Here's a related question. So, we have $H=\{(x,y)\in\mathbb{R}^2: y>0\}$ and define the metric $g=\frac{1}{y^2}(dx^2+dy^2)$. I know that the circle $x^2+y^2=1, y>0$ is the image of a geodesic in $H$. An obvious parametrization of this curve is $\gamma(t)=(\cos t,\sin t), t\in (0,\pi)$. So, computing the length of this curve in $H$, we get $$L(\gamma)=\int_0^{\pi} (g_{\gamma(t)}(\dot{\gamma},\dot{\gamma}))^{1/2}=\int_0^{\pi} \frac{(-\sin t)^2+(\cos t)^2}{y\circ\gamma(t)}dt=\int_0^{\pi} \csc tdt$$but this latter integral doesn't converge.

This came up as I was trying to compute the arc-length parametrization $s(t)=\int_0^t(g_{\gamma(\tau)}(\dot{\gamma},\dot{\gamma}))^{1/2}$ to find $t(s)$, but was finding that $s(t)$ contained a divergent term.

What am I doing wrong here?

Edit: So it's been point out to me that this curve has infinite length, which is what the calculation indicates. But my question is this:

The arc length parameter here is $$s(t)=\int_0^t(g_{\gamma(\tau)}(\dot{\gamma},\dot{\gamma}))^{1/2}=\int_0^t \csc\tau d\tau$$which also doesn't converge. I guess the idea is that the length from any point on the geodesic toward the $x$-axis is infinite, since that goes off to infinity, which I suppose makes sense as $y\to 0$.

So my question is then, how do we calculate $s(t)$ and its inverse? Do we just pick any $t_0$ for the lower bound of the integral?

  • $\begingroup$ Why would you expect it to have finite length? $\endgroup$ – Chappers Sep 16 '15 at 4:45
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    $\begingroup$ Probably nothing. The length of that geodesic is $\infty$ $\endgroup$ – Thomas Sep 16 '15 at 4:45
  • $\begingroup$ @Chappers Thomas Oh. Well that would explain that question. Thanks! Now, the question I had was that, when calculating the arc length parametrization, you still get a divergent term. $\endgroup$ – Moya Sep 16 '15 at 4:48

The semicircular geodesic in this model is the analog of an infinitely extended straight line in Euclidean space. Attempting to parametrize it starting at an endpoint is the same as trying to parametrize a Euclidean line starting at infinity - it doesn't really make sense. What you need to do is choose your parameter to be zero somewhere in between ($(x,y) = (0,1)$ makes the most sense) and integrate from there, so that the two endpoints of the semicircle correspond to your parameter going to $\pm \infty$.

  • $\begingroup$ Thank you! So, with this in mind, I calculated $s(t)=\ln|\csc t-\cot t|$, but I'm having trouble calculating the inverse function $t(s)$. Any suggestion? $\endgroup$ – Moya Sep 16 '15 at 5:10
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    $\begingroup$ $\rm csc(t) - cot(t) = tan(t/2)$. $\endgroup$ – Anthony Carapetis Sep 16 '15 at 5:17
  • $\begingroup$ Trig identities, of course. Thank you for all your help! $\endgroup$ – Moya Sep 16 '15 at 5:20

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