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The CMA (Clemmow - Mullaly - Allis) diagram for a plasma of electrons and immobile ions is shown below:

enter image description here

Source: Gurnett & Bhattacharjee, Introduction to Plasma Physics with Space, Laboratory and Astrophysical applications page 133

While in some sense I understand the diagram where the topological shapes show how the $R$ and $L$ modes (right polarized and left polarized modes), corresponding to wave propagation at an angle $\theta = 0$ to the magnetic field, transit to the $O$ and $X$ modes (ordinary and extraordinary modes), when $\theta$ changes from $0$ to $\frac{\pi}{2}$

Earlier in the book, on page 129, it is stated that the combinations of the topological shape for each transition is purely determined by the value of the refractive index $n$ at $\theta = 0$ and $\theta = \frac{\pi}{2}$. Basically, notice that in each region we have two topological shapes, such as 2 ellipsoids in the bottom left triangle. This corresponds to the two modes $R$ and $L$ and how they transit as $\theta$ changes from $0$ to $\frac{\pi}{2}$

For example, if we look at the topological shape at the bottom left triangle near the origin, bounded by the $R = 0$ line and the axes, I understand that the ellipsoid means that the refractive index does not change sign when transiting from $\theta = 0$ to $\theta = \frac{\pi}{2}$ for BOTH the $R$ and $L$ modes.

What I don't understand is:

(a) In order to draw the topological shapes, we need to know whether the refractive index $n$ changes sign. How do we determine this from the CMA diagram?

(b) In some cases, the $R$ mode transits to the $X$ mode while in others, the $R$ mode transits to the $O$ mode. How can we tell whether the $R$ mode transits to the $O$ or the $X$ mode? How do we know is not possible that both the $R$ and $L$ mode transits to $X$?

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  • $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Jun 16, 2021 at 5:27
  • $\begingroup$ @honeste_vivere Hi! Pardon me, but I don't really understand your comment. Are you saying that the L mode always transits to O mode and R mode always transits to X? $\endgroup$
    – D. Soul
    Jun 20, 2021 at 7:21
  • $\begingroup$ @D.Soul - 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. I retract my previous comment as it seems there are cases where the polarization of the O- and X-modes can be slightly different from what I stated. I almost always work with plasmas that satisfy $\omega_{pe} \gg \Omega_{ce}$, so it greatly limits the possibilities for free modes. Also, some polarizations are not accessible from free-space, which is important to remember. $\endgroup$ Jun 21, 2021 at 22:16
  • $\begingroup$ @honeste_vivere Thanks for your reply! Yes, I understand the meaning of the contours, however, i was wondering exactly how these contours were determined. How do they know whether the L mode changes to an X or and O mode? How do they know what the topological shapes look like? It doesn't seem to be a trivial matter. Is there more complicated technicalities involved? $\endgroup$
    – D. Soul
    Jun 22, 2021 at 8:51
  • $\begingroup$ @honeste_vivere did Clemmow, Mullaly and Allis really went to compute the refractive index for each region? Or were they determined experimentally? Also how did they know whether the L mode at zero degrees become the X or the O mode at 90 degrees? For example in the bottom left near the origin, it shows the L mode become the O mode. How did they know this? Was it also determined experimentally? Also, would you like to post your comments as an answer so i can accept it? $\endgroup$
    – D. Soul
    Jun 22, 2021 at 12:59

1 Answer 1

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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.

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  • $\begingroup$ Thank you, for the part where you mentioned "In the case of free space modes, the O-mode is left-handed and the X-mode is right-handed" may i ask how this is determined? Or rather, is there any literature I can read? $\endgroup$
    – D. Soul
    Jun 24, 2021 at 2:24
  • $\begingroup$ It's from the full index of refraction in the high frequency limit (i.e., $\omega \gg \omega_{pe}$ and $\omega \gg \Omega_{ce}$). The R and L terms survive but because it's such high frequency they don't care about the magnetic field direction anymore (aside from Faraday rotation, of course). Gurnett and Bhattacharjee's book discuss this. $\endgroup$ Jun 24, 2021 at 12:30
  • $\begingroup$ Thank you for your patience! $\endgroup$
    – D. Soul
    Jun 25, 2021 at 7:11

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