geometric interpretation of $\vec v \cdot \operatorname{curl} \vec v = 0 $

Question

There is a family of surfaces orthogonal to the vector field $\vec v \in \mathbb R^3$ iff $\vec v \cdot \operatorname{curl} \vec v = 0 $. Now the necessity part is trivial, but the proof of sufficiency I have seen in physics textbooks, e.g. Kestin: Thermodynamics, is kind of mindless integration similar to the one usually offered to prove Poincare's lemma as in Flanders, or just asserted as in Born&Wolf: Optics. Is there an intuitive and geometric interpretation of this condition that would make it obvious why its sufficiency must be true?

Answer 1

This is the Frobenius integrability condition.

Consider the dual one-form $\alpha$. Then your statement $\mathbf{v} \cdot \operatorname{curl}{\mathbf{v}} = 0$ is equivalent to $\alpha \wedge d\alpha = 0$.

Notice that, at each point $p$, the plane defined by $\ker \alpha_p$ is the plane orthogonal to $\mathbf{v}_p$.

I'll roughly sketch the proof of this theorem in the three-dimensional case below. Perhaps you can interpret the messy proof in your textbook in the framework I suggest (it's probably essentially the way they did it). The full proof, in arbitrary dimension, is accessible if you know a decent amount of differential topology, and involves less calculation.

To clarify the frobenius integrability condition:

Given a smoothly varying family of planes at each point in 3D, consider, locally, any one-form $\alpha$ such that these planes are given by $\ker \alpha$ (equivalently, consider an orthogonal vector field $\mathbf{v}$ and take its dual one-form).

Then the family of planes is integrable, i.e. there exists a family of surfaces with those planes as their tangent planes, if and only if $\alpha \wedge d\alpha = 0$.

Rough sketch:

One proof is essentially that you can "integrate" the planes by projecting in some direction, drawing a curve there, and "lifting" it back so that it stays tangent to the planes (this becomes, in coordinates, solving an ODE, i.e. "integrating"). If you draw multiple curves with the same starting and ending points, or equivalently go around a loop, you need to show that you end up in the same place (i.e. you've stayed on one surface in your family of surfaces). This is like a curvature condition, and the two-form $d\alpha$ helps measure the discrepancy between the lift at the start and the end of the loop, just like curvature does.

The condition $\alpha \wedge d\alpha = 0$ implies that $d\alpha$ is zero on the two-planes, i.e. the kernels of $\alpha$ (i.e. "there's no curl on the orthogonal planes to $\mathbf{v}$"). This then implies that this discrepancy I described in the previous paragraph, coming from $d\alpha$, is in fact zero.

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Another view on necessity, citing Darboux's theorem in contact geometry:

You can use Stokes' theorem to get necessity. Here's another view, analyzing what happens when $\alpha \wedge d\alpha \neq 0$.

If $\alpha_p \wedge d\alpha_p \neq 0$ then this is true in the neighborhood of $p$, say an open ball around $p$. Without loss of generality, suppose this is positive, i.e. a positive multiple of the standard volume form $dvol = dx \wedge dy \wedge dz$ (otherwise swap your $x$ and $y$ coordinates).

This is the contact condition in 3D. In this case, on this open ball around $p$, the one-form $\alpha$ defines a contact structure given by the planes orthogonal to $\mathbf{v}$, equivalently the planes given by $\ker \alpha$.

There is a theorem, due to Darboux which states that all contact structures are locally equivalent. That is, there is some change of coordinates near $p$ which makes this locally the standard contact structure, which is the collection of planes given by the kernels of $\alpha = dz + x\, dy$.

Assuming the Darboux theorem, let's keep going and prove that this is non-integrable:

The wikipedia page for contact structure has a nice picture of the standard three-dimensional contact structure. The contact case is sometimes called the maximally non-integrable case. It looks pretty non-integrable, but how do we prove it?

Claim: This standard contact structure is not integrable, i.e. cannot be "integrated" into a collection of surfaces.

There are various proofs. Here's a simple one: Show that given any two points $p$ and $q$, there's a curve $\gamma(t)$ tangent to the planes (equivalently, for which $\alpha(\gamma_*(dt)) = 0$ is identically zero) connecting these two. This is called a legendrian curve.

(The situation is actually quite flexible and there are many curves between any $p$ and $q$.)

This then implies it can't be integrable, since such curves would have to stay on one surface in the family.

Here's a hint on one way to do this: Consider the standard contact form $\alpha = dz + xdy$. Project to the $x-y$ plane. To figure out how to "lift" a drawing of a curve on this plane into 3D, notice that for $\alpha$ to be zero, we simply need $dz/dy = -x$. This will allow you to draw many such curves between any pair of points $p$ and $q$.

Notice that this is roughly the opposite of what I mentioned above in the sketch of the proof of the Frobenius integrability theorem.

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I'll note, finally, that in dimensions higher than 3 there are situations between the contact condition (sometimes also called "maximally non-integrable") and the frobenius integrability condition ("integrable" i.e. there exists a family of surfaces with the specified tangent planes).

Answer 2

Here's a more basic answer, involving no differential forms, trying to get at the intuition more:

The condition $\mathbf{v} \cdot \operatorname{curl}(\mathbf{v})$ is essentially stating that $\mathbf{v}$ has no curl on the planes orthogonal to $\mathbf{v}$.

What does it mean to restrict curl to a plane? Well, parametrize the plane by $r(s,t)$. Then we have $r_s \times r_t$ is normal to the plane.

It turns out that $(\operatorname{curl}{\mathbf{v}}) \cdot (r_s \times r_t) = \frac{d}{ds} (\mathbf{v}\cdot r_t) - \frac{d}{dt} (\mathbf{v}\cdot r_s)$. This is precisely the scalar two-dimensional curl of the vector field $(\mathbf{v}\cdot r_s,\mathbf{v}\cdot r_t)$, which is what shows up in the two-dimensional Stokes's theorem. (In fact you can use this fact to deduce Green's theorem from the 2D Stokes's theorem.)

So this dot product of $\mathbf{v}$ with $\operatorname{curl}{\mathbf{v}}$ being zero represents the fact that $\mathbf{v}$ has no curl on its orthogonal planes.

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Now, if you scale $\mathbf{v}$ by a function, the statement $\mathbf{v} \cdot \operatorname{curl}{\mathbf{v}} = 0$ is unchanged. Why? Intuitively, this is saying there's no curl on orthogonal planes, and scaling $\mathbf{v}$ scales the curl, but otherwise only adds curl in directions perpendicular to $\mathbf{v}$, i.e. on planes not orthogonal to $\mathbf{v}$.

Consider some function $f: \mathbb{R}^3 \rightarrow \mathbb{R_+}$, some function to scale by, and consider a new, scaled, vector field $f\mathbf{v}$.

Then $\operatorname{curl}(f\mathbf{v}) = \operatorname{grad}(f) \times \mathbf{v} + f \operatorname{curl}(\mathbf{v})$. Thus $$(f\mathbf{v})\cdot\operatorname{curl} (f\mathbf{v}) = (f\mathbf{v}) \cdot \operatorname{grad}(f) \times \mathbf{v} + (f\mathbf{v}) \cdot f \operatorname{curl}(\mathbf{v}) = f^2 (\mathbf{v} \cdot \operatorname{curl}{\mathbf{v}})$$

[An alternative method at this point: Show that if $\mathbf{v}\cdot \operatorname{curl}{\mathbf{v}} = 0$ then there's some $f$ such that $\operatorname{curl}(f\mathbf{v}) = 0$. Then the vector field $f\mathbf{v}$ is conservative and may be integrated as $\operatorname{grad}(h)$ for some function $h$. Then the level sets of $h$ are your surfaces.]

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Now here's the idea of the proof:

Suppose that at some point, $\mathbf{v} \cdot \hat{\mathbf{z}}$ is nonzero (else change your coordinate names so this is true). This is then true on some open set near the point by continuity.

Scale $\mathbf{v}$ such that $\mathbf{v} \cdot \hat{\mathbf{z}} = 1$ on an entire region near the point. This new vector field still satisfies the condition, by the previous section. So the notation is simple, let's just keep calling this scaled version $\mathbf{v}$.

Now given a curve $(x(t),y(t))$ on the $x-y$ plane and a starting $z$ value $z(0)$, solve for $z(t)$ such that $(x(t),y(t),z(t)) \cdot \mathbf{v} = 0$. (This is just an ODE on $z(t)$, and always solvable given the initial condition $z(0)$.) Call the resulting curve $c(t) = (x(t),y(t),z(t))$.

We'd like these curves to fill out the family of surfaces we're interested in. For example, do this with all the curves in the $x-y$ plane parallel to the $x$-axis and again with all the curves in the $x-y$ plane parallel to the $y$-axis. These should be coordinate lines on our surfaces.

What could go wrong? Maybe these lines don't meet up when you go around a square. In fact, this is the only thing that could go wrong. How do we show this doesn't happen if $\mathbf{v} \cdot \operatorname{curl}{\mathbf{v}} = 0$?

The basic idea is that the lack of curl on the planes orthogonal to $\mathbf{v}$ means that such a closed curve, going around a loop, can't "spin up" to a different ending $z$ value, due to this lack of curl.

Here's the argument:

Work by contradiction. Suppose it does happen. Then we'll have a square that, when we go around the square as our curve in the $x-y$ plane, the final $z$ value $z(T)$ when we get back to where we started doesn't match up with the starting $z(0)$.

Finish off the curve $c(t) = (x(t),y(t),z(t))$ having it go completely vertical to close up (i.e. between the final and initial z values).

Then the line integral of $\mathbf{v}$ along $c$ is precisely $z(0) - z(T)$ because the only contribution is from the vertical portion (since otherwise $c$ is orthogonal to $\mathbf{v}$) and there we've got $\mathbf{v} \cdot \hat{\mathbf{z}} = 1$.

But it turns out this is a contradiction. We should be able to use Stokes' theorem to show that the integral around this closed loop is zero, using $\mathbf{v} \cdot \operatorname{curl}{\mathbf{v}} = 0$. (Actually, the setup here isn't quite right for proving this. I was trying to translate how all this works to vector fields and do it without changing coordinates, just scaling $\mathbf{v}$. I don't think I've quite got the right setup for the proof. The intuition I mention is still basically right. I'll stop for now.)

In any case, hopefully this has some intuition in here. Please ask if it all sounds greek.