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

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.

• 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
• To confirm zyx's comment about planar sections: If the ambient metric is Lorentzian instead of Euclidean, the induced metric is hyperbolic (and homogeneous under the ambient isometry group), so the geodesics are plane sections. The respective metrics (hyperbolic and hyperboloid) are$$ds^{2} = dt^{2} + \sinh^{2} t\, d\phi^{2},\qquad ds^{2} = \cosh(2t)\, dt^{2} + \sinh^{2} t\, d\phi^{2}.$$Expressing $t$ as a function of $\phi$ along a geodesic, hyperboloid geodesics "grow more slowly than" hyperbolic geodesics (by a standard formula for Clairaut patches), and therefore are not plane sections. – Andrew D. Hwang May 6 '15 at 17:07

You might want to look at A. Pressley, Differential Geometry, text problem 8.1, he calculates the geodesics for the hyperboloid of one sheet $$x^2+y^2-z^2=1$$ and finds that they are actually four, two straight lines, one circle and one hyperbola. But he keeps things very simple and finds them just by using some basic definition.

The shortest path between the points $p$ and $q$ on the upper sheet where $z>0$ on the hyperboloid $\{x^2+y^2-z^2=-1\}$ is the curve where the hyperboloid intersects the plane that contains the origin and the points $p$ and $q$.

(As for how to compute the Euclidean distance, the other posters seem to know their differential geometry better than I do. If you wanted the hyperbolic distance on the other hand, induced by the quadratic form used in defining the hyperboloid, there is a simpler way. But that was not your question and it's easier to look up.)

If you use cylindrical coordinates $(r,\theta, z)$ since $(r,z)$ relation is known, only $(r,\theta)$ relation needs to be found out.

The Clairaut's Law is especially suited to finding geodesics on surfaces of revolution.

The procedure can be used to find geodesics on any surface of revolution when

meridian is given.

Choose any one of the two convex sheets of the hyperboloid given.

$z^2 - r^2 = 1$

Differentiate with respect to r; $z = r* \tan(\phi).$ (1*)

Choose Clairaut's constant $a = r \sin(\psi).$ (2*)

$(r=a)$ is where all lines are tangent at minimum radius.

From differential geometry, $dr/ \sin(\phi) = r d\theta * \cot(\psi).$ (3*)

Equns (1*) to (3*) are adequate to find $r=f(\theta)$, after eliminating $z,\phi,\psi$ and integrating.