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Billiards are a class of dynamical systems in which a point particle moves uniformly in a domain $D\subset \mathbb{R}^d$ except for mirror-like reflections from the boundary. Varying $D$ leads to examples satisfying many ergodic properties. Billiards enhance visual explanations of dynamical concepts to students and the general public. There are many applications in physics and image processing. The free motion and/or reflection rule may be generalized.

5 votes
1 answer
187 views

Unfoldings of trajectories on the Veech triangle $V_4$

$V_4$ is a Veech triangle, so the dynamics of billiards on it are very well understood. Above is the unfolding of $V_4$, with edge identification indicated by numeral. …
Alex Becker's user avatar
3 votes
0 answers
141 views

Reference for standard lemmas in polygonal billiards

I'm trying to track down references for a couple of well-known results in the theory of polygonal billiards for a paper I'm working on. … This is stated for example in Section 2.1 of "Obtuse Triangular Billiards II: 100 Degrees Worth of Periodic Trajectories" by R. E. …
Alex Becker's user avatar
1 vote
1 answer
268 views

Computing saddle connections in flat structures

Background: A polygonal billiards table $P$ with rational angles gives rise to a flat structure $S(P)$ in a standard way, described here. …
Alex Becker's user avatar
4 votes
0 answers
83 views

Do identical orbit tiles imply identical combinatorial types?

Given a periodic trajectory on a triangle, we can associate to this trajectory a sequence of integers $1,2$ and $3$ by labeling the edges of the triangle and taking the sequence of edges the trajector …
Alex Becker's user avatar
10 votes
1 answer
446 views

Does the $n$-gonal billiards conjecture follow from the $m$-gonal conjecture when $m>n$?

I.e., does the $3$-gonal billiards conjecture follow from the $m$-gonal billiards conjecture for any $m>3$? … Furthermore, does the $n$-gonal billiards conjecture follow from the $m$-gonal billiards conjecture for any $n$ and $m>n$? There is one case for which this is easily seen. …
Alex Becker's user avatar