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The Cold-Plasma Dielectric Tensor

Source lecture(s): PC368 Lec 13

Intuition

A magnetised plasma is an anisotropic dielectric. Push the electrons along \(\mathbf{B}\) and they respond freely; push them across and they gyrate instead, responding differently — and with a component at right angles to the push, because \(\mathbf{v}\times\mathbf{B}\) turns the motion. Three independent responses, so three independent numbers. Those numbers are the Stix parameters, and every named wave in a cold plasma is a corner of the parameter space they span.

Deriving it

Each species obeys the linearised cold momentum equation (no pressure — that is what "cold" buys):

\[-i\omega m_\sigma \mathbf{u}_\sigma = q_\sigma(\mathbf{E} + \mathbf{u}_\sigma\times\mathbf{B}_0)\]

Solve for \(\mathbf{u}_\sigma\) parallel and perpendicular to \(\hat{\mathbf{z}}\) separately, sum the currents \(\mathbf{J} = \sum_\sigma n_\sigma q_\sigma\mathbf{u}_\sigma\), and fold the result into Ampère's law. With signed cyclotron frequency \(\omega_{c\sigma} = q_\sigma B_0/m_\sigma\) (negative for electrons):

\[\overleftrightarrow{K} = \begin{pmatrix} S & -iD & 0\\ iD & S & 0\\ 0 & 0 & P\end{pmatrix}\]

Stix parameters

\[S = 1 - \sum_\sigma\frac{\omega_{p\sigma}^2}{\omega^2-\omega_{c\sigma}^2},\qquad D = \sum_\sigma\frac{\omega_{c\sigma}}{\omega}\frac{\omega_{p\sigma}^2}{\omega^2-\omega_{c\sigma}^2},\qquad P = 1 - \sum_\sigma\frac{\omega_{p\sigma}^2}{\omega^2}$$ $$R = S + D,\qquad L = S - D\]

The names are mnemonic: Sum and Difference, Parallel, Right and Left. \(P\) is the unmagnetised answer — no \(\omega_c\) appears, because motion along \(\mathbf{B}\) does not feel it. The off-diagonal \(D\) is the gyrotropic term, and it is the reason a magnetised plasma rotates the polarisation of light passing through it.

The dispersion relation

Substituting into the wave equation gives a quadratic in \(n^2 = c^2k^2/\omega^2\):

\[An^4 - Bn^2 + C = 0$$ $$A = S\sin^2\theta + P\cos^2\theta,\quad B = RL\sin^2\theta + PS(1+\cos^2\theta),\quad C = PRL\]

Two roots: two modes at every frequency and angle.

The two limits worth memorising

Parallel propagation (\(\theta = 0\)): \(A = P\), and the roots collapse to

\[n^2 = R \quad\text{and}\quad n^2 = L\]

Right- and left-hand circularly polarised waves. Because \(R \ne L\) they travel at different speeds, and a linear polarisation rotates — Faraday rotation, the standard remote measurement of astrophysical magnetic fields. The R wave resonates at \(\omega_{ce}\) (electrons gyrate the same way the wave rotates, so they stay in phase and absorb) and the L wave at \(\omega_{ci}\).

Perpendicular propagation (\(\theta = \pi/2\)): \(A = S\), and

\[n^2 = P \quad (\text{\textbf{O}-mode}) \qquad n^2 = \frac{RL}{S} \quad (\text{\textbf{X}-mode})\]

The ordinary mode has \(\mathbf{E}\parallel\mathbf{B}_0\) and does not notice the field at all — its cutoff is plain \(\omega_{pe}\), which is what makes it the mode of choice for interferometry. The extraordinary mode has \(\mathbf{E}\perp\mathbf{B}_0\), is partly longitudinal, and resonates where \(S = 0\) — the hybrid resonances:

\[\omega_{UH}^2 = \omega_{pe}^2 + \omega_{ce}^2, \qquad \frac{1}{\omega_{LH}^2} \approx \frac{1}{\omega_{ce}\omega_{ci}} + \frac{1}{\omega_{pi}^2}\]

Cutoffs versus resonances

Condition \(n^2\) Physical behaviour
Cutoff \(P=0\), \(R=0\), or \(L=0\) \(\to 0\) wavelength → ∞, wave reflects
Resonance \(A = 0\) \(\to \infty\) wavelength → 0, wave slows and is absorbed

Cutoffs are angle-independent; resonances are not. This asymmetry is the whole art of RF heating: you must find a path through frequency and angle that reaches the resonance layer without first hitting a cutoff.

Common mistakes

  • Dropping the sign of \(\omega_{ce}\). Electrons have negative charge, so their signed cyclotron frequency is negative. Getting this wrong swaps R and L and inverts the Faraday rotation.
  • Assuming resonance means the wave stops existing. It means \(n^2\to\infty\), so \(\lambda\to0\); the cold model then breaks down and warm-plasma effects (finite Larmor radius, Landau damping) take over and absorb the energy.
  • Forgetting the ions. They contribute negligibly near \(\omega_{pe}\) but they own everything below \(\omega_{ci}\) — including Alfvén waves and the lower hybrid resonance.

Knowledge graph position

Prerequisites: Plasma frequency, Larmor radius, plasma waves. Leads to: resonance cones, RF heating and current drive, magnetospheric wave propagation.

Quiz

Q1 (conceptual). Why does the O-mode not care about the magnetic field?

Answer

Its electric field is parallel to \(\mathbf{B}_0\), so the electron motion it drives is along the field — where \(\mathbf{v}\times\mathbf{B} = 0\). The response is the unmagnetised one, \(n^2 = P = 1 - \omega_{pe}^2/\omega^2\), which is exactly why its cutoff cleanly measures density.

Q2 (computational). A tokamak has \(B = 5\) T and \(n_e = 10^{20}\) m⁻³. Find \(f_{ce}\), \(f_{pe}\) and the upper hybrid frequency.

Answer

\(f_{ce} = eB/2\pi m_e = 1.4\times10^{11}\) Hz = 140 GHz. \(f_{pe} = 8980\sqrt{10^{14}} = 8.98\times10^{10}\) Hz ≈ 90 GHz. \(f_{UH} = \sqrt{f_{pe}^2+f_{ce}^2} = \sqrt{90^2+140^2} \approx 166\) GHz. (Which is why ECRH systems are built around 140–170 GHz gyrotrons.)

Q3 (MCQ). Faraday rotation of a signal crossing a magnetised plasma happens because:

  • (a) the plasma is birefringent: \(R \neq L\), so the two circular polarisations travel at different speeds
  • (b) the magnetic field exerts a torque on the photons
  • (c) the plasma absorbs one polarisation
  • (d) the wave frequency shifts
Answer

(a). A linear polarisation is a sum of R and L circular components; unequal phase velocities accumulate a relative phase, which rotates the plane. The rotation is proportional to \(\int n_eB_\parallel\,dl\) — which is how the Galactic magnetic field is mapped.