Pinball on the infinite plane

The problem is essentially the Sinai billiard. That takes place on a finite square table with a circular hole removed (=peg added). There are standard bounces off the straight edges as well as off the peg.

There is a standard procedure of "unfolding" across flat edges: you just take a reflected copy of the table across any flat edge. There is a correspondence between trajectories in the unfolded table and the original table (a reflection in the original table across a flat edge just becomes a straight trajectory in the unfolded table).

Repeating this, you obtain exactly the model in the question. Sinai gave a statistical analysis of the properties of the trajectory which must imply that the set of angles for which the trajectory remains bounded has measure 0.

This is an infinite horizon Lorentz gas. Ergodicity in the extended space has been shown for this model, but comparatively recently: D. Szasz and T. Varju, J. Stat. Phys. 129 59-80 (2007). But as noted already the set of initial conditions has zero measure, so this is not sufficient in itself. The initial conditions are a smooth one dimensional set in the full two dimensional collision space, so one would expect that if the set of orbits in the collision space with a bound $r$ has dimension $d(r)>1$, the desired set will have dimension $d(r)-1$, approaching unity as $r\to\infty$.

This is a reckless guess, but I wonder if the set of angles might be countable (so consider this a request for a discussion of why that might not be so if you wish.) Let the code of a trajectory be the sequence of posts hit. My wild intuition is that

• The entire infinite code should reveal the starting angle, while an initial segment of a possible code will only confine the starting angle to a sector.
• A bounded trajectory will have to have an eventually periodic code.

If those two things are true, then my guess will follow from there being only countably many eventually periodic codes. Note that I do not say that the actually bounded trajectory has to be eventually periodic (although I am saying that it will eventually be asymptotic to a periodic trajectory with another starting point.)

There will be some purely periodic trajectories (starting at the origin) but only countably many. A very boring one has code $(0,0),(1,0),(0,0),(1,0),\cdots$ . Given another starting post there would be another countable cohort of purely periodic trajectories starting there. I suspect from answers to other questions that (for fixed $r$) there is a unique magic angle from the origin which with results in the code $(0,0),(1,1),(0,1),(1,1),(0,1),(1,1),\cdots$ The center of the ball follows a path asymptotic to the segment from $(r,1).$ to $(1-r,1).$ Of course much flashier things would be possible.

Side question: Is it any different if we say that the ball is a point and the posts have radius $r?$ I'll assume not and speculate on: "could it be that with radius $r=0.48$ (say) the ball would end up arbitrarily far from the start (except for a set of starting angles of measure 0)?" Think of the squares between the posts as rooms. The ball would generally bounce around in any give room for a long time but I am thinking that it would follow a 2 dimensional drunkards walk with a small but positive transition probability.