Consider $H^2$ to be the hyperbolic $2$-space or radius $1$ (for instance, the upper half plane model) with its hyperbolic metric (coming from the corresponding Riemannian metric). Now, consider two distinct points $p, q \in H^2$, let $d = d(p, q)$ (where this is the hyperbolic distance) and denote by $c$ the unit-speed geodesic from $p$ to $q$. Let $\gamma_1$ be a unit-speed geodesic starting at $p$ and let $\gamma_2$ be a unit-speed geodesic starting at $q$, and assume that:
- $\gamma_1$ is perpendicular to $c$ at $p$.
- $\gamma_2$ is perpendicular to $c$ at $q$.
- $\gamma_1(t)$ and $\gamma_2(t)$ are on the "same side" of the geodesic $c$ for $t > 0$.
For $t > 0$, how can we compute $$d(\gamma_1(t), \gamma_2(t))$$ in terms of $d$? I have been told that $$d(\gamma_1(t), \gamma_2(t)) = d \cosh(t),$$ and that one can prove this using Jacobi fields, but I have no idea how to even start proving this question. I know that the length of a Jacobi field can measure this "geodesic deviation" for two geodesics starting at the same point, but here the geodesics start at different points.
Moreover, is there a similar formula for the distance between two geodesics satisfying the same conditions as above in the unit sphere $\mathbb{S}^2 \subset \mathbb{R}^3$ with the angular metric (coming from the restricted Euclidean metric)? By "similar", I mean $$d(\gamma_1(t), \gamma_2(t)) = d \cos(t), $$ for small positive $t$?
Best Answer
The points $p, q, \gamma_1(t), \gamma_2(t)$ span a Saccheri quadrilateral $Q$ in the hyperbolic plane, where $d$ is the base and $t$ is the common length of the legs of $Q$. The hyperbolic distance between $\gamma_1(t), \gamma_2(t)$ is called the summit $s$ of $Q$. The correct formula for the summit is not $d\cosh(t)$ but $$ \cosh(s)= \cosh(d) \cosh^2(t) - \sinh^2(t) $$
or $$ \sinh(s/2)= \cosh(t) \sinh(d/2). $$ See references in the link (primarily, Greenberg's book). Or, you can look here for a self-contained proof.