9

I want to see if the Stokes theorem can be applied to a given flux integral of the vector field $F$. To do so, I need to determine if the vector field $F$ is the curl of some other vector field. How would I determine this?

Chappers
  • 67,606
Neil Philip
  • 159
  • 6
    Check its divergence? A divergence-free field is the curl of another field. – Cameron Williams Dec 15 '15 at 01:15
  • Yep, that makes sense Cameron. If div F = 0, F is the curl of another field, Stokes theorem applies. See my answer for an explanation. – Neil Philip Dec 15 '15 at 01:33
  • However it does not give you the explicit expression of the field, which you need for the Stokes theorem. – Kuifje Dec 15 '15 at 01:33
  • No, I suppose it doesn't. Actually finding which vector field G F is the curl of might require some integration. – Neil Philip Dec 15 '15 at 01:34
  • As it happens, I answered a question on how to invert the curl operator fairly recently, here. – Chappers Dec 15 '15 at 01:37
  • 2
    Look at the wiki page of Helmholtz decomposition, it tell you how to decompose a vector field into a curl-free and divergence-free components. For a vector field to be curl of something, it need to be divergence-free and the wiki page also have the formula for building the corresponding vector potentials. – achille hui Dec 15 '15 at 01:40
  • 1
    Contra @Cameron Williams, a divergence-free field (in three dimensions, say) is not necessarily the curl of another field. That implication holds only if the field is defined in a region with vanishing second cohomology. – symplectomorphic May 02 '17 at 06:23
  • @CameronWilliams: In Newtonian mechanics the gravitational field of a point mass is divergence free but is not the curl of any vector field – Dan Fox Mar 26 '20 at 14:08

2 Answers2

1

You can determine whether a vector field can be written as the curl of another vector field (in $\mathbb{R}^3$) by looking at it's divergence. Assume a vector field $F$ can be written as the curl of another vector field, call it $G$. Then $F = \text{curl}~G$. Take the divergence of $F$, and say $\text{div}~F \not= 0$. Then, as implied by Clairaut's Theorem, $\text{div}~F = \text{div}(\text{curl}~G) = 0$, which contradicts $\text{div}~F \not= 0$. So therefore $F$ is not the curl of another vector field if $\text{div}~F \not= 0$. So if $\text{div}~F = 0$, then the field is the curl of another field. It also means the vector field is incompressible (solenoidal)!

S/O to Cameron Williams for making me realize the connection to divergence there.

Neil Philip
  • 159
  • 6
    Note that this does require that the domain is simply connected, else cohomology comes and bites you. – Chappers Dec 15 '15 at 01:39
  • ^Right! In my multivariate calculus we typically only deal with simply connected domains, but yeah, always check for that, also when deciding on properties of conservation based solely on the curl. – Neil Philip Dec 15 '15 at 01:44
  • See this page for how to typeset math on this site using mathjax/latex. It's really easy to learn. – Winther Dec 15 '15 at 01:48
  • Okay, new to StackExchange, thanks Winther! – Neil Philip Dec 15 '15 at 01:49
  • 7
    Your argument only shows that if $F$ is the curl of $G$ then $\text{div}~F = 0$. It does not show that if $\text{div}~F = 0$ then $F = \text{curl}~G$ which is what you want. See this page for this. – Winther Dec 15 '15 at 01:50
  • 1
    @Chappers: you mean contractible, not simply connected. The obstruction to a solenoidal field (in three dimensions) having a vector potential deals with second cohomology, not first cohomology. I see this error (saying "simply connected" where "contractible" or "vanishing $k$-th cohomology" for $k\neq1$ is meant) all the time on the internet. A nice counterexample of a solenoidal (divergence-free) field that is not the curl of another field even in a simply connected domain is given on page 126 of Counterexamples in Analysis. – symplectomorphic May 02 '17 at 06:18
  • 1
    @symplectomorphic You're right, of course. And thank you for the reference, which I shall no doubt end up waving at undergraduates! – Chappers May 02 '17 at 17:24
  • 2
    This answer is wrong. The vector field $G(x) = |x|^{-3}x$ is divergence free but is not the curl of any vector field. The answer is right if it is made very clear that the divergence free vector field is defined on all of $R^{3}$. – Dan Fox Mar 26 '20 at 14:10
  • As others have said, this answer is wrong. Even following the original argument, this line doesn't follow at all from what comes before it: "So if div F=0, then the field is the curl of another field." This shouldn't be accepted as the answer. – Paul Zhang Mar 14 '22 at 10:03
-1

I found an algorithm that seems to work out, but perhaps someone can advise me where the divergence free comes into-play to power this algorithm.

Let $\textbf F$ be a Euclidean vector-field such that $\nabla \cdot \textbf{F}=0$. If there exists a vector-field $\textbf A$ such that $F=\nabla \times A,$ then we have the following set of equations. Additionally, I have found that they have the corresponding solutions, where $\oplus $ is the exclusive or symbol: $$\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=F_x: A_z=\int F_x dy \oplus A_y=-\int F_x dz $$

$$-\left(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z}\right)=F_y: A_z=-\int F_ydx \oplus A_x=\int F_y dz$$

$$\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=F_z: A_y=\int F_zdx \oplus A_x=-\int F_zdy$$

These exclusive or $\oplus $ solutions yield two choices for $\textbf A$

Andrew Dynneson
  • 93