# Normal vector fields

1. Aug 26, 2011

### WannabeNewton

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.

2. Aug 27, 2011

### HallsofIvy

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.

3. Aug 27, 2011

### WannabeNewton

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?
Thank you so very much by the way. Cleared my head up; I think my problem was that I kept thinking of the function everywhere outside the hypersurface and not just what happens on it.

Last edited: Aug 27, 2011