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 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 γ and v being a unit vector pointing inwards. We then define the billiard map T:M->M. To understand the map T, we consider a point mass traveling from x in direction v. Let x1 be the first point on γ that this point mass intersects and suppose that v1 is the new direction of the mass upon incidence. Then T maps (x,v) to (x1,v1). We now introduce an alternate ''coordinate system'' describing M. Parametrize gamma by arc-length t and fix a point (x,v)isM. We can find t such that x=gamma(t) and

How to show that the Billiard flow is invariant with respect to the area form $\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})d\mathrm{\alpha <!--\; \alpha \; -->}\wedge <!--\; \wedge \; -->dt$
Consider a plane billiard table $D\subset <!--\; \subset \; -->{\mathbb{R}}^2$ (i.e. a bounded open connected set) with smooth boundary $\mathrm{\gamma <!--\; \gamma \; -->}$ being a closed curve. Next, let M denote the space of tangent unit vectors (x,v) with x on $\mathrm{\gamma <!--\; \gamma \; -->}$ and v being a unit vector pointing inwards. We then define the billiard map
$T:M\to <!--\; \to \; -->M.$
To understand the map T, we consider a point mass traveling from x in direction v. Let $x_1$ be the first point on $\mathrm{\gamma <!--\; \gamma \; -->}$ that this point mass intersects and suppose that v1 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 $\mathrm{\gamma <!--\; \gamma \; -->}$ by arc-length t and fix a point $(x,v)\in <!--\; \in \; -->M$. We can find t such that $x=\mathrm{\gamma <!--\; \gamma \; -->}(t)$ and let $\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->(0,\mathrm{\pi <!--\; \pi \; -->})$ be the angle between the tangent line at x and v. The tuple $(t,\mathrm{\alpha <!--\; \alpha \; -->})$ 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
$\mathrm{\omega <!--\; \omega \; -->}:=\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{d}\mathrm{\alpha <!--\; \alpha \; -->}\wedge <!--\; \wedge \; -->\mathrm{d}t$
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 $\omega =\sin <!--\; \; -->\alpha d\alpha \wedge <!--\; \wedge \; -->dt$ is T-invariant.
Proof. Define $f(t,t_1)$ to be the distance between $\mathrm{\gamma <!--\; \gamma \; -->}(t)$ and $\mathrm{\gamma <!--\; \gamma \; -->}(t_1)$. The partial derivative $\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}{t}_1}$ is the projection of the gradient of the distance $\left|\mathrm{\gamma <!--\; \gamma \; -->}(t)\mathrm{\gamma <!--\; \gamma \; -->}(t_1)\right|$ on the curve at point $\mathrm{\gamma <!--\; \gamma \; -->}(t_1)$. This gradient is the unit vector from $\mathrm{\gamma <!--\; \gamma \; -->}(t)$ to $\mathrm{\gamma <!--\; \gamma \; -->}(t_1)$ and it makes angle ${\mathrm{\alpha <!--\; \alpha \; -->}}_1$ with the curve; hence $\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}{t}_1=\cos <!--\; \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_1$. Likewise, $\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}t=-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}$. Therefore,
$\mathrm{d}f=\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}t}\mathrm{d}t+\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}{t}_1}\mathrm{d}{t}_1=-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{d}t+\cos <!--\; \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_1\mathrm{d}{t}_1$
and hence
$0={\mathrm{d}}^{2}f=\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{d}\mathrm{\alpha <!--\; \alpha \; -->}\wedge <!--\; \wedge \; -->\mathrm{d}t-<!--\; -\; -->\sin <!--\; \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_1\mathrm{d}{\mathrm{\alpha <!--\; \alpha \; -->}}_1\wedge <!--\; \wedge \; -->\mathrm{d}{t}_1.$
This means that $\mathrm{\omega <!--\; \omega \; -->}$ 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 $M\times <!--\; \times \; -->M$?
In the proof, are we specifically considering $(t,\mathrm{\alpha <!--\; \alpha \; -->})$ and $(t_1,{\mathrm{\alpha <!--\; \alpha \; -->}}_1)$ such that $T(t,\mathrm{\alpha <!--\; \alpha \; -->})=(t_1,{\mathrm{\alpha <!--\; \alpha \; -->}}_1)$?
I am having a hard time understanding how the author obtains $\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}{t}_1=\cos <!--\; \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_1$ and $\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}t=-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}$. The explanation given feels mostly heuristic, how could I go about constructing a rigorous proof?