In The Dispersion Relation of a Magnetized Plasma we found that a magnetic field alters the dispersion relation differently for left and right polarized light passing through a plasma. But what about a linear polarized wave? The emerging effect is termed the "Faraday Rotation" and has numerous applications. Find out what we can learn about the cosmos and graphene using this effect.

## Problem Statement

In the mentioned problem we assumed an electron density ne and a constant magnetic field B = B0·ez.

We wanted to know the dispersion relation for circularly polarized light,

$\mathbf{E}^{l/r}\left(\mathbf{r},t\right) = E_{0}\exp\left(\mathrm{i}\left(kz-\omega t\right)\right)\left[\mathbf{e}_{x}\pm\mathrm{i}\mathbf{e}_{y}\right]\ .$

We found that

$\epsilon_{r}^{l/r}\left(\omega\right) = 1-\frac{\omega_{p}^{2}}{\omega\left(\omega\mp\Omega\right)}$

with the plasma and cylcotron frequencies

$\omega_{p}^{2}=\frac{n_{e}e^{2}}{\epsilon_{0}m_{e}} , \Omega=eB_{0}/m_{e}\ .$

In conclusion the response of the electron plasma was strongly depending on the polarization of the wave.

Find the rotation of a plane wave polarized in x direction passing through a magnetized plasma of thickness d  for frequencies much greater than both ωp and Ω!

## Hints

What is the relation of a circular and a linear polarized plane wave?

How can we know the rotation from the dispersion relation?

## Solution

We know that a linear polarized plane wave can be seen as a superposition of circular polarized ones. From

$\begin{eqnarray*}\mathbf{E}^{l/r}\left(\mathbf{r},t\right) & = & E_{0}\exp\left(\mathrm{i}\left(kz-\omega t\right)\right)\left[\mathbf{e}_{x}\pm\mathrm{i}\mathbf{e}_{y}\right]\end{eqnarray*}$

we can see that

$\begin{eqnarray*}\mathbf{E}^{x}\left(\mathbf{r},t\right) & = & \frac{1}{2}\left(\mathbf{E}^{l}\left(\mathbf{r},t\right)+\mathbf{E}^{r}\left(\mathbf{r},t\right)\right)\ .\end{eqnarray*}$

Now we may treat the plasma as a medium with the derived dispersion relation. The dispersion relation is given by:

$\begin{eqnarray*}k_{l,r}\left(\omega\right) & = & \sqrt{\epsilon_{r}^{l,r}\left(\omega\right)}k_{0}\ .\end{eqnarray*}$

Since the high-frequency case is considered we may use a Taylor expansion of the relative permittivity. This will yield the difference between both polarizations. It is

$\begin{eqnarray*}\sqrt{\epsilon_{r}^{l/r}\left(\omega\right)} & = & \sqrt{1-\frac{\omega_{p}^{2}}{\omega\left(\omega\mp\Omega\right)}}\\ & = & 1-\frac{\omega_{p}^{2}}{2\omega^{2}}\mp\frac{\omega_{p}^{2}\Omega}{2\omega^{3}}+\mathcal{O}\left(\omega^{-4}\right)\ . \end{eqnarray*}$

Putting the latter result into the electric field we find an equation for the electric field inside the plasma:

$\begin{eqnarray*}\mathbf{E}^{x}\left(\mathbf{r},t\right) & = & \frac{E_{0}}{2}\exp\left(-\mathrm{i}\omega t\right)\left(\exp\left(\mathrm{i}k_{l}z\right)\left[\mathbf{e}_{x}+\mathrm{i}\mathbf{e}_{y}\right]+\exp\left(\mathrm{i}k_{l}z\right)\left[\mathbf{e}_{x}-\mathrm{i}\mathbf{e}_{y}\right]\right)\\& = & \frac{E_{0}}{2}\exp\left[\mathrm{i}\left\{ \left(1-\frac{\omega_{p}^{2}}{2\omega^{2}}\right)k_{0}z-\omega t\right\} \right]\\& & \cdot\left(\exp\left({\color{red}-}\mathrm{i}\frac{\omega_{p}^{2}\Omega}{2\omega^{3}}k_{0}z\right)\left[\mathbf{e}_{x}+\mathrm{i}\mathbf{e}_{y}\right]+\exp\left({\color{red}+}\mathrm{i}\frac{\omega_{p}^{2}\Omega}{2\omega^{3}}k_{0}z\right)\left[\mathbf{e}_{x}-\mathrm{i}\mathbf{e}_{y}\right]\right)\ .\end{eqnarray*}$

Now we can see that after a certain propagation distance d both polarizations have accumulated different phases. To find how much the incident wave has been rotated we just have to look for the prefactors in front of ey, the calculation for ex would be similar:

$\begin{eqnarray*} \mathrm{i}\left(\exp\left(-\mathrm{i}\frac{\omega_{p}^{2}\Omega}{2\omega^{3}}k_{0}d\right)-\exp\left(+\mathrm{i}\frac{\omega_{p}^{2}\Omega}{2\omega^{3}}k_{0}d\right)\right)\mathbf{e}_{y} & = & 2\sin\left(\frac{\omega_{p}^{2}\Omega}{2\omega^{3}}k_{0}d\right)\mathbf{e}_{y}\ . \end{eqnarray*}$

Hence we find a rotation of

$\begin{eqnarray*} \varphi & = & \frac{\omega_{p}^{2}\Omega}{2\omega^{3}}k_{0}d\ . \end{eqnarray*}$

For some applications for this effect please have a look at the following background!

## Background: The Faraday Effect and its Applications

It is clear that any time more or less free charged particles are passed by an electromagnetic wave and a magnetic field is involved, one might observe the Faraday Effect we are about to calculate. One tool designed from the effect is the Faraday Rotator changing the angle of polarization due to the studied effect.

If one knows the rotation and the plasma density, one can in principle measure the magnetic field. This is also true for the biggest laboratory one can imagine - the universe! See for example Faraday Rotation of Microwave Background Polarization by a Primordial Magnetic Field on how the effect is used to unravel the last secrets of space and time.

We might very well change the length scales from lightyears to Ångström. Here we find the very promising material graphene which is a monolayer of carbon atoms. Graphene has an extremely high conductivity because of hybridized electrons. Thus, when a light wave is passing through this approximately five Ångström thick layer one might still expect a measurable Faraday Rotation. An experiment was done in 2010 and show the extraordinary properties of graphene, see Giant Faraday rotation in single- and multilayer graphene.

## Latest Articles

### The 7 Essential Physics Equations

Published in Blog

Explore the historical significance and real-world applications of the 7 most important physics equations, from Newton to Einstein and beyond.

### Quantum Electrodynamics for Quantum Computation

Explore the indispensable role of Quantum Electrodynamics in advancing quantum computation, from foundational principles to practical applications.

## Discover

### Total Internal Reflection in a Prism - iFDTD Tutorial

Published in Blog
Tags:

In this second tutorial we will show how to define a refractive index profile n(r) within the interactive FDTD toolbox. We will simulate the total internal reflection in a glass prism using a Gaussian shaped beam.

### The Disadvantages of Solar Energy

Published in Blog

Before we can all admire the new and exciting green future, let us have a clear and analytic look on what the disadvantages of solar energy are. Let us also find out what these disadvantages mean practically.