hawatajwizp

Answered

2022-06-25

Consider the hyperbolic 3-space $\left({\mathbb{H}}^{3},d{s}^{2}\right)$ with
${\mathbb{H}}^{3}:=\left\{\left(x,y,z\right)\in {\mathbb{R}}^{3}|z>0\right\},\phantom{\rule{1em}{0ex}}d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}+d{z}^{2}}{{z}^{2}}$
Geodesics for this space are circular arcs normal to $\left\{z=0\right\}$ and vertical rays normal to $\left\{z=0\right\}$ .

Answer & Explanation

Cristopher Barrera

Expert

2022-06-26Added 24 answers

Step 1
A clean way to do this is to work with the hyperboloid model of the hyperbolic space where ${H}^{n}$ . Then, given two points p, q in the (upper) hyperboloid ${H}^{n}\subset {R}^{1,n}$,n, their midpoint is
$\frac{p+q}{\sqrt{}}.$
If you know how to go back and forth between the upper half space and the hyperboloid model, you can translate this formula into the upper half space model. (I think it will not be pretty.) See also this question.

anginih86

Expert

2022-06-27Added 8 answers

Step 1
Assume that the geodesic passing through the given points ${p}_{1}$ and ${p}_{2}$ is the circle with center at $p=\left(a,b,0\right)$ and radius r. Then
$\left(x-a{\right)}^{2}+\left(y-b{\right)}^{2}+{z}^{2}={r}^{2}$
The vectors $u=p{p}_{1}$ and $v=p{p}_{2}$ satisfy
$\left(u×v\right).k=0$
Using these two equations, you can evaluate a, b and r. I think(I am not sure) the rest is easy.
I remembered that the Euclidean bisector of the segment ${p}_{1}{p}_{2}$ intersects the plane $z=0$ at the center of the required geodesic.

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