[Physics] The geodesic line on Poincare half plane

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I was calculating the geodesic lines on Poincare half plane but I found I somehow missed a parameter. It would be really helpful if someone could help me find out where my mistake is.

My calculation is the following:

Let $ds^2=\frac{a^2}{y^2}(dx^2+dy^2)$, then we could calculate the nonvanishing Christoffel symbols which are $\Gamma^x_{xy}=\Gamma^x_{yx}=-\frac{1}{y}, \Gamma^y_{xx}=\frac{1}{y}, \Gamma^y_{yy}=-\frac{1}{y}$. From these and geodesic equations, we have $$\ddot{x}-y^{-1}\dot{x}\dot{y}=0$$ $$\ddot{y}+y^{-1}\dot{x}^2=0$$ $$\ddot{y}-y^{-1}\dot{y}^2=0$$

From the last equation, it's straightforward that $y=Ce^{\omega\lambda}$, where $C$ and $\lambda$ are integral constants. Then substitute the derivative of $y$ into the first equation, we have, $$\ddot{x}-\omega\dot{x}=0$$ Therefore we have $x=De^{\omega\lambda}+x_0$ where $D, x_0$ are integral constants. However, by the second equation, we have, assuming $C$ is nonzero, $$C^2+D^2=0$$ And this leads to a weird result which is $$(x-x_0)^2+y^2=0$$
But the actual result should be $(x-x_0)^2+y^2=l^2$, where $l$ is another constant.

Best Answer

You say $C,\lambda$ are constants of integration, but that gives $\ddot{x}-\lambda\dot{x} = 0$ instead. Since your followup would be inconsistent, I will assume you meant $C,\omega$ are constants of integration.

You should not have three components to the geodesic equation, but rather two: $$\ddot{y} + \Gamma^y_{xx}\dot{x}\dot{x} + \Gamma^y_{xy}\dot{x}\dot{y} + \Gamma^y_{yx}\dot{y}\dot{x} +\Gamma^y_{yy}\dot{y}\dot{y} = 0\text{.}$$ You are also missing a factor of $2$ for your $\ddot{x}$ equation.

I will give you a further hint to say that since the $x$-coordinate is cyclic, $\dot{x} = Ey^2$ for some constant $E$. If you're not familiar with Killing vector fields, you can see this from the Euler-Lagrange equations on $L = \frac{1}{2}g(u,u)$, where $u^\mu = (\dot{x},\dot{y})$, which is also a nice way to find geodesics.

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