Billiards

September 24, 2009

Amie Wilkinson asked me the following question some time ago:

Given a smooth convex Jordan curve J \subseteq \mathbb{R}^2, consider the billiard map \varphi: J \times (0, \pi) \rightarrow J \times (0, \pi), let \pi_\theta: J \times (0, \pi) \rightarrow (0, \pi) be the projection.

a) If \forall (p, \theta) \in J \times (0, \pi), \pi_\theta \circ \varphi (p, \theta) = \theta, does this imply J is a circle?

(Yes, in fact we only need \varphi to fix the second component of points in \{p, q \} \times (0, \pi) for a chosen pair of points p, q. Classical geometry)

b) If \forall p \in J, \pi_\theta \circ \varphi (p, \pi/2) = \pi/2, does this imply J is a circle?

(No, my example was a cute construction that attaches six circular arcs together)

c) What’s the smallest set S \subseteq (0, \pi) s.t. if \varphi fixes the second component on J \times S then J has to be a circle?

I am still thinking about c)…My guess is that any sub interval would work, and of course any dense subset inside a given set works equally well as the whole set…

But is it possible to have only finitely many angles? Maybe even two angles?

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: