The Virial Theorem and Long-Range Repulsion

Question

I am working on a problem (the 12th problem in the third chapter of the third edition of Goldstein's Classical Mechanics) that asks the reader to consider certain long-range interactions between atoms in a gas in the form of central forces derivable from a potential

\begin{equation*} U(r) = \frac{k}{r^{m}} \end{equation*}

where $r$ is the distance between any pair of atoms and $m\in \mathbb{N}$. In particular, it asks the reader to determine the addition to the virial of Clausius, i.e.,

\begin{equation*} -\frac{1}{2}\overline{\sum_{i}{\mathbf{F}_{i}\cdot\mathbf{r}_{i}}} \end{equation*}

that is due to such interactions. To that end, I wrote the force acting on the $i^{\mathrm{th}}$ particle as

\begin{equation*} \mathbf{F}_{i}' = \mathbf{F}_{i} + \sum_{j\neq i}{\frac{km}{r_{ij}^{m+1}}\hat{\mathbf{r}}_{ij}} = \mathbf{F}_{i} + \mathbf{f}_{ij} \end{equation*}

where $\mathbf{r}_{ij}$ is the vector $\mathbf{r}_{j}-\mathbf{r}_{i}$. The thing, then, is to consider the additional term

\begin{equation*} \sum_{\substack{i,j\\i\neq j}}{\mathbf{f}_{ij}\cdot\mathbf{r}_{i}} \end{equation*}

in the average. My instinct is that it should be zero, but I'm having trouble showing it. It is clear that $\mathbf{f}_{ij} = -\mathbf{f}_{ji}$, whereby the terms in the sum can be written as sums of pairs of the form

\begin{equation*} \mathbf{f}_{ij}\cdot\mathbf{r}_{i} + \mathbf{f}_{ji}\cdot\mathbf{r}_{j} = \mathbf{f}_{ij}\cdot\mathbf{r}_{ji}. \end{equation*}

But $\mathbf{f}_{ij}$ is parallel to $\mathbf{r}_{ji}$, so these don't, in general, vanish. Is my instinct correct (is there no additional contribution to the virial), and so is it possible to show that the terms vanish? Otherwise, is it only in the time average that they vanish? If they don't vanish at all, what's a productive way forward? (Please forgive my over-liberal use of the word vanish.)

Answer 1

The term you are having problems with does certainly not vanish as it represents the average kinetic energy. If you have a potential

$$ U_i = \frac{k}{r_i^{m}} $$

due to particle $i$ acting on the test particle, then this is equivalent to the force

$$ F_i = -\frac{\partial{U_i}}{{\partial{r_i}}}=\frac{km}{r_i^{m+1}} $$

Now

$$ -\frac{1}{2}\overline{\sum_{i}{F_i\cdot r_i}} $$

is nothing but the average kinetic energy $\overline{T}$ of the system so

$$ \overline{T}= -\frac{1}{2}\overline{\sum_{i}{F_i\cdot r_i}} = -\frac{1}{2}\overline{\sum_{i}{\frac{km}{r_i^{m+1}}\cdot r_i}} = -\frac{m}{2}\overline{\sum_{i}{\frac{k}{r_i^{m}}}} = -\frac{m}{2}\overline{U} $$

which is the Virial Theorem for a potential of (inverse) power $m$ (note that sometimes this would be referred to in this case as a potential of power $-m$, so the sign in the equation would then be positive, with $m$ taken as negative).