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.)


1 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

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


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

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).

  • $\begingroup$ Somehow, I though the vector $\mathbf{r}_{i}$ (the radius vector of the $i^{\mathrm{th}}$ atom) in the virial product was different than the vector $\mathbf{r}_{ij}$ (the radius vector separating the $i^{\mathrm{th}}$ and $j^{\mathrm{th}}$ atoms) the magnitude of which shows up in the power law force under consideration. In your calculation, you seem to consider them as the same--are they? $\endgroup$
    – kandb
    Commented Mar 12 at 8:09
  • $\begingroup$ @kandb You can leave out the vector notation because we are dealing with central forces i.e. force vector between the particles is always along the line connecting them.. That means the $r_i$ in the virial expression cancels with one of $r_i$s from the potential. $\endgroup$
    – Thomas
    Commented Mar 12 at 8:23
  • $\begingroup$ Ok, so leaving out the vector notation, the potential varies as $|r_{i}-r_{j}|^{-m}$ while the radius in the virial product is just $r_{i}$, right? $\endgroup$
    – kandb
    Commented Mar 12 at 8:34
  • $\begingroup$ @kandb If you use a relative vectors for one, you also have to use it for the other, so you might just as well use scalars in the first place. In the Wikipedia article section 'Connection with the potential energy between particles' and following, this is worked out in detail en.wikipedia.org/wiki/… $\endgroup$
    – Thomas
    Commented Mar 12 at 19:26

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