Length of parametrized path

Question

Can someone guide me through how to solve this problem?

Let $P = (0,1)$ and $Q = (1,1)$, and let $\gamma$ be the following parametrized path in $\mathbb H^2$ from $P$ to $Q$: $\gamma(t) = (t,1)$.

Find the length of $\gamma$. Using this length find the length of the geodesic $PQ$.

Thank you!

Answer 1

I assume that you're using the Poincaré half plane model. The Wikipedia article for that has a section titled Metric which states:

The metric […] is given by $$(\mathrm ds)^2=\frac{(\mathrm dx)^2+(\mathrm dy)^2}{y^2}$$

So you'd compute

$$s=\int_{t=0}^1\mathrm ds=\int_{t=0}^1\sqrt{\frac{(\mathrm dt)^2+0}{1^2}}=\int_{t=0}^1\mathrm dt=1$$

The actual length computation is surprisingly simple. That's because the path is actually part of a horocycle, and horospheres are Euclidean subspaces of any hyperbolic space. With the right coordinate system (and the $t$ in your case is right), you get simple Euclidean behavior there.

What does this tell you about the geodesic? Well, it sounds like you've learned some connection between horocycle arc lengths and geodesic lengths. I can't think of such a connection right now, so instead I'd integrate to compute the geodesic length.

What's your geodesic path? It's an arc of a circle with center on the $y=0$ line passing through $P$ and $Q$. So it satisfies

$$\left(x-\tfrac12\right)^2+y^2=\tfrac54$$

and you could parametrize it using the $x$ coordinate, with

$$y = \sqrt{\tfrac54-\left(x-\tfrac12\right)^2}$$

Find the directional derivatives, plug them into the metric formula, integrate and you are done.

Answer 2

HINT:

By using the half-plane model find why or why not, the length of path in can be given by:

$$ \cosh^{-1} \left( 1+ \dfrac{1^2 + 0^2 }{2 \cdot 1 \cdot 1 } \right)$$