# Pinball on the infinite plane

Imagine pinball on the infinite plane, with every lattice point $\mathbb{Z}^2$ a point pin. The ball has radius $r < \frac{1}{2}$. It starts just touching the origin pin, and shoots off at angle $\theta$ w.r.t. the horizontal. It reflects from pins in the natural manner:

What happens? More specifically,

Q. For a given $r$, what is the measure of the set of angles $\theta$ for which the pinball remains a finite distance from the origin forever?

Many other questions suggest themselves, but let me leave it at that basic question for now.

This problem seems superficially similar to Polya's Orchard problem (e.g., explored in the MO question, "Efficient visibility blocker s in Polya’s orchard problem"), but the reflections produce complex interactions. It is also similar to Pach's enchanted forest problem, mentioned in the MO question, "Trapped rays bouncing between two convex bodies", but it seems simpler than that unsolved problem, due to the lattice regularity. Has it been considered before, in some guise? If so, pointers would be appreciated. Thanks!

Addendum. My question can be rephrased in terms of the "Sinai billiard," as Anthony Quas explains: I am asking for the fate of radial rays in the situation illustrated:

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I'd suggest putting the origin pin back in, and starting the ball in the same direction, but just touching the origin pin (which can be done since $r< {1\over 2}$). – Jeff Strom Feb 23 '12 at 15:17
Please add the [billiards] tag. – Zsbán Ambrus Feb 23 '12 at 15:33
@Jeff: Nice suggestion! Now followed. @Zsbán: Thanks; done. – Joseph O'Rourke Feb 23 '12 at 15:37

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.

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Applying ergodic theory here is problematic. Indeed Sinai's ergodicity implies that the set of trapped trajectories has zero measure. But unfortunately the set of trajectories in the question has zero measure itself (essentially, they all start from one point). – Sergei Ivanov Feb 23 '12 at 20:03
As with the trapped rays question, I'll comment here that this is a (Sinai billiard =) Lorentz gas. – Steve Huntsman Feb 23 '12 at 23:00
@ Sergei: If we allow the ball to just touch the original pin and then allow for arbitrary initial momentum. We then get all the trapped trajectories. – Liviu Nicolaescu Feb 24 '12 at 14:13

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 $r\to\infty$.

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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.

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@Aaron: Just on your side question: The two circumstances are equivalent: shrink the ball, grow the pins correspondingly. Here is one way to see it. Consider the situation as I described it, and focus on when the ball just contacts a pin. Pick up the disk and center it on the pin. Now think of the pin as this disk, and what used to be the center of the ball as a point particle. One obtains the same reflection. – Joseph O'Rourke Feb 23 '12 at 18:39
That is how I saw it. For some reason the second point of view makes it easier to imagine asymptotic trajectories. – Aaron Meyerowitz Feb 23 '12 at 18:44
@Aaron: The bounded orbit should not be periodic. You may take two periodic orbits say A and B which are close to each other at some part and produce a nonperiodic orbit say which goes 7 times around A switches and goes 19 times around B then switches back and goes 27 times around A and so on. – Anton Petrunin Feb 23 '12 at 23:03

Denote by $\Theta_r$ the set of such angles. I am inclined to believe that the set $\Theta_r$ satisfies a zero-one law, with a possible threshold $r_0$. (If $r< r_0$, then set $\Theta_r$ has measure zero, while if $r>r_0$ the complement of $\Theta_r$ has measure zero.)

Here is my "argument". Fix $r\in (0,\frac{1}{2})$. Observe that we have a map

$$T_r: S^1\to S^1$$

defined as follows. Shoot a ball touching the pin at the origin in the direction $\theta$. Denote by $P_r(\theta)\in\mathbb{Z}^2$ the location of the first pin touched by the ball. After it touches $P_r(\theta)$, the ball will continue traveling along a ray of angle $T_r(\theta)$.

Clearly $\theta\in \Theta_r \Rightarrow T_r(\theta)\in \Theta_r$. The map $T_r$ is bijective and the set $\Theta_r$ is $T$-invariant. I am inclined to believe that $T_r$ is ergodic with respect to a measure absolutely continuous with respect to the arclength measure on $S^1$. If this is the case then the zero-one phenomenon above holds. (Do not ask me why I have this ergodic belief.)

The threshold statement seems harder to "argue", but observe that if $r=0$ then no irrational angle belongs to $\Theta_r$

Oops! $T_r$ is indeed not injective and in fact it is the wrong map. Here is the correct map. First fatten the pins to disk of radii $r$, and reduce the ball to a point particle. Consider the cylinder $C=S^1\times [-\pi/2,\pi/2]$, where $S^1$ is the boundary of a fat pin. A particle leaves the boundary of this fat pin with a velocity making an angle $\phi\in [-\pi/2,\pi/2]$ with the outer normal to the boundary at the departure point.

Then $T_r: C\to C$. This is almost injective (problems do appear when the particle grazes the boundary of a pin.) However is measure preserving for a Liouville-type measure. associated to the geodesic flow on the plane with the fat pins removed. The ergodicity is not obvious, but the dynamics of maps such as $T_r$ are being investigated by ergodic theorists. (I'm not one.)

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Why is $T_{r}$ injective? – Peter Sarkoci Feb 23 '12 at 17:20
Why do you have that ergodic belief? – Sean Eberhard Feb 23 '12 at 17:44
Yeah, there's no way that $T_r$ is injective. If you look at a ball of... "small" radius moving straight up, and then look what happens as you let $\theta$ vary, you can see that you span most of the circle with that. If you then consider the same setup except with the ball heading straight down, right, and left, you see that this map can not be injective for small values of $r$. – Simon Rose Feb 23 '12 at 18:26