Nonlinear Boltzmann collision operator ($Q(f,f)(v)=\int B(g,n)(f^{\prime }{f}_{\ast }^{\prime }-f{f}_{\ast })d{v}_{\ast }dn$)

For equal-mass elastic collisions, one reflection parametrization is $v^{\prime }=v-((v-v_{\ast })\cdot n)n$ and $v_{\ast }^{\prime }=v_{\ast }+((v-v_{\ast })\cdot n)n$. The fixed-$n$ transformation preserves pairwise momentum, kinetic energy and velocity-pair measure. A nonnegative kernel invariant under particle interchange and collision reversal yields a gain-loss operator. Its weak Boltzmann collision identity proves conservation and its entropy identity proves the Boltzmann H theorem. Unlike the Maxwell molecule collision operator, this general kernel may depend on relative speed.

The nonnegative weight describing the rate and angular dependence of binary elastic collisions in the Boltzmann equation. In a scattering-angle representation with solid-angle measure $d\Omega$, it contains relative speed $g$ times differential scattering cross-section. A parametrization by collision normals has a different Jacobian. Any symmetry factor from integrating over a hemisphere instead of a sphere must be included consistently.

Write $f=M(1+h)$ around a fixed Maxwellian and expand the nonlinear Boltzmann collision operator to first order. The weak Boltzmann collision identity gives the nonnegative dissipation form in $L^2(Mdv)$, and polarization proves symmetry on an appropriate common domain. Its zero modes are the collision invariants when the scattering kernel is nondegenerate. A spectral gap, and self-adjointness of an unbounded realization, require additional domain and kernel hypotheses.

The angular collision rate is integrable at each nonzero relative velocity, with any needed speed-growth bounds stated separately. This makes the separate gain and loss integrals available under suitable velocity integrability. Non-cutoff kernels can have infinite angular collision rate and require cancellation rather than the same naive gain-loss split.

Here $\Delta \psi =\psi (v^{\prime })+\psi (v_{\ast }^{\prime })-\psi (v)-\psi (v_{\ast })$ is the pairwise collision change, not a Laplace operator. Average over the two particle labels and use collision reversal to obtain the formula. Every collision invariant has $\Delta \psi =0$, giving mass, momentum and kinetic energy conservation. A second symmetrization replaces $f{f}_{\ast }\Delta \psi /2$ by $(f^{\prime }{f}_{\ast }^{\prime }-f{f}_{\ast })(-\Delta \psi )/4$.