# Compute geodesic between two points in 2 dimensions

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 $$$$ds^2= \frac{l^2}{z^2} \Big(dx^2+ dz^2 \Big).$$$$ The geodesic equation is given by $$\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$$ into account. So either one cannot take the approach I am taking, or I am doing some mistakes along the way. Any input on how I am doing or thinking about this wrong is very welcomed. • Did you try substituting the proposed solutions forx$and$y$into your differential equation? – Sal Commented Nov 22, 2021 at 13:37 • Yes and unfortunately, they don't. Commented Nov 22, 2021 at 13:44 • Would Mathematics be a better home for this question? Commented Nov 22, 2021 at 15:13 • @Sal thanks very much again. Just to clarify. Can$x(\lambda)$and$z(\lambda)$be obtained from Euler Lagrange equation, if we treat the integrand as the Lagrangian? Commented Nov 23, 2021 at 11:36 • @JohanHansen No problem. You can obtain an equation for$x(z)$that describes circles using the given Lagrangian$\mathcal{L}(x,x',z)$– Sal Commented Nov 24, 2021 at 2:16 ## 1 Answer To start, this is the Poincaré half-plane, which is a well known space. The given solution is a geodesic: it's easy to see that it extremizes the path length given by Eq. (21.9). However, it won't fulfill the geodesic equations because $$\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. • Thank you very much for clarifying that a geodesic does not necessarily solve the geodesic equation if it is not expressed in terms of an affine parameter. That is one mistake I did. Another source of confusion is probably that I don't have a clear understanding of how to compute the metric along a semi circle. Do you know where I can find an explicit calculation on this? Commented Nov 22, 2021 at 14:29 • @JohanHansen You just take the metric$ds = (\ell/z)\sqrt{dx^2+dy^2}$and plug in$x$and$z$as functions of$\lambda$(and take their differentials). Commented Nov 23, 2021 at 13:36 • should$dy$be a$dz$? I see how I can take their differentials. But I don't see how we get$x(\lambda) = \frac{l}{2}cos(\lambda)$. That I would love to understand. Commented Nov 23, 2021 at 14:08 • Yes, sorry, that's$dz$. And what I'm saying is that you take the solution that is given to you and plug it into the metric, not the other way around. All I'm doing is showing that, along the curve given by$(x(\lambda), z(\lambda))$,$\lambda\$ is not proportional to arclength. Commented Nov 23, 2021 at 15:04
• Thanks for clarifying. That makes sense. Commented Nov 23, 2021 at 16:56