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In Poincare half-plane model,

$M = \{(x,y) \in \mathbb R^2 \mid y > 0\}$ and $ds = \frac{1}{y^2}(dx^2 + dy^2)$.

In wiki (https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model), it says that geodesic (straight line) "looks" like semicircles. Can anyone please show me an explicit derivation how to find a geodesic? I have been looking for examples, but could find none.

Arctic Char
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James C
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  • @ArcticChar Almost. With a few more explanation, I agree that this post may be closed. Why is it the case that geodesic equation requires $d/dt(\dot x/y^2) = 0$? – James C May 28 '21 at 08:04
  • @ArcticChar I really need a step-by-step derivation. If $\gamma(t)$ is a geodesic, then
    1. Is $\gamma'(t) = \dot{x} \partial_x + \dot{y} \partial_y$?
    2. Why does geodesic equation require $d/dt(\dot x/y^2) = 0$?
     – James C May 28 '21 at 08:10
  • https://math.stackexchange.com/questions/2063328/calculating-the-geodesic-hyperbolic-plane – Hans Lundmark May 28 '21 at 08:35
  • Download my (free) differential geometry text, linked in my profile. Read Section 2 of Chapter 3. – Ted Shifrin May 28 '21 at 19:09
  • I did not do the calculation, but I suppose you have to calculate all the Christoffel symbols and the geodesic equation would give you that. @PaulPogba – Arctic Char May 29 '21 at 04:08

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