 Weak Boltzmann collision identity

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 $(f^{\prime }{f}_{\ast }^{\prime }-f{f}_{\ast })(-\Delta \psi )/4$.