# How to show that the Billiard flow is invariant with respect to the area form $\sin(\alpha)d\alpha\wedge dt$

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?

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.
• Thank you! If $f$ is only depends on $t, t_1$, why does it make sense to write $\mathrm{d}^2f = \sin{\alpha}\mathrm{d}\alpha \wedge \mathrm{d}t - \sin{\alpha_1} \mathrm{d}\alpha_1 \wedge \mathrm{d}t_1$? – Quoka Feb 11 '20 at 8:46
• 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... – frog Feb 11 '20 at 8:56
• Thanks again. last question: why is it that we are getting $\alpha_1$ and not $\pi - \alpha_1$? – Quoka Feb 11 '20 at 9:08
• 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. – frog Feb 11 '20 at 9:42