced by this temperature gradient is,

$$\vec{j}_{\text{elec}}= \frac{e}{4\pi^3\hbar^2}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\nabla_{\vec{k}}E(\vec{k})\left( \nabla_{\vec{k}}E(\vec{k})\cdot\left(\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T\right) \right) d^3k.$$

The relationship between the electrical current density and the temperature gradient can be written as a matrix,

$$$\left[ \begin{array}{cccc} j_{x} \\ j_{y} \\ j_{z} \\ \end{array} \right]= \left[ \begin{array}{cccc} \kappa_{xx} & \kappa_{xy} & \kappa_{xz} \\ \kappa_{yx} & \kappa_{yy} & \kappa_{yz} \\ \kappa_{zx} & \kappa_{zy} & \kappa_{zz} \\ \end{array} \right]\left[ \begin{array}{cccc} \frac{\partial T}{\partial x} \\ \frac{\partial T}{\partial y} \\ \frac{\partial T}{\partial z} \\ \end{array} \right].$$$

The thermoelectric coefficients $$$\kappa_{ij}=\frac{e}{4\pi^3\hbar^2T}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu} \nabla_{\vec{k}}E(\vec{k})\cdot\hat{e}_i \left( E(\vec{k}) - \mu \right) \nabla_{\vec{k}}E(\vec{k})\cdot\hat{e}_j d^3k.$$$

Here $\hat{e}_i$ are the unit vectors $i = [x,y,z]$. For cubic crystals the thermoelectric coefficient is a constant,

$$$\kappa=\frac{e}{4\pi^3\hbar^2T}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu} \left( E(\vec{k}) - \mu \right) \left( \nabla_{\vec{k}}E(\vec{k})\cdot\hat{z}\right)^2 d^3k.$$$ $$\kappa=\frac{e}{3\pi^2T}\int\limits_0^{\infty} \tau(\vec{k}) \frac{\partial f_0}{\partial \mu} \left( E(\vec{k}) - \mu \right) \left|\vec{v}_{\vec{k}}\right|^2 k^2 dk.$$

In steady state the current density must be constant along the wire. This implies that $\nabla T/T$ is also a constant and that the temperature falls exponentially along the wire. The electrons carry heat as well as charge. The heat current is,

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

Generally, the relationship between the thermal current density and the temperature gradient is described by the thermal conductivity matrix,

$$$\left[ \begin{array}{cccc} j_{Qx} \\ j_{Qy} \\ j_{Qz} \\ \end{array} \right]= - \left[ \begin{array}{cccc} K_{xx} & K_{xy} & K_{xz} \\ K_{yx} & K_{yy} & K_{yz} \\ K_{zx} & K_{zy} & K_{zz} \\ \end{array} \right]\left[ \begin{array}{cccc} \frac{\partial T}{\partial x} \\ \frac{\partial T}{\partial y} \\ \frac{\partial T}{\partial z} \\ \end{array} \right].$$$

The thermal conductivity matrix can be calculated from the dispersion relation as,

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

Here $\hat{e}_i$ are the unit vectors $i = [x,y,z]$. For cubic crystals the thermal conductivity is a constant,

$$$K=\frac{1}{4\pi^3\hbar^2T}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu}\left( \left( E(\vec{k}) - \mu \right) \nabla_{\vec{k}}E(\vec{k})\cdot\hat{z}\right)^2 d^3k.$$$ $$K=\frac{1}{3\pi^2T}\int\limits_0^{\infty} \tau(\vec{k}) \frac{\partial f_0}{\partial \mu} \left( E(\vec{k}) - \mu \right)^2 \left|\vec{v}_{\vec{k}}\right|^2 k^2 dk.$$