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I have been following the mentioned paper ("Aspects of Holographic Entanglement Entropy" hep-th/0605073 by Shinsei Ryu and Tadashi Takayanagi), trying to get into AdS/CFT. In section 6.2 they go on to calculate the geodesic's length using the ambient space coordinates $\Re^{2,2}$. The geodesics are given by the following expression:

$\vec{X}=\frac{R}{\sqrt{a^2-1}}\sinh(\frac{\lambda}{R})\cdot\vec{x}+R\left[\cosh(\frac{\lambda}{R})-\frac{a}{\sqrt{1-a^2}}\sinh(\frac{\lambda}{R})\right]\cdot\vec{y}\quad$ (6.4)

Where :

  • $\vec{x}=\left(\cosh(\rho_0) \cos(t), \cosh(\rho_0) \sin(t), \sinh(\rho_0) , 0\right)$
  • $\vec{y}=(\cosh(\rho_0) \cos(t), \cosh(\rho_0) \sin(t), \sinh(\rho_0) \cos(2πl/L), \sinh(\rho_0) \sin(2πl/L))$
  • $a=1 + 2 \sinh^2 \rho_0 \sin^2 (πl/L)$

Since we are in AdS3 the coordinates relevant are $\rho$ which has the cut off point $\rho_0$, t and $\theta\in\left[0,\frac{2\pi l}{L}\right] $. $\lambda$ is an affine parameter and R is the AdS3 radius.

The issue is that for the life of me I can't understand how one derives (6.4). Any help is appreciated.

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  • $\begingroup$ I'm not presently able to go through the paper and do an explicit check, but if you take the metric and calculate the geodesic distance this should come out by integrating $ds$. I'll try to check this, and if not may thy Mathematica chip and calculate. $\endgroup$
    – VaibhavK
    Commented Apr 30 at 4:58

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