# Normal vector fields

Hi guys, I was wondering if anyone could post or point me to a proof of the statement that given a hypersurface $\Sigma$, specified by setting a function $f(x) = const.$, the vector field $\xi ^{\mu } = \triangledown ^{\mu }f = g^{\mu \nu }\triangledown _{\nu }f$ will be normal to $\Sigma$ in the sense that $\xi ^{\mu }$ will be orthogonal to all $u\in T_{p}(\Sigma )$ for some $p\in \Sigma$. I tried to visualize it for trivial manifolds but I really couldn't. If there isn't really a proof of any kind could someone at least make the statement more intuitive. Thanks in advance.

HallsofIvy
Homework Helper
The simplest way of looking at it is to use the "directional derivative". If f(X) is a real valued function (X is a point in some space) and $\vec{v}$
is a vector in that space, then the "derivative of f in the diretion $\vec{v}$", the rate at which f is increasing in that direction, is given by $D_\vec{v} f= \nabla f\cdot\vec{v}$. In particular, if $\vec{v}$
is tangent to the surface f(X)= constant then f does not change: $D_\vec{v} f= \nabla f\cdot \vec{v}= 0$ which says directly that $\nabla f$ and $\vec{v}$
are perpendicular. Since $\vec{v}$
could be any vector in the tangent plane, $\nabla f$ is perpendicular to the tangent plane, i.e. normal to the surface.

Oh, the fact that I saw it in component form was confusing me. If I could just restate what you said, for my own clarification: since $\Sigma$ is specified by $f(x) = const.$ then on $\Sigma$, $\triangledown _{\nu }f = 0$ for all $\nu$ so $\xi ^{\mu } = g^{\mu \nu }\triangledown _{\nu }f = 0$ identically as well on the hypersurface. Then, for all $\mathbf{v}\in T_{p}(\Sigma )$, $\xi ^{\mu }v_{\mu } = 0$ and since $\mathbf{v}$ lies on the tangent space to $\Sigma$ at p, $\boldsymbol{\xi}$ must be orthogonal to it so it is then normal to $\Sigma$. Is this basically the gist of what you said?