Consider a system in which the total force acting on the system consists of conservative forces $F'$ and frictional forces $f^i$, where $f^i$ depends on velocity. It is to show that for such a system the virial theorem holds in the form $$\overline{T}~=~-\frac{1}{2}\sum\overline{{F_i}'\cdot r_i}. \tag{1}$$

I defined a quantity $$G=\sum_i p_i\cdot r_i. \tag{2}$$ The total force on the system $$\dot{p_i}=F=F_i'+f_i.\tag{3}$$

Taking time derivative of $G$ and time averaging over limit $0$ to $\tau$. If the Time period is taken to be too large or periodic the time average of $\frac{dG}{dt}$ vanishes. To show the virial theorem holds, all I have to prove that the time average of $\sum{f_i.\dot{r_i}}$ over time period which goes to infinity or the system is periodic in $\tau$ is zero.

i.e, $$\frac{1}{\tau}\int_{0}^{\tau}{f_i(\dot{r_i})}.\dot{r_i}dt\tag{4}$$ is zero in the limit $\tau\to\infty$ or $\tau$ is periodic. But how can I prove it. For the simple case, If I take $f_i(\dot{r_i})=\dot{r}_i$, so the integral becomes, $$\frac{1}{\tau}\int_{0}^{\tau}{\dot{r_i}}^2dt\tag{5}$$ I don't know how to proceed from here.


1 Answer 1


In this answer we just recall the standard approach, which works for a friction force of the form $$\vec{f}_i~=~-k_i \vec{v}_i,\tag{A}$$ where $k_i$ is a constant (which may depend on the $i$'th particle), i.e. Stokes' drag.

  1. Firstly, improve/change/replace the virial quantity $G$ in eq. (2) into $$G~:=~\sum_i \left\{\vec{p}_i\cdot \vec{r}_i+\frac{1}{2}k_i r_i^2\right\},\tag{B}$$ so that the time-derivative becomes independent of friction force: $$\begin{align} \dot{G}~\stackrel{(B)}{=}~&\sum_i \left\{\vec{p}_i \cdot \vec{v}_i+(\dot{\vec{p}}_i+k_i \vec{v}_i)\cdot \vec{r}_i\right\}\cr ~\stackrel{(A)+(3)}{=}&2T+\sum_i \vec{F}^{\prime}_i\cdot \vec{r}_i.\end{align}\tag{C}$$
  2. Secondly, invoke reasonable assumptions about the improved $G$ (periodicity or boundedness, so that the left-hand side of eq. (C) vanishes in time average) to show the virial theorem (1).

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