Advanced Solid State Physics

Outline

Electrons

Magnetic effects and
Fermi surfaces

Magnetism

Linear response

Transport

Crystal Physics

Electron-electron
interactions

Quasiparticles

Structural phase
transitions

Landau theory
of second order
phase transitions

Superconductivity

Quantization

Photons

Exam questions

Appendices

Lectures

Books

Course notes

TUG students

      

Nerst Effect

In 1886, professor Albert von Ettingshausen and his PhD student Walther Nernst were studying the Hall effect in bismuth at the TU Graz. They discovered that if a temperature gradient was applied perpendicular to a magnetic field, a voltage could be measured perpendicular to both the temperature gradient and the magnetic field. This is known as the Nernst effect. As usual, we start wth the expression for the electronic current density is,

\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 +\frac{e}{\hbar}\nabla_{\vec{k}}E(\vec{k})\times\vec{B}\right) \right) d^3k. \end{equation}

The sample is electrically isolated so $\vec{j}_{\text{elec}}=0$.

\begin{equation} 0 = \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 +\frac{e}{\hbar}\nabla_{\vec{k}}E(\vec{k})\times\vec{B}\right) \right) d^3k. \end{equation}

This equation can be solved for the relationship between $\nabla_{\vec{r}}\tilde{\mu}$, $\nabla_{\vec{r}}T$, and $\vec{B}$. The Nernst coefficient is defined as,

\begin{equation} N_{lmn}= \frac{\nabla \tilde{\mu}_l}{e\nabla T_mB_n}. \end{equation}

The Nernst coefficients can be determined by substituting unit vectors for $\nabla T$ and $\vec{B}$ to determine $\nabla \tilde{\mu}$ and then iteratively solving equations such as,

\begin{equation} 0 = \frac{e}{4\pi^3\hbar^2}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\nabla_{\vec{k}}E(\vec{k})\cdot\hat{e}_i\left( \nabla_{\vec{k}}E(\vec{k})\cdot\left(\left( \begin{array}{cccc} eN_{xyz} \\ eN_{yyz} \\ eN_{zzz} \end{array} \right)+\frac{E(\vec{k})-\mu}{T}\hat{y} +\frac{e}{\hbar}\nabla_{\vec{k}}E(\vec{k})\times\hat{z}\right) \right) d^3k. \end{equation}

This represents three equations for the three unknown coefficients since all three components of the current density must be zero. Here $\hat{e}_i$ are the unit vectors $i = [x,y,z]$.