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Let $ds^2 = \frac{dx^2+dy^2}{y^2}$ for $y>0$. i.e., Poincare half plane model.

I want to derive the distance between two points given by the wikipedia \begin{align} \operatorname{dist}(\langle x_1, y_1 \rangle, \langle x_2, y_2 \rangle) = \operatorname{arcosh} \left( 1 + \frac{(x_2-x_1)^2+(y_2-y_1)^2}{2y_1y_2} \right) \end{align} There are some posts in stackexchange like

Going from Metric to Distance Function in the Poincaré Half Plane

So my trial is solving geodesic equations explicitly.

\begin{align} \frac{d^2 x^\mu}{ds^2} + \Gamma^{\mu}{}_{\nu\rho} \frac{dx^\nu}{ds} \frac{dx^\rho}{ds} =0 \end{align}

The metric is given by

\begin{align} g=(g_{ij}) = \begin{pmatrix} y^{-2} & 0 \\ 0 & y^{-2} \end{pmatrix} \end{align}

and the non-trivial Christoffel symbol reads \begin{equation} \Gamma^1_{12}=\Gamma^1_{21}=-\frac{1}{y},\\ \Gamma^2_{12}=\Gamma^2_{21}=0,\\ \Gamma^1_{11}=\Gamma^1_{22}=0,\\ \Gamma^2_{11}=\frac{1}{y},\\ \Gamma^2_{22}=-\frac{1}{y}. \end{equation} Then the geodeisc equation reads \begin{align} &\frac{d^2x}{ds^2} - \frac{2}{y} \frac{dx}{ds} \frac{dy}{ds}=0 \\ &\frac{d^2y}{ds^2} + \frac{1}{y} \left( \frac{dx}{ds}\right)^2 - \frac{1}{y} \left(\frac{dy}{ds}\right)^2 =0 \end{align} I am trying to solve this with mathematica but fails. Nonetheless, for me solving this geodesic equation does not produce the distance formula...

Assuming $\frac{dx}{ds}\neq 0$, one get further and obtains

\begin{align} \frac{d}{ds} \left( y \frac{dy}{ds} \left(\frac{dx}{ds}\right)^{-1} + x \right) =0 \end{align} ...

phy_math
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    You should really prove that the metric is $SL_2(\Bbb{R})$ invariant, so up to a $SL_2(\Bbb{R})$ transformation you can assume that one of your points is $i$ and the other $b$ is purely imaginary, the geodesic is the portion of the imaginary axis and the distance is $|\log \Im(b) |$. That is $i e^t,t\in \Bbb{R}$ is a arc-length parametrized geodesic and all the others are image of it by a $SL_2(\Bbb{R})$ transformation. – reuns Feb 15 '22 at 13:33
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    The question of deriving the distance formula in the hyperbolic plane (upper half-plane model) was asked many times before on MSE. Your approach is the most direct and the most ineffective one. I do not doubt, by working very hard and for a long time you can eventually succeed, but why? You already realize that your first obstacle is to solve the geodesic equation, which is a nontrivial task unless you know what the answer is. – Moishe Kohan Feb 16 '22 at 00:53

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