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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.

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Have you ever read this: en.wikipedia.org/wiki/Geodesic May be of interest for you –  Ilya Oct 25 '11 at 16:12
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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

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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 $$ \frac{\mathrm{d}}{\mathrm{d} s} \left( \phi^\prime(s) \sinh^2(t(s)) \right) = \sinh^2(t(s)) \left( \phi^{\prime\prime}(s) + \frac{2}{\tanh(t(s))} t^\prime(s) \phi^\prime(s) \right) \left. = \right|_{\text{eq. of motion}} = 0 $$

Unfortunately this does not get us any closer to the geometric interpretation of the geodesic. Is it an interesection of the hyperboloid with a plane containing two points ? I suspect so, but I do not see how to prove it.

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Probably the integral of motion is just what is implied by rotational symmetry, i.e, Clairaut's equation for geodesics on a surface of revolution, specialized to a hyperboloid. I don't think the geodesic on a quadric surface is contained in a plane, at least for an ellipsoid integrating the geodesic flow was an illustrious problem first solved by Jacobi using elliptic integrals. –  zyx Oct 26 '11 at 7:15

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