3
$\begingroup$

I got the following problem: Let $F$ be a continously differentiable vector field with $div(F)=0$. We define an integral curve as a curve that is tangent to the vector field F.

Now let us take a curve C that is never tangent to F. When we look at the integral curves of F through C we get a surface, call it S.

Prove that the flux of F through every section of S is equal.

Any help? I tried to find a general parametrization to the surface F and compute the flux "by hand". I got that the flux is zero through every section but that seems odd, so I am probably mistaken. Plus, I haven't used that fact that the field is divergence free.

Please help :)

$\endgroup$
3
$\begingroup$

I think the OP is right, regardless of the value of $\nabla \cdot F$.

Why? Recall that the flux through a surface $S$ is

$\displaystyle \int_S F \cdot \vec n \; dS, \tag 1$

where $dS$ is the area element of $S$, and $\vec n$ is a unit normal field; but here we have

$F \cdot \vec n = 0 \tag 2$

on $S$, since $F$ is tangent to the surface. Thus the integral (1) must vanish, independently of $\nabla \cdot F$.

Note that we required no special parametrization to see this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.