I'm currently studying the effects of Faraday rotation - specifically high frequency EM waves in a cold magnetised plasma, the external magnetic field is constant and uniform.

I am considering EM waves parallel to the external field so from the dielectric tensor we can retrieve dispersion relations for left hand and right hand circularly polarised waves:

Left hand: $$k_L=\frac{\omega}{c}(1-\frac{\omega_{pe}^2}{\omega(\omega_{ce}+\omega)})^{0.5}$$

Right hand: $$k_R=\frac{\omega}{c}(1+\frac{\omega_{pe}^2}{\omega(\omega_{ce}-\omega)})^{0.5}$$

Where $\omega$ is the angular frequency of the EM wave, $\omega_{pe}$ the angular frequency of electron plasma oscillations and $\omega_{ce}$ the electron cyclotron frequency

Now my university notes make the following claim:

$$k_L - k_R = \Delta k \approx \frac{\omega_{ce}}{2\omega}\frac{\omega_{pe}^2}{c\omega}$$

I have the following working:

$$\Delta k = \frac{\omega}{c}((1-\frac{\omega_{pe}^2}{\omega(\omega_{ce}+\omega)})^{0.5}-(1+\frac{\omega_{pe}^2}{\omega(\omega_{ce}-\omega)})^{0.5})$$

Now using the Taylor expansion approximation(as we are in the high frequency limit $\omega >> \omega_{pe}$) $$(1+x)^{0.5} = 1+\frac{x}{2}$$

After some algebra we arrive at:

$$\Delta k \approx \frac{\omega_{ce}\omega_{pe}^2}{c(\omega^2 -\omega_{ce}^2)}$$

Any pointers would be greatly appreciated!


So first you should rearrange things to normalize them properly to give: $$ \begin{align} k_{R} & = \frac{ \omega }{ c } \sqrt{ 1 - \frac{ \omega_{pe}^{2} }{ \omega^{2} \left( 1 + \frac{ \Omega_{ce} }{ \omega } \right) } } \tag{0a} \\ k_{L} & = \frac{ \omega }{ c } \sqrt{ 1 - \frac{ \omega_{pe}^{2} }{ \omega^{2} \left( 1 - \frac{ \Omega_{ce} }{ \omega } \right) } } \tag{0b} \end{align} $$ where $\Omega_{ce}$ is the electron cyclotron frequency, $\omega_{pe}$ is the electron plasma frequency, and $\omega$ is the frequency of the mode in question.

Now if you are in the limit where $\omega \gg \omega_{pe}$ and not near a star or strongly magnetized body, then it is very likely that $\omega \gg \Omega_{ce}$ too. In fact, it is likely that $\omega_{pe} \gg \Omega_{ce}$ but that's not critical at the moment. First we perform a Taylor expansion on the $\tfrac{ \Omega_{ce} }{ \omega }$ terms to get: $$ \begin{align} k_{R} & \approx \frac{ \omega }{ c } \sqrt{ 1 + \left( \frac{ \omega_{pe} }{ \omega } \right)^{2} \left( \frac{ \Omega_{ce} }{ \omega } \right) } \tag{1a} \\ k_{L} & \approx \frac{ \omega }{ c } \sqrt{ 1 - \left( \frac{ \omega_{pe} }{ \omega } \right)^{2} \left( \frac{ \Omega_{ce} }{ \omega } \right) } \tag{1b} \end{align} $$ where we have used: $$ \left( 1 \pm \frac{ 1 }{ x } \right)^{-1} \approx \pm x - x^{2} \tag{2} $$

The next step involves expanding terms of the form: $$ \left( 1 - a \ x^{2} \right)^{1/2} \approx 1 - a \frac{ x^{2} }{ 2 } \tag{3} $$ such that our approximation for $k_{R(L)}$ goes to: $$ \begin{align} k_{R} & \approx \frac{ \omega }{ c } \left[ 1 + \frac{ 1 }{ 2 } \left( \frac{ \Omega_{ce} }{ \omega } \right) \left( \frac{ \omega_{pe} }{ \omega } \right)^{2} \right] \tag{4a} \\ k_{L} & \approx \frac{ \omega }{ c } \left[ 1 - \frac{ 1 }{ 2 } \left( \frac{ \Omega_{ce} }{ \omega } \right) \left( \frac{ \omega_{pe} }{ \omega } \right)^{2} \right] \tag{4b} \end{align} $$

The rest is just arithmetic so I will leave it as an exercise.

  • 1
    $\begingroup$ Thanks so much for this alternate method, it turns out that there is an erroneous factor of 1/2 in my notes - and yes you are correct, we can make the other assumption you have made. Cheers. $\endgroup$ Dec 22 '20 at 10:35

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