# Orthogonal lines to the plane $z=0$ in hyperbolic space are geodesics

Consider the hyperbolic space as the upper half space of $\mathbb{R}^3$ $$\mathbb{H}^3=\{(x,y,z)\in\mathbb{R}^3:z>0\}$$ equipped with the metric $$ds^2=\frac{1}{z^2}(dx^2+dy^2+dz^2)$$ Prove that straight lines perpendicular to the plane $z=0$ are geodesics if they are arc-length parameterized.

Of course, such lines can be parameterized by $$\alpha(t)=(x,y,t)$$ where $(x,y)$ is an arbitrary fixed point of the plane. But, even though this is arc-length parameterized under the usual metric, it's not under the defined metric, and indeed it's not geodesic.

My problem is conceptually serious too, as I don't know how to even check it. For example, assume the parameterization given by $$\alpha(t)=(x,y,t^2+t^3)$$ Then we would have $$\alpha'(t)=(0,0,2t+3t^2)$$ Then the module is given by $$|\alpha'(t)|^2=\frac{1}{z^2}(2t+3t^2)^2$$ Now is this arc-length parameterized? I don't know, since different variables appear.

• Is $Z$ supposed to be the same as $z$? In mathematical notation, uppercase and lowercase letters are considered different symbols. – Jack Lee Jan 24 '17 at 21:01
• Oh, it's the same thing indeed, I'll correct that thanks. – F.Webber Jan 24 '17 at 21:04
• Hint: Write $\alpha(t) = (x,y,f(t))$, and figure out what $f$ has to be in order to make $\alpha$ unit-speed. – Jack Lee Jan 24 '17 at 21:05
• I have the same doubt with that too, though, which is that both variables $t$ and $z$ appear. I'll tell you what I've done, I arrive at $\frac{1}{z^2}f'(t)^2$, if I want that to be one, I need $f'(t)=z$ and hence $f(t)=\frac{z^2}{2}$, this looks weird though since that is not a function of $t$! – F.Webber Jan 24 '17 at 21:12
• You have to evaluate $z$ at the point $\alpha(t)$. – Jack Lee Jan 24 '17 at 21:13

To arc-length parametrize you may integrate: $ds = dz/z$ to get $s-s_0=\pm \log(z/z_0)$ or $z= z_0 \exp( \pm (s-s_0))$.
In order to see that a vertical segment minimizes a distance (i.e. is a geodesic) consider a $C^1$ path $t\in [a,b]\mapsto \gamma(t)=(x(t),y(t),z(t))$ from $(x,y,z_0)$ to $(x,y,z_1)$ with $0<z_0<z_1$. Since $ds\geq dz/z$, with equality iff $x(t),y(t)$ are constant, you deduce that the projection of the path to the vertical segment gives the minimal length (and provided $z_0\leq z(t)\leq z_1$ for $t\in [a,b]$).