# How are the topological shapes determined in a CMA diagram?

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

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 $$X$$?

• Consider to spell out acronyms. Jun 16, 2021 at 5:27
• @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? Jun 20, 2021 at 7:21
• @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. Jun 21, 2021 at 22:16
• @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? Jun 22, 2021 at 8:51
• @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? Jun 22, 2021 at 12:59

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.

• 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? Jun 24, 2021 at 2:24
• 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. Jun 24, 2021 at 12:30
• Thank you for your patience! Jun 25, 2021 at 7:11