Can a non-zero vector field have zero divergence and zero curl?

Question

I don't see how. Curl and divergence are essentially "opposites" - essentially two "orthogonal" concepts. The entire field should be able to be broken into a curl component and a divergence component and if both are zero, the field must be zero.

I'm visualizing it like a vector in $\mathbb{R}^2$. A vector cannot have a zero $x$ component and a zero $y$ component and still be non-zero.

EDIT: Here's a slightly more formal formulation of my thoughts: The way I see it, the curl and divergence form a "basis" - they are essentially orthogonal vectors. So how can a non-zero vector not be in their span?

Please don't just give me a counterexample. Please explain why my logic is incorrect.

Answer 1

That «the curl and divergence form a basis» does not really mean anything. The curl and the divergence are operators acting on vector fields, and they do not form a basis in any sense.

The contemplation of any counterexample to your claim should provide ample food for thought...

Answer 2

You've had some complex analysis, so you know what a harmonic function is. Take the gradient of any harmonic function. They also have harmonic functions in three dimensions, same example.

You said not to do that. Life is tough.

Two dimensional, we can take harmonic function $x^2-y^2,$ which is the real part of $(x+yi)^2,$ to get vector field $$ (2x, -2y). $$ This has divergence zero and "curl" (as used in Green's Theorem) zero. It really is the curl, we just write it as a scalar.

No more difficult in three dimensions, we may take function $x^2 + y^2 - 2 z^2,$ giving vector field $$ (2x, 2y,-4z). $$ Again, zero divergence and zero curl.

Answer 3

Using geometric calculus--the calculus of clifford algebra--we can write any vector field $F$ in terms of its value on a boundary curve $\partial M$ and its divergence and curl within a region $M$.

$$iF(p) = \oint_{\partial M} G(p-p') \, d\ell' \, F(p') + \int_M G(p-p') \, dA' \, \nabla F|_{p'}$$

where $G(p) = p/2\pi p^2$ is the 2d Green's function for $\nabla$. If $\nabla \cdot F = 0$ and $\nabla \wedge F= 0$, then $\nabla F = 0$ everywhere, and the area integral goes to zero.

But the line integral still remains, and $F$ is totally determined by its values on that bounding curve. A holomorphic function is determined by its values on some closed curve, is it not? This is just the 2d vector version of that concept.

So you can see, there are three parts to any decompsition of a vector field: a divergence-full part that is curl-free, a curl-full part that is divergence-free, and a divergence and curl-free part from the closed line integral (one word for this in the geometric calculus literature is monogenic, which is used to distinguish from the weaker condition of being harmonic).

Answer 4

Any vector field over $\mathbb{R}^2$ can be represented by a curl-free and divergence-free component provided the magnitude of the vectors decay suitably fast as one approaches infinity. So your intuition is partially correct, but the full theorem requires one extra condition. (As others have said, see the Helmholtz Theorem.) There are certainly vector fields which are non-constant and have zero curl and divergence everywhere in $\mathbb{R}^2$--namely, those which are unbounded at infinity. Notice the examples provided above have zero divergence and curl and are unbounded for large $(x,y)$.

Answer 5

One way your logic fails is that both curl and divergence are differentials of the field, and differentials don't "see" constant terms. And consequently, the simplest counterexample to your claim is a non-zero constant field: It has zero curl and zero divergence everywhere, yet it is nowhere zero.

Answer 6

I've recently discovered Maxwells equations, which are fascinating and I can think of two examples with real world analogs.

For an electric field to satisfy div(E) = 0, in a region, there must be no point charges within the region. Similarly for an electric field to satisfy curl(E) = 0, in a region, there must a static magnetic field in the region (because change in magnetic field is proportional to curl(E).

I can think of a few cases which satisfy this. The first is trivial E=constant (where constant is a vector). Div(E) = 0 because there are no sources or sinks and curl(E) = 0 because there are no gradients. This would correspond to an electric field between two infinite charged plates (or at least two plates much larger than the gap between them).

The second case is consider a region in space R and outside R there are static charges which generate a static electric field. This field can be quite complicated in shape, but as long as it is static and the magentic field is static, it must satisfy the conditions div(E) = 0 and curl(E) = 0.