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advanced27 min studyLesson 355 in Physics

Relax toward a Maxwellian without losing conservation

Use moment matching and separate physical relaxation from numerical stepping.

Before you begin
On this page

The question for today

What you’ll understand

  • Construct a conserving BGK relaxation operator.
  • Solve a homogeneous nonconserved perturbation exactly.
  • Distinguish physical stability, numerical stability, and positivity.

The full collision integral contains detailed angular and relative-speed information. A relaxation model replaces that complexity with motion toward a local equilibrium over one timescale. This simplification is useful only if the target equilibrium shares the moments that collisions must conserve. The resulting BGK model also teaches an important numerical distinction: an exact relaxing equation can have a poorly chosen time step that oscillates or makes a distribution negative.

The target must share the protected moments

The BGK operator is the difference between a local Maxwellian and the actual distribution divided by a positive relaxation time. The Maxwellian is recomputed from the actual density, mean velocity, and kinetic temperature at that same point. Its five conserved moments therefore match those of the distribution. Taking invariant moments of the difference gives zero exactly in the continuous model, preserving the defining conservation laws of elastic gas collisions.

Relaxation toward an externally fixed Maxwellian is a different model. It can describe a system exchanging momentum and heat with a bath, but it generally does not conserve the gas’s own momentum and energy. Likewise, a velocity-dependent relaxation time placed directly under the BGK difference can spoil moment cancellation unless the construction is modified. The attractive phrase “relaxes to equilibrium” is insufficient to establish either conservation or the appropriate physical environment.

CBGK[f]=M[f]−fτ,∫(M[f]−f){1,v,v2} d3v=0,τ>0C_{\rm BGK}[f]=\frac{M[f]-f}{\tau},\qquad \int (M[f]-f)\{1,\mathbf v,v^2\}\,d^3v=0,\qquad \tau>0
τ is relaxation time in s. M[f] is the Maxwellian with the same number, momentum, and kinetic-energy moments as f; the displayed braces denote the separate invariant moments.↔ Scroll wide equations horizontally.

Homogeneous relaxation is an exact benchmark

In a force-free spatially uniform gas, the conserved moments cannot change through BGK collisions. Its target Maxwellian is consequently constant in time. The remaining difference from equilibrium decays exponentially, with every nonconserved perturbation assigned the same relaxation time in this simple model. The exact solution is a convex combination of the initial distribution and its matched Maxwellian, so nonnegative initial data stay nonnegative.

The single-timescale assumption is not an exact property of the Boltzmann operator. Different angular and energy distortions generally relax at different rates. In particular, basic monatomic BGK ties viscosity and thermal conductivity so that its Prandtl number is one, whereas dilute monatomic gases have values near two thirds. Matching a viscosity by choosing the relaxation time therefore does not also guarantee an accurate heat conductivity, high-order kinetic spectrum, or rarefied boundary response.

f(t)=M+[f(0)−M]e−t/τ,a(t)=a0e−t/τf(t)=M+[f(0)-M]e^{-t/\tau},\qquad a(t)=a_0e^{-t/\tau}
a is the amplitude of a zero-invariant-moment perturbation; its units depend on the chosen perturbation basis. This exact formula assumes homogeneous, force-free relaxation with fixed conserved moments.↔ Scroll wide equations horizontally.

A stable equation can have a bad update

Explicit Euler stepping replaces exponential relaxation by a multiplier one minus the time step divided by the relaxation time. A perturbation decays in magnitude when this ratio lies strictly between zero and two. Ratios above one alternate its sign, and the full distribution update is no longer a convex combination. Thus linear amplitude stability alone does not guarantee nonnegative populations. A conservative velocity discretization must also approximate the Maxwellian moments consistently; the analytic operator cannot rescue an inconsistent quadrature.

Use the optimization lab to inspect this multiplier without calling it a gas. Reset and set step size to 0.5 and initial position to 1, then Step once; predict 0.5. Reset and set step size to 1.5 and initial position to 1, then Step once; predict −0.5. Both shrink in magnitude, but the latter changes sign. The instrument is gradient descent on a quadratic, not a BGK distribution or a test of particle-population positivity.

ak+1=(1−Δtτ)ak,0<Δt/τ<2 for amplitude decay,0≤Δt/τ≤1 for a convex updatea_{k+1}=\left(1-\frac{\Delta t}{\tau}\right)a_k,\qquad 0<\Delta t/\tau<2\ \text{for amplitude decay},\qquad 0\leq\Delta t/\tau\leq1\ \text{for a convex update}
Δt and τ are in s. The first condition concerns the linear relaxation mode; the second is a sufficient positivity condition for the homogeneous explicit BGK mixture when both distributions are nonnegative.↔ Scroll wide equations horizontally.

Work it through

Compare exact and stepped decay

A homogeneous BGK perturbation has amplitude 0.20 initially and relaxation time 2 ms. Find its amplitude after 2 ms exactly, and compare one explicit Euler step of 2 ms with two steps of 1 ms.

  1. The elapsed time equals one relaxation time, so exact amplitude is 0.20/e≈0.07358. This exponential follows the continuous equation and preserves the sign of the initial amplitude.

  2. One Euler step has Δt/τ=1, so its multiplier is zero and it removes the perturbation completely. That is stable and nonnegative as a distribution mixture, but it is inaccurate for this finite time interval.

  3. Two 1 ms steps each have multiplier 1−1/2=1/2, giving 0.20×(1/2)²=0.0500. It is closer to the exact value but still underestimates it. Numerical decay rate approaches the physical one only with a controlled time resolution.

The result

The exact amplitude is 0.07358; one coarse Euler step gives zero and two half-size steps give 0.0500. Stability alone does not guarantee accuracy.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

Inspect a relaxation multiplier

Reset and set η=0.5, x₀=1, then Step once; predict x₁=0.5. Reset and set η=1.5, x₀=1, then Step once; predict x₁=−0.5.

Inspect a relaxation multiplierGradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Current readings: Position x0: 2; Objective x²/2: 2; Iteration behavior: Stable step. All parameters are adjustable with the labeled controls below.−4−202402.174.356.528.7Parameter xf(x) = x²/2n = 0
ObjectiveDiscrete iterate
Position x0
2
Objective x²/2
2
Iteration behavior
Stable step
n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis quadratic gradient-descent recurrence matches an Euler relaxation multiplier algebraically. It does not track gas moments, Maxwellians, or distribution positivity.

Pause & reason

Check your understanding

What makes the continuous basic BGK operator conserve momentum and energy?

Your turn

Try it yourself

Take a homogeneous perturbation amplitude 0.10 and choose Δt=1.5τ. Compute the amplitude after two Euler steps, compare the exact value at 3τ, and assess what stability does and does not establish.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The Euler multiplier is −0.5 each step, while exact decay uses exp(−3).

Compare with the solution

Two Euler steps give 0.10×(−0.5)²=0.025. Exact evolution gives 0.10exp(−3)≈0.004979. The discrete amplitude shrinks but alternates in sign, so it is linearly stable yet inaccurate here. Because the full update has a negative coefficient multiplying the old distribution, this step size does not guarantee positivity for arbitrary nonnegative initial distributions.

Keep the idea

Take it with you

  • BGK conserves the gas moments only when its target is matched appropriately.
  • Single-time relaxation is a model with restricted transport predictions.
  • Numerical stability, accuracy, and positivity are distinct requirements.

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