 Solution

To distinguish the moment quantity from the gravitational constant, write it as $\mathcal{G}={\sum }_i{\mathbf{p}}_i\cdot {\mathbf{r}}_i$. Newton's equations give ${\dot{\mathbf{p}}}_i={\mathbf{F}}_i$ and ${\dot{\mathbf{r}}}_i={\mathbf{p}}_i/m_i$. Differentiating provesAverage from $0$ to $\tau$, divide by $\tau$, and apply the supplied long-time boundary condition. The result isThe PDF's intermediate formula omits this minus sign. It cannot follow from the preceding derivative identity: an attractive inverse-square circular orbit already has $\mathbf{F}\cdot \mathbf{r}<0$ and $T>0$, contradicting the printed positive-sign version. The later homogeneous-potential relation is consistent with the corrected sign.

Interpret $W$ as the total pair potential energy, with each pair counted once, for example $W={\sum }_{i<j}C|{\mathbf{r}}_i-{\mathbf{r}}_j{|}^{p+1}$. The printed pair sum leaves one particle index unspecified; an ordered-pair convention with the corresponding factor absorbed into $C$ has the same degree of homogeneity. Under a common dilation of all positions,Differentiating with respect to $\lambda$ at one is the Euler theorem for homogeneous functions. Since ${\mathbf{F}}_i=-{\nabla }_{i}W$, it givesSubstitution into the correctly signed averaged identity proves the virial theorem for a homogeneous pair potential:For $p\ne -1$, the conserved total energy can be evaluated using these averages:Use the kinetic temperature convention $⟨T⟩=\kappa \Theta$, where $\kappa =3N{k}_B/2>0$ for three-dimensional thermalized particles, or the same positive coefficient as a kinetic-temperature definition. At fixed particle number and interaction parameters,Thus the equilibrium heat capacity is negative precisely whensubject to existence of the bounded equilibrium sequence assumed by the virial theorem. At $p=-3$ the virial energy is zero and this derivative vanishes. At $p=-1$ the written power-law potential is constant, its force vanishes, and the bounded-motion virial condition excludes nonzero kinetic energy; division by $p+1$ is not permissible. The Newtonian case $p=-2$ recovers $E_{\mathrm{tot}}=-⟨T⟩$.