Simplifying the search for a geodesic

Question

How can calculations for the geodesics on the surface $U=\{(x,y,z): c(x^2+y^2)-z^2=0, z>0\}$ be simplified by noting that is locally Euclidean? I can see that the property means that when we open up the sector, the geodesics should appear as straight lines, but how can simplify the process of finding them explicitly? Hopefully there is no need for variational calculus?

Answer 1

Geodesics are preserved under isometry. Let $\beta = \sqrt{c^2 + 1}$. Let $\Omega := \{ (r,\theta) | r\in\mathbb{R}_+, \theta \in (0,2\pi/\beta) \}$. Then the map from $\Omega$ to $U$ given by

$$ (r,\theta) \to (r\cos\theta / \beta, r\sin\theta/\beta, cr/\beta ) $$

is an isometry. Straight lines in polar coordinates are given by $r(\theta) = r_0 \sec(\theta - \phi)$ for $\theta \neq \phi$ and also $theta = \phi$ for lines through the origin. Any line through the origin is radial and can be easily seen to map to a radial ray from the apex of the cone. The other lines can all be locally expressed as the images of $r(\theta) = r_0 \sec(\theta-\phi)$ under the isometry given above.