The following intro is taken from this answer: https://physics.stackexchange.com/a/264526/59023
If we consider the case of a cold uniform plasma with only linear waves, then one can show the dielectric tensor has the form:
$$
\begin{align}
\overleftrightarrow{\mathbf{K}} & = \left[
\begin{array}{ c c c }
S & -i \ D & 0 \\
i \ D & S & 0 \\
0 & 0 & P
\end{array} \right] \tag{0}
\end{align}
$$
where the $S$, $D$, and $P$ terms are defined as:
$$
\begin{align}
S & = 1 - \sum_{s} \frac{\omega_{ps}^{2}}{\omega^{2} - \Omega_{cs}^{2}} \tag{1a} \\
& = \frac{1}{2} \left( R + L \right) \tag{1b} \\
D & = \sum_{s} \frac{ \Omega_{cs} \ \omega_{ps}^{2} }{\omega \left( \omega^{2} - \Omega_{cs}^{2} \right)} \tag{1c} \\
& = \frac{1}{2} \left( R - L \right) \tag{1d} \\
P & = 1 - \sum_{s} \frac{ \omega_{ps}^{2} }{ \omega^{2} } \tag{1e} \\
R & = 1 - \sum_{s} \frac{ \omega_{ps}^{2} }{ \omega \left( \omega + \Omega_{cs} \right) } \tag{1f} \\
L & = 1 - \sum_{s} \frac{ \omega_{ps}^{2} }{ \omega \left( \omega - \Omega_{cs} \right) } \tag{1g}
\end{align}
$$
where $\Omega_{cs}$ is the gyrofrequency (or cyclotron frequency) of species $s$, $\omega_{ps}$ is the plasma frequency of species $s$, and $\omega$ is the wave frequency. These parameters are defined as:
$$
\begin{align}
\Omega_{cs} & = \frac{ Z_{s} \ e \ B_{o} }{ \gamma \ m_{s} } \\
\omega_{ps} & = \sqrt{\frac{ n_{s} \ Z_{s}^{2} \ e^{2} }{ \varepsilon_{o} \ m_{s} }}
\end{align}
$$
where $Z_{s}$ is the charge state of species $s$ (e.g., +1 for protons), $e$ is the elementary charge, $B_{o}$ is the quasi-static magnetic field magnitude, $\gamma$ is the relativistic Lorentz factor, $m_{s}$ is the mass of species $s$, $n_{s}$ is the number density of species $s$, and $\varepsilon_{o}$ is the permittivity of free space.
Note that the $S$, $D$, $P$, $R$, and $L$ terms do have physical significances, e.g., $R$ and $L$ terms correspond to right- and left-hand polarized modes, respectively, called the R- and L-modes. Solutions for $S = 0$ correspond to the so called hybrid resonances, e.g., the lower hybrid resonance frequency. Similarly, $P = 0$ can correspond to a linearly polarized Langmuir wave propagating parallel to the quasi-static magnetic field. The $D = 0$ solution is relevant for regions of space with multiple ion species, corresponding to the so called crossover frequencies.
The dispersion relation, $D(\mathbf{k}, \omega)$, is derived from the equation:
$$
\mathbf{n} \times \left( \mathbf{n} \times \mathbf{E} \right) + \overleftrightarrow{\mathbf{K}} \cdot \mathbf{E} = 0
$$
where we rewrite this in the tensor form $\overleftrightarrow{\mathbf{D}} \cdot \mathbf{E} = 0$. If $\overleftrightarrow{\mathbf{D}}$ has a determinant that goes to zero, then there is a non-trivial solution for $\mathbf{E}$. This solution is the dispersion relation, $D(\mathbf{k}, \omega)$, which can be simplified down if we assume the index of refraction, $\mathbf{n}$, is parallel to the wave vector, $\mathbf{k}$, then:
$$
D\left( \mathbf{k}, \omega \right) = A \ n^{4} - B \ n^{2} + R \ L \ P = 0 \tag{2}
$$
where the terms $A$ and $B$ are defined by:
$$
\begin{align}
A & = S \ \sin^{2}{\theta} + P \cos^{2}{\theta} \tag{3a} \\
B & = R \ L \ \sin^{2}{\theta} + P \ S \ \left( 1 + \cos^{2}{\theta} \right) \tag{3b}
\end{align}
$$
where $\theta$ is the angle between $\mathbf{k}$ and the quasi-static magnetic field. The dispersion relation in Equation 2 has the unique solutions of:
$$
n^{2} = \frac{B \pm F}{2 \ A} \tag{4}
$$
where $F$ is defined by:
$$
F = \left( R \ L - P \ S \right)^{2} \sin^{2}{\theta} + 4 \ P^{2} \ D^{2} \ \cos^{2}{\theta} \tag{5}
$$
We can see that $\Im{ [F] } = 0$, always. Since the terms $A$ and $B$ are real, then we can say that $n^{2}$ must either be purely real ($n^{2}$ $>$ 0) or purely imaginary ($n^{2}$ $<$ 0). If $n^{2}$ $<$ 0, then the wave becomes evanescent. Evanescence occurs – purely imaginary with no finite real part – because a finite real part implies that energy is lost to the plasma. In a cold, collisionless plasma, this cannot occur as there is no energy absorption mechanism.
Note that $n \rightarrow \infty$ corresponds to resonance and $n \rightarrow 0$ corresponds to a cutoff.
Propagation Parallel to Magnetic Field
If we assume that $\theta \rightarrow 0$ in Equation 4 above, then there are three roots: $P = 0$, $n^{2} = R$, and $n^{2} = L$. As mentioned before the latter two are called the R- and L-modes, respectively, where the $R$ and $L$ designate the polarization of the electric field oscillations with respect to the quasi-static magnetic field.
Propagation Perpendicular to Magnetic Field
If we assume that $\theta \rightarrow \pi/2$ in Equation 4 above, then there are two roots: $n^{2} = P$ and $n^{2} = R L/S$. For the $n^{2} = P$ case, the electric and magnetic field oscillations are orthogonal to the wave vector but the electric field oscillations can be parallel to the quasi-static magnetic field. This mode is linearly polarized and is called the ordinary mode or O-mode.
The $n^{2} = R L/S$ has both longitudinal (i.e., parallel to wave vector) and transverse (i.e., perpendicular to wave vector) electric field oscillations. It also has magnetic field oscillations parallel to the quasi-static magnetic field (but still orthogonal to the wave vector). Thus, this mode does care about the quasi-static magnetic field. It is called the extraordinary mode or X-mode. The X-mode has cutoffs at $R = 0$ and $L = 0$ and resonances at $S = 0$ (i.e., hybrid resonances mentioned above).
Polarization of O- and X-modes
This is more complicated as it can depend on whether $\Omega_{cs} > \omega_{ps}$ and from what direction one takes the frequency limit (i.e., start from infinity and reduce $\rightarrow$ called free space modes). In the case of free space modes, the O-mode is left-handed and the X-mode is right-handed.
Clemmow-Mullaly-Allis (CMA) Diagrams
CMA diagrams are typically labeled in the following way. Each line/contour has two labels, the polarization at wave normal angle of zero degrees and the mode of propagation at ninety degrees. The $R$, $L$, $S$, and $P$ terms change sign at the bounding surfaces defined by $R = 0$, $L = 0$, $S = 0$, $P = 0$, $R = \infty$, and $L = \infty$.
Closed contours indicate a continuous index of refraction – index of refraction remains real and is continuous in its change across a boundary. Open contours indicate a change to an imaginary index of refraction. Since this is a cold plasma approximation, the only option of imaginary indices of refraction is evanescence since the waves cannot transfer energy to the particles directly.
It doesn't seem to be a trivial matter. Is there more complicated technicalities involved?
– D. Soul Jun 22 '21 at 08:51