 Solution

A nonzero vector field has hypersurface orthogonality when it is locally normal to a family of hypersurfaces. In terms of its associated differential one-form $K^{♭}=g_{\mu \nu }{K}^{\nu }d{x}^{\mu }$, the Frobenius theorem gives the equivalent conditionslocally with nonzero $f$ and $dh$; square brackets denote tensor index antisymmetrization.

All coefficients of the Reissner-Nordstrom metric are independent of $t$, and $K={\partial }_t$ has constant coordinate components. Thus the Lie derivative obeys ${\mathcal{L}}_{K}g={\partial }_{t}g=0$: $K$ is a Killing vector field. Moreover,Thus $K$ has hypersurface orthogonality, with the $t=\text{constant}$ hypersurfaces providing the orthogonal family wherever this chart is regular. It is timelike where $F>0$, since $g(K,K)=-F$. Its hypersurface orthogonality also holds in a regular extension across a horizon; singular static coordinates do not invalidate the Frobenius theorem test.