# Virial theorem for many-body systems under closed systems

$$2\langle T\rangle=\left\langle\sum_i \vec r _i \cdot \nabla_i V\right\rangle$$ The right-hand side is the so-called virial of the forces. The virial theorem (3.33) tells us that under the mentioned assumptions the time average of the kinetic energy is equal to one half of the virial of the system. Special statements can be derived for closed systems $$V\left( \vec r _1, \ldots, \vec r _N\right)=\frac{1}{2} \sum_{i, j} V_{i j}\left(r_{i j}\right)$$ if the internal potential $$V_{i j}$$ can be written as $$V_{i j}=\alpha_{i j} r_{i j}^m ; \quad m \in Z$$ Then it follows (proof?): $$\sum_i \vec r _i \cdot \nabla_i V=m V,$$ whereby the virial theorem simplifies to $$2\langle T\rangle=m\langle V\rangle$$
My confusion is: how exactly do I get from $$V_{i j}=\alpha_{i j} r_{i j}^m ;\quad m \in Z$$ that. $$\sum_i \vec{r}_i \cdot \nabla_i V=m V$$
1. Convince yourself that $$\frac{\partial}{\partial \, \vec{r}_k} |\vec{r}_i-\vec{r}_j|= \delta_{ki} \frac{\vec{r}_k-\vec{r}_j}{|\vec{r}_k-\vec{r}_j|}+\delta_{kj}\frac{\vec{r}_k-\vec{r}_i}{|\vec{r}_k-\vec{r}_i|}=(\delta_{ki}-\delta_{kj})\frac{\vec{r}_i-\vec{r}_j}{|\vec{r}_i-\vec{r}_j|} \quad \text{for }\, i\ne j.$$
2. Use this formula to compute $$(\ast)=\sum\limits_{k=1}^N \vec{r}_k \cdot \frac{\partial}{\partial \, \vec{r}_k} V(\vec{r}_1,\ldots, \vec{r}_N)$$ for the potential $$V(\vec{r}_1, \ldots, \vec{r}_N)=\frac{1}{2}\sum\limits_{i \ne j}^N \alpha_{ij} \, (r_{ij})^m \quad \text{with} \; r_{ij}=|\vec{r}_i-\vec{r}_j|,$$ leading to the desired result \begin{align}(\ast) &=\frac{m}{2}\sum\limits_{k=1}^N \sum\limits_{i\ne j}^N\alpha_{ij}(r_{ij})^{m-1}(\delta_{ki}-\delta_{kj})\, \vec{r}_k\cdot(\vec{r}_i-\vec{r}_j)/r_{ij}\\ &=\frac{m}{2}\sum\limits_{i\ne j}^N\alpha_{ij}(r_{ij})^m \\&=mV(\vec{r}_1,\ldots,\vec{r}_N). \end{align}