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m →Statement and derivation: MOS:MATH#PUNC, fmt., style |
m →Connection with the potential energy between particles: punct., style, fmt. |
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=== 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:
where {{math|'''F'''<sub>''jk''</sub>}} is the force applied by particle {{mvar|j}} on particle {{mvar|k}}. Hence, the virial can be written▼
<math display="block">
</math>
-\frac12\,\sum_{k=1}^N \sum_{j=1}^N \mathbf{F}_{jk} \cdot \mathbf{r}_k \,.▼
▲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 =
</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
<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} \sum_{k=2}^N \sum_{j=1}^{k-1} \mathbf{F}_{jk} \cdot \end{align}</math>
where we have
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} =
-
</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
&= -\sum_{k=2}^N \sum_{j=1}^{k-1} \frac{dV_{jk}}{dr_{jk}} \frac{|
& =-\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> === Special case of power-law forces ===
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