# Hyperbolic Two Body Problem

## Math
$$G= PSL(2, \mathbb{R})$$ and the action on $\mathbb{H}$ is

$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
z = \frac{az + b}{cz+d}
$$

The Lie algebra is $\mathfrak{g} := \mathfrak{sl}(2, \mathbb{R})$ (i.e. traceless 2 by 2 real matrices). We coordinatize $\mathfrak{g}$ via the global chart
$$
(\zeta, \eta, \xi) \mapsto \begin{bmatrix} \zeta & \eta-\xi \\
\eta+\xi & -\zeta
\end{bmatrix}
$$

Let $Q:= \mathbb{H}^2 \setminus \Delta$. Using the natural action on $Q$ induced by the action of $G$ on There is an isomorphism $\Psi: \mathbb{R}^+ \times G \to Q$ defined by
$$
\Psi(r,g) = (g \cdot \mathbb{i} , g \cdot (e^r \mathbb{i})).
$$

for proof that $\Psi$ is an isomorphism???

## Mechanics
In words, we have a Kinetic minus Potential energy Lagrangian, and configuration space is two copies of the hyperbolic plane. There is a symmetry given by the isometries of the hyperbolic plane (this is $SL(2) / \mathbb{R}+$)

$$
Q = \mathbb{H}^2 \\
L(z_1, z_2, u_1, u_2) =
\quad \frac{m_1}{2 \Im(z_1)^2} \|u_1\|^2 + \frac{m_2}{2\Im(z_2)^2} \|u_2\|^2 - V(r)
$$
where $r$ is the hypobolic distance between $z_1$ and $z_2$.

We can pull-back our Lagrangian to $\mathbb{R}^+ \times G$
$$
\Psi^*L(r,g,\dot{r}, \dot{g}) = \frac{1}{2} (\dot{r}, \dot{g})^i M_{ij} (\dot{r}, \dot{g})^j - V(r)
$$

Since we are ultimately going to do $G$ reduction, and this kinetic energy is just a inner product on a Lie algebra $\mathbb{sl}(2, \mathbb{R})$. So the reduced Lagrangian is on $T\mathbb{R}^+ \times \mathfrak{g}$.

$$
\ell(r,v,\xi) = \frac{1}{2} (v,\zeta, \eta, \xi)^T M_{ij} (v, \zeta, \eta, \xi) - V(r)
$$

$$
M_{ij}(r,e) = \begin{bmatrix}
m_2 & 2m_2 & 0 & 0 \\
2m_2 & 4(m_1+m_2) & 0 & 0 \\
0 & 0 & 4m_1 + m_2 \frac{(1 + e^{2r})^2}{e^{2r}} & \frac{1-e^{4r}}{e^{2r}} \\
0 & 0 & \frac{1-e^{4r}}{e^{2r}} & \frac{m_2(1-e^{2r})^2}{e^2r}
\end{bmatrix}
$$

## Discrete Mechanics
It's actually easier to create the discrete Lagrangian first by
$$
\ell_d: \mathbb{R}^+ \times \mathbb{R}^+ \times G \to \mathbb{R}
$$
by the simple relation
$$
\ell_d(r_1, r_2, g) := \ell(r_1, r_2 - r_1, \log(g))
$$
Then we can define the discrete Lagrangian, $L_d: Q \times Q \to \mathbb{R}$ by
$$
L_d(r_1,g_1, r_2, g_2) := \ell_d(r_1, r_2, g_2 g_1^{-1})
$$

Therefore our discrete Lagrangian is expressible as a composition of elementary matrix functions.

The discrete equations of motion are found by considering the function
$$
A_d(q_1, q_2, q_3) = L_d(q_1, q_2) + L_d(q_2, q_3)
$$

then solving for $q_3$ given $q_1, q_2$ and the constraint $\frac{\partial A_d}{\partial q_2} (q_1,q_2, q_3) = 0.$

Discrete Lagrange-Poincare dynamics amounts to solving one-step of Euler-Lagrange with the initial condition $(r_1, I, r_2, g_2)$ to get a point $(r_3, g_3)$, then translating backwards by $g_2$ to get the updated initial condition $(r_2, I), (r_3, g_3g_2^{-1})$ which serves as the input in the next time-step.

$$
A_d(r_1,r_2,r_3, g_1, g_2, g_3) = L_d(r_1, g_1, r_2, g_2) + L_d(r_2, g_2, r_3, g_3) \\
= L_d(r_1, I, r_2, g_1^{-1} g_2) + L_d(r_2, I, r_3, g_2^{-1} g_3) \\
= \ell_d(r_1, r_2, g_1^{-1} g_2) + \ell_d(r_2, r_3, g_2^{-1} g_3)
$$

The Langrage-Poincare equations amount to solving for $(r_3, g_3)$ for fixed $r_1, r_2, g_2$ in the equations
$$
\p \ell_d(r_1, r_2, g_2) + \ell_d(r_2, r_3, g_2^{-1} g_3)
$$