Why is the acceleration vector the spatial gradient of the lapse function?

Question

If we have a Lorentzian manifold $(M, g)$ with a foliation by spacelike surfaces $\Sigma_t$ with unit-normal vector field $n$, we can define the lapse function $N$ by $$ \partial_t = N n + X $$ where $X$ is the shift vector. I have seen several claims that, for any $Y$ tangent to $\Sigma_t$, we have $$ g(\nabla_n n, Y) = \nabla_Y \ln N = N^{-1} \nabla_Y N = N^{-1} Y(N), $$ but I cannot find a proof for this. It is often stated as a trivial consequence of the definitions but I cannot derive it myself. Is there something obvious I am missing?

Answer 1

Write \begin{align*} \langle \nabla_n n, Y\rangle &= -\langle n, \nabla_Y n\rangle-\langle n, [n,Y]\rangle \\ &=-\frac 12 Y \langle n,n\rangle - \langle n, N^{-1} Y(N)\rangle \\ &=N^{-1} Y(N). \end{align*} You can get the formula for $[n,Y]$ by writing $n=\frac 1N(\partial_t-X)$ and using the coordinate expression for the Lie bracket.