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

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.

• It is impossible to explain why your logic is incorrect because you did not give us any «logic» to justify your claim. – Mariano Suárez-Álvarez Nov 24 '14 at 5:21
• In any case both the divergence and the curl are computed in terms of the derivatives of the components of your field, so it is enough that these derivatives all vanish for both curl and divergence be zero. Can you think of a non-zero vector field whose components all have vanishing derivatives? – Mariano Suárez-Álvarez Nov 24 '14 at 5:23
• Divergence and curl are opposites? More like apples and oranges. – JohnD Nov 24 '14 at 5:50
• @dfg: Well, the explanation as to why you are incorrect is because we can produce vector fields like $\mathbf{F}=\langle 1, 1, 1\rangle$. – JohnD Nov 24 '14 at 6:03
• Perhaps the right thing to say is this: the kernels of divergence and curl do span a $C^2$ functions (hence the Hemholtz decomposition), but their intersection is non-trivial, since it consists of all Laplacian functions. – Ben Grossmann Jul 15 '16 at 19:24

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.

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).

• So is plain old Helmholtz theorem just a special case of the Hodge Decomposition, with the contribution from the line integral being essentially a constant? – user_of_math Nov 24 '14 at 8:39
• Sorry, your answer was quite a bit over my head. I don't know any clifford algebra or what Green's Function is. From what I understood from your answer, a vector field can be decomposed into 3 parts: a divergence-full curl-free part, a curl-full and divergence free part and a "divergence and curl free part from the closed line integral". I'm not entirely sure what you mean by the third part - could you please expand on this? – dfg Nov 24 '14 at 15:00
• @user_of_math Eh, I'm going to retract that. Helmholtz is more involved because it uses the Green's function for $\nabla^2$ instead of $\nabla$, so I shouldn't relate the two decompositions so trivially. – Muphrid Nov 24 '14 at 17:17
• @dfg The line integral I wrote has zero divergence and zero curl. But, even when the volume integral is zero at every $p$--because $F$ has zero divergence and zero curl, say--the line integral can give you nonzero $F$. – Muphrid Nov 24 '14 at 17:31
• I always find answers like this to questions like this curious. (Not a criticism per se, just an observation.) The OP doesn't seem to have a grasp on divergence and curl, but (s)he might grasp an answer reliant on clifford algebra? – JohnD Nov 24 '14 at 17:44

Since you asked not just for a counterexample but also for an explanation of what is wrong with your reasoning, I would like to focus on what aspects of your intuition are correct, and where (I think) they fall short of capturing the full situation.

First, you are not entirely wrong in thinking that

the curl and divergence... are essentially orthogonal vectors

More precisely, I would say that you have noticed an important and useful property of vector fields: any vector field $$\vec{F}$$ can be decomposed as a sum of two fields $$\vec{F} = \vec{A} + \vec{B}$$, where $$\vec{A}$$ is "irrotational" (i.e. $$\operatorname{curl} \vec{A} = 0$$) and $$\vec{B}$$ is "incompressible" (i.e. $$\operatorname{div} \vec{B} = 0$$). Another way to say this is that $$\vec{A}$$ gets all of $$\vec{F}$$'s divergence and none of its curl, while $$\vec{B}$$ gets all of $$\vec{F}$$'s curl and none of its divergence.

This decomposition is, in some respects, analogous to decomposing a vector into its $$x$$ and $$y$$ components. An even better analogy might be to a force exerted on an object moving along a circular track; such a force can be decomposed into a radial component (pointing outward) and a tangential component (pointing along the direction of motion). In many respects it is reasonable to think of the divergence and curl of a vector field as being sort of like the radial and tangential components (respectively) of a vector.

But note that I said "sort of like", not "exactly like". There is one crucial difference, and it gets to the heart of your question. The difference is this: the decomposition of $$\vec{F}$$ into curl-free and divergence-free components is not unique. That is, we might have two different decompositions $$\vec{F} = \vec{A}_1 + \vec{B}_1$$ and $$\vec{F} = \vec{A}_2 + \vec{B}_2,$$ where $$\vec{A}_1$$ and $$\vec{A}_2$$ are both curl-free, and $$\vec{B}_1$$ and $$\vec{B}_2$$ are both divergence-free. How can this be? Well, we would need $$\vec{A}_1 + \vec{B}_1 = \vec{A}_2 + \vec{B}_2$$ which is equivalent to $$\vec{A}_1 - \vec{A}_2 = \vec{B}_2 - \vec{B}_1$$ Notice that the left-hand side of this equation is curl-free, and the right-hand side is divergence-free. So in order to satisfy it, we would need to find a nonzero vector field $$\vec{H}$$ satisfying both $$\operatorname{curl}\vec{H} = 0$$ and $$\operatorname{div}\vec{H} = 0$$. This is precisely what you are asking about! If we had such a vector field $$\vec H$$, then from any decomposition $$\vec{F} = \vec{A} + \vec{B}$$ we could form another one, namely $$\vec{F} = (\vec{A} + \vec{H}) + (\vec{B} - \vec{H})$$

Several of the other answers have given you examples of vector fields $$\vec{H}$$ having the required property, so you know they exist, and that means that the decomposition of a vector field into irrotational and incompressible components is not unique; it is in precisely this respect that the analogy breaks down, because the decomposition of a vector into its radial and tangential components is unique.

To go back to the wording of your question:

The way I see it, the curl and divergence form a "basis"

I would say that this intuition is half-right. To have a basis, you need vectors that both span a space, and are linearly independent. The "all-curl" (divergence-free) vector fields and the "all-divergence" (curl-free) vector fields do "span" the set of vector fields, in the sense that any vector field can be written as a sum of fields of those two types. But they are not "independent" because we do not have a unique way to write a vector field as a sum.

• Great answer, should be the top one and should be accepted. The only answer to actually address what the OP asked! – Thomas Aug 18 '20 at 22:02
• I agree with Thomas, this was excellent. It helped to clear my confusion as well. – novawarrior77 Jan 5 at 1:54

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...

• Perhaps this should go in the comments of the post, instead of presenting this as an answer to the question. It is not. – novawarrior77 Jan 5 at 1:48

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.

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 $(x,y)$.

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.

Think of a beam of parallel ray vectors of constant magnitude in evry point in 3d space. It has both divergence and curl values 0. That means partial derivative in all dimensions are 0. Still the vector exist in constant form, i.e. everywhere with same magnitude and direction. For example uniformly charged plane surfaces of limited infinite dimensions do have set of constant electric field vectors such as parallel and independent of coordinates. This field does have neither divergence nor curl.