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Collision Operators

Source: PHY653B Ch. 1

Intuition

The collision term \(\mathcal{C}[f]\) is the right-hand side of the Boltzmann equation and the part that makes it irreversible. In a plasma it is also, usually, small — and knowing how small is what licenses the collisionless models this course spends most of its time on.

Why plasma collisions are weak

A plasma with \(N_D \gg 1\) has many particles inside a Debye sphere, so each one is nudged by many distant neighbours rather than struck by a few close ones. The ratio of collisional to collective effects scales as the plasma parameter \(g = 1/N_D\), and for a tokamak \(N_D \sim 10^8\). Formally, \(\mathcal{C}[f] \to 0\) is legitimate precisely when \(N_D \gg 1\) — the same condition that defines an ideal plasma.

This is the licence for the Vlasov equation: not that collisions never happen, but that they act on a timescale far longer than the wave physics of interest.

Three operators, three prices

Landau (Fokker–Planck). The correct one for Coulomb collisions, derived from accumulated small-angle scattering:

\[\mathcal{C}_L[f] = \frac{\partial}{\partial v_\alpha}\int d\mathbf{v}'\, U_{\alpha\beta}\left(f'\frac{\partial f}{\partial v_\beta} - f\frac{\partial f'}{\partial v'_\beta}\right)\]

It conserves particles, momentum and energy exactly, and drives \(f\) to a Maxwellian. It is also a nonlinear integro-differential operator in a five-dimensional velocity convolution — the single most expensive object in computational plasma physics. Codes that need it (edge, runaway electrons, fast-ion slowing) pay dearly.

BGK. Relax toward local equilibrium at a single rate:

\[\mathcal{C}_{\rm BGK}[f] = -\frac{1}{\tau}\left(f - f^{\rm eq}\right)\]

Crude — one relaxation time for every moment, so it gets the Prandtl number wrong — but it conserves the right things by construction if \(f^{\rm eq}\) is built from the local moments, and it costs almost nothing. BGK is not a toy. It is the entire collision model of the lattice Boltzmann method, where it produces correct Navier–Stokes behaviour with a viscosity you can predict and measure.

Collisionless. \(\mathcal{C} = 0\). Exact for the timescales of most wave and instability problems, and the reason Landau damping — a collisionless damping — is the flagship benchmark of Chapter 2.

The catch with setting it to zero

Removing collisions removes the only thing that smooths velocity space. Nothing then stops structure in \(f\) from getting finer and finer — phase mixing runs forever, and on a grid it runs until it hits \(\Delta v\) and produces recurrence. Physically, real collisions eventually erase filaments below the collisional scale; numerically, a small velocity-space diffusion does the same job. Adding it is defensible, but it must be stated: you have chosen to break reversibility to buy well-behaved late-time results.

Common mistakes

  • Believing collisionless means "no dissipation of wave energy". Landau damping is collisionless and irreversible-looking. Energy moves from the wave to the resonant particles with \(\mathcal{C} = 0\).
  • Using BGK where the Prandtl number matters. A single relaxation time ties viscosity and thermal conductivity together. Multi-relaxation-time schemes exist precisely to break that link.
  • Adding numerical velocity diffusion silently. It is a physics change, not a numerical detail, and it sets a floor on the damping rates you can measure.

Knowledge graph position

Prerequisites: Coulomb collisions, ideal plasma. Leads to: Vlasov solvers, LBM.

Quiz

Q1 (conceptual). What condition licenses setting \(\mathcal{C}[f] = 0\)?

Answer

\(N_D \gg 1\): many particles per Debye sphere, so collective long-range forces dominate binary collisions. Equivalently the plasma parameter \(g = 1/N_D \ll 1\). It is the same condition that defines an ideal plasma.

Q2 (conceptual). Why is BGK acceptable for lattice Boltzmann but not for a runaway-electron calculation?

Answer

LBM only needs the correct hydrodynamic limit, and Chapman–Enskog shows single-time relaxation delivers Navier–Stokes with \(\nu = c_s^2(\tau-\frac12)\). Runaway electrons depend on the velocity dependence of the drag (\(\nu \propto v^{-3}\)), which a single relaxation time destroys — the entire runaway mechanism disappears.

Q3 (MCQ). The Landau collision operator is expensive mainly because it is:

  • (a) implicit in time
  • (b) a nonlinear convolution over the full velocity space
  • (c) only stable at small time steps
  • (d) not conservative
Answer

(b). Each point in velocity space couples to every other through the tensor kernel, so a naive evaluation is quadratic in the velocity-grid size — and it is nonlinear in \(f\) as well.