What vector field property means “is the curl of another vector field?”

Question

I'm an undergraduate mathematics educator and I teach a lot of multivariable calculus. I posed this question on MSE over four years ago and I haven't gotten any definitive answers (despite 12 upvotes and a bounty posted). It could be there's no answer, but someone suggested I ask on this forum.

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I know that a vector field $\mathbf{F}$ is called irrotational if $\nabla \times \mathbf{F} = \mathbf{0}$ and conservative if there exists a function $g$ such that $\nabla g = \mathbf{F}$. Under suitable smoothness conditions on the component functions (so that Clairaut's theorem holds), conservative vector fields are irrotational, and under suitable topological conditions on the domain of $\mathbf{F}$, irrotational vector fields are conservative.

Moving up one degree, $\mathbf{F}$ is called incompressible if $\nabla \cdot \mathbf{F} = 0$. If there exists a vector field $\mathbf{G}$ such that $\mathbf{F} = \nabla \times \mathbf{G}$, then (again, under suitable smoothness conditions), $\mathbf{F}$ is incompressible. And again, under suitable topological conditions (the second cohomology group of the domain must be trivial), if $\mathbf{F}$ is incompressible, there exists a vector field $\mathbf{G}$ such that $\nabla \times\mathbf{G} = \mathbf{F}$.

It seems to me there ought to be a word to describe vector fields as shorthand for “is the curl of something” or “has a vector potential.” But a google search didn't turn anything up, and my colleagues couldn't think of a word either. Maybe I'm revealing the gap in my physics background. Does anybody know of such a word?

TL;DR: gradient is to conservative as curl is to ___?

Answer 1

I think it’s just called a solenoidal field (incompressible field), because by definition, if we have $\mathbf{\nabla}\times \mathbf{A}= \mathbf{V}$, $$\mathbf{\nabla}\cdot(\mathbf{\nabla}\times\mathbf{A})= \mathbf{\nabla}\cdot \mathbf{V }=0$$ because the divergence of the curl is 0. Because of this, any field that can be derived from a vector potential is automatically incompressible. Since every incompressible field can be expressed as the curl of some potential, they are precisely equivalent. Therefore, we already have a name for it, and it doesn’t need a new one.

Answer 2

In the general case -- i.e in any number of dimensions, the analogue of $\nabla\times(\nabla \phi)=0$ and $\nabla\cdot(\nabla\times {\bf A})=0$ is $d^2=0$ where $d$ is the exterior derivative anding on $p$-forms. This means that if $\omega=d\eta$ then $d\omega=0$. A p-form $\omega$ such that $d\omega=0$ is said to be closed. If $\omega= d\eta$ then $\omega$ is said to be exact.

You mentioned cohomology, so probably you know what I have just written, and are instead asking what are "closed" and "exact" are called ordinary vector calculus. The answer is that I think that there is no standard name for this situation because most vector calculus is done in contractable spaces where closed $\Rightarrow$ exact. Certainly I have never seen a name.

Answer 3

If the domain is topologically trivial, then, as explained in the other answers, "is a curl" is the same as "incompressible," i.e., has zero divergence. So that's your answer.

In examples like the electric field of a point charge, the domain has a hole in it. This breaks the equivalence between incompressibility and is-a-curl. However, you asked this on a physics site, so you need to realize that for a physicist, features like the singularity of this electric field are unphysical idealizations. Classically, we would think of this as an idealization of the field of some charge distribution like a uniformly charged sphere. This is why physicists don't need a different name for the is-a-curl property. Space isn't a swiss cheese, and we don't have fundamental physical fields that fail to be defined at certain points.