Cold Magnetized Plasma Dispersion Relation #

Now that we have a pretty good handle on the unmagnetized case, let’s include the background magnetic field \( \vec B_0 \) in the linearized equation of motion \[m_\alpha \pdv{\vec v_{\alpha, 1}}{t} = q_\alpha (\vec E_1 + \vec v_{\alpha, 1} \cross \vec B_0)\] We define \( \vec \omega_{c, \alpha} \) to be the vector gyrofrequency \[\vec \omega_{c, \alpha} \equiv \frac{q_\alpha}{m_\alpha} \vec B_0\] \[\pdv{\vec v_{\alpha, 1}}{t} + \vec \omega_{c, \alpha} \cross \vec v_{\alpha, 1} = \frac{q_\alpha}{m_\alpha} \vec E_1\]

For a linear system like this, where \( Av = F \), in general the generic solutions are perturbations to the solutions of the homogeneous equation \( Av = 0 \). For example, if there is a basis of solutions \( v_{0n} \) to \( Av = 0 \), then the generic solution to \( Av = F \) will be

where \( v_F \) satisfies the non-homogeneous equation \( Av = F \).

Let’s start with the homogeneous case, setting \( \vec E_1 = 0 \):

We already kind of know what the trajectories are going to be. We expect to see helical motion due to the static magnetic field \( \vec B_0 \). But let’s follow the full mathematical method to solve it because things will get more complicated when we include the electric field. For simplicity, we’ll drop the \( \alpha \) subscripts from here on, but they’re still implied. We can represent the cross product by means of a matrix operator \( A \): \[A = \begin{bmatrix} 0 & \omega_z & - \omega_y \\ - \omega_z & 0 & \omega_x \\ \omega_y & - \omega_x & 0 \end{bmatrix} \qquad \vec \omega \cross \vec B = A \cdot \vec B\] \[\pdv{\vec v_1}{t} = A \vec v_1\] We know that the solution to this is just \[\vec v_1 = e^{ A t} \vec v_0\] We can write down the power series representation of \( e^{At} \) in powers of the matrix \( A \) \[e^{At} = \sum _{n=0} ^\infty \frac{A^n t^n}{n!}\]

For our specific matrix operator (the cross product), we can write higher order powers in terms of lower powers by making use of the vector identity: \[A = \omega \cross (\ldots) \qquad A^2 = \omega \cross (\omega \cross (\ldots)) = \omega (\omega \cdot (\ldots)) - \omega^2 (\ldots)\] \[A^3 = - \omega^2 \omega \cross (\ldots) = - \omega ^2 A\] We can write out the pattern like this: \[A = A \qquad A^3 = - \omega ^2 A \qquad A^5 = \omega ^4 A\] \[A^2 = A^2 \qquad A^4 = - \omega^2 A^2 \qquad A^6 = \omega^4 A^2\] We can now expand the summation and group the even and odd components \[e^{At} = I + \sum_{k = 0} ^\infty \frac{(-1)^k}{(2k + 1)!} \omega^{2k + 1} \left( \frac{A}{\omega} \right)t^{2k + 1} + \sum_{m = 1} ^{\infty} \frac{(-1)^m}{(2m)!} \omega^{2m} \left( \frac{A^2}{\omega^2} \right) t^{2m}\] The two even/odd series are exactly the power series representations for \( \sin(\omega t) \) and \( \cos (\omega t) \): \[e^{At} = I + \sin (\omega t) \frac{A}{\omega} + (1 - \cos (\omega t)) \frac{A^2}{\omega}\] We’ve ended up with a closed form of the exponential of the cross product. This is called the Euler-Rodriguez rotation operator. If we sum cross products continuously, we get a rotation \[\vec v(t) = e^{At} \vec v_0 \] \[= \vec v_0 + \sin (\omega_c t) \frac{A \vec v_0}{\omega_c} + (1 - \cos (\omega_c t)) \frac{A^2 \vec v_0}{\omega_c ^2}\] \[\frac{A \vec v_0}{\omega_c} = - \frac{\vec \omega_c \cross \vec v_0}{\omega_c} = - \hat b_0 \cross \vec v_0 \qquad \hat b_0 = \frac{\vec B_0}{B_0}\] \[\frac{A^2 \vec v_0}{\omega_c ^2} = \frac{\vec \omega_c \cross \vec \omega_c \cross \vec v_0}{\omega_c ^2} = \hat b_0 \cross \hat b_0 \cross \vec v_0\] Now we split the velocity into components parallel to the magnetic axis \( \vec v_{0, \parallel} \) and perpendicular \( \vec v_{0, \perp} \) \[\vec v_{0, \parallel} = (\vec v_0 \cdot \hat b_0) \hat b_0\] \[\vec v_{0, \perp} = \vec v_0 - \vec v_{0, \parallel} = \vec v_0 - (\vec v_0 \cdot \hat b_0) \hat b_0 = - \hat b_0 \cross \hat b_0 \cross \vec v_0\] We can identify the \( A/\omega \) and \( A^2 / \omega \) terms, and we’re left with \[\vec v(t) = \vec v_0 - (1 - \cos (\omega_c t)) \vec v_{0, \perp} - \sin (\omega_c t) \hat b_0 \cross \vec v_{0, \perp}\] \[= \vec v_{0, \parallel} + \cos (\omega_c t) \vec v_{0, \perp} - \sin (\omega_c t) \hat b_0 \cross \vec v_{0, \perp}\] Now we’ve got the solution we expected: a rotation about the \( \hat b_0 \) axis with frequency \( \omega_c \), where the parallel component of velocity is unaffected by the magnetic field.

If we re-introduce the \( \vec E_1 \) forcing term, we perturb the helical trajectory \[\pdv{\vec v}{t} = A \vec v + \vec E_1\] In the inhomogeneous case, we usually solve the differential equation using an integrating factor \( e^{At} \) \[v(t) = e^{At} \int _{-\infty} ^{t} e^{- A t'} E_1 (t') \dd t' = \int _{-\infty} ^{t} e ^{A(t - t')} \vec E_1 (t') \dd t'\]

Because this has the form of a convolution, the solution will be easier in spectral variables. Transforming the differential equation into Fourier space, \[\pdv{v}{t} = A \vec v + \vec E_1 \rightarrow - i \omega \hat{\vec v} = A \hat{\vec v} + \hat{\vec E}\] Since the operator \( A \) is constant, it doesn’t transform. \[\rightarrow - i \omega( I - i\frac{A}{\omega}) \hat{\vec v} = \hat{\vec E}\] How do we solve this now? In the unforced case, we just get eigenvalues from \( \det(I - i \frac{A}{\omega}) = 0) \). But for the forced case we need to invert the whole operator \[\hat{\vec v} = \frac{i}{\omega} (I - i \frac{A}{\omega}) ^{-1} \hat{\vec E}\] The operator \( (I - i \frac{A}{\omega}) \) isn’t the easiest to invert, but we can perform an eigendecomposition of \( A \) to get \[A = Q \Lambda Q^{-1} \rightarrow (I + A)^{-1} = (Q Q^{-1} + Q \Lambda Q^{-1})^{-1}\] \[= (Q(I + \Lambda) Q^{-1})^{-1}\] \[= Q(I + \Lambda)^{-1} Q^{-1}\] So our strategy here is to

• Perform an eigendecomposition of \( A \)
• Find eigenvalues of \( I + A \)
• Invert the operator \( I - \frac{i}{\omega} A \) to find \( \frac{i}{\omega} (I - \frac{i}{\omega})^{-1} \vec E \)

We choose our coordinate system such that \( \vec \omega_c = \omega_c \hat b = \omega_c \hat z \). Then, \[A = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \omega_c\] \[\rightarrow \lambda = 0, \pm i \omega_c\] \[\vec v = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad \frac{1}{\sqrt 2}\begin{bmatrix} -i \\ 1 \\ 0 \end{bmatrix}, \quad \frac{1}{\sqrt 2}\begin{bmatrix} i \\ 1 \\ 0 \end{bmatrix}\] \[\rightarrow Q = \begin{bmatrix} - i / \sqrt{2} & i/ \sqrt{2} & 0 \\ 1/ \sqrt{2} & 1/ \sqrt{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}\] \[Q^{-1} = \begin{bmatrix} i / \sqrt{2} & 1/ \sqrt{2} & 0 \\ -i/ \sqrt{2} & 1/ \sqrt{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}\] \[\Lambda = \begin{bmatrix} i \omega_c & 0 & 0 \\ 0 & - i \omega c \\ 0 & 0 & 0 \end{bmatrix}\] \[I - \frac{i}{\omega} A \rightarrow I - \frac{i}{\omega} \Lambda = \begin{bmatrix} 1 + \frac{\omega_c}{\omega} & 0 & 0 \\ 0 & 1 - \frac{\omega_c}{\omega} & 0 \\ 0 & 0 & 1 \end{bmatrix}\] To double-check our work, if \( \vec E_0 = 0 \), then \( \det(I - \frac{i}{\omega} A) = 0 \) so \( \omega = \pm \omega_c \), so that’s a good sign. Now we can solve for \[\hat \vec v = Q(I - \frac{i}{\omega} \Lambda)^{-1} Q^{-1} (\frac{i}{\omega} \hat{\vec E})\] where \( \Lambda \) are the eigenvalues of \( A \) and \( Q, Q^{-1} \) are the eigenvectors of \( A \). For our case where \( A = \vec \omega_c \cross (\ldots) \), \[(I - \frac{i}{\omega} \Lambda) ^{-1} = \begin{bmatrix} (1 + \omega_c / \omega)^{-1} & 0 & 0 \\ 0 & (1 - \omega_c / \omega)^{-1} & 0 \\ 0 & 0 & 1 \end{bmatrix}\] We can just focus our attention on the rotational part, ignoring the parallel component in the \( \hat z \) direction for now \[\rightarrow \begin{bmatrix} (1 + \omega_c/\omega)^{-1} & 0 \\ 0 & (1 - \omega_c/\omega)^{-1} \end{bmatrix}\] We define the convenience terms \( P \) and \( N \) to be \[P \equiv \frac{\omega}{\omega + \omega_c}\] \[N \equiv \frac{\omega}{\omega - \omega_c}\] \[\rightarrow Q(I - \frac{i}{\omega} \Lambda) ^{-1} Q^{-1} = \frac{1}{2} \begin{bmatrix} -i & i \\ 1 & 1 \end{bmatrix} \begin{bmatrix} P & 0 \\ 0 & N \end{bmatrix} \begin{bmatrix} i & 1 \\ -i & 1 \end{bmatrix}\] Writing it all out… \[\frac{i}{\omega} (\ldots) = \frac{i}{2} \begin{bmatrix} -i & i \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1/(\omega + \omega_c) & 0 \\ 0 & 1/(\omega - \omega_c) \end{bmatrix} \begin{bmatrix} i & 1 \\ -i & 1 \end{bmatrix}\] \[= \frac{i}{2 \omega} \begin{bmatrix} P + N & -i (P - N) \\ i (P - N) & P+N \end{bmatrix}\] Note that

Permittivity Tensor in Uniformly Magnetized Cold Plasma #

Let’s call the big matrix above \( R_{\alpha, i, j} \). This lets us pull out the conductivity tensor \( \hat{\vec j} = \hat \sigma \hat {\vec E} \) \[\hat{ \vec j} = \sum_\alpha n_\alpha q_\alpha \frac{q_\alpha}{m_\alpha} \frac{i}{\omega} R_{\alpha} \vec E = (\sum_\alpha \omega_{p, \alpha} ^2 R_\alpha) \frac{i}{\omega} \epsilon_0 \hat{\vec E}\] So the conductivity tensor for a cold uniformly magnetized plasma is \[\hat \sigma = \frac{i \epsilon_0}{\omega} \sum_\alpha \omega_{p, \alpha} ^2 R_\alpha\]

As a good check, we can compare this with the conductivity for the unmagnetized case. If \( B_0 = 0 \), then \( R_\alpha \) is just the identity matrix and we recover \( \sigma = \frac{i \epsilon_0}{\omega} \omega_p ^2 \).

We can also get the dielectric and susceptibility tensors, so that we can write out the dispersion relation and start to analyze the different wave species allowed. From \( \epsilon = \epsilon_0 + \frac{i}{\omega} \sigma \) we get \[\epsilon = \epsilon_{ij} = \epsilon_0 (I _{ij} - \sum_\alpha \frac{\omega_{p, \alpha}^2}{\omega^2} R_{\alpha, ij})\] \[\rightarrow \chi_{ij} = - \sum_\alpha \frac{\omega_{p, \alpha}^2}{\omega^2} R_{\alpha, ij}\]

Trajectories of Particles in Magnetized Plasma #

Let’s take a minute to look at what happens to a particle if we suppose that a transverse, linearly polarized wave is a solution. \[\vec E_1 = \begin{bmatrix} e^{i kz - \omega t} \\ 0 \\ 0 \end{bmatrix} E_0\] \[\rightarrow \hat v_x = i \frac{\omega}{\omega^2 - \omega_c ^2} \frac{q}{m} \hat E_x\] \[\hat v_y = \frac{\omega_c}{\omega^2 - \omega_c ^2} \frac{q}{m} \hat E_x\] If we use a test field \( \hat E(\omega) = \delta(\omega - \omega_0) E_0 \) to “pick out” a frequency \( \omega_0 \), then \[v_x (t) = i \frac{\omega_0}{\omega_0 ^2 - \omega_c ^2} \frac{q}{m} E_0 e^{i (kz - \omega_0 t)} \qquad v_y(t) = \frac{\omega_c}{\omega_0 ^2- \omega_c^2 } \frac{q}{m}E_0 e^{i(k z - \omega_0 t)}\] To simplify the situation, define an amplitude \[A_v = \frac{\omega_0}{\omega_0 ^2 - \omega_c ^2}\frac{q E_0}{m}\] \[\rightarrow v_x = i A_v e^{i (kz - \omega_0 t)} \quad v_y = \frac{\omega_c}{\omega_0} A_v e^{i (kz - \omega_0 t)}\] Integrating velocity to get position, \[x = - \frac{A_v}{\omega_0} e^{i (kz - \omega_0 t)} \qquad y = \frac{\omega_c}{\omega_0} \frac{A_v}{\omega_0} i e^{i (kz - \omega_0 t)}\] Defining some more simplifying constants, \[A_x \equiv \frac{1}{\omega_0 ^2 - \omega_c ^2} \frac{q E_0 }{m} \qquad A_y \equiv \frac{\omega_c}{\omega_0} A_x\] \[\rightarrow x(t) = - A_x \cos (kz - \omega_0 t) \qquad y(t) = - A_y \sin(kz - \omega_0 t)\] How do the response amplitudes \( A_x \), \( A_y \) change with \( \omega_0 \)? There are obviously linear resonances at \( \omega = \pm \omega_c \) where the amplitudes grow indefinitely. Of course, the amplitude doesn’t actually go to \( \infty \) at resonance, we just leave the linear region where our equation of motion is valid.

Note that if you flip the sign of \( q \), the sign of \( A_x / A_y \) flips, meaning the response is asymmetric between particle species.

We can see from the resonance locations that the linear approximation fails near \( \omega \rightarrow \omega_c \). Depending on the value of \( \omega \), there is a wide variety of response by the particles which is also asymmetric in their charge. \[\frac{A_x}{A_y} = \frac{\omega_c}{\omega_0}\]

• In the case \( \omega \gg \omega_c \), then \( A_y \ll A_x \). Here, we say that the response is essentially unmagnetized, because the response is basically the same as we would see if there was no background magnetic field.
• If \( \omega < \omega_c \), then \( A_y > A_x \) and the response is definitely magnetized.

Particles have very different responses based on their charge-to-mass ratio \[\omega_{c, i} \ll \omega_{c, e}\] Since the charge-to-mass ratio of the electron is 1836 times greater than that of fully ionized ions, the difference is significant. In most cases, we’ll get \( \omega_{ci} \ll \omega \ll \omega_{ce} \), where the ions are effectively unmagnetized and the electrons are magnetized.

That’s what the response looks like for a transverse linear polarized wave, leading to a twisting elliptical orbit.

Dispersion Tensor for Cold Magnetized Plasma #

To get the dispersion relation, we start by writing the dielectric tensor \( \epsilon / \epsilon_0 \) in a more conventional format due to Stix: \[\frac{\epsilon}{\epsilon_0} = \begin{bmatrix} S & - i D & 0 \\ + i D & S & 0 \\ 0 & 0 & P \end{bmatrix}\] where we have some mnemonic terms defined as:

We refer to the “left” and “right” handedness of these solutions because different species with different \( \omega_{c, \alpha} \) respond differently when moving with or against the magnetic loop direction defined by the magnetizing field. In a non-magnetized plasma, there are no differences between \( R \) and \( L \) and we only see the ordinary plasma wave.

Now we can reduce everything to a dispersion tensor \( D(\omega, k) \) such that \[\hat D_{ij} \hat E_j = 0\] Starting with the Helmholtz wave equation \[\vec n \cross \vec n \cross \hat{\vec E} + \frac{\hat \epsilon}{\epsilon_0} \hat{\vec E} = 0\]

We rotate our coordinate system such that \( \vec k \) lies in the \( x \)-\( z \) plane, so \( \vec n = (n_x, 0, n_z) \). \[\vec n \cross \vec n \cross \vec E = (n_i n_j - n^2 I_{ij})E_j = \begin{bmatrix} n_x ^2 - n^2 & 0 & n_x n_z \\ 0 & - n^2 & 0\\ n_z n_x & 0 & n_z ^2 - n^2 \end{bmatrix} \] \[\rightarrow \vec n \cross \vec n \cross \hat{\vec E} + \frac{\hat \epsilon}{\epsilon_0} \cdot \hat{\vec E} = 0\] Let \( \theta \) be the angle between \( \vec k \) and the \( z \) axis \[n_x = n \sin \theta \qquad n_z = n \cos \theta\] \[\rightarrow 0 = \begin{bmatrix} S - n^2 \cos ^2 \theta & -i D & n^2 \cos \theta \sin \theta \\ i D & S - n^2 & 0 \\ n^2 \cos \theta \sin \theta & 0 & P - n^2 \sin ^2 \theta \end{bmatrix} \begin{bmatrix} \hat E_x \\ \hat E_y \\ \hat E_z \end{bmatrix}\]

We take the determinant of the big matrix and get a bi-quadratic equation

\[A n^4 - B n^2 + C = 0 \] \[A = S \sin ^2 \theta + P \cos ^2 \theta\] \[B = R L \sin ^2 \theta + P S (1 + \cos ^2 \theta)\] \[C = PRL\] Solving the quadratic, we get \[n^2 = \frac{B \pm F}{2A} \qquad F^2 = B^2 - 4AC = (RL - PS)^2 \sin^4 \theta + 4 P^2 D^2 \cos ^2 \theta\] Because we’re solving in terms of \( n^2 \), we will get a forward and backwards propagating mode for each solution, but also the \( \pm F \) gives us two modes for each angle \( \theta \). In the unmagnetized case, we only had one mode, so we say that the magnetization has split the isotropy of our previously isotropic plasma.

We have \( n^2 = n^2 (\theta, \omega) \). We can invert to get an expression for \( \theta \) in terms of \( n^2 \) and \( \omega \) \[\tan ^2 \theta = - \frac{P (n^2 - R)(n^2 - L)}{(Sn^2 - RL)(n^2 - P)}\] This is a pretty complicated expression with a lot of possible solution modes. The next section covers each of the solution types in detail, and puts the various “families” of solution into a topographical diagram called the CMA diagram.