# Expression of fields in Faraday rotation

1. Jan 15, 2014

### EmilyRuck

Hello!
Talking about propagation of an electro-magnetic field in a non-isotropic medium, I've got some troubles with the expression in object, used to show the Faraday rotation of the polarization of a field.

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

An electro-magnetic field enters a particular medium, propagating along the $\hat{\mathbf{u}}_z$ direction. In $z = 0$, its electric field is $\mathbf{E} = E_0 \hat{\mathbf{u}}_x$. It could also be written as a superposition of two circularly polarized waves:

$\mathbf{E} = \displaystyle \frac{E_0}{2} (\hat{\mathbf{u}}_x + j \hat{\mathbf{u}}_y) + \displaystyle \frac{E_0}{2} (\hat{\mathbf{u}}_x - j \hat{\mathbf{u}}_y)$

The two components have different propagation constants, $β_-$ and $β_+$, in the medium. In a generic $z$ position we could write

$\mathbf{E} = \displaystyle \frac{E_0}{2} (\hat{\mathbf{u}}_x + j \hat{\mathbf{u}}_y)e^{-j β_- z} + \displaystyle \frac{E_0}{2} (\hat{\mathbf{u}}_x - j \hat{\mathbf{u}}_y)e^{-j β_+ z}$

2. Relevant equations

Rearranging the last expression (this is done in several books, like Pozar), we obtain:

$\mathbf{E} = E_0 e^{-j (β_+ + β_-) \frac{z}{2}} \left\{ \hat{\mathbf{u}}_x \cos \left[ \left( β_+ + β_- \right) \displaystyle \frac{z}{2} \right] - \hat{\mathbf{u}}_y \sin \left[ \left( β_+ - β_- \right) \displaystyle \frac{z}{2} \right] \right\}$

This is done to show that the polarization is still linear like in the original field $\mathbf{E} = E_0 \hat{\mathbf{u}}_x$, but its "orientation" has changed with the position $z$.
But could this still be called a wave? Its dipendence from $z$ is no more only in the exponential $e^{-j β z}$, but is also contained in the cosine and sine terms.

3. The attempt at a solution

It apparently does no more satisfy the Helmholtz wave equation, because deriving the $x$ component with respect to $z$ gives a completely different result that that obtained deriving the same component with respect to time (assuming that we are using phasors).

So, how can I interpret this expression? Shouldn't it still satisfy the Helmholtz equation? Shouldn't it be still a wave?

Emily

2. Jan 16, 2014

### TSny

Hello.

You started with a superposition of two circularly polarized that propagate with different wavenumbers $\beta$.
Each of these satisfy a Helmholtz equation. But $\beta$ appears in the form of the Helmholtz equation. So, the two individual circularly polarized waves satisfy different Helmholtz equations.

Try to show that the sum of the two circularly polarized waves cannot satisfy a Helmholtz equation.

Does the superposition of the two circularly polarized waves represent a wave in the medium? That depends on the definition of a wave. If you define a wave loosely as "a disturbance in a medium that propagates" or something like that, then I would say you have a wave. If you define a wave to be "a mathematical expression that satisfies the wave equation $\psi_{,xx} - \frac{1}{v^2}\psi_{,tt} = 0$ for some $v$", then you do not have a wave. That's because each component satisfies a different wave equation (with different value of $v$), so the sum of the components does not satisfy the wave equation. But that would be true anytime you have a superposition of waves of different wavelength propagating in a dispersive medium.