Transport theory and the Boltzmann equation

Two opposite populations, with no bulk flow

Imagine a small, spatially uniform sample of dilute gas. Half its particles tend to move to the right and half to the left. The average velocity is zero, but motion along that direction is stronger than motion across it. Can collisions change this distribution while leaving particle number, total momentum and total energy unchanged?

Yes. Conservation fixes only particular averages, not the whole distribution of velocities. Collisions can redistribute energy between directions without creating energy or starting a bulk flow. Our supplied example begins with directional pressures in the ratio 5:1:1. Its equilibrium ratio will be 1:1:1, with the same sum of 7. The numerical values are dimensionless model values, not measurements of a named material.

We will first describe the distribution, identify what collisions preserve, and solve a deliberately simple relaxation model. Then we will see why a flowing gas needs more than a local equilibrium formula. The advanced return path derives the collision entropy inequality, the linearized invariant modes, and the conditions on a fluid approximation. The example is not an exact solution of the full Boltzmann collision integral.

Define the distribution before taking its averages

Let f(x,v,t) d³x d³v be the expected number of particles in a small position-and-velocity element. Here x is position, v is particle velocity and t is time. Integrating over velocity gives the local number density n=∫f d³v. Dividing by n gives a normalized local velocity probability density. The factor 1/n matters: u=(1/n)∫v f d³v is the mean flow velocity, not the unnormalized particle flux.

A particle’s velocity relative to that mean is c=v−u, called its peculiar velocity. Then ∫c f d³v=0. This is an average cancellation; individual particles still move. A distribution can have u=0 while being anisotropic, meaning different directions have different statistics. It can also have zero heat flux without being Maxwellian.

In SI units, n has units m⁻³ and f has units s³ m⁻⁶ when particle count is treated as dimensionless. A velocity marginal integrates over the other velocity components. Its graph is a density per velocity interval, not the probability of one exact velocity and not a picture of particle positions. The measure must stay the same when comparing curves or defining an entropy.

Separate travel, acceleration and collisions

For a specified velocity-independent force F(x,t) on each particle of mass m, the kinetic equation is ∂_t f+v·∇_x f+(F/m)·∇_v f=C[f]. Free streaming moves particles between positions. The force moves their velocities. The collision operator C redistributes velocities through encounters at a position. These are different mechanisms, even when they act simultaneously.

For example, a distribution that varies in space can change through streaming even where its collision term is zero. Conversely, our uniform force-free example has no streaming or acceleration term, so its change comes entirely from the assigned collisions. Uniformity here does not describe two spatial clouds separated by a hidden interface.

The displayed convective force term uses zero divergence of F/m in velocity space. A more general acceleration law needs the conservative phase-space form ∂_t f+∇_x·(v f)+∇_v·(a f)=C[f]. One must not silently use the simpler equation for an arbitrary compressible phase-space flow.

Read collision gain and loss

For the classical, dilute, elastic, binary-collision description, a standard gain-minus-loss form is C[f](v)=∫d³v₁ dΩ B(v−v₁,Ω)[f(v′)f(v₁′)−f(v)f(v₁)]. All distributions in this expression refer to the same position and time. The two velocity pairs are linked by the stated elastic scattering rule. Microreversibility allows the inverse encounter to supply the gain term in this common notation.

The nonnegative kernel B contains relative speed and the differential scattering cross section in the chosen angular convention. The loss term removes particles from a small velocity neighbourhood around v through encounters with other velocities. The gain term adds particles whose incoming velocities scatter into that neighbourhood. A collision does not remove a particle from the gas merely because it removes it from one velocity interval.

The product of two one-particle distributions is a closure assumption for the incoming pair statistics, often called molecular chaos. It is not the claim that collisions create no correlations, or that every exact many-particle state always factors into independent particles. The dilute binary regime and suitable scale separation matter. A measured cross section belongs in a quantitative material prediction; none has been assigned to our example.

Identify what collisions cannot change

An elastic encounter between identical particles preserves the pair’s total momentum and kinetic energy, as well as its particle count. A collision invariant ψ satisfies ψ(v)+ψ(v₁)=ψ(v′)+ψ(v₁′). For the ordinary nondegenerate monatomic elastic problem the independent invariants are 1, the three velocity components, and |v|². Multiplying by fixed mass factors gives the corresponding physical momentum and energy.

The collision integral therefore obeys ∫C d³v=0, ∫m v C d³v=0 and ∫(m|v|²/2)C d³v=0. These are local collision statements. Streaming can still change the number, momentum and energy in a small region by carrying them across its boundary, and an external force can exchange momentum and energy with the gas.

A directional second moment such as ∫v_x² f d³v is not separately invariant. Neither is a fourth moment. Their relaxation is possible precisely because only the appropriate total energy is constrained. Additional species, internal states, reactions or inelastic encounters change which invariants and energy bookkeeping are appropriate.

Match a local Maxwellian

A local Maxwellian with the same number density, flow and translational energy is M=n[m/(2πk_B T)]^(3/2) exp[−m|v−u|²/(2k_B T)]. The Boltzmann constant k_B converts temperature to energy. For this monatomic model, the matching kinetic temperature is defined by (m/2)∫|c|² f d³v=(3/2)n k_B T. Away from equilibrium this energy-based definition need not have every meaning of an equilibrium thermometer reading.

Under the usual elastic collision-connectivity conditions, a positive collision-equilibrium distribution has ln M equal to a linear combination of the invariants. The normalizable isotropic form is the Maxwellian above. Thus C[M]=0 for the full classical operator in this regime. That equation alone does not say that a spatially varying local Maxwellian solves the streaming equation.

Local equilibrium and global equilibrium are different. Nearby particles may have almost Maxwellian velocity statistics while density, temperature or mean flow varies across an apparatus. Those slowly varying conserved fields drive transport. A gas can therefore be close to local equilibrium and still conduct heat or dissipate shear.

Solve a matching-moment relaxation model

To make the opening question exactly solvable, assign the BGK model C_BGK=(M[f]−f)/τ, with a positive constant relaxation time τ. The Maxwellian M[f] must share f’s number, momentum and total peculiar kinetic energy. Relaxation toward an arbitrarily chosen Gaussian could violate conservation and would be a different, generally unsuitable model.

In a homogeneous force-free sample these matching moments are constant. Consequently M[f] is constant too, and the equation has the exact solution f(t)=e^(−t/τ)f(0)+(1−e^(−t/τ))M. It is a positive convex mixture, not a polynomial correction that can turn negative in a tail. The equality is exact for this assigned BGK problem, not for every Boltzmann kernel or every nonequilibrium flow.

Use σ₀=√(k_B T₀/m), dimensionless velocity w=v/σ₀ and time s=t/τ, with positive symbolic T₀ and density n₀. Write Φ=σ₀³f/n₀, so ∫Φ d³w=1. Let G_θ(w)=(2πθ)^(−3/2)exp[−|w|²/(2θ)]. Our initial distribution is Φ₀=[G₁(w−2e_x)+G₁(w+2e_x)]/2, where e_x is a unit vector along x.

The two components each have variance 1 in every direction, but their opposite drift velocities contribute to energy. Their combined mean is zero; their second moments are ⟨w_x²⟩=5 and ⟨w_y²⟩=⟨w_z²⟩=1. The conserved sum is 7, so the matching isotropic variance is θ_eq=7/3, not 1. Therefore Φ_s=e^(−s)Φ₀+(1−e^(−s))G_(7/3).

See directional pressure relax at fixed energy

Define the pressure tensor P_ij=m∫c_i c_j f d³v. A diagonal component measures the corresponding directional momentum flux; its unit is the pascal. The scalar kinetic pressure is p=(P_xx+P_yy+P_zz)/3. Outside an isotropic state, that average does not replace the individual tensor components.

In the example, divide P by n₀k_B T₀. The exact components are P_xx=7/3+(8/3)e^(−s) and P_yy=P_zz=7/3−(4/3)e^(−s), in these normalized units. Their sum stays 7 while their difference P_xx−P_yy=4e^(−s) decays. At one relaxation time, s=1, the difference is about 1.4715178: one time constant is not complete equilibration.

Every off-diagonal component and the heat-flux vector vanish here by reflection symmetry. Nevertheless the finite-time distribution is generally not Maxwellian. Its fourth x moment is ⟨w_x⁴⟩=49/3+(80/3)e^(−s), beginning at 43 and approaching 49/3. These are shape changes at fixed number, mean and total second moment.

At the final supplied time s=8, the pressure difference is still about 0.0013418505, strictly positive. Exact isotropy belongs to the separately labelled s→∞ limit. A plotted curve may look converged at finite resolution; the accompanying values must not teach that a finite endpoint is exactly the limit.

Keep the distribution outside the plotting window

The x marginal initially has two peaks associated with the two opposite drift populations. The y marginal starts as a narrower, single Gaussian. The common equilibrium marginal has variance 7/3. Comparing both directions shows redistribution that an x-only plot could hide. These curves are ensemble velocity densities, not trajectories of individual molecules.

The supplied samples run from w=−8 to w=8 in steps of 1/20. The Gaussian mixtures themselves extend over the whole real line and are normalized there. Their positive tails outside this plotting window are not cut off physically, and the sampled curves are not renormalized to make the displayed area one.

This distinction also matters for moments: the exact pressure and fourth-moment formulas use the full infinite-domain distribution. A finite sum of plotted density samples is neither that integral nor a numerical entropy calculation. An honest finite canvas can show the important shape while its text and tables retain the complete model and its limits.