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Classical statistical mechanics question

  1. Jan 8, 2011 #1
    Hello. I read this assertion in a book: if we take at an initial time t_0 a constant density distribution [itex]\rho(p,q,t_0)[/itex]
    in phase space, then this implies that \rho wil remain independent of time for all [itex]t>t_0[/itex] because by Liouville's theorem
    [tex]\frac{\partial\rho}{\partial t}=-\sum(\frac{\partial\rho}{\partial q}\dot{q}+\frac{\partial\rho}{\partial p}\dot{p})=0[/tex]
    I don't understand this because we take [itex]\frac{\partial\rho}{\partial q}=\frac{\partial\rho}{\partial p}=0[/itex] only at t=t_0 which means that [itex]\frac{\partial\rho}{\partial t}=0[/itex] only at t=t_0 it doesn't mean that [itex]\frac{\partial\rho}{\partial t}=0[/itex] for t>t_0 therefore we can not assert that [itex]\frac{\partial\rho}{\partial q}=0[/itex] and [itex]\frac{\partial\rho}{\partial q}=0[/itex] for [itex]t>t_0[/itex]
  2. jcsd
  3. Feb 21, 2011 #2
    I just happened across this post and thought I'd respond (though its a bit late). Why do you take the partials of rho to be zero, as they are not necessarily. Liouville's theorem says that density is constant in time, not in space (over q's and p's)
  4. Feb 21, 2011 #3
    Yes, a uniform distribution would correspond to the high-temperature Botlzmann distribution. Such a distribution is stationary.

    You can use analyticity and induction to prove this. As you say only [tex]\dot{\rho}(t_0) = 0[/tex] immediately follows. Therefore at an infinitesimal time later you have [tex]\rho(t_0+dt) = \rho(t_0) + dt \dot{\rho}(t_0) = \rho(t_0)[/tex]. You can probably see the rest.
  5. Feb 21, 2011 #4
    Liouville's theorem does not say that rho is constant in time it says that the materilal(substancial) derivate of rho is zero. If you stay in one place in phase space then rho may vary in time.
  6. Feb 21, 2011 #5
    I think you are simply restating Liuville's theorem which says [itex]\frac{d\rho}{dt}=0[/itex] at all times, that is not what I asked.
  7. Feb 21, 2011 #6
    By constant in time I meant that the total derivative [tex] d\rho/dt = 0 [/tex]. But the partials with respect to position need not be, that is: [tex] \partial \rho / \partial q \neq 0 [/tex], ditto with respect to momenta. Are you asking for a proof of the theorem? Many books don't really prove the theorem but sort of wave their hands and throw up a continuity equation for the density.
  8. Feb 21, 2011 #7
    You have a 1st order partial differential equation for the function rho (the velocities can be written in terms of the coordinates through derivatives of the Hamiltonian).

    What is the boundary condition for a 1st order partial differential equation? The answer is values of the function on the hypersurface.

    You are given the values on the hypersurface t=0.

    So that should be enough to determine the solution completely.

    Hence rho=constant is the solution for all time and all phase space.

    Now there's always the question whether these Cauchy conditions are too much and no solution exists (for 1st order differential equation the Cauchy conditions will be just the value of the function on a hypersurface). However, for this problem you don't have that problem, since rho=constant is obviously a solution.

    So rho=constant is the only solution that satisfies the boundary conditions and the differential equation.
    Last edited: Feb 21, 2011
  9. Feb 21, 2011 #8
    I meant partial derivatives with respect to time. Replace all [tex]\dot{\rho}[/tex] with [tex]\frac{\partial \rho}{\partial t}[/tex] and consider fixed coordinates. This is not Liouville's theorem, it's just calculus. If the distribution cannot change at the initial time, and it cannot change a moment later, and it cannot change a moment later, ..., then it cannot change at all (barring non-analyticity).
  10. Feb 22, 2011 #9
    No, I'm not asking for a proof of the theorem, maybe I did not express myself clearly. I'm sorry but my english is not good
  11. Feb 22, 2011 #10
    this sounds convincing, thank you
  12. Feb 22, 2011 #11
    Yes, I agree that what you are saying is just calculus and you don't need to perform a numerical integration to get that rho is constant in time in the sense your are taking it because we know by one of the fundamental theorems of calculus that a null derivative implies a constant function.

    Consider this, rho is constant along a trayectory in phase space and that's what Liouville's theorem is all about. However this doen not mean that if you fix your position in phase space and let the time flow rho will remain constant, neither it means that if you freeze time at same instant t and wonder arround phase space rho wil remain constant as you move.

    I think the argument given by RedX could be correct
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