 ## Electrochemical potential

The electric current density can be expressed in terms of the band structure $E(\vec{k})$, the externally applied electric field $\vec{E}$, and the temperature gradient $\nabla_{\vec{r}}T$.

\begin{equation} \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(e\vec{E}+ \nabla_{\vec{r}}\mu+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T \right) \right) d^3k. \end{equation}

The quantity $e\vec{E} + \nabla_{\vec{r}}\mu$ is the gradient of the electrochemical potential. The electrochemical potential for electrons is $\tilde{\mu}= -e\phi+\mu$ where $\phi$ is the electrostatic potential and $\mu$ is the chemical potential. This is the appropriate potential energy to use in a problem where there is a concentration gradient of electrons and an electrostatic field present. Generally, the force is minus the gradient of the potential energy, $\vec{F} = -\nabla \tilde{\mu} = e\nabla_{\vec{r}}\phi - \nabla_{\vec{r}}\mu= -e\vec{E} - \nabla_{\vec{r}}\mu$. This force has an electrostatic component $-e\vec{E}$ that is in the opposite direction of the electric field and a component $- \nabla_{\vec{r}}\mu$ corresponding to the driving force for diffusion that points from high concentration to low concentration. The work required to move an electron from $\vec{r}_1$ to $\vec{r}_2$ in this force field is $\tilde{\mu}(\vec{r}_2)- \tilde{\mu}(\vec{r}_1)$.

A voltmeter with two contacts (red and black) measures the difference in electrochemical potential divided by the elementary charge, $(\tilde{\mu}(\text{red})-\tilde{\mu}(\text{black}))/e$. If a voltmeter is used to measure the voltage across a charged capacitor, it measures the difference in electrostatic potential. If a voltmeter is used measure the voltage on a battery, it measures the difference in chemical potential. Often there are both electric fields and concentration gradients present so a voltmeter measures a combination of both the difference in electrostatic potential and the difference in chemical potential. For experiments where a voltmeter is used, or experiments where a voltage source is used, it is convenient to write the electric current density in terms of the electrochemical potential.

\begin{equation} \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. \end{equation}

More generally, a voltmeter measures an electromotive force $\mathcal{E}$. Unfortunately, the terminology is not very clear. Electromotive force is measured in volts and is the potential energy of an electron at the red contact minus the potential energy of an electron at the black contact divided by $e$. In addition to electric fields and concentration gradients, changing magnetic fields can produce an electromotive force. A difference in the electrostatic potential $\phi(\text{red})-\phi(\text{black})$ is an electromotive force, a difference in the chemical potential $(\mu(\text{red})-\mu(\text{black}))/e$ is an electromotive force, and a difference in the electrochemical potential $(\tilde{\mu}(\text{red})-\tilde{\mu}(\text{black}))/e$ is an electromotive force.

In terms of the electrochemical potential, the current densities are,

$$\vec{j}_{\text{elec}}= \frac{e}{4\pi^3}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\vec{v}_{\vec{k}}\left( \vec{v}_{\vec{k}}\cdot\left(\nabla_{\vec{r}}\tilde{\mu}+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T \right) \right) d^3k.$$ $$\vec{j}_n= -\frac{1}{4\pi^3}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\vec{v}_{\vec{k}}\left( \vec{v}_{\vec{k}}\cdot\left(\nabla_{\vec{r}}\tilde{\mu}+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T \right) \right) d^3k.$$ $$\vec{j}_U= -\frac{1}{4\pi^3}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}E(\vec{k})\vec{v}_{\vec{k}}\left( \vec{v}_{\vec{k}}\cdot\left(\nabla_{\vec{r}}\tilde{\mu}+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T \right) \right) d^3k.$$ $$\vec{j}_Q= -\frac{1}{4\pi^3}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\left( E(\vec{k}) - \mu \right)\vec{v}_{\vec{k}}\left( \vec{v}_{\vec{k}}\cdot\left(\nabla_{\vec{r}}\tilde{\mu}+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T\right) \right) d^3k.$$