Editing Virial theorem (section)

=== Connection with the potential energy between particles ===

The total force {{math|'''F'''<sub>''k''</sub>}} on particle {{mvar|k}} is the sum of all the forces from the other particles {{mvar|j}} in the system:
<math display="block">
 \mathbf{F}_k = \sum_{j=1}^N \mathbf{F}_{jk},
</math>
where {{math|'''F'''<sub>''jk''</sub>}} is the force applied by particle {{mvar|j}} on particle {{mvar|k}}. Hence, the virial can be written as
<math display="block">
 -\frac12\,\sum_{k=1}^N \mathbf{F}_k \cdot \mathbf{r}_k =
 -\frac12\,\sum_{k=1}^N \sum_{j=1}^N \mathbf{F}_{jk} \cdot \mathbf{r}_k.
</math>

Since no particle acts on itself (i.e., {{math|1='''F'''<sub>''jj''</sub> = 0}} for {{math|1 ≤ ''j'' ≤ ''N''}}), we split the sum in terms below and above this diagonal and add them together in pairs:
<math display="block">\begin{align}
 \sum_{k=1}^N \mathbf{F}_k \cdot \mathbf{r}_k
  &= \sum_{k=1}^N \sum_{j=1}^N \mathbf{F}_{jk} \cdot \mathbf{r}_k =
     \sum_{k=2}^N \sum_{j=1}^{k-1} \mathbf{F}_{jk} \cdot \mathbf{r}_k + \sum_{k=1}^{N-1} \sum_{j=k+1}^{N} \mathbf{F}_{jk} \cdot \mathbf{r}_k \\
  &= \sum_{k=2}^N \sum_{j=1}^{k-1} \mathbf{F}_{jk} \cdot \mathbf{r}_k + \sum_{j=2}^N \sum_{k=1}^{j-1} \mathbf{F}_{jk} \cdot \mathbf{r}_k = 
     \sum_{k=2}^N \sum_{j=1}^{k-1} (\mathbf{F}_{jk} \cdot \mathbf{r}_k + \mathbf{F}_{kj} \cdot \mathbf{r}_j) \\
  &= \sum_{k=2}^N \sum_{j=1}^{k-1} (\mathbf{F}_{jk} \cdot \mathbf{r}_k - \mathbf{F}_{jk} \cdot \mathbf{r}_j) =
     \sum_{k=2}^N \sum_{j=1}^{k-1} \mathbf{F}_{jk} \cdot (\mathbf{r}_k - \mathbf{r}_j),
\end{align}</math>
where we have used [[Newton's laws of motion|Newton's third law of motion]], i.e., {{math|1='''F'''<sub>''jk''</sub> = −'''F'''<sub>''kj''</sub>}} (equal and opposite reaction).

It often happens that the forces can be derived from a potential energy {{mvar|''V''<sub>''jk''</sub>}} that is a function only of the distance {{math|''r''<sub>''jk''</sub>}} between the point particles {{mvar|j}} and {{mvar|k}}. Since the force is the negative gradient of the potential energy, we have in this case
<math display="block">
 \mathbf{F}_{jk} = -\nabla_{\mathbf{r}_k} V_{jk} =
 -\frac{dV_{jk}}{dr_{jk}} \left(\frac{\mathbf{r}_k - \mathbf{r}_j}{r_{jk}}\right),
</math>
which is equal and opposite to {{math|1='''F'''<sub>''kj''</sub> = −∇<sub>'''r'''<sub>''j''</sub></sub>''V''<sub>''kj''</sub> = −∇<sub>'''r'''<sub>''j''</sub></sub>''V''<sub>''jk''</sub>}}, the force applied by particle {{mvar|k}} on particle {{mvar|j}}, as may be confirmed by explicit calculation. Hence,
<math display="block">\begin{align}
 \sum_{k=1}^N \mathbf{F}_k \cdot \mathbf{r}_k
  &= \sum_{k=2}^N \sum_{j=1}^{k-1} \mathbf{F}_{jk} \cdot (\mathbf{r}_k - \mathbf{r}_j) \\
  &= -\sum_{k=2}^N \sum_{j=1}^{k-1} \frac{dV_{jk}}{dr_{jk}} \frac{|\mathbf{r}_k - \mathbf{r}_j|^2}{r_{jk}} \\
  & =-\sum_{k=2}^N \sum_{j=1}^{k-1} \frac{dV_{jk}}{dr_{jk}} r_{jk}.
\end{align}</math>

Thus
<math display="block">
 \frac{dG}{dt} = 2 T + \sum_{k=1}^N \mathbf{F}_k \cdot \mathbf{r}_k =
 2 T - \sum_{k=2}^N \sum_{j=1}^{k-1} \frac{dV_{jk}}{dr_{jk}} r_{jk}.
</math>