A Solve GHY Boundary Term Problem for Calculations

Zitter
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Hi everybody ! In one of my papers I need to add Gibbons-Hawking-York boundary term in order to calculate everything properly. I found a paper (https://www.sciencedirect.com/science/article/pii/S0370269316306530 ) in which authors included this term into the action. My problem is: I tried to calculate this term as a practice work, but I didn't obtain the same result as authors of paper. There is no point in posting my result, because my formula is much bigger than compact form from paper. I checked my hand-made calculations in Mathematica and they are the same, so probably there is a problem at the beginning of my thinking. Could somebody tell where is the error ?

1) Firstly, I calculated the trace of extrinsic curvature ##K=-\frac{1}{\sqrt{-g}}\partial_{\mu} (\sqrt{-g} n^{\mu})##, where ##n_{\mu}## is the unit vector normal to the boundary. The hyper-surface boundary is given by ##r=R_0##, where ##R_0## is constant. I obtained ##n_t=0,\ n_r=\sqrt{\frac{B}{A}},\ n_{\varphi}=0## and ##n^t=0,\ n^r=\sqrt{\frac{A}{B}},n^{\varphi}=0 ##.

2) In order to add GHY term into Einstein-Hilbert action I used Stokes theorem to change "surface" integral into "volume" integral. One can rewrite ##K## as ## \bar{K}^{\mu}n_{\mu}##, where ##\bar{K}^{\mu}## is vector ##(\bar{K}^t,\bar{K}^r,\bar{K}^{\varphi})=(0,\sqrt{\frac{A}{B}}K,0)##. By using Stokes theorem expression ##d^2x \sqrt{-h} K## is replaced by ##d^3x \sqrt{-g} \bar{K}^{\mu}_{;\mu}##, where ##\bar{K}^{\mu}_{;\mu}## is divergence of ##\bar{K}^{\mu}## vector.

3) Now, inside action integral we have ##\sqrt{-g}\left(\frac{1}{2\kappa}(R-2\Lambda)+\frac{1}{\kappa}\bar{K}^{\mu}_{;\mu}\right) ## plus ##\sigma##-model part.

Is this reasoning correct ? Thank you in advance for any help. Few months ago I changed my field of study from QM to GR and I have gaps in knowledge, which of course I try to reduce as hard as possible.
 
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Zitter said:
1) Firstly, I calculated the trace of extrinsic curvature ##K=-\frac{1}{\sqrt{-g}}\partial_{\mu} (\sqrt{-g} n^{\mu})##, where ##n_{\mu}## is the unit vector normal to the boundary. The hyper-surface boundary is given by ##r=R_0##, where ##R_0## is constant. I obtained ##n_t=0,\ n_r=\sqrt{\frac{B}{A}},\ n_{\varphi}=0## and ##n^t=0,\ n^r=\sqrt{\frac{A}{B}},n^{\varphi}=0 ##.

Are you want to check extrinsic curvature of the formula (4) in the paper of Harms and Stern? Dis you calculate the inverse metric ##g^{\mu\nu}##? can you post it?
 
Yes. I want to calculate ##K## by using ansatz on metric given by formula (4) in the paper. The inverse metric I calculated is
$$\begin{pmatrix}
-\frac{1}{A+3r^2\Omega^2 }& 0 & \frac{2\Omega}{A+3r^2 \Omega^2} \\
0 & \frac{A}{B} & 0 \\
\frac{2 \Omega}{A+3r^2 \Omega^2} & 0 & \frac{A-r^2\Omega^2}{r^2A+3r^4 \Omega^2}
\end{pmatrix}$$
 
Zitter said:
Yes. I want to calculate ##K## by using ansatz on metric given by formula (4) in the paper. The inverse metric I calculated is
$$\begin{pmatrix}
-\frac{1}{A+3r^2\Omega^2 }& 0 & \frac{2\Omega}{A+3r^2 \Omega^2} \\
0 & \frac{A}{B} & 0 \\
\frac{2 \Omega}{A+3r^2 \Omega^2} & 0 & \frac{A-r^2\Omega^2}{r^2A+3r^4 \Omega^2}
\end{pmatrix}$$
The metic ##g_{\mu\nu}## can be written as
$$
\left[
\begin{array}{ccc}
-A+r^2\Omega^2 & 0 &0\\
0&\frac{B}{A} & r^2\Omega\\
0&r^2\Omega&r^2
\end{array} \right]
$$
Am I right?
 
The ##r^2\Omega## term should be in positions (1,3) and (3,1), but thanks to you I found error in my calculations. I forgot that it should be ##r^2\Omega## and not ##2r^2\Omega## :D. Now everything is ok and calculations are correct .
 
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