 Solution

Use the correct Gaussian entropy decompositionThe last coefficient is $1/2$, as follows from $-\log {\gamma }_N=N\log (2\pi )/2+|\mathbf{v}{|}^2/2$; the printed hint omits it. Under the allowed differentiability and integrability assumptions, the supplied collision invariants conserve mass and energy. Differentiating therefore giveswhere the extra derivative term $\int {\partial }_{t}F$ vanishes by mass conservation. Thus the Kac entropy production isFor one pair, call the double integral $A_{ij}$. The measure-preserving substitution $(\mathbf{v},\theta )\mapsto (R_{ij,\theta }\mathbf{v},-\theta )$ interchanges $F$ and $F\circ R_{ij,\theta }$, with angles taken modulo $2\pi$. Averaging the original and substituted expressions givesSince $N/(4\pi C_N)=1/[2\pi (N-1)]$, the desired Kac entropy dissipation formula isFor positive values $a,b$, $(a-b)(\log a-\log b)\ge 0$ because the logarithm is increasing. At two zeros use value zero; at one zero and one positive value use the nonnegative extended value $+\infty$. One may first use positive densities and then regularize by $(F+\epsilon {\gamma }_N)/(1+\epsilon )$; rotation invariance of ${\gamma }_N$ preserves the formula and permits the usual limit at zeros under the stated assumptions.

Thus relative entropy is nonincreasing along the evolution. When the dissipation is finite, zero dissipation means pairwise rotation invariance and hence radiality, by the preceding kernel argument. Radial normalized densities other than ${\gamma }_N$ can be stationary with positive relative entropy: vanishing dissipation is not a claim that the unique stationary density is Gaussian.