In the static spacetime, the extrinsic curvature of hypersurface $t=constant$ is zero How can I prove that in the static spacetime, the extrinsic curvature of hypersurface $t=constant$ is zero? My efforts all are failed. Any hint would be greatly appreciated.
 A: Ref : Padmanabhan, Gravitation, Cambridge, p $531-534$
In ADM formalism, we may write : 
$ds^2=g_{mn}dx^mdx^n= -N^2dt^2+h_{\alpha\beta}(dx^\alpha+N^\alpha dt)(dx^\beta+N^\beta dt)$
$N$ is called the lapse function, and $N^\alpha$ are called shift functions. Here the latin letters are for $4$-metrics, while greek letters are for the induced $3$- metrics. 
We may write the induced $3-$ metrics $h_{\alpha\beta}$, in a four-dimensional notation with $h_{mn} = g_{mn}+n_mn_n$, where $n (n_o=-N, n_\alpha=0)$ is the normal to the hypersurfaces $t$ = Constant. This gives $h_{oo}= N^\gamma N_\gamma$,  $h_{0\alpha}= N_\alpha$, $h_{\alpha\beta} = g_{\alpha\beta} $
The extrinsic curvature is $K_{mn}= -h^a_m\nabla_an_n = -(\nabla_mn_n+n^an_m\nabla_an_n)$, it can be shown that this curvature is symmetric in $m,n$. For the spatial components of the extrinsic curvature, one get, with some calculus :
$K_{\alpha\beta}= -\nabla_\beta n_\alpha=\frac{1}{2N} (D_\beta N_\alpha + D_\alpha N_\beta-\partial_0 h_{\alpha\beta})$ 
where $D_mV_n= h^a_mh^b_n \nabla_a V_b$ is the spatial covariant derivative of a vector $X$ orthogonal to $n$ ($X$ is tangential to the hypersurface $t$=Constant)
In the case of the static space-time, we have $h_{\alpha\beta} = h_{\alpha\beta} (x^\gamma)$, and $N^\alpha=0$, so we have $K_{\alpha\beta}=0$. The components $K_{oj}=0$ too.
A: Look at the ADM equations.  Apply the staticity condition to them.
A: The unit normal to the 3-surfaces is proportional to the Killing field. The coefficient being the norm of the Killing filed. You can compute the derivative using the Killing equations a couple of times to get the result.  
