= Solution
For a time-independent function $F$, <Hamilton's equations> give
$$
\frac{dF}{dt}=\sum_k(F_{q_k}H_{p_k}-F_{p_k}H_{q_k})=[F,H].
$$
Thus a constant of motion has \b[$[F,H]=0$]. For $F=\sum_kp_k$ and the given <Hamiltonian>, $[F,H]=-\sum_kV_{q_k}$. Simultaneous translation $q_k\mapsto q_k+s$ leaves every difference $q_i-q_j$, and hence $V$, unchanged. Differentiation with respect to $s$ gives $\sum_kV_{q_k}=0$, so \b[$[F,H]=0$ also in this example]. This is conservation of total momentum from translational symmetry.
Back to article page
