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:
- Is the domain of $f$ the set $M\times M$?
- In the proof, are we specifically considering $(t, \alpha)$ and $(t_1, \alpha_1)$ such that $T (t, \alpha) =(t_1,\alpha_1)$?
- 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?
Best Answer
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.
Late edit to adress the comment by IMHackingXD: If you're looking at the following picture, you see a positively oriented curve $\gamma$ and a red vector say $\gamma(t_1)-\gamma(t)$ and a blue vector $\gamma(t_1)-\gamma(t+\varepsilon)$
Now $\gamma(t+\varepsilon) \approx \gamma(t) + \varepsilon \gamma'(t)$
so that going from $f(t,t_1)$ to $f(t+\varepsilon,t_1)$, the value of $f(t,t_1)$ is roughly reduced by $2\varepsilon\cos(\alpha)\|\gamma'(t)\|$. A bit more precise: $\gamma(t_1)-\gamma(t+\varepsilon) \approx \gamma(t_1)-\gamma(t)-\varepsilon \gamma'(t)$ Then $$ \|\gamma(t_1)-\gamma(t+\varepsilon)\|^2 \approx f(t,t_1)(f(t,t_1)-2\varepsilon\cos(\alpha)\|\gamma'(t)\|)+\varepsilon^2\|\gamma'(t)\|^2 $$ Does that help?