Acceleration and extrinsic curvature

Question

I am currently reading a textbook on numerical relativity (Numerical Relativity, M.Shibata), and in the preliminary work is defined the extrinsic curvature $K_{ab}$. That is okay, but then they derive a relation which involve a term called an acceleration :

$a_{b}=n^{c}\nabla_{c}n_{b}$.

It indicates that it is a spatial vector, but I really can't figure it out. ($n^{\mu}$ denote the vector field normal to the hypersurface $\Sigma_{t}$)

Can anyone tell me why ? Thanks in advance.

Answer 1

To show that $\mathbf{a}$ is purely spatial one could show that it is orthogonal to $\mathbf{n}$: $$ n^\mu a_\mu=n^\mu n^\nu \nabla_\nu n_\mu=\frac 1 2 n^\nu \nabla_\nu (n^\mu n_\mu)=\frac 1 2 n^\nu \nabla_\nu (-1)=0. $$

So $\mathbf{a}$ is orthogonal to the time-like vector $\mathbf{n}$ and tangent to the spatial hypersurface $\Sigma_t$, therefore it is a purely spatial vector.