# Shortest path on hyperboloid

On the sphere $S^2$, the shortest path between two points is the great circle path. How about $H^2$, the hyperboloid $x^2+y^2-z^2=-1, z\ge 1$, with the Euclidean distance? Is there a formula for the shortest path between two points on the surface? And what is the length of the shortest path?

Note that it is not the hyperbolic distance; it is the Euclidean distance.

-
Have you ever read this: en.wikipedia.org/wiki/Geodesic May be of interest for you –  Ilya Oct 25 '11 at 16:12
There are a few differential equations you need to solve to obtain equations for the geodesic... note that you have a surface of revolution, so the task is somewhat easier. See formulae 31-35 here. –  Ｊ. Ｍ. Oct 25 '11 at 16:21
Even for a hyperboloid which is not axially symmetric, it should be possible to find the geodesics using Jacobi's ellipsoidal coordinates (which were invented for computing the geodesics on an ellipsoid). I've never tried to go through all the details, though. –  Hans Lundmark Oct 25 '11 at 18:47
The Euclidean metric in $\mathbb{R}^3$ induces a metric on a hyperboloid, and the shortest path will be shortest with respect to this distance. Standard coordinates on a hyperboloid $z^2 - x^2 - y^2 = 1$ would be $$x = \sinh(t) \sin \phi \qquad y = \sinh(t) \cos \phi \qquad z = \cosh(t)$$ with the interval: $$\mathrm{d}s^2 = \cosh(2 t) \mathrm{d} t^2 + \sinh^2(t) \mathrm{d} \phi^2$$ That means that non-zero Christoffel symbols are $$\Gamma^t_{tt}=\tanh(2t) \qquad \Gamma^t_{tt}=-\frac{1}{2}\tanh(2t) \qquad \Gamma^\phi_{t\phi} = \Gamma^\phi_{\phi t} = \frac{1}{\tanh(t)}$$ Geodesic equations are readily obtained: $$t^{\prime\prime}(s) + \tanh(2 t(s)) \left( (t^\prime(s))^2 - \frac{1}{2} (\phi^\prime(s))^2 \right) = 0 \qquad \phi^{\prime\prime}(s) + \frac{2}{\tanh(t(s))} t^\prime(s) \phi^\prime(s) = 0$$ It is not hard to see, that the latter equation admits an integral of motion, i.e. $\phi^\prime(s) \sinh^2(t(s)) = \mathcal{L}$, because $$\frac{\mathrm{d}}{\mathrm{d} s} \left( \phi^\prime(s) \sinh^2(t(s)) \right) = \sinh^2(t(s)) \left( \phi^{\prime\prime}(s) + \frac{2}{\tanh(t(s))} t^\prime(s) \phi^\prime(s) \right) \left. = \right|_{\text{eq. of motion}} = 0$$