Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 6 / 4 / c

The printed hint with unchanged $\sigma$ is not a valid change of variables: at fixed $\sigma$, the outgoing velocities forget the direction of $v-v_{\ast }$. A correct proof uses the angular exchange for elastic collisions.

Set $C=(v+v_{\ast })/2$ and $v-v_{\ast }=r\omega$, where $r>0$ and $\omega \in {\mathbb{S}}^2$. Then $dvd{v}_{\ast }=dC{r}^{2}drd\omega$, while $v^{\prime }=C+r\sigma /2$ and $v_{\ast }^{\prime }=C-r\sigma /2$. The gain integral becomesExchange the two independently integrated angular variables $\omega$ and $\sigma$. The measure is unchanged, and reverting to $v=C+r\omega /2$, $v_{\ast }=C-r\omega /2$ givesThe measure-zero set $r=0$ causes no difficulty. This proves the requested identity without invoking the false fixed-$\sigma$ Jacobian assertion.