Account for gains and losses in velocity space
Build the dilute-gas collision operator from relative speed and elastic scattering.
On this page
The question for today
What you’ll understand
- Interpret the gain and loss terms of the collision integral.
- Identify the units and role of a differential cross section.
- Verify elastic collision invariants without assuming individual velocities are conserved.
A collision changes velocities while leaving the collision point essentially fixed on the macroscopic scale. The Boltzmann collision operator counts particles entering and leaving a chosen velocity cell through binary encounters. Its compact gain-minus-loss structure contains strong physical assumptions: dilute particles, short interactions, uncorrelated incoming partners, and reversible elastic scattering. We will keep those assumptions visible and distinguish a collision rate for one tagged particle from the number of collision events in an entire gas.
Streaming and collisions act differently
Between encounters a particle streams through space and is accelerated by any external force. Along that trajectory the distribution changes only through the collision term. A force term differentiates with respect to velocity and includes acceleration, force divided by mass, when the velocity measure is used. This is a common convention trap when translating a formula originally written in momentum coordinates. The same physical equation must preserve its units after that change of variables.
At a given position, binary collisions couple four velocities: two incoming and two outgoing. Elasticity preserves their total momentum and kinetic energy but generally preserves neither particle’s velocity. The interaction range and duration must be small compared with macroscopic lengths and the mean time between collisions. Dense liquids, strongly correlated recollisions, and long-range cumulative deflections can require different kinetic descriptions; calling every interaction a Boltzmann collision does not establish those approximations.
Relative speed supplies the incident flux
For a particle at velocity v, partners with velocity v₁ approach at relative speed g. Multiplying g by the differential cross section gives the collision kernel for scattering into a small solid angle. Its product with the partner velocity density and velocity volume gives a rate. Integrate over partner velocities and scattering directions to count losses from the selected cell. Reversed elastic encounters supply the corresponding gain term through the product of the associated primed distributions.
The symmetric notation uses microscopic reversibility and the collision-map Jacobian to express gain and loss with the same nonnegative kernel. It should not be read as a claim that the primed and unprimed particle populations are always equal. Only detailed balance at equilibrium makes each paired product equal. There is no extra one-half in the tagged-particle loss frequency; a half appears when counting total unordered collision events, because summing tagged encounters counts each pair twice.
Check kinematics before rates
For equal masses, introduce center-of-mass velocity and relative velocity. An elastic collision retains the center velocity and rotates the relative vector without changing its magnitude. This construction gives an immediate momentum and energy audit for any proposed numerical collision. Merely preserving speed for each particle separately is too restrictive, while independently drawing two outgoing velocities usually violates the shared conservation laws. A valid collision sampler must enforce the correct pair constraints.
The vector lab isolates one part of this geometry. Reset and set angle to 0° and magnitude to 2; predict components (2,0). Reset and set angle to 90° and magnitude to 2; predict (0,2). A fixed-length relative vector can rotate without changing relative kinetic energy. The instrument contains one vector only, so it does not reconstruct the two outgoing particles, use a differential cross section, or sample collision events.
Work it through
Rotate an elastic encounter
Two equal-mass particles initially have velocities (2,0,0) and (0,0,0) m/s. Rotate their relative velocity by 90° into +y without changing its length. Find the outgoing velocities and verify momentum and energy.
The center-of-mass velocity is (1,0,0) m/s. The incoming relative vector is (2,0,0) m/s, so the rotated relative vector is (0,2,0) m/s. Keeping its length fixed implements elastic relative motion.
Add and subtract half that rotated vector from the center velocity. The outgoing velocities are (1,1,0) and (1,−1,0) m/s. Their sum remains (2,0,0) m/s, preserving total momentum after multiplication by the common mass.
The incoming sum of squared speeds is 4+0=4 m²/s². The outgoing sum is 2+2=4 m²/s². Multiplying either by m/2 gives the same pair kinetic energy. Individual particle energies have changed, which is allowed.
The outgoing velocities are (1,1,0) and (1,−1,0) m/s; pair momentum and kinetic energy are conserved.
The living notebook
Explore the model
Interactive lab · vectors
Rotate a relative velocity
Reset and set angle=0°, magnitude=2; predict (x,y)=(2,0). Reset and set angle=90°, magnitude=2; predict (0,2).
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 single vector illustrates preservation of relative-speed magnitude under rotation. It does not implement collisions or calculate cross sections and rates.
Pause & reason
Check your understanding
Your turn
Try it yourself
In a representative-speed dilute-gas estimate, take n=2×10²⁰ m⁻³, total cross section 10⁻¹⁹ m², and relative speed 250 m/s. Find the tagged collision frequency, mean waiting time, and total event rate per volume.
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 ν=nσg and divide nν by two only when counting pair events per volume.
Compare with the solution
The tagged frequency is 5000 s⁻¹, hence the mean waiting time is 2×10⁻⁴ s under this constant-rate approximation. The unordered event rate is nν/2=5×10²³ m⁻³s⁻¹. These are rate estimates using one representative relative speed, not an evaluation of a general velocity-dependent collision integral.
Keep the idea
Take it with you
- The collision kernel combines incident relative flux with a differential cross section.
- Elastic scattering conserves pair momentum and energy.
- Tagged encounter rates and unordered event rates use different counting factors.
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