# Boltzmann Transport Equation : Hall Effect in the Relaxation Time Approximation

1. Oct 31, 2011

### gougoune5

Hello everyone!

I have been on this website for quite a while, and found some interesting answers to many questions, and I decided to create an account to seek you help with a particular issue I encountered in my assignment.

1. The problem statement, all variables and given/known data

I need to show that the Boltzmann Transport Equation (BTE), in the relaxation time approximation, for electrons in a semiconductor with constant and uniform electric (E) and magnetic (B) fields gives a steady-state electron distribution function of the form $g=g_B+f$, with f the typical Fermi distribution function and

$g_B=-\frac{\partial f}{\partial \epsilon}\textbf{v}\cdot\textbf{X}_B$ is the distribution perturbation

where

$\textbf{X}_B (1+\phi^2)=\textbf{X}+\textbf{X}\times\phi+( \textbf {X}\cdot\phi)\phi$

$\textbf{X}=-q\tau\textbf{E}$

$\phi=-\frac{q\tau}{m}\textbf{B}$

We consider the spatial gradients of g and of temperature to be zero. q is the electronic charge (in absolute value), τ is the relaxation time, v is the particle velocity, m is the electron effective mass and $\epsilon$ is the energy.

2. Relevant equations

The BTE, including the aforementioned approximations and in the relaxation time framework, boils down to

$-\frac{q}{\hbar}( \textbf{E} + \textbf{v} \times \textbf{B} )\vec{\nabla_k} g = -\frac{g-f}{\tau}$

3. The attempt at a solution

When I work out the equation relative to $g_B$ rather than g, I can cancel out some terms due to the vector orthogonalities arising from the cross product.

I have worked out that $\vec{\nabla}_k f=\hbar \textbf{k} \frac{\partial f}{\partial \epsilon}$

Also, knowing that $\hbar\textbf{k}=m\textbf{v}$, we can change the gradient on k by a gradient on v.

Finally, come up with

$-q\frac{\partial f}{\partial \epsilon} \textbf{v}\cdot \textbf{E}-\frac{q}{\hbar}(\textbf{E}+\textbf{v} \times \textbf{B} ) \vec{\nabla}_k g_B = \frac{g-f}{\tau} = \frac{g_B}{\tau}$

or

$-q\frac{\partial f}{\partial \epsilon} \textbf{v}\cdot \textbf{E}-\frac{q}{m}(\textbf{E}+\textbf{v} \times \textbf{B} ) \vec{\nabla}_v g_B = \frac{g-f}{\tau} = \frac{g_B}{\tau}$

However, I have absolutely no idea how to isolate $g_B$ in this equation. In our course, we did a similar treatment for the case with no magnetic field, and we approximated $\vec{\nabla}_k g \simeq \vec{\nabla}_k f$ , which is equivalent to saying $\vec{\nabla}_k g_B \simeq 0$.

However, this poses a problem here since we oversee the effect of the magnetic field in this approximation due to the gradient of f being parallel to v, thus perpendicular to $\textbf{v} \times \textbf{B}$

I have seen in another work someone postulating that we have a solution of the form $g_B=-\textbf{v} \cdot\textbf{P}(\epsilon)\frac{\partial f}{\partial \epsilon}$, where $\textbf{P}(\epsilon)$ is a vectorial quantity dependent solely on the energy.

They managed to work out that $(\textbf{v} \times \textbf{B})\cdot\vec{\nabla}_k g \simeq -\textbf{v}\cdot(\textbf{B} \times \textbf{P})\frac{\partial f}{\partial \epsilon}$ but I can't for the life of me figure out how they swapped around the vectors in this product...

Also, their final solution was obtained by assigning orthogonal axes to the fields and writing out the individual scalar components of P. However, I have to work out the solution for arbitrary fields and keep everything in vectorial notation. Even if the approximation above holds true, I don't know how to isolate P in the resulting equation, how can I invert the cross product? :

$\textbf{P}-\frac{q\tau}{m}(\textbf{B} \times \textbf{P}) + q\tau \textbf{E}=0$

I am not very familiar with manipulating vectorial equations such as these and I have been turning this upside down for a few days now, so any help would be appreciated!