Solve GHY Boundary Term Problem for Calculations

[Zitter]

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.

 

[craigthone]

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?

 

[Zitter]

Yes. I want to calculate $K$ by using ansatz on metric given by formula (4) in the paper. The inverse metric I calculated is
$$\begin{pmatrix}
-\frac{1}{A+3r^2\Omega^2 }& 0 & \frac{2\Omega}{A+3r^2 \Omega^2} \\
0 & \frac{A}{B} & 0 \\
\frac{2 \Omega}{A+3r^2 \Omega^2} & 0 & \frac{A-r^2\Omega^2}{r^2A+3r^4 \Omega^2}
\end{pmatrix}$$

 

[craigthone]

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?

 

[Zitter]

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 .