Dtyjlui

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:
$Gamma_{22}^2gamma_2'(t)gamma_2'(t) neq 0$

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 :

${ dx^2+dy^2over y^2 }geq {dy^2 over y^2}$ 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.