Compute geodesic between two points in 2 dimensions

Question

So I am trying to follow Tom Hartmans notes, page 190, and understand the example when he computes the entanglement entropy in 2d. In order to do this, we need the minimal "surface" connecting two points, $\frac{-L}{2}, \frac{L}{2}.

I don't really understand his approach, so I tried to verify the result by computing the geodesic equation. I don't see what I do wrong, but I don't get the same result as Tom.

We have the metric \begin{equation} ds^2= \frac{l^2}{z^2} \Big(dx^2+ dz^2 \Big). \end{equation}

The geodesic equation is given by $$ \frac{d^2x^\mu}{d\lambda^2} + \Gamma^{\mu}_{\sigma \nu} \frac{dx^\sigma}{d\lambda} \frac{dx^\nu}{d\lambda} =0$$

The non-zero Christoffel symbols are: $$ \Gamma^{x}_{zx}= \frac{-1}{z}, \quad \Gamma^{z}_{xx} = \frac{1}{z}, \quad \Gamma^{z}_{zz} = \frac{-1}{z} . $$

By using the geodesic equation I get: \begin{align} & \frac{d^2x}{d\lambda^2} -\frac{1}{z} \frac{dx}{d\lambda} \frac{dz}{d\lambda} =0 \\ & \frac{d^2z}{d\lambda^2} +\frac{1}{z} \Big(\frac{dx}{d\lambda}\Big)^2 -\frac{1}{z} \Big(\frac{dz}{d\lambda}\Big)^2=0. \end{align} Now, I guess I can solve these differential equations for the explicit form for $x(\lambda)$ and $z(\lambda)$. But I took the solutions from Tom's notes: $$ x= \frac{L}{2}\cos(\lambda), \quad z= \frac{L}{2}\sin(\lambda), \quad \lambda \in (\frac{\epsilon}{L}, \pi - \frac{\epsilon}{L} \Big) .$$

I don't see how these are solutions to my geodesic equations so I am doing something wrong. I don't see where the $L$'s come from in his solution for $x$ and $z$. By computing the geodesic equations in this manner I am also unsure about where to take the cutoff $\epsilon$ into account.

So either one cannot take the approach I am taking, or I am doing some mistakes along the way. Any input on how I am doing or thinking about this wrong is very welcomed.

Answer 1

To start, this is the Poincaré half-plane, which is a well known space.

The given solution is a geodesic: it's easy to see that it extremizes the path length given by Eq. (21.9). However, it won't fulfill the geodesic equations because $\lambda$ is not arclength (aka an affine parameter) - it's just a parameter. You can see this explicitly by computing the metric along the semicircle and seeing that

$$ds = \frac{\ell}{\sin\lambda} d\lambda,$$

so that

$$s = \ell \log \tan \frac{\lambda}{2}$$

is a good parameter for the geodesic equation. In principle you could rewrite the solutions in terms of $s$ and they should satisfy the geodesic equation - though I'm not sure it's worth the trouble! If you really want to check that the semicircle satisfies the original geodesic equations, a (probably) better way to do it is to start with the equation

$$x^2 + z^2 = \left(\frac{L}{2}\right)^2,$$

take two derivatives, plug in the geodesic equations, and hope that everything checks out.