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  How to calculate the extrinsic curvature of boundary of $AdS_2$

+ 1 like - 0 dislike

I have a simple but technical problem:

How to calculate the extrinsic curvature of boundary of  $AdS_2$?

The boundary of $AdS_2$ metric
is given by $(t(u),z(u))$.
The induced metric on the boundary is
$$ds^2_{bdy}=g_{\alpha\beta}dx^{\alpha}dx^{\beta}=g_{\alpha\beta}\frac{\partial x^{\alpha}}{\partial y^a}\frac{\partial x^{\beta}}{\partial y^b} dy^ady^b==g_{\alpha\beta}e^{\alpha}_ae^{\beta}_bdy^ady^b=h_{ab} dy^ady^b $$
For $Ads_2$ case, $ds^2_{bdy}= h_{uu}dudu$ where $h_{uu}= \frac{z'^2+t'^2}{z^2}$

My calculation is the following:

1) compute normal vecotr $(n^t,n^z)$

From the orthogonal relation $e^\alpha_a n_\alpha=o $ and unit norm condition $g_{\alpha\beta}n^{\alpha}n^{\beta}=1$, we have $n^t=\frac{zz'}{\sqrt{t'^2+z'^2}} , n^z=-\frac{zt'}{\sqrt{t'^2+z'^2}} $

2) compute the extrinsic curvature $K=\nabla_\alpha n^{\alpha}$

$$K=\nabla_\alpha n^{\alpha}=\frac{1}{\sqrt g}[\partial_t(\sqrt g n^t)+\partial_z(\sqrt g n^z)]=\frac{1}{\sqrt g}[\frac{1}{t'}\partial_u(\sqrt g n^t)+\frac{1}{z'}\partial_u(\sqrt g n^z)]$$

I tried some times but I can not reprodue the result in the paper.
My question is whether there are some mistakes in the formulas I used above.

asked Jul 15, 2017 in Theoretical Physics by C Thone (110 points) [ revision history ]
edited Jul 17, 2017 by C Thone

1 Answer

+ 0 like - 0 dislike

Obviouly, the last step is not correct.

You can see 


for the correct calculation of K. 

answered Jan 3, 2021 by anonymous [ no revision ]

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