Geodesic in hyperbolic plane

Question

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.

Answer 1

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.

Answer 2

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))

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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.