0

It is known that if a vector field is divergence free, thus for sure it is the curl of a suitable vector field.

My question is: if a vector field is not divergence free, one can aspect that it is anyway curl of a vector field?

Thank you.

C. Bishop
  • 3,418
  • 1
    Your first statement is “for sure” only true if the vector field is (nice and) defined on all of space. If, for example, it has a singularity at one point, your claim may fail. The theorem is that (again with assumptions about continuous second-order partial derivatives), the divergence of the curl of a vector field is always $0$. Does this answer your second question? – Ted Shifrin Mar 01 '23 at 00:24
  • 1
    For a worked out example where even in the divergence free case things can become complicated please see my answer to your own question. @TedShifrin is the father of all of my knowledge. Thanks again! – Kurt G. Mar 01 '23 at 05:29
  • @TedShifrin, thank you for your comment. You say that the divergence of the curl of a vector field is always $0$, but is still not clear to me what happens if, e.g. it has a singularity. I mean, assume that $F$ is a vector field with a singularity at a point and such that is not divergence free. May I conclude that does not exist $G$ such that $F= curl(G)$? I hope you could help again. Thank you. – C. Bishop Mar 01 '23 at 09:32
  • That's a no brainer. When there is a VF $G$ s.t. $F=\nabla\times G$ in an open set then $\nabla\cdot F=0$ in that set. At the singularity of $F$ the question cannot even be asked. – Kurt G. Mar 01 '23 at 10:31

1 Answers1

1

The answer to your question is a trivial no, simply by taking the contrapositive of the following obvious theorem, which I state in excruciating detail so there is no ambiguity:

Let $U\subset\Bbb{R}^3$ be an open set, $G:U\to\Bbb{R}^3$ be a twice Frechet-differentiable vector field, and define $F:U\to\Bbb{R}^3$, $F:=\text{curl}(G)$. Then, $\text{div}(F):U\to\Bbb{R}$ is the zero function.

The proof is by equality of mixed partials (which holds due to twice Frechet-differentiability of $G$).

peek-a-boo
  • 55,725
  • 2
  • 45
  • 89
  • peek-a-boo, thank you for the answer. Does the same hold if $F$ has a singularity at a point? – C. Bishop Mar 01 '23 at 09:18
  • @C.Bishop was my answer not clear? if the divergence is not zero, then $F$ cannot be the curl of a vector field. Anyway, I just edited my answer to be as precise as possible. If that still doesn't answer your question then you're going to have to be more explicit about what exactly you're trying to ask. – peek-a-boo Mar 01 '23 at 09:56