4
$\begingroup$

Consider a plane billiard table $D \subset \mathbb{R}^2$ (i.e. a bounded open connected set) with smooth boundary $\gamma$ being a closed curve. Next, let $M$ denote the space of tangent unit vectors $(x,v)$ with $x$ on $\gamma$ and $v$ being a unit vector pointing inwards. We then define the billiard map $$ T : M \to M. $$ To understand the map $T$, we consider a point mass traveling from $x$ in direction $v$. Let $x_1$ be the first point on $\gamma$ that this point mass intersects and suppose that $v_1$ is the new direction of the mass upon incidence. Then $T$ maps $(x,v)$ to $(x_1, v_1)$.

We now introduce an alternate ''coordinate system'' describing $M$. Parametrize $\gamma$ by arc-length $t$ and fix a point $(x,v) \in M$. We can find $t$ such that $x = \gamma(t)$ and let $\alpha \in (0, \pi)$ be the angle between the tangent line at $x$ and $v$. The tuple $(t, \alpha)$ uniquely determines the point $(x,v)$ in $M$, and thus offers and alternative description of this space.

My question is as follows: I want to show that the area form given by $$ \omega := \sin{\alpha}\,\mathrm{d}\alpha \wedge \mathrm{d}t $$ is invariant under $T$.

I found a proof of this invariance property proof in S. Tabachnikov's Geometry and billiards but I'm having some trouble understanding a critical part of the proof.

If anyone can explain the proof to me (or provide me with another proof) I would highly appreciate it. An intuitive explanation is also appreciated, but I am looking for a rigorous proof if possible. We restate this theorem formally below and provide the proof as given by Tabachnikov.

Theorem 3.1. The area form $ω = \sin α \,dα \wedge dt$ is $T$-invariant.

Proof. Define $f(t, t_1)$ to be the distance between $\gamma(t)$ and $\gamma(t_1)$. The partial derivative $\frac{\partial f}{\partial{t_1}}$ is the projection of the gradient of the distance $\left\vert{\gamma(t)\gamma(t_1)}\right\vert$ on the curve at point $\gamma(t_1)$. This gradient is the unit vector from $\gamma(t)$ to $\gamma(t_1)$ and it makes angle $\alpha_1$ with the curve; hence $\partial f/\partial t_1 = \cos{\alpha_1}$. Likewise, $\partial f/\partial t = -\cos{\alpha}$. Therefore, $$ \mathrm{d}f = \frac{\partial f}{\partial t} \mathrm{d}t + \frac{\partial f}{\partial t_1}\mathrm{d}t_1 = -\cos{\alpha}\,\mathrm{d}t + \cos{\alpha_1}\,\mathrm{d}t_1 $$ and hence $$ 0 = \mathrm{d}^2f = \sin{\alpha}\mathrm{d}\alpha \wedge \mathrm{d}t - \sin{\alpha_1} \mathrm{d}\alpha_1 \wedge \mathrm{d}t_1. $$ This means that $\omega$ is a $T$-invariant form.

The above proof is copied directly from the book. I have the following questions about his method:

  1. Is the domain of $f$ the set $M\times M$?
  2. In the proof, are we specifically considering $(t, \alpha)$ and $(t_1, \alpha_1)$ such that $T (t, \alpha) =(t_1,\alpha_1)$?
  3. I am having a hard time understanding how the author obtains $\partial f/\partial t_1 = \cos{\alpha_1}$ and $\partial f/\partial t = -\cos{\alpha}$. The explanation given feels mostly heuristic, how could I go about constructing a rigorous proof?
$\endgroup$
1
$\begingroup$

As you explain in your question, the set $M$ consists of points $(x,v)$, where $x$ is a point of $\partial D$ and $v$ is an inward pointing unit vector. Now you chose to parametrize the set of points in $M$ differently: If $\gamma:[0,\ell)\to\partial D$ is the parametrization of $\partial D$, this gives rise to a map $f:[0,\ell)\times[0,\ell)\to (0,\infty)$, where actually $f$ is bounded by $\operatorname{diam}(D)$. I hope this helps for the first question. For the third question: Since $f$ is defined to be the distance between $\gamma(t)$ and $\gamma(t_1)$ you may write $$ f(t,t_1)=|\gamma(t_1)-\gamma(t)|. $$ In this way you can compute $$ \partial_{t_1}f(t,t_1) = \frac{\langle \gamma(t_1)-\gamma(t),\dot\gamma(t_1)\rangle}{|\gamma(t_1)-\gamma(t)|}. $$ Since $\gamma$ is a unit speed curve, you can write $$ \partial_{t_1}f(t,t_1) =\frac{\langle \gamma(t_1)-\gamma(t),\dot\gamma(t_1)\rangle}{|\gamma(t_1)-\gamma(t)||\dot\gamma(t_1)|}=\cos\left(\sphericalangle(\gamma(t_1)-\gamma(t),\dot\gamma(t_1))\right). $$ According to your definition, the angle $$ \sphericalangle(\gamma(t_1)-\gamma(t),\dot\gamma(t_1))=\alpha_1. $$ Finally, for the second question, the answer is yes.

$\endgroup$
  • $\begingroup$ Thank you! If $f$ is only depends on $t, t_1$, why does it make sense to write $\mathrm{d}^2f = \sin{\alpha}\mathrm{d}\alpha \wedge \mathrm{d}t - \sin{\alpha_1} \mathrm{d}\alpha_1 \wedge \mathrm{d}t_1$? $\endgroup$ – Quoka Feb 11 '20 at 8:46
  • $\begingroup$ This is just expressing $f$ by means of other coordinates. The map $f$ actually depends on the angles in an implicit way: Once you choose $t$ and $t_1$ there is a unique angle $\alpha$ such that $\gamma(t)$ is "sent" to $\gamma(t_1)$ by the map $T$ and this in turn defines also an angle $\alpha_1$. But I agree that it is a little confusing since the two systems of coordinates are mixed up... $\endgroup$ – frog Feb 11 '20 at 8:56
  • $\begingroup$ Thanks again. last question: why is it that we are getting $\alpha_1$ and not $\pi - \alpha_1$? $\endgroup$ – Quoka Feb 11 '20 at 9:08
  • $\begingroup$ Yes, you're right of course. You get $\pi-\alpha_1$ but If you reverse the orientation of $\gamma$, you really get $\alpha_1$. In any case, the form $\omega$ will be invariant. $\endgroup$ – frog Feb 11 '20 at 9:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.