advanced27 min studyLesson 354 in Physics

Identify the five slow fields of a monatomic gas

Use collision invariants to derive conservation and a local Maxwellian closure.

On this page

What you’ll understand

  • List the independent elastic collision invariants.
  • Recover local density, flow velocity, and temperature from a Maxwellian.
  • Distinguish local collision equilibrium from a global stationary solution.

Collisions can rapidly reshape a velocity distribution while leaving several moments unchanged. Those protected moments become the slow variables of hydrodynamics. For a simple three-dimensional monatomic gas they are particle number, three momentum components, and kinetic energy. The local Maxwellian is the distribution compatible with detailed balance for those constraints. It closes the ideal-fluid equations, but it does not eliminate streaming or make every inhomogeneous gas globally equilibrated.

Conserved pairs give vanishing moments

A collision invariant is a function whose sum over the two particles is unchanged by every allowed encounter. For elastic equal-mass particles, constants count particles, linear functions of velocity count momentum, and the squared speed counts kinetic energy. Symmetrizing the collision integral shows that its moment against any such invariant vanishes. This cancellation is local in position: an ordinary binary collision transfers velocity-space population without moving particles across a macroscopic spatial distance.

Taking these moments of the full kinetic equation therefore leaves spatial fluxes and external-force terms, but no collisional source of mass, momentum, or energy in the isolated gas. Internal exchanges between species require more careful bookkeeping; one species’ momentum can change while total momentum stays fixed. Reactions and inelastic internal excitation alter which individual moments are protected. Our five-field account assumes one nonreacting classical species with translational energy as its only internal energy.

∫C[f](1mv12mv2)d3v=(000)\int C[f]\begin{pmatrix}1\\m\mathbf v\\\tfrac12mv^2\end{pmatrix}d^3v=\begin{pmatrix}0\\\mathbf0\\0\end{pmatrix}
The scalar, vector, and scalar rows express five independent local conservation statements. Their physical moments are number density, momentum density, and kinetic energy density for elastic single-species collisions.↔ Scroll wide equations horizontally.

Detailed balance selects a Gaussian

Zero collision production requires the logarithm of a positive equilibrium distribution to be a linear combination of collision invariants. Complete the velocity square in that expression. Normalizability fixes the coefficient of squared speed to be negative, while the linear term shifts the Gaussian’s center to the local flow velocity. The multiplicative constant fixes number density. In three dimensions the remaining width is conventionally expressed through temperature and Boltzmann’s constant.

The resulting Maxwellian has equal random variances in all directions, isotropic pressure, and zero heat flux in the local rest frame. It maximizes kinetic entropy among distributions with the same five moments. The gas may have a large mean speed in a laboratory frame without changing these local random moments. Galilean translation shifts the center of the velocity distribution; it does not broaden the distribution or heat the gas by itself.

M=n(m2πkBT)3/2exp⁡[−m∣v−u∣22kBT],Pij=nkBTδij,q=0M=n\left(\frac{m}{2\pi k_BT}\right)^{3/2}\exp\left[-\frac{m|\mathbf v-\mathbf u|^2}{2k_BT}\right],\qquad P_{ij}=nk_BT\delta_{ij},\quad \mathbf q=0
M has the same number-distribution units as f, n is in m⁻³, m in kg, T in K, and kB in J/K. These Maxwellian moment identities apply to a classical monatomic gas.↔ Scroll wide equations horizontally.

A local closure is not a static universe

Each spatial point can have its own density, velocity, and temperature while its collision operator vanishes. Streaming between neighboring points generally makes the full kinetic left-hand side nonzero, so such a local Maxwellian is the leading approximation in a slowly varying flow. Taking its moments gives Euler-level fluxes. Viscous stress and heat conduction require departures from this zeroth-order shape; adding those fluxes without changing the closure would be inconsistent.

The heat lab illustrates slow variation after a constitutive closure has already been chosen. Reset and set spatial mode to 1 and diffusivity to 0.5, then Step once; predict amplitude exp(−0.25). Reset and set mode to 2 and diffusivity to 0.5, then Step once; predict exp(−1). These are exact periodic heat-equation modes. They display a continuum consequence of local transport, not collision invariants, a Maxwellian velocity distribution, or a derivation of the heat equation.

eint=32nkBT,etotal=12mn∣u∣2+einte_{\rm int}=\frac32nk_BT,\qquad e_{\rm total}=\frac12mn|\mathbf u|^2+e_{\rm int}
The energies are per volume in J/m³. This split assumes three translational degrees of freedom; molecular rotation or vibration would change the internal caloric relation.↔ Scroll wide equations horizontally.

Separate thermal and bulk energy

A local Maxwellian has n=10²⁵ m⁻³, particle mass 4×10⁻²⁶ kg, thermal energy kBT=4×10⁻²¹ J, and flow speed 10 m/s. Find mass density, pressure, random energy density, and total kinetic energy density.

  1. Mass density is mn=(10²⁵)(4×10⁻²⁶)=0.4 kg/m³. The ideal-gas pressure is n kBT=4×10⁴ Pa, regardless of the common bulk velocity in this nonrelativistic description.

  2. Random energy density is three halves of pressure, giving 6×10⁴ J/m³. This follows from three equal Gaussian velocity variances; it is not obtained by inserting the bulk flow speed into equipartition.

  3. The bulk contribution is ½×0.4×10²=20 J/m³. Adding it gives total kinetic energy density 60020 J/m³. The much smaller bulk contribution is still a distinct flux-bearing part of the laboratory energy.

The result

The gas has density 0.4 kg/m³, pressure 40000 Pa, random energy density 60000 J/m³, and total kinetic energy density 60020 J/m³.

Explore the model

Interactive lab

Interactive lab · heat

Inspect a closed continuum mode

Reset and set n=1, κ=0.5, then Step once; predict amplitude exp(−0.25)≈0.779. Reset and set n=2, κ=0.5, then Step once; predict exp(−1)≈0.368.

Inspect a closed continuum modeThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. 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 forThe exact periodic heat mode presupposes a continuum diffusion equation. It does not calculate collisions or establish local Maxwellian equilibrium.

Check your understanding

Why can a spatially varying local Maxwellian have C[M]=0 yet fail to solve the complete stationary kinetic equation?

Try it yourself

An isolated uniform gas has zero mean flow and diagonal pressure tensor (2,3,4) Pa with no off-diagonal components. Elastic collisions relax it to an isotropic Maxwellian without changing density or total energy. Find the final pressure and internal energy density.

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

Give me a hint

Use the trace of the pressure tensor; collisions redistribute directional energy while conserving its total.

Compare with the solution

Initially the trace is 9 Pa, so random energy density is ½×9=4.5 J/m³. The final equal diagonal components must each be 3 Pa. Thus the final scalar pressure is 3 Pa and internal energy remains 4.5 J/m³. The original anisotropic state had the same scalar kinetic temperature but was not already a Maxwellian.

Take it with you

  • Elastic collisions protect five local moments of a monatomic gas.
  • A Maxwellian fixes isotropic pressure and zero local-rest-frame heat flux.
  • Local collision equilibrium is a leading closure, not automatically global stationarity.

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