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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.


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 ___?

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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.

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  • $\begingroup$ Being derived from a vector potential is sufficient but not necessary for incompressibility, right? $\endgroup$ – Matthew Leingang Nov 19 at 1:00
  • $\begingroup$ Wolfram MathWorld says it’s both: mathworld.wolfram.com/SolenoidalField.html . $\endgroup$ – Tesseract Nov 19 at 1:15
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    $\begingroup$ @MatthewLeingang: Your field F has a divergence at the origin, which is a delta-function. If you like, you can say that the origin isn't part of its domain, but then you can't apply Gauss's theorem to it. MathWorld doesn't spell out assumptions about whether the domain has to be all of 3-space or something, but I would be inclined to believe their claim unless you can come up with a counterexample that doesn't have this defect. Of course it would be nice to see a proof. $\endgroup$ – Ben Crowell Nov 19 at 2:02
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    $\begingroup$ @MatthewLeingang: Sorry, I don't understand how this relates to Stokes' theorem. Stokes' theorem would involve a line integral around the boundary of an open surface, but I don't see any of that in what you wrote. Your F has a nonzero divergence, so I don't understand what is the point of discussing F. $\endgroup$ – Ben Crowell Nov 19 at 3:33
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    $\begingroup$ Somehow the mathworld article is ignoring the domain. In fact the divergence free and having a vector potential are mathematically equivalent iff the Betti number $b_2$ of the domain is 0. Intuitively speaking this means that the domain has no holes. So $r/|r|^3$ is indeed a counterexample since it is defined on $\mathbb{R}^3 \setminus \{0\}$. However if you treat it as a function on $\mathbb{R}^3 \setminus [0,\infty) \times \mathbb{R}^2$ there is in fact a potential. The general proof is a bit involved, but on a star-shaped domain it's just Poincare's lemma. $\endgroup$ – mlk Nov 19 at 11:55
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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.

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  • $\begingroup$ So far answers here are the same as the ones on MSE. So I'm starting to think there's no phrase for it besides “is a curl.” $\endgroup$ – Matthew Leingang Nov 19 at 1:30
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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.

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  • $\begingroup$ Thanks. I was indeed asking for the physicist's perspective! $\endgroup$ – Matthew Leingang Nov 19 at 16:21
  • $\begingroup$ "Space isn't a swiss cheese" bummer $\endgroup$ – user2723984 Nov 19 at 18:35

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