Past exam of the kinetic theory course of the University of Oxford / 2023 / 1 / a

Write $K_{ij}=K({\mathbf{x}}_i,{\mathbf{x}}_j)$ and $\mathbf{P}={\sum }_i{\mathbf{v}}_i$. The unit particle masses make this sum the total momentum. Exchange the two indices and use the symmetry $K_{ij}=K_{ji}$:Thus every interacting pair transfers equal and opposite momentum, even though the velocity alignment dissipates kinetic energy.

For the total kinetic energy $E_N=\frac{1}{2}{\sum }_i|{\mathbf{v}}_i{|}^2$, apply the same exchange of indices:The dissipated kinetic energy is that of relative motion. Since the total momentum is constant, $E_N-|\mathbf{P}{|}^2/(2N)=\frac{1}{2}{\sum }_i|{\mathbf{v}}_i-\mathbf{P}/N{|}^2$ also decreases. Equality holds when every pair with $K_{ij}>0$ has the same velocity; the hypotheses permit disconnected interacting groups, so they alone do not guarantee eventual global velocity alignment.