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Liouville equation - equilibrium

  1. Oct 7, 2013 #1
    Hi!
    I have problem with uderstanding of Liouville equation. Which sais that if we have a
    Hamiltonian system (energy is conserved), then the the volume of phase space is
    conserved, or equivalently the probability density is conserved (the total derivative
    of probability density per time is 0).
    On the other hand, and isolated hamiltonian system is going to an equilibrium state,
    where the probability density is the same for all state in phase space.

    How can system reach the equilibrium, when the probaility density is on the develpment
    trajectory conserved and at the begining is not the same for all state?

    Thaks a lot!
     
  2. jcsd
  3. Oct 8, 2013 #2

    Jano L.

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    Gold Member

    Isolated Hamiltonian system cannot be "going to an equilibrium state".

    Let the energy of such system be ##E##.

    In general almost any microstate on the energy hypersurface ##E## remains accessible to the system, it is only a question of time when the system will get there.

    This can be easily seen from the Liouville equation: the probability behaves as incompressible liquid, it does not accumulate in some hypothetical microstate of equilibrium (nor in a region of equilibrium) in phase space.

    The concept of "equilibrium state" is a notion from thermodynamics, used to describe a body whose macroscopic description - macrostate ##S##(pressure, temperature) has a meaning and whose time variation can be neglected.

    In statistical physics, knowledge of the macrostate implies probability distribution function ##\rho(q,p)## on the phase space. Since the macrostate of the body in equilibrium can be considered as time-independent, it is most natural to require that the value of ##\rho## for any microstate ##q,p## be also time-independent.

    From this assumption and the Liouville theorem it follows that the Poisson bracket of ##\rho## and ##H## vanishes. This can happen if ##\rho## is a function of ##H##.

    Th equilibrium value of the distribution ##\rho## is the same only for the microstates that correspond to the energy ##E##, i.e. for ##q,p## such that ##H(q,p) = E##.

    It is not the system what can reach the equilibrium. It is the function ##\rho(q,p)## that may reach the equilibrium value ##\rho(q,p) = \rho_0(E)##. In order to accomplish this, however, it has to evolve differently than according to the above Liouville equation, since as you have noticed, the latter preserves the initial values of density ##\rho## (it only changes ##q,p## they correspond to.)

    The system has to be brought to interaction with other system (thermal reservoir) and then disconnected. During the interaction, the above Liouville equation for ##\rho## does not apply; there will be new forces due to the reservoir. As a result of these external forces, the resulting ##\rho## may get close to equilibrium value ##\rho_0(E')## for the newly acquired value of energy ##E'##.
     
  4. Oct 23, 2013 #3

    Jano L.

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    Gold Member

    Was the above helpful? Now that I'm reading the posts again, I am not sure whether we understood each other.
     
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