Normalize the distribution before taking moments
Separate particle number, mean motion, and random momentum transport.
On this page
The question for today
What you’ll understand
- Declare the velocity-space measure and normalization.
- Compute density, mean velocity, and peculiar-velocity moments.
- Distinguish pressure from bulk momentum flux.
A one-particle distribution compresses enormous microscopic information into a density over position and velocity. It does not specify which particular atom follows which path. Its moments yield measurable quantities such as number density, flow velocity, and stress, but only after a normalization convention is declared. Dividing by number density in the wrong place changes a velocity into a particle current, while using laboratory velocities in a pressure moment mixes thermal motion with bulk flow.
Counting particles in phase space
Use f(x,v,t) so that f times a small position volume and a small velocity volume is the expected number of particles in that cell. Integrating over velocity gives the local number density, and integrating over both spaces gives the total particle number. A normalized probability density would instead integrate to one. These are related conventions, but their formulas for conditional averages have different prefactors. Always state which object a symbol represents before applying a familiar integral.
Velocity and momentum coordinates also use different densities. For particles of one fixed mass, momentum equals mass times velocity, so the three-dimensional measure acquires a factor m³. Equality of particle counts then requires the velocity density to be m³ times the momentum density evaluated at the corresponding momentum. A missing Jacobian silently alters every moment. Quantum occupation numbers using phase-space cells divided by Planck’s constant require another convention, which we do not adopt here.
Subtract the local flow
The local mean velocity is the first velocity moment divided by number density. Subtract it from each velocity to define the peculiar velocity. Its first moment vanishes by construction. The second peculiar moment defines the kinetic pressure tensor: particles crossing an imagined surface carry fluctuating momentum relative to the local flow. Its diagonal components need not be equal away from equilibrium, so a general distribution cannot be summarized by one scalar temperature without discarding information.
For a classical monatomic gas, one may define kinetic temperature from the trace of this tensor even before full equilibrium. The resulting scalar describes total random translational energy, but isotropic pressure and a Maxwellian velocity shape are additional claims. Two counterstreaming narrow beams can have the same density and scalar kinetic temperature as a thermal gas while differing strongly in higher moments. Moment agreement alone therefore does not establish local equilibrium or justify a fluid closure.
Transport includes more than pressure
The full momentum flux consists of bulk transport plus the pressure tensor. Likewise, total kinetic energy splits into bulk kinetic energy and random internal energy. The heat flux uses a third peculiar moment, not the total laboratory energy flux. This distinction lets conservation equations describe moving gases without treating an ordinary advected energy current as heat conduction. Closing those equations requires relating these higher moments to density, velocity, temperature, and possibly their gradients.
The probability lab gives a finite counting illustration. Reset and set success probability to 0.75 and trials to 8; predict mean count 6 and variance 1.5. Reset and set probability to 0.5 and trials to 8; predict mean 4 and variance 2. These are exact binomial probabilities for independent trials. They illustrate normalization and moments, not a measured molecular velocity distribution, collision correlations, or an anisotropic pressure tensor.
Work it through
Two velocity populations
A one-dimensional illustrative distribution has number density 8 m⁻³ and particle mass 2 kg. Three quarters of particles move at +2 m/s and one quarter at −2 m/s. Compute mean velocity, longitudinal pressure, and the bulk/random kinetic energy split.
The weighted mean velocity is 0.75×2+0.25×(−2)=1 m/s. The two peculiar velocities are therefore +1 m/s and −3 m/s, whose weighted mean is zero.
The peculiar second moment is 0.75×1²+0.25×(−3)²=3 m²/s². Multiplying by mn=16 kg/m³ gives longitudinal pressure Pxx=48 Pa. It is not the full laboratory momentum flux, which is 64 Pa.
Total kinetic energy density is ½mn times the laboratory second moment 4, hence 32 J/m³. Bulk energy is ½mn u²=8 J/m³ and random energy is Pxx/2=24 J/m³, giving the same total. This toy has no transverse degrees of freedom.
The mean velocity is +1 m/s and longitudinal pressure is 48 Pa; total kinetic energy density separates into 8 J/m³ bulk plus 24 J/m³ random motion.
The living notebook
Explore the model
Interactive lab · probability
Compute moments of a normalized count law
Reset and set p=0.75, n=8; predict mean 6 and variance 1.5. Reset and set p=0.5, n=8; predict mean 4 and variance 2.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 bars are exact binomial masses for independent trials. They illustrate moment bookkeeping, not molecular velocity samples or a gas collision model.
Pause & reason
Check your understanding
Your turn
Try it yourself
In the same one-dimensional toy, use density 5 m⁻³, mass 2 kg, and equal populations at +3 and −3 m/s. Find mean velocity, Pxx, and random kinetic 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
Equal populations cancel the first moment but reinforce the second moment.
Compare with the solution
The mean velocity is zero. The second peculiar moment is 9 m²/s², so Pxx=(2 kg)(5 m⁻³)(9 m²/s²)=90 Pa. Random energy density is Pxx/2=45 J/m³ and the bulk contribution vanishes. A zero mean does not imply a motionless or pressureless gas.
Keep the idea
Take it with you
- The distribution and its phase-space measure must be declared together.
- Peculiar moments separate random motion from bulk flow.
- A few matching moments do not guarantee a Maxwellian distribution.
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