Liouville's Theorem: Sketching Rectangle Motion in px-x Plane

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SUMMARY

Liouville's theorem asserts that the motion of phase-space points governed by Hamilton's equations conserves phase-space volume. The Hamiltonian for a particle under a constant force F is given by H(x,p_{x}) = p_{x}^2/(2m) - Fx. The discussion focuses on visualizing the motion of a phase space rectangle defined by 0 < x < A and 0 < p < B over time. Participants suggest solving the equations of motion for x and p_{x} using the corners of the rectangle as initial conditions to trace the evolution of the phase space volume.

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  • Understanding of Hamiltonian mechanics
  • Familiarity with Liouville's theorem
  • Knowledge of phase-space representation
  • Ability to solve differential equations
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  • Study Hamilton's equations of motion in detail
  • Explore phase-space volume conservation principles
  • Learn about the implications of constant forces in Hamiltonian systems
  • Investigate graphical representations of phase-space trajectories
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Students and researchers in physics, particularly those studying classical mechanics and Hamiltonian dynamics, will benefit from this discussion.

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Homework Statement


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

H(x,p_{x}) = \frac{p_{x}^2}{2.m} - F.x
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 p_{x}-x plane.

Homework Equations


\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.

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...
 
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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.
 
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.
 
andrew1982 said:
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.

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.
 
Thank you, guys!

So, I'm using the following equations:

\dot{x}=\frac{dH(x,p_{x})}{dp_{x}} = \frac{p_{x}}{m}

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

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)

\dot{x}=0

\dot{p}_{x}=F

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.
 
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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