The Geometric Viewpoint | Periodic Billiard Paths

Periodic Billiard Paths

Posted on April 25, 2014 8:30 am by Dan Medici Comment

“When are we actually going to use this in real life?” A student complains to his or her math teacher. It is the question that we’ve all heard, at one point or another, in our high school years. When we are young, it can be difficult to see how mathematics is related to the ‘real world’. As we grow older, those of us who continue to study mathematics realize that one of the beautiful things about mathematics (and geometry in particular) is that it is present all over the place. This can be seen, for example, in the fact that the golden ratio and the Fibonacci numbers show up, well, almost everywhere. It should be no surprise, then, that almost anything can be studied using mathematics. Here we will see how to take a simple game of skill, billiards, and use geometry to study the game mathematically.

Let’s get acquainted with the ‘rules’ of mathematical billiards, which are somewhat different from the game with which many of us are familiar. First off, we play with only one infinitesimally small billiard ball. For this reason, our goal is not to hit other billiard balls into pockets; rather, it’s to see how our ball travels once we’ve set it in motion. We also play on a frictionless table and require all collisions to be elastic (this means that energy is conserved during each collision; i.e. the ball doesn’t ever slow down). So, once our ball starts moving, it doesn’t stop. The only exception is that we say the ball has stopped if it lands precisely in a corner. Our final rule change is that we can make our billiard table whatever shape we want; later on we will discuss what shapes behave ‘nicely’ as billiard tables.

There is also an important law of physics that our ball must obey; in fact, this law is also present in real billiards. We require that when the billiard ball strikes an edge of our table that

angle of incidence = angle of reflection

That is, the angle the ball’s path (prior to collision) and the edge of the table make is equal to the angle the ball’s path (after the collision) and the edge of the table make.

One of the interesting open questions in the study of mathematical billiards is, given a particular shape as our billiard table, whether it is possible to find a billiard path that is periodic; that is, one which repeats itself over and over again. A billiard path is simply the path that a billiard ball takes once it is set in motion. Here are some simple examples of periodic billiard paths. Notice that the table does not necessarily have to have straight edges.

In the case of the circle, notice that the path is periodic because it bounces perpendicularly off the edges; this is a common way to find periodic billiard paths, as we will see. We will, however, consider only polygonal billiard tables from here on. It is unknown whether every polygonal table has a periodic billiard path, and the solution to this question is an active area of research. R. Schwartz has recently proved that every triangular table with largest angle no greater than 100 degrees has a periodic billiard path. So, what other polygons are known to have periodic billiard paths? We will look at a particular class of polygons called rational polygons and prove the following theorem:

Theorem: Every rational polygon has a periodic billiard path.

A rational polygon is one whose angles are all rational multiples of $\pi$ . It’s easy to get lost in the subtleties of a full proof of this theorem, so we will not discuss every detail. If you are interested in seeing the details and a more rigorous discussion, most of what follows can be found in chapter 17 of Richard Schwartz’ book, Mostly Surfaces.

The first detail we need to discuss is what we mean by a translation surface. We will demonstrate this concept with a simple example. Consider an isosceles triangle $T$ with angles $\frac{\pi}{4}, \frac{\pi}{4},$ and $\frac{\pi}{2}$ . Let’s glue the congruent edges together. We call the ‘glued up’ space $\bar{T}$ . $\bar{T}$ is an example of what we call a cone surface. Now, pick a point $x \in T$ that is a vertex of $T$ . Then we call $\bar{x}$ (the point of $\bar{T}$ that $x$ became after gluing) a cone point. In our example, we have two cone points since two vertices of our triangle were glued together. For each cone point, we define the cone angle to be

cone angle of $\bar{x}$ = $\sum\limits_{x \in \bar{x}}$ angle at $x$ .

The summation is over all points in $T$ which are glued together to form $\bar{x}$ . So, in our example, the cone angle of each cone point is $\frac{\pi}{2}$ because we glued the two vertices that have angle $\frac{\pi}{4}$ together, and the vertex with angle $\frac{\pi}{2}$ is not glued to anything but itself. Now, we can finally give a definition of a translation surface: A translation surface is a cone surface such that each cone angle is an integer multiple of $2\pi$ . We will not be able to prove this here, but what we need to know in order to proceed is that for any rational polygon $P$ , we can take finitely many copies of $P$ and come up with gluings of pairs of edges so we obtain a translation surface. In general, we cannot do this for polygons which are not rational; as we will see shortly, this is the reason why we can only prove that rational polygons have a periodic billiard path.

Here’s an easy example of how we can construct a translation surface from a rational polygon. Let’s begin with one of the simplest examples of a rational polygon: a square. Call our square $S$ . Let $e_i$ , $i = 1,2,3,4$ be the edges of $S$ as shown to the left.

For each $e_i$ , let $R_i$ be the reflection over that edge. If $e_i$ and $e_j$ are parallel, then $R_i = R_j$ (this isn’t exactly true, but hold on!). So, in our example, $R_1 = R_3$ and $R_2 = R_4$ . Let $G$ be the group consisting of all possible compositions of $R_1$ and $R_2$ . We can see that $G = \left\{ {I, R_1, R_2, R_1 \circ R_2} \right\}$ since any other composition gives us something already in $G$ . Note that the order of $G$ is 4; in particular, $G$ is a finite group! This is very important: going through this process with any rational polygon will yield a finite group. If we generate a group this way for a polygon which is not rational, then the group will not necessarily be finite, and if that happens we cannot continue. We make ‘copies’ of $S$ by applying each of our group elements to $S$ . We also shift our copies of $S$ so that they are all disjoint from each other. The fact that we shift is what allows us to say two reflections are equal if the corresponding edges are parallel. We now have four copies of our original square.

We begin with the obvious gluings: $e_1$ to $R_1(e_1)$ , $e_2$ to $R_2(e_2)$ , $R_2(e_1)$ to $R_1 \circ R_2(e_1)$ , and $R_1(e_2)$ to $R_1 \circ R_2(e_2)$ .

Now we have a bigger square whose edges we need to glue together. To do this, we will glue opposite edges together. For example, we glue $e_3$ to $R_1(e_3)$ , $R_2(e_3)$ to $R_1 \circ R_2(e_3)$ , etc. (For general polygons, there are equations for determining which gluings to make, but we won’t go into that). It is easy to verify that all cone angles are integer multiples of $2\pi$ . The resulting surface is a torus, which is a doughnut or ring shape, and this is our translation surface. It’s important to note that we still think of our translation surface as being flat. Although our translation surface does look like the surface below, we prefer to visualize it as a flat polygon with with the gluings that we’ve defined.

We need two lemmas before we can tackle our theorem. Before we introduce our first lemma, however, we need a little background information. Define a path $\gamma \in \bar{P}$ (where $\bar{P}$ is a translation surface) to be straight if when we ‘unglue’ our translation surface, the path becomes a straight line. For example, the equator is a straight path on the Earth; although the equator is curved, it looks like a straight line on a map.

We need a way to go between our polygon $P$ and the translation surface $\bar{P}$ that we build from $P$ . To accomplish this, we define a function $\pi : \bar{P} \rightarrow P$ . $\pi$ takes as an input some point $\bar{p}$ and outputs the point on $P$ corresponding to $\bar{p}$ after we construct our translation surface.

Lemma 1: Let $\hat{\gamma}$ be a straight path on $\bar{P}$ which does not go through any cone points of $\bar{P}$ . Then $\gamma = \pi(\hat\gamma)$ is a billiard path on $P$ .

We won’t go through a full proof here. However, it is not so difficult to convince yourself of this result: take a piece of paper and fold it in half, creating a crease. Unfold it and draw a straight line which crosses the crease. This straight line is analogous to a straight path which crosses an edge from one copy of $P$ to another. When you refold the paper (folding the paper is analogous to hitting $\bar{P}$ with our $\pi$ function), you’ll see that the straight line now ‘bounces’ off the crease in accordance with the rule angle of incidence = angle of reflection. In the figure below, the dotted line is the image of the straight path after refolding the paper.

Now, for any $\bar{x} \in \bar{P}$ , and for any direction on $\bar{P}$ , define $f(\bar{x})$ to be the point of $\bar{P}$ obtained by starting at $\bar{x}$ and moving one unit in the chosen direction. $f$ is defined everywhere except when the path between $\bar{x}$ and $f(\bar{x})$ hits a cone point. If we think of our path as traveling at unit speed, we can say that $\pi(f^n(\bar{x}))$ is the location of our billiard ball on $P$ after $n$ seconds. We will now prove our second lemma:

Lemma 2: Pick a point $p \in \bar{P}$ . Then there is a point $q \in \bar{P}$ that is very close to $p$ and $f^n(q)$ is very close to $p$ (for some $n$ ).

Proof: Let $D_0$ be a small disk around $p$ , and for each $n \in \mathbb{N}$ let $D_n$ be a disk centered around $f^n(p)$ of the same size as $D_0$ . Since the area of $\bar{P}$ is finite, the disks $D_0, D_1, D_2, \ldots$ are not disjoint. Hence we can find two disks, $D_i$ and $D_j$ that intersect at a point. Call this point $x_n$ and assume $i < j$ . Then we can see that $D_{i-1}$ and $D_{j-1}$ intersect at the point $f^{-1}(x_n) = x_{n-1}$ . We can continue this process until we see that $D_0$ and $D_{j-i}$ intersect at the point $x_0$ , which is obtained by composing $f^{-1}$ with itself $(j-i)$ times and evaluating it at $x_n$ . But then $x_0$ is very close to $p$ and also $f^{j-i}(x_0)$ is very close to $p$ (since $x_o$ is contained in the disks $D_0$ and $D_{j-i}$ ). This completes the proof.

We now have all the tools we need to prove the theorem. We let $P$ be any rational polygon, and we build a translation surface $\bar{P}$ from $P$ . Pick a point $p \in \bar{P}$ and a direction that is perpendicular to and heading towards a nearby edge of $P$ . $\hat{\gamma}$ will be a straight path which starts at $p$ and in the chosen direction. By our first lemma, there is a billiard path $\gamma$ in $P$ which corresponds to $\hat{\gamma}$ . By our choice of direction, $\gamma$ is traveling perpendicularly to an edge of $P$ at time 0. By our second lemma, there is a $q$ very close to $p$ such that $f^n(q)$ is also very close to $p$ (for some $n$ ). Let $\beta$ be the path starting at $q$ and in our chosen direction. We make our choice of “very close” to mean that $q$ and $p$ are on the same copy of $P$ used in the construction of $\bar{P}$ . Recall that $\beta$ is a straight path, and so $\beta$ is traveling in the same direction at times 0 (when it is at $\pi(q)$ ) and $n$ (when it is at $\pi(f^n(q))$ ). But at time 0, $\beta$ is traveling perpendicularly to an edge of $P$ , since it is traveling in the same direction as $\gamma$ . Hence, $\beta$ is traveling perpendicularly to an edge of $P$ at times 0 and $n$ . Therefore, $\beta$ hits the edge perpendicularly twice, and so it periodic, since it just retraces its path each time it hits the edge perpendicularly. Such a path might look like this:

There are two natural questions to ask now. The first is “What about polygons that are not rational? Do these polygons not have periodic billiard paths?” The answer here is ‘no’, there are some non-rational polygons which do have periodic billiard paths. To convince yourself of this, notice that any parallelogram has a periodic path. In fact, there are no known examples of non-rational polygons which do not have a periodic billiard path. The second, and more difficult, question to ask is “Does this result hold in three dimensions as well? Is it true that rational polyhedra have periodic billiard paths?” The intuitive answer might be ‘yes’ because none of our rules change when we move up in dimension. It has been proven that the answer is in fact ‘yes’ for tetrahedra, but the result is unknown for more complicated polyhedra, like this one.

Sources

Mostly Surfaces by Richard Evan Schwartz

Obtuse Triangular Billiards II: 100 Degrees Worth of Periodic Trajectories by Richard Evan Schwartz

Images were either hand drawn or found using a Google Image search

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