Why does the curl of a function provide this particular amount of information?

In a classical electrodynamics textbook (Griffiths), it is mentioned that even though the electric field function, $E:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$, is a (3D) vector valued function, the amount of information needed to fully describe it is equivalent to the amount needed to describe a scalar valued function, and this is because it is the gradient of an electric potential function, $U:\mathbb{R}^{3}\rightarrow \mathbb{R}$ and so is completely determined by it. The author goes on to explains that this is so because $E$ has the property that its curl is zero evevrywhere, and this is what restricts the freedom in determining $E$ (and what enables it to be the gradient of a scalar function in the first place).

This is all fine (and fun), but I find myself unable to answer this question: why does the imposition of zero curl provide an amount of information exactly equivalent to two scalar-valued functions (no more, no less)?

An example of the sort of reasoning that leads me to other conclusions: since zero curl is equivalent to equating three pairs of partial derivatives ($\partial E_{x}/\partial y = \partial E_{y}/\partial x$ and so on), leaving 6 out of 9 partial derivatives "free", it seems that "one third of the amount of information" required to describe $E$ has been taken up, as opposed "two thirds"...
I am of course assuming here that the first order partial derivatives determine the behavior of the function (perhaps up to a constant as in the 1D case?), and that these partial derivatives are independent of each other and thus provide equal amounts of information. Is either of these assumptions wrong? Is my whole reasoning off?

Any insight into this question and a possible answer would be appreciated.

*The question is, of course, a general one about vector-valued functions, with $E$ just being a particular case.

**I have not mentioned all the obvious assumption of smoothness necessary for everything to be defined..

***I realise that I'm playing fast-and-loose with the word "information", and that my whole question is very un-formal. Any reference to an area of mathematics which may perhaps put such intuitions on rigorous footing are more then welcome.

Edit: I am not asking why (or when) a vector field having zero curl is equivalent to it being a gradient of some scalar field. I am asking about the amount of information we get about a vector field when we determine its curl.

• It is not totally true that $\nabla \times \vec{E} = \vec{0}$ everywhere. See the Maxwell–Faraday equation. – md2perpe Sep 16 '18 at 20:02
• Welcome New contributor Evan! I recommend that you take a look at this question regarding cross-posting on Stack Exchange sites: Is cross-posting a question on multiple Stack Exchange sites permitted if the question is on-topic for each site? – Alfred Centauri Sep 16 '18 at 22:43
• @AlfredCentauri I did indeed ask this question in math-SE too. Is this not considered to be ok? If not I will, of course, remove this question from one of the sites – Evan Sep 16 '18 at 22:44
• @AlfredCentauri Thank you for pointing out that questions should be asked in only one SE. Unfortunately, I see I can not delete my question here so I'm not sure as to how to proceed – Evan Sep 16 '18 at 22:48
• Evan, since two volunteers here have taken the time to write an answer, I think deleting the question might be disrespectful. I wonder if there's a way to merge the answers here to the Math.SE question? I could flag the question for moderator attention if you would like for me to do that. – Alfred Centauri Sep 16 '18 at 22:56

If the curl were an arbitrary vector field, it would “contain as much information as the field itself”. (I'm using scare quotes because this is all on the same hand-waving level of rigorosity that your arguments address.)

However, the curl satisfies $\nabla\cdot(\nabla\times E)=0$. Thus, we have one scalar constraint on the curl, which reduces the number of free scalar functions in the curl from $3$ to $2$, so specifying the curl reduces the number of free scalar functions in $E$ from $3$ by $2$ to $1$.

Your considerations about the partial derivatives don't work out because the partial derivatives aren't independent. If they were, that would replace one scalar function by three scalar functions. The partial derivatives are subject to $\frac{\partial^2}{\partial x\partial y}=\frac{\partial^2}{\partial y\partial x}$, and likewise for the other two pairs of coordinates.

• Thank you for taking the time to answer my question and having the patience to wave hands with me. Pointing out the restriction on the curl and the dependencies in the derivatives made this whole matter much clearer. May I ask if there is a rigorous way that you know of in which I could have asked my question? – Evan Sep 16 '18 at 22:30
• @Evan: There isn't, but the key phrase here is "that I know of" :-) I think hand waving works reasonably well for these sorts of things. – joriki Sep 17 '18 at 4:11

The correct context for these kinds of question is de Rham cohomology. This is the natural cohomology associated with differential forms and these, in turn, are the natural generalisations of the integrands we find in 3d vector analysis to manifolds of arbitrary curvature and dimension. Unfortunately, this is not yet a part of the standard toolkit of every physicist in the way that vector analysis is.

In this context, what you are asking is when we are given a differential form $\alpha$ when is there a differential form $\beta$ such that $d\beta=\alpha$. Such a differential form is called exact.

Say this is true, then since it is always true that $d^2=0$ (which is the correct generalisation of vector analysis identities such as $curl(grad f) = 0)$ then we must have $d\alpha=d^2\beta=0$ and hence it must hold that $d\alpha=0$. Now such differential forms are closed. So our first condition is that for differential form to be exact, it is neccessary that it is closed. However, this condition is not sufficient. Further conditions rely upon the global topology of the space in question. If this space is like a Euclidean space then it is possible. Now, by definition, space is taken to be always locally Euclidean. Hence, locally speaking, a closed form is always exact.

• Thank you for taking the time to write an answer. It is very helpful to have a reference to the relevant area of mathematics. I will say that it seems you have provided me with a description of the conditions under which a differential form being closed is equivalent to it being exact (which in vector analysis terms, if I understand correctly, are conditions under which a vector field having zero curl is equivalent to it being a gradient of a scalar field). Why does this provide insight into the amount of information we get about the vector field by making it have zero curl (i.e. be closed)? – Evan Sep 16 '18 at 21:32
• @evan: Perhaps you might want to consider what you mean by 'amount of information'? I'm not sure really what you mean by this and nor what you are looking for. Also, you might want to consider why you say that 'the first order behaviour of partial derivatives determine a function' - in fact you need all orders - and even then that's not sufficient; in those cases in which it is sufficient, the function is called analytic. – Mozibur Ullah Sep 16 '18 at 21:53
• I realize that my question is very informal and wave-handy, making me all the more grateful that you are trying to help. What I talk about the "amount of information" needed to describe a function, I am alluding to the fact that in a sense, an arbitrary vector field in R^3 is "built" of 3 independent scalar fields, and therefore requires "3 times the amount of information" to describe. continuation: – Evan Sep 16 '18 at 22:00
• By (perhaps mistakingly) saying that the "first derivatives determine the behavior of the function, I was generalizing from the 1D case in which a C^1 function is determined up to a constant (i.e. in behavior) by it's derivative via integration. – Evan Sep 16 '18 at 22:05
• @evan: Saying a function is C1 is equivalent to saying that it is determined by its first derivatives; there's no new information there. You're simply saying the same thing in two different ways and sowing misinformation and confusion. – Mozibur Ullah Sep 16 '18 at 22:12

What you are asking about is known as Helmholtz decomposition.

Probably the easiest place to see the decomposition is in Fourier space where the condition of zero curl means that $\vec k\times \vec E[\vec k]=0$ which directly implies that $\vec E = f(\vec k)\vec k$ as a statement that $\vec E$ points in the $\hat k$ direction. All of the standard requirements for a Fourier transform to exist then also need to be applying to the electric field.

I suspect that the real tension is coming from a question more like this: given a (co)vector field $F_\alpha$ it is not in principle hard to form a Jacobian $\nabla_\alpha F_\beta$ and this clearly can be decomposed into a symmetric and antisymmetric part. The question is, it is clear that $\nabla_\alpha \phi$ is sufficient to generate a symmetric matrix, but how could we show it is necessary (with the various "assuming the field is nice" that we always have in physics)? In other words how can we prove that a nice symmetric Jacobian is a Hessian?

We might just appeal to the Fourier space again but when it's phrased like that I think there might be another clarifying insight. A real symmetric matrix is a special case of a Hermitian matrix and those are diagonalizable, but a diagonal matrix could just be integrated directly into a solution as the sum of all the double integrals of its diagonal components.