Faraday Rotation - $\Delta k$ Approximation 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!
 A: 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.
