Integrate a velocity distribution into directional pressure
A zero mean velocity does not imply zero pressure. Consider independent uniform velocity components on intervals −ux..ux, −uy..uy and −uz..uz, with ux=100a, uy=100b and uz=100 m/s. Mass density is 0.6 kg/m³. This original anisotropic packet is not asserted to be thermal equilibrium: collisions would normally relax unequal directional moments unless the distribution were maintained.
Learning outcome: Retain two anisotropic recipes and the isotropic a=b=1 case. Derive the positive-half-space collision integral, compare mean velocity with mean square, and distinguish adding energy from redistributing it.
Starting recap
Review definite integrals, probability densities, mean versus mean square, temperature units and the first-law work convention. State the domain and distinguish calibration data from unused validation.
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At a=2,b=1, the normal pressures are 8000,2000,2000 Pa. Translational energy density is 6000 J/m³. If an isolated relaxation preserves that energy and reaches an isotropic state at the same density, its scalar pressure is 4000 Pa, not the original x pressure.
Open the experimentPredict
Which moment determines the x-normal kinetic pressure in this even distribution?
Experiment
Retain two anisotropic recipes and the isotropic a=b=1 case. Derive the positive-half-space collision integral, compare mean velocity with mean square, and distinguish adding energy from redistributing it.
Integrate directional momentum flux
Worked teaching model · synthetic values · Independent uniform velocity components on ±100a, ±100b and ±100 m/s; zero mean and mass density 0.6 kg/m³. Elastic wall reflection. This supplied anisotropic distribution is a transient or maintained nonequilibrium packet, not a Maxwell equilibrium gas. No physical gas handling.
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Cumulative x-component second moment
Graph axes rescale independently; read units and values, not screen angles. 1: uniform component distribution (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Cumulative x-component second moment
uniform component distribution
| upper component limit (m/s) | integral of vx squared times density (m²/s²) |
|---|---|
| -200 | 0 |
| -195 | 487.6042 |
| -190 | 950.8333 |
| -185 | 1390.313 |
| -180 | 1806.667 |
| -175 | 2200.521 |
| -170 | 2572.5 |
| -165 | 2923.229 |
| -160 | 3253.333 |
| -155 | 3563.438 |
| -150 | 3854.167 |
| -145 | 4126.146 |
| -140 | 4380 |
| -135 | 4616.354 |
| -130 | 4835.833 |
| -125 | 5039.063 |
| -120 | 5226.667 |
| -115 | 5399.271 |
| -110 | 5557.5 |
| -105 | 5701.979 |
| -100 | 5833.333 |
| -95 | 5952.188 |
| -90 | 6059.167 |
| -85 | 6154.896 |
| -80 | 6240 |
| -75 | 6315.104 |
| -70 | 6380.833 |
| -65 | 6437.813 |
| -60 | 6486.667 |
| -55 | 6528.021 |
| -50 | 6562.5 |
| -45 | 6590.729 |
| -40 | 6613.333 |
| -35 | 6630.938 |
| -30 | 6644.167 |
| -25 | 6653.646 |
| -20 | 6660 |
| -15 | 6663.854 |
| -10 | 6665.833 |
| -5 | 6666.563 |
| 0 | 6666.667 |
| 5 | 6666.771 |
| 10 | 6667.5 |
| 15 | 6669.479 |
| 20 | 6673.333 |
| 25 | 6679.688 |
| 30 | 6689.167 |
| 35 | 6702.396 |
| 40 | 6720 |
| 45 | 6742.604 |
| 50 | 6770.833 |
| 55 | 6805.313 |
| 60 | 6846.667 |
| 65 | 6895.521 |
| 70 | 6952.5 |
| 75 | 7018.229 |
| 80 | 7093.333 |
| 85 | 7178.438 |
| 90 | 7274.167 |
| 95 | 7381.146 |
| 100 | 7500 |
| 105 | 7631.354 |
| 110 | 7775.833 |
| 115 | 7934.063 |
| 120 | 8106.667 |
| 125 | 8294.271 |
| 130 | 8497.5 |
| 135 | 8716.979 |
| 140 | 8953.333 |
| 145 | 9207.188 |
| 150 | 9479.167 |
| 155 | 9769.896 |
| 160 | 10080 |
| 165 | 10410.1 |
| 170 | 10760.83 |
| 175 | 11132.81 |
| 180 | 11526.67 |
| 185 | 11943.02 |
| 190 | 12382.5 |
| 195 | 12845.73 |
| 200 | 13333.33 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| Mean vx | 0 | m/s |
| Mean squared vx | 13333.33 | m²/s² |
| Normal pressure on x wall | 8000 | Pa |
| Normal pressure on y wall | 2000 | Pa |
| Normal pressure on z wall | 2000 | Pa |
| Translational energy density | 6000 | J/m³ |
| Scalar pressure after hypothetical energy-preserving isotropization | 4000 | Pa |
Equal mean velocity does not imply equal directional momentum flux.
For an even distribution, collision flux on an x wall integrates 2m vx times number flux vx f(vx) over positive vx. Thus Px=rho integral vx² f(vx) dvx. A uniform component on −u..u has second moment u²/3. The directional pressures differ unless all widths agree. Half their trace gives translational energy density; only an isotropic equilibrium justifies one scalar kinetic pressure. Hypothetical relaxation conserves energy here, not each component moment.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
Equal mean velocity does not imply equal directional momentum flux.
For an even distribution, collision flux on an x wall integrates 2m vx times number flux vx f(vx) over positive vx. Thus Px=rho integral vx² f(vx) dvx. A uniform component on −u..u has second moment u²/3. The directional pressures differ unless all widths agree. Half their trace gives translational energy density; only an isotropic equilibrium justifies one scalar kinetic pressure. Hypothetical relaxation conserves energy here, not each component moment.
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