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The Weibel Instability

Source: PHY653B Ch. 3

Intuition

The Weibel instability is the standard benchmark for electromagnetic PIC, in the same way Landau damping is the benchmark for electrostatic Vlasov. Its driver is remarkable: a plasma with no beams, no gradients, no currents, and no free energy other than a temperature anisotropy spontaneously generates magnetic field out of nothing.

The mechanism

Take a plasma hotter along \(x\) than along \(y\): \(T_x > T_y\). Seed a small magnetic perturbation \(B_z(y)\). The Lorentz force sorts the fast \(x\)-moving particles into current filaments — those moving \(+x\) deflect one way, those moving \(-x\) the other. Those filaments carry current, and their current reinforces the seed field.

Positive feedback. The field grows exponentially, drawing energy from the anisotropy until the distribution is isotropic, at which point growth stops. It is a purely kinetic, purely electromagnetic instability with no fluid counterpart, and it converts thermal free energy directly into magnetic energy.

Why it is the right benchmark

  • It requires the full Maxwell solver. An electrostatic code cannot produce it at all — the physics is entirely in the transverse magnetic field. It therefore tests the parts of an EM PIC code an electrostatic benchmark leaves untouched.
  • It has an analytic growth rate in the linear phase for a bi-Maxwellian, so there is a number to hit.
  • It saturates physically, at a level set by isotropisation rather than by numerics — so it also tests the nonlinear stage.
  • It is genuinely important. Weibel is the leading candidate for generating the magnetic fields in collisionless astrophysical shocks — gamma-ray-burst afterglows, supernova remnants — and hence for the field that makes those shocks work at all. It also matters in fast-ignition inertial fusion, where it filaments the hot-electron beam.

What to measure

  1. Linear growth rate. Fit \(\ln \mathcal{E}_B(t)\) over a window after the initial transient and before saturation. As always the field energy grows at \(2\gamma\) — the same factor of two that catches everyone.
  2. Filament wavelength. The fastest-growing mode has a characteristic \(k\); the filaments should appear at that spacing, visible directly in a current-density map.
  3. Saturation level. Compare the final magnetic energy against the anisotropy energy available. It should account for the free energy, not exceed it.
  4. Isotropisation. \(T_x\) and \(T_y\) should converge. If they do not, the run stopped early or something is injecting energy.

Common mistakes

  • Insufficient markers. Weibel grows from noise, and PIC has plenty. With too few markers the instability starts from an unphysically large seed and the "linear phase" is contaminated from the outset. Run at 2× and 4× \(N_{\rm ppc}\) and check the growth rate is unchanged; the saturation level is much more forgiving than the growth phase.
  • Fitting through the initial transient. The seed is broadband noise; only after the fastest mode dominates is the growth a clean exponential.
  • Confusing Weibel with two-stream. Two-stream is electrostatic and longitudinal, driven by relative drift; Weibel is electromagnetic and transverse, driven by anisotropy. Both filament, and both are easy to mislabel in a plot.

Knowledge graph position

Prerequisites: PIC, Maxwell/FDTD solvers, Boris pusher. Leads to: collisionless shock simulation, astrophysical magnetogenesis, fast-ignition physics.

Quiz

Q1 (conceptual). What is the free energy source for the Weibel instability?

Answer

Temperature anisotropy — the plasma is hotter in one direction than another. No drift, no gradient, no current is needed. The instability converts that anisotropy into magnetic energy and saturates when the distribution becomes isotropic.

Q2 (conceptual). Why can an electrostatic PIC code not reproduce Weibel?

Answer

The instability lives entirely in the transverse electromagnetic field: the growing quantity is \(B_z\), generated by transverse current filaments. An electrostatic code solves only Poisson for a longitudinal \(\mathbf{E}\) and has no magnetic field to grow.

Q3 (MCQ). In a PIC simulation of Weibel, the linear growth phase is most sensitive to:

  • (a) the saturation level
  • (b) the marker count, because the instability grows out of PIC noise
  • (c) the box length
  • (d) the shape-function order alone
Answer

(b). Weibel amplifies whatever seed it is given, and in PIC that seed is statistical noise. Too few markers means an unphysically large starting amplitude and a contaminated linear phase — so the growth rate must be shown to be independent of \(N_{\rm ppc}\).