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Liouville's theorem

  1. Apr 11, 2007 #1
    1. The problem statement, all variables and given/known data
    According to Liouville's theorem, the motion of phase-space points defined by Hamilton's equations conserves phase-space volume. The Hamiltonian for a single particle in one dimension, subjected to a constant force F, is

    [tex]H(x,p_{x}) = \frac{p_{x}^2}{2.m} - F.x[/tex]
    Consider the phase space rectangle of initial points defined by
    0 < x < A and 0 < p < B

    Let the points in the rectangle move according to Hamilton's equations for a time t and sketch how the rectangle changes with time in the [tex]p_{x}[/tex]-x plane.


    2. Relevant equations
    [tex] \frac{d\rho}{dt}= \frac{\partial\rho}{\partial t} +\sum_{i=1}^d\left(\frac{\partial\rho}{\partial q^i}\dot{q}^i +\frac{\partial\rho}{\partial p_i}\dot{p}_i\right)=0. [/tex]


    3. The attempt at a solution
    Substituting the Hamiltonian from the problem inside the Liouville's equation I can see that the density of particles of this volume is constant.
    But, I don't know how to show the movement of this rectangle with time.
    I guess that there is no difference...
     
    Last edited: Apr 11, 2007
  2. jcsd
  3. Apr 11, 2007 #2

    Dick

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    You can be more explicit about picturing the motion of the rectangle for the harmonic oscillator. You know p^2/2m+k*x^2/2=E which is a constant of motion. So the points in phase space move on concentric ellipses. That should make your sketch a little more expressive.
     
  4. Apr 13, 2007 #3
    But it was a constant force, not an oscillator, right?

    It was some time ago that I did these things, but an approach could be to solve the equations of motion for x and px and then use the corners of the rectangle in phase space as starting conditions for 4 different trajectories. Then you can see where the corners are at time t later and how the phase space volume has evolved.... and all points that started inside the rectangle will still be there. Just an idea.
     
  5. Apr 13, 2007 #4

    Dick

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    Ooops, you are right! It's not a oscillator. Tracing the motion of the corners is pretty much what I was suggesting - except the trajectories will no longer be ellipses. Be careful not to assume that the boundaries of the region remain straight lines.
     
  6. Apr 13, 2007 #5
    Thank you, guys!

    So, I'm using the following equations:

    [tex]\dot{x}=\frac{dH(x,p_{x})}{dp_{x}} = \frac{p_{x}}{m}[/tex]

    [tex]\dot{p}_{x}=-\frac{dH(x,p_{x})}{dx} = F[/tex]

    Now I thinking to substitute inside these equations the points of the corners.
    (0,0), (A,0), (A,B) and (0,B).

    For instance:
    (0,0)

    [tex]\dot{x}=0[/tex]

    [tex]\dot{p}_{x}=F[/tex]

    So, there is a variation in the p-axis, but there's no variation in the x-axis (I don't know if this is the right interpretation).

    Do you think that I'm going in the right way?
    Thanks a lot for your comments.
     
  7. Apr 13, 2007 #6

    Dick

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    You have p increasing linearly in time. As p becomes non-zero then the derivative of x becomes non-zero and x becomes nonzero. So saying xdot is zero is only true at a particular time. The physics here is SAME as an object falling in a uniform gravitational field. You know how to solve that, right?
     
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