 Solution

For fixed $\mathbf{R}$ and $t$, define $h(\mathbf{x},s)=n_2(\mathbf{a},\mathbf{b},t)$ with $\mathbf{a}=\mathbf{x}+(s-1)\mathbf{R}$ and $\mathbf{b}=\mathbf{x}+s\mathbf{R}$. The chain rule givesso $\boxed{{\partial }_{s}h=\mathbf{R}\cdot {\nabla }_{\mathbf{x}}h}$. The derivative with respect to $\mathbf{x}$ here translates both arguments of the two-particle number density; it is not a derivative of its first argument alone.

Take the spatial divergence of the proposed interaction pressure tensor. Its pair potential factor is independent of $\mathbf{x}$, so the preceding identity givesExchange symmetry of the two-particle distribution function implies $n_2(\mathbf{x}-\mathbf{R},\mathbf{x})=n_2(\mathbf{x},\mathbf{x}-\mathbf{R})$. In the second term, substitute $\mathbf{R}\mapsto -\mathbf{R}$. Since $R_i{\varphi }^{\prime }(R)/R$ is odd, both terms become equal:On the other hand, at fixed ${\mathbf{x}}_{\star }=\mathbf{x}+\mathbf{R}$, the first-argument derivative of the pair potential is ${\nabla }_{\mathbf{x}}\varphi =-\mathbf{R}{\varphi }^{\prime }(R)/R$. Thus the interaction force density in b is $-{\partial }_{x_j}{P}_{ij}^{\varphi }$. Moving it to the left yields local conservation of momentum:The integral over $s$ distributes each pair's momentum flux along the straight segment joining the particles. The factor $1/2$ removes the double counting of ordered pairs. Central forces also make $P_{ij}^{\varphi }=P_{ji}^{\varphi }$.