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Virial Theorem and Frictional Forces

  1. Sep 25, 2008 #1
    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!)
     
  2. jcsd
  3. Sep 25, 2008 #2

    olgranpappy

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    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
    [tex]
    \langle{\vec v}\cdot{\vec r}\rangle\;,
    [/tex]
    where [itex]\langle\ldots\rangle[/itex] 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).
     
  4. Sep 25, 2008 #3
    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.
     
  5. Sep 25, 2008 #4

    olgranpappy

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    I'll expand and expound:

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

    The particle travels along a trajectory
    [tex]
    {\vec r}(t)
    [/tex]
    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
    [tex]
    r(t)<R\;,
    [/tex]
    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
    [tex]
    \langle\ldots\rangle\equiv
    \lim_{T\to\infty}\frac{1}{T}\int_0^Tdt(\ldots)
    [/tex]

    So then
    [tex]
    \langle \frac{d r^2}{dt}\rangle
    =\lim_{T\to\infty}\frac{1}{T}\int_0^T\frac{d r^2}{dt}
    =\lim_{T\to\infty}\frac{r^2(T)-r^2(0)}{T}\;.
    [/tex]
    But that last quantity is necessarily less than
    [tex]
    lim_{T\to\infty}
    \frac{R^2}{T}=0\;.
    [/tex]
    (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.
     
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