A How to calculate the extrinsic curvature of boundary of AdS_2

Tags:
1. Jul 15, 2017

craigthone

I have a simple but technical problem:

How to calculate the extrinsic curvature of boundary of AdS_2?
I am not very familiar with this kind of calculation.

$$ds^2=\frac{dt^2+dz^2}{z^2}$$
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 (nt,nz)

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. Thanks in advance.

2. Jul 20, 2017

PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.