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Completing Knowledge in Mechanics (virial)

  1. Nov 8, 2005 #1
    Hi everybody,

    I would like to know: 1)what is the so-called VIRIAL in physics, 2) what is the VIRIAL THEOREM and 3) comments on its importance in physics.
    It always sounded too much mysterious to me.

    Thank you all,

  2. jcsd
  3. Nov 8, 2005 #2


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    There's nothing particularly mysterious about the concept except the word itself which is derived from the Latin vis or viris (force).

    The physics concept simply relates the total kinetic energy of a stable system to the total potential energy. E.g. in a gravitational system such as a galaxy, the total kinetic energy of all the stars is negative one half times the total gravitational potential energy. Similar statements apply to systems of charged particles.
  4. Nov 8, 2005 #3
    let [tex] G = \sum (p_i\prime \cdot r_i [/tex]
  5. Nov 8, 2005 #4
    whoa, sorry about that. I'm trying to learn the LaTex codes.

    let [tex] G = \sum (p_i \cdot r_i) [/tex]
    then we are interested in [tex] \frac {dG}{dt} [/tex]
    perform this derivative and you will see that:
    [tex] \frac {dG}{dt} = 2T + \sum F_i \cdot r_i [/tex]
    take the time average of this quantity:
    [tex] \frac {1}{\tau} \int \frac {dG}{dt} dt = \overline {2T} + \overline{\sum(F_i \cdot r_i)} [/tex]

    if the motion is periodic, then the time average of the left side will be zero. If it is not periodic, then we choose tau sufficiently long as to make the time average zero. then it follows that
    [tex] \overline T = - \frac{1}{2} \overline{\sum(F_i \cdot r_i)} [/tex]

    Now, If we're dealing with conservative forces, [tex] F_i = -\nabla V [/tex] which would be the case in, for instance, gravitation. What the equation means, in that context, is that the average kinetic energy is half the average potential energy.

    It is worth noting that even if there are non conservative forces present, say frictive forces proportional to the velocity of the particles, that the virial theorem still holds. There is a caveat, though. You must periodically put energy into the system, otherwise all time averages will go to zero (Goldstein, 86).
    Last edited: Nov 8, 2005
  6. Nov 8, 2005 #5
    Dear Ptabor,

    I would ask you to make just one rapid revision in your last post, for I am not sure to have correctly filled some blanks.

    Thank you,

  7. Nov 8, 2005 #6
    to expound a little further on the topic, I'll apply this to thermodynamics (adapted from an argument in Goldstein).
    In the case of a gas with N atoms within a container of volume V, the equipartition theorem holds and the average kinetic energy is given by
    [tex] \overline T = \frac {3}{2}Nk_b T [/tex] where the T on the right side is the temperature, and the T on the left side is the average kinetic energy of the atoms.
    As for the [tex] F_i [/tex], these are the forces not only of constraint (provided by the walls of the container) but also the inter-atom interactions. For our discussion here, we assume that collisions between atoms are rare compared with collisions with the walls (indeed, this is the definition of an ideal gas).
    Given this premise, what we are really looking at is a pressure being exerted on the container by the particles. we can then write
    [tex] dF_i = -P ndA [/tex] where n is the normal to the surface of the container.
    To put it another way: [tex] \frac {1}{2} \sum(F_i \cdot r_i) = - \frac{P}{2} \int {(n \cdot r) dA} [/tex]
    Recall that gauss's theorem says [tex] \int {(n \cdot r) dA} = \int {(\nabla \cdot r) dV} = 3V [/tex]
    thus we may write: [tex] \frac{3}{2}Nk_bT = \frac{3}{2}PV [/tex]
    cancelling the 3/2 factor we have recovered the ideal gas law.
    I hope this helps.
  8. Nov 8, 2005 #7


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