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Electromagnetic Waves in an Unmagnetised Plasma

Source lecture(s): PC368 Lec 11

Intuition

Light entering a plasma has to push the electrons around, and pushing electrons costs the wave something. The result is the simplest and most useful dispersion relation in the subject: a plasma is transparent to high frequencies and a mirror to low ones, with a hard boundary at the plasma frequency.

The dispersion relation

Linearise the electron fluid against a transverse wave (\(\mathbf{k}\perp\mathbf{E}\), so no density perturbation and no pressure), feed the current into Ampère's law, and:

EM waves in an unmagnetised plasma

\[\omega^2 = \omega_{pe}^2 + c^2k^2\]

Everything follows from reading this equation carefully:

  • \(\omega < \omega_{pe}\): \(k^2 < 0\). The wave is evanescent — it does not attenuate by absorption, it simply cannot propagate, and is reflected. The skin depth is \(c/\omega_{pe}\).
  • \(\omega > \omega_{pe}\): propagates, with refractive index \(n = ck/\omega = \sqrt{1 - \omega_{pe}^2/\omega^2} < 1\).
  • \(\omega = \omega_{pe}\): the cutoff. Since \(\omega_{pe} \propto \sqrt{n_e}\), cutoff frequency is a direct proxy for density.

The refractive index is less than one

\(n < 1\) means the phase velocity \(v_p = c/n\) exceeds \(c\). No relativity is harmed: the group velocity

\[v_g = \frac{d\omega}{dk} = \frac{c^2k}{\omega} = c\,n < c\]

carries the energy and the information. Note the tidy result \(v_pv_g = c^2\). A plasma is the standard classroom example of superluminal phase velocity precisely because the effect is large and unambiguous.

Where you meet this

The ionosphere. Peak \(n_e \sim 10^{12}\) m⁻³ gives \(f_{pe} \approx 9\) MHz. Below that, radio reflects — which is how shortwave broadcasting reaches the other side of the planet, and why AM stations carry much further at night (the absorbing D layer recombines after sunset). Above it, signals pass through: FM at 100 MHz, GPS at 1.5 GHz, and every satellite link. The critical frequency is measured routinely by ionosondes, sweeping frequency and timing the echo.

Density interferometry. Since the phase accumulated across a plasma depends on \(n_e\), a microwave or laser beam crossing a tokamak returns a line-integrated density. This is the standard density diagnostic in every fusion device, and it is nothing more than this dispersion relation used backwards.

Radio bursts from the Sun. As a disturbance travels outward through the falling coronal density, the local \(\omega_{pe}\) falls with it, and the emission drifts down in frequency. Type III bursts sweep from hundreds of MHz to kHz in minutes; the drift rate measures the speed of the electron beam exciting them.

Metals. The same equation with conduction electrons puts \(\omega_{pe}\) in the ultraviolet, which is why metals reflect visible light and go transparent in the far UV.

Common mistakes

  • Calling the reflection "absorption". Below cutoff, \(k\) is purely imaginary: the wave field decays evanescently and the energy goes back out. Nothing is dissipated in the collisionless limit.
  • Worrying about \(v_p > c\). Only \(v_g\) transports energy, and \(v_g < c\) always.
  • Assuming this holds when there is a magnetic field. It does not. Adding \(\mathbf{B}\) splits this single branch into the whole zoo of cold-plasma modes — R, L, O and X — with several cutoffs and resonances. Explore them in the cold plasma wave explorer.

Knowledge graph position

Prerequisites: Plasma frequency, plasma waves. Leads to: magnetised waves, cold-plasma dispersion, RF heating and diagnostics.

Quiz

Q1 (conceptual). Why does a plasma reflect rather than absorb radio below \(\omega_{pe}\)?

Answer

Below cutoff \(k^2 < 0\), so the solution is evanescent, not oscillatory. The field decays over a skin depth \(c/\omega_{pe}\) and the energy is returned — with no collisions there is no dissipation channel. It is total reflection, exactly like light beyond the critical angle.

Q2 (computational). An ionosonde finds a critical frequency of 8 MHz. What is the peak electron density?

Answer

\(f_{pe} = 8980\sqrt{n_e[\text{cm}^{-3}]}\) Hz, so \(\sqrt{n_e} = 8\times10^6/8980 = 891\), giving \(n_e \approx 7.9\times10^5\) cm⁻³ \(= 7.9\times10^{11}\) m⁻³.

Q3 (MCQ). GPS signals at 1.5 GHz reach the ground because:

  • (a) they are strong enough to burn through
  • (b) 1.5 GHz is far above the ionospheric plasma frequency of ~9 MHz
  • (c) the ionosphere is transparent to all frequencies
  • (d) satellites are above the ionosphere
Answer

(b). Well above cutoff the plasma is nearly transparent — though not perfectly: the residual \(n < 1\) delays the signal by a density-dependent amount, and correcting that ionospheric delay is a routine part of precision GPS.