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I am trying to derive equation 2.4 in the article https://arxiv.org/pdf/hep-th/0603001.pdf The AdS3 metric in global coordinates $(t,\rho,\theta)$ is given by \begin{align} ds^2=R^2(-\cosh^2\rho dt^2+d\rho^2+\sinh^2\rho d\theta^2) \end{align} where $R$ is the AdS radius. They introduce a cut-off $\rho_0$, s.t. $\rho\leq\rho_0$, with the relation for the total system length $L$ and lattice cut-off is $e^{\rho_0}=\frac{L}{a}$. At the boundary $\rho_0$, they geometry is identified with a cylinder $(t,\theta)$, where $0\leq\theta\leq\frac{2\pi l}{L}$. The static geodesic length $L_\gamma$ connecting the two points $0$ and $\frac{2\pi l}{L}$ is found to be \begin{align} \cosh(\frac{L_\gamma}{R})=1+2\sinh^2\rho_0\sin^2\frac{\pi l}{L}. \end{align} I have trouble obtaining this relation, and wonder if there is some trick i am unaware of? Any hints/help are appreciated.

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I found the answer in another article they've written https://arxiv.org/pdf/hep-th/0605073.pdf, section 6.2.

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