= Solution
The <Faraday law> and <Ampere-Maxwell law> give $\delta\mathbf B=(c/\omega)\mathbf k\times\delta\mathbf E$ and
$$
\mathbf k\times(\mathbf k\times\delta\mathbf E)+\frac{\omega^2}{c^2}\boldsymbol\epsilon\cdot\delta\mathbf E=0.
$$
With <refractive index> $n=kc/\omega$, the <linear system> is therefore
$$
\begin{pmatrix}
S-n^2\cos^2\theta&ig&n^2\sin\theta\cos\theta\\
-ig&S-n^2&0\\
n^2\sin\theta\cos\theta&0&P-n^2\sin^2\theta
\end{pmatrix}\delta\mathbf E=0.
$$
Set
$$
\boxed{R=S-g,\qquad L=S+g,\qquad RL=S^2-g^2.}
$$
Here $R$ labels the <cold-plasma right-handed wave> and $L$ the <cold-plasma left-handed wave>, with their <polarizations> fixed explicitly in <1/d>. In the <electron>-only response,
$$
R=1-\frac{\omega_{pe}^2}{\omega(\omega-\Omega_e)},\qquad
L=1-\frac{\omega_{pe}^2}{\omega(\omega+\Omega_e)}.
$$
Writing $N=n^2$, expansion of the <determinant> yields the <dispersion relation>
$$
A N^2-BN+C=0,\quad
A=S\sin^2\theta+P\cos^2\theta,\quad
B=RL\sin^2\theta+PS(1+\cos^2\theta),\quad C=PRL.
$$
Equivalently,
$$
P(N-R)(N-L)\cos^2\theta+(SN-RL)(N-P)\sin^2\theta=0.
$$
Division gives
$$
\boxed{\tan^2\theta=-\frac{P(n^2-R)(n^2-L)}{(Sn^2-RL)(n^2-P)}.}
$$
The undivided <determinant> equation remains valid when this rational form has simultaneous zero factors.
Back to article page
