Graduate Classical particle in a 2D box

[jordi]

I am trying to understand ergodic theory, i.e. how simple systems reach equilibrium.

I consider a classical particle in a 2D (or 3D) box. Funnily, I have never seen this example in books (probably due to lack of knowledge). Instead, in QM, the particle in a box is a prototypical example.

My question is: is a classical particle in a 2D box ergodic? In other words: given any point in the 4D phase space (position and momentum), will the particle visit any other point in the 4D phase space, and when doing so, with the same probability for all states?

From inspection, it seems clear there are some points in the phase space for which ergodicity fails: for example, when the particle moves parallel to any of the walls (the particle keeps bouncing up and down through the same line). Also, there are some other "cycles", such as four rebounds, for example hitting with 45 degrees, which results in a cycle parallel and perpendicular to the diagonals of the box.

However, it could be that these cycles are a "measure zero" of all points in the phase space, so ergodicity could still hold "almost always" (in probabilistic terms).

Or maybe not? Could it be that for any starting point in the 4D phase space, eventually there is a cycle, that returns to the original point in the phase space after a finite amount of rebounds on the wall?

Is this classical example, apparently so simple (no interactions!) been solved?

 

[DrClaude]

Check out
http://www.scholarpedia.org/article/Dynamical_billiards

 

[jordi]

Thank you for the reference.

It seems clear that this system is not ergodic: when an incoming classical particle rebounds on a wall with X degrees with respect to the perpendicular (say 30 degrees), the outcoming particle also has X degrees with respect to the perpendicular. Since the box is composed of four 90 degrees corner, the next rebound will have the incoming classical particle with a 90 degrees - X to the perpendicular. The outcoming particle is also at 90 degrees - X to the perpendicular. And now again, the incoming particle is at X degrees with respect to the new perpendicular, and so on.

What it is not clear to me yet is if all the points in the box are "visited" by the particle, or if there is a closed cycle in a finite number of rebounds.

But clearly, a particle will only have very specific movement directions, so unless there is an interaction that "bumps" the particle, there will be no ergodicity. But just a minimum of interaction will do, I assume.

Does anybody know if the particle visits all the points in the box?

 

[jordi]

I have found an answer: "The dynamics on an individual invariant torus depends on whether k is rational or irrational: in the former case the geodesic flow is periodic, and in the latter it is ergodic, and in fact, uniquely ergodic. In particular, a billiard trajectory with a rational slope is periodic, while the one with an irrational slope is dense in the square."

https://www-fourier.ujf-grenoble.fr/~lanneau/references/masur_tabachnikov_chap13.pdf

So, in a box, if the slope is rational, there is a cycle. If the slope is irrational, the trajectory "bumps" on all points of the box, eventually. But the directions are limited to just two. So there is no ergodicity.

For sure, we are talking about differences between rationals and irrationals, and due to quantumness, we cannot distinguish between the two.

If there is a minimum of interaction among the particles, then it seems we should be closer to ergodicity. The question then is: when two particles bump on each other, are there "preferred" directions? (in the probabilistic sense). It seems that this is getting closer to molecular chaos.

 

[jordi]

Also, it is surprising how different QM and classical mechanics are, for this system.