# Geodesic in hyperbolic plane

I'm trying to show that $\gamma(t) = (0,t)$ is a geodesic in the hyperbolic plane, that is for $\mathbb{R}^2$ equipped with the metric $g_{11}=g_{22} = \frac{1}{y^2}$, $g_{12}=0$.

The way I was trying to do this was by computing the associated Christoffel symbols and then show that $\gamma$ satisfies the geodesic equation. The Christoffel symbols I computed are: $\Gamma_{11}^1 = \Gamma_{12}^2 = \Gamma_{22}^1 = 0$, $\Gamma_{11}^2 = \frac{1}{y}$, $\Gamma_{12}^1 = \Gamma_{22}^2 = \frac{-1}{y}$.

Clearly $\gamma'(t) = (0,1), \gamma''(t) = (0,0)$, so for the second geodesic equation I believe reduces to: $\Gamma_{22}^2\gamma_2'(t)\gamma_2'(t) \neq 0$

I'm sure there's just a sign error or something in there but I can't spot it at all.

You can prove this without complicated calculation :

${ dx^2+dy^2\over y^2 }\geq {dy^2 \over y^2}$ implies that given two point on the vertical line $x=0$ and a path $c(t)=(x(t),y(t))$ between these two points, the length of this path is greater that the length of the path $d(t)=(0,y(t))$, which contains the vertical segment between these points. This segment is therefore the unique shortest path between these points, and is a geodesic.

I don't understand how, from the generic second geodesic equation

$$\ddot{y}+\Gamma_{11}^2(\dot{x})^2+\Gamma_{12}^2(\dot{x})^2(\dot{y})^2+\Gamma_{21}^2(\dot{x})^2(\dot{y})^2+\Gamma_{22}^2(\dot{y})^2=0$$

you obtain this single term ; in this equation, the central terms vanish and it remains:

$$\ddot{y}=\tfrac{(\dot{x})^2}{y}-\tfrac{(\dot{y})^2}{y}$$

(Take also a look at (https://physics.stackexchange.com/q/91113))

I take this opportunity to explain a simple physical model that I have never seen explained very clearly.

This model provides the geodesics of $$\mathbb{H}$$ , i.e., the half circles orthogonal to the real axis as optical shortest paths in a medium with a variable optical index: $$n=\tfrac{1}{y}$$ at point $$(x,y)$$ (this index is thus constant along horizontal lines).

Question: What is the trajectory followed by a light ray starting in $$(x_0,y_0)$$ with an incidence angle $$i_0$$ with respect to a vertical reference (see figure below) ?

Solution : Let us consider $$\mathbb{H}$$ as a ''stratified medium'' with an infinity of infinitesimal diopters separating medias with respective indices $$\dfrac{1}{y+dy}$$ and $$\dfrac{1}{y}$$. For such a diopter, the Snell's refraction law (https://en.wikipedia.org/wiki/Snell%27s_law) gives :

$$\tag{1}\dfrac{1}{y+dy}\sin{(i+di)}=\dfrac{1}{y}\sin{(i)} \ \iff \ \sin{(i+di)}=\left(1+\dfrac{dy}{y}\right)\sin{(i)}.$$

Let us expand the LHS of (1) up to the first order:

$$\sin(i)+\cos{(i)} di=\sin{(i)}+\dfrac{dy}{y}\sin(i).$$

$$\dfrac{\cos{(i)}}{\sin{(i)}} di=\dfrac{dy}{y}.$$

This differential equation can be integrated as follows:

$$\ln(\sin{(i)})=\ln(y)+K.$$

Let $$K=-\ln(R)$$. The previous relationship is equivalent to:

$$\sin{(i)}=\dfrac{y}{R} \$$ with initial conditions $$\sin{(i_0)}=\dfrac{y_0}{R}$$ giving

$$R=\dfrac{y_0}{\sin(i_0)}.$$

Out of which, finally, we get $$y=R \sin{i}$$ : it's, as awaited, a circular arc with radius $$R$$ centered on the $$x$$ axis.

Remark 1: It is interesting to see that this law $$n=\tfrac{1}{y}$$ is a kind of "potential" with respect to the law $$d=\tfrac{1}{y^2}$$ expressing the hyperbolic distance to the $$x$$ axis.

Remark 2: We have not considered specifically here the particular case of the vertical lines. • the reason, I believe the equation reduced to one term was that all others vanished. Either because the Christoffel symbol was 0 or because $\gamma'' = 0$. Thanks for the extensive explanation on the intuition behind these geodesics, I'm sure it will help my visualisation in the future. I'd also love to know where/what my mistake is! – user291678 Mar 31 '17 at 16:50
• The equation you give $\ddot{y}=\tfrac{(\dot{x})^2}{y}-\tfrac{(\dot{y})^2}{y}$ is of course true, but I have a specific curve I wish to check is a geodesic. Namely $(x(t),y(t)) = (0,t)$, it was substituting into the equation you've given which gave me the single non-zero term. So surely this then says that $(0,t)$ is in fact not a geodesic? – user291678 Mar 31 '17 at 16:57
• @user291678: You need to reparametrize the vertical rays, but, yes, they're geodesics. See the reference I gave you above. – Ted Shifrin Mar 31 '17 at 17:57
• @TedShifrin the reference you provided was useful in my understanding! I think I am getting confused because of the question itself. It explicitly asked to show that $\gamma(t) = (0,t)$ is a geodesic, so I assumed all that needed to be done was to essentially plug it into the geodesic equation and show that they do vanish. – user291678 Mar 31 '17 at 18:10
• Most texts will call it a pre-geodesic if the curve (reparametrized appropriately) becomes a geodesic. Some of us are a bit sloppier. They should have just described it as a vertical ray in words. :) – Ted Shifrin Mar 31 '17 at 18:28