Virial Theorem and Frictional Forces

[Old Guy]

In the virial theorem, why do velocity-dependent frictional forces disappear? I've seen this stated a number of times, but never any explanation. (And, obviously, haven't been able to figure it out myself!)

 

[olgranpappy]

In the virial theorem, why do velocity-dependent frictional forces disappear? I've seen this stated a number of times, but never any explanation. (And, obviously, haven't been able to figure it out myself!)

I don't know if it is true in general, but for forces which are directly proportional to the velocity the virial is
proportional to
$$\langle{\vec v}\cdot{\vec r}\rangle\;,$$
where $\langle\ldots\rangle$ is time averaging.

But, that virial is the time average of a total derivative w.r.t time (i.e., of r^2/2) and so is zero if the quantity r is bounded and the averaging is done over a long time. (Or if the motion is periodic and the averaging is done over a period of the motion).

 

[Old Guy]

Thanks very much; can you expand a bit on your answer? Specifically, what does it mean for r to be "bounded", and how does it make the time average of a total time derivative go to zero? I understand about the "long time" bit, although even there it would seem to approach a limit of zero, and not actually equalling zero. Am I thinking about this right? Thanks again.

 

[olgranpappy]

Thanks very much; can you expand a bit on your answer? Specifically, what does it mean for r to be "bounded", and how does it make the time average of a total time derivative go to zero? I understand about the "long time" bit, although even there it would seem to approach a limit of zero, and not actually equalling zero. Am I thinking about this right? Thanks again.

I'll expand and expound:

First of all, I'm discussing a single classical particle.

The particle travels along a trajectory
$${\vec r}(t)$$
under the influence of some forces.

When I say that r is bounded I mean that I can choose a "bound"--some (possibly large) number (R) such that I always have
$$r(t)<R\;,$$
for all t. Put more simply, the particle never escapes off to infinity and is always within some finite (though perhaps large) distance from the origin. Also, Because r is bounded so too is r^2 bounded.

Next, I define the "time averaging" as
$$\langle\ldots\rangle\equiv<br /> \lim_{T\to\infty}\frac{1}{T}\int_0^Tdt(\ldots)$$

So then
$$\langle \frac{d r^2}{dt}\rangle<br /> =\lim_{T\to\infty}\frac{1}{T}\int_0^T\frac{d r^2}{dt}<br /> =\lim_{T\to\infty}\frac{r^2(T)-r^2(0)}{T}\;.$$
But that last quantity is necessarily less than
$$lim_{T\to\infty}<br /> \frac{R^2}{T}=0\;.$$
(it's also necessarily greater than -R^2/T)
Thus, because the quantity r^2 was bounded the time-average of the derivative of r^2 was zero. This is true for any bounded quantity.