I asked this question first on physics.SE but I got no complete answer so I thought maybe someone here could help.
I'm trying to understand how to derive the extrinsic curvature (in order to understand some calculation on fluid/gravity dynamics). But I hit a wall in my progress. I stuck at trying to verify the extrinsic curvature of a slowing-rotating Kerr solution ( page 94, E.Poisson: A relativistic Toolkit: The mathematics of black holes ): $$ds_+^2=g_{ab}\;dx^a dx^b=-f dt^2+f^{-1}dr^2+r^2 d\Omega^2-\frac{4Ma}{r}\;\sin^2(\theta)\;dr\;d\phi$$ where $f=1-2M/r$ and cut of $r=R$ which defines the hypersurface $\Sigma$ on which we'll work on.
I derive the induced metric (with $\psi=\phi-\Omega t$ and $\Omega=\frac{2Ma}{R^3}$) : \begin{equation} h_{ab}\;dy^ady^b=-f dt^2+R^2(d\theta^2+\sin^2\theta d\psi^2) \end{equation}
But from here on, I stuck. So my stucking points are:
- How can I derive the form of the unit normal vector $n_a$ here and in not so obvious metrics ?(The answer for this problem is $n_a=f^{-1/2}\partial_a\;r$, this means that $n_a=f^{-1/2}\delta^r_a$) ?
- The extrinsic curvature is defined as
$$K_{AB}=n_{a;b}\frac{\partial x^a}{\partial y^A}\frac{\partial x^b}{\partial y^B} \;\;\;\;\;\;\; (1)$$
where $x^a=x^a(y^A)$.
Edit: Until now I have compute the $n_a=f^{-1/2}\delta^{r}_{a}$ but if I try to compute the components of the extrinsic curvature, my answers are in complete disagreement with the answers from the book. I found for example $$K_{\theta\theta}=1/2$$ which is obviously wrong. The right answer is $$K_{\theta\theta}=f^{1/2}/R$$
Any suggestions? Thank you.