# Normal vector field to the hypersurface

Consider a hypersurface $\Sigma$ in a manifold $M$ specified by setting a single function to a constant: $$f(x)=f_{*}$$

Define the vector field $$\zeta^{\nu}=g^{\mu\nu}\nabla_{\mu}f$$

How to show that $\zeta^{\mu}$ is orthogonal to all vectors $V^{\nu} \in T_p\Sigma$?

That is:

$$g_{\mu\nu}\zeta^{\mu}V^{\nu}=0$$

• How is this a physics question? Commented Sep 18, 2016 at 20:39
• @ACuriousMind I'm pretty sure he's studying Carroll's GR book and he thought it would be more appropriate here. Commented Sep 18, 2016 at 20:45
• yes and would prefer a physics perspective because math people use hard to follow notation.
– user169903
Commented Sep 18, 2016 at 20:48

$V^\nu \nabla_\mu$ is the derivative along the integral curve of $V$. This curve lies on the hypersurface. If $f(x)$ is constant on the hypersurface, this means that

$V^\nu \nabla_\mu f= 0$

that is what you wanted to prove.

• Thanks. Perfect and to the point!
– user169903
Commented Sep 18, 2016 at 21:00