Helmholtz decomposition, the process for splitting a vector field into parts which have vanishing divergence and curl, plays a central role in our ability to quantize the electromagnetic field because it permits us to separate the gauge invariant solenoidal part of the vector potential, $\mathbf{A}_{\mathrm{sol}}$, from its gauge dependent irrotational part, $\mathbf{A}_{\mathrm{irrot}}$. The formulae for these components of the field can be defined from the Wikipedia article as \begin{align} \mathbf{A}_{\mathrm{irrot}}(\mathbf{x}) &= -\frac{\nabla}{4\pi} \int_V \frac{\nabla' \cdot \mathbf{A}(\mathbf{x}')}{\left|\mathbf{x}-\mathbf{x}'\right|} \operatorname{d}V' + \frac{\nabla}{4\pi} \oint_{\partial V} \hat{n}'\cdot \frac{\mathbf{A}(\mathbf{x}')}{\left|\mathbf{x}-\mathbf{x}'\right|} \operatorname{d}S' \\ &= -\frac{\nabla}{4\pi} \int_V \mathbf{A}(\mathbf{x}') \cdot \nabla \frac{1}{\left|\mathbf{x}-\mathbf{x}'\right|} \operatorname{d}V',\ \mathrm{and}\\ \mathbf{A}_{\mathrm{sol}}(\mathbf{x}) &= \mathbf{A}(\mathbf{x}) - \mathbf{A}_{\mathrm{irrot}}(\mathbf{x}). \end{align} Now, for a field that already satisfies $\nabla\times\mathbf{A}=0$ the first operator is the identity and the second is $0$, which both act locally, and the same is true for the swapped versions. Equivalently, if a field is either solenoidal or irrotational then that can be determined locally by application of the divergence and curl and seeing which one is zero. Given a mixed field, though, the operator that constructs its irrotational component looks highly non-local, especially if $V$ is expanded to cover all of space.

Can this determination be made locally? That is, does going the other direction with $V$, towards zero size by, say, partitioning $V$ into compact sub-volumes, work? My instinct is to say that the partition is inherently non-local because of the existence of magnetic scalar potential techniques.

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    $\begingroup$ I don't understand what you mean by "inherently" nonlocal, or what sort of evidence would be needed to show it. $\endgroup$ Commented Dec 24, 2017 at 20:01
  • $\begingroup$ @EmilioPisanty No way to do it with an integration kernel that has a support on a set of measure $0$ (i.e. $\delta(\mathbf{x}-\mathbf{x}')$ and a finite number of derivatives thereon). $\endgroup$ Commented Dec 24, 2017 at 20:04
  • $\begingroup$ More on Helmholtz decomposition. $\endgroup$
    – Qmechanic
    Commented Dec 25, 2017 at 13:01

2 Answers 2


Yes, it is non-local.

A useful way to think about the decomposition is that $\mathbf{A}_{\text{irrot}}$ is the gradient vector field closest to $\mathbf{A}$, with the closeness of two fields measured by the average norm-squared difference of the vectors in the fields. This averaging operation makes the decomposition non-local.

For example, if we have a vector field $$ \mathbf{A}(x,y) = \begin{bmatrix} 0 \\ x \end{bmatrix}, $$ then the closest gradient is the constant field equal to $\mathbf{A}$ at the center of the domain, and thus is not locally calcuable from $\mathbf{A}$.

  • $\begingroup$ I would have accepted your answer, but the argument presented only showed that one particular way of constructing the components was non-local, not that all ways must be. $\endgroup$ Commented Dec 28, 2017 at 3:25
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    $\begingroup$ I'm not sure that I understand your objection: if one method for splitting out the components depends on the volume over which the decomposition is performed (here because of the averaging), then any other method for performing the decomposition also depends on the volume (or we're talking about a non-uniquely-defined decomposition, which I do not believe to be the case here). $\endgroup$
    – RLH
    Commented Dec 28, 2017 at 4:20
  • $\begingroup$ That's a good point, though it doesn't appear to be explicit in the post, just an implication taken from a particular example that is everywhere harmonic. I did still upvote, for what it's worth. $\endgroup$ Commented Dec 29, 2017 at 19:58
  • $\begingroup$ Ah. The key idea from my post was "This averaging operation makes the decomposition non-local." The part after "For example..." was a minimal-working-example of the principle. $\endgroup$
    – RLH
    Commented Jan 2, 2018 at 7:14

Yes, Helmholtz decomposition is inherently a non-local process because it is only unambiguous when the integrals are performed over all of a simply connected space that contains all relevant sources of the field (where a "source" is defined as any region in the field where the divergence or curl is non-zero). As the usage of the magnetic scalar potential demonstrates, it fails on sub-spaces because of the ambiguity inherent in the classification. In standard Helmholtz decomposition, divergenceless and solenoidal are taken as synonymous, likewise with divergent and irrotational (curl free). Fields that have both zero divergence and zero curl are assumed to be excluded by the boundary condition at infinity that the fields vanish there.

When you limit your considerations to finite sized regions, though, the assumption breaks down because a source that is outside of your bounded region can produce a field that is both divergenceless and irrotational everywhere in the region of interest. For that reason, local categorization is only unambiguous if it is done via positive properties. In other words, a vector field can be locally described unambiguously by breaking it down into three components: \begin{align} \mathbf{F}(\mathbf{x}) & = \mathbf{F}_{\mathrm{div}}(\mathbf{x}) + \mathbf{F}_{\mathrm{curl}}(\mathbf{x}) + \mathbf{F}_{\mathrm{har}}(\mathbf{x}),\ \mathrm{where} \\ \nabla\cdot\mathbf{F}_{\mathrm{div}}(\mathbf{x}) &\neq 0\ \quad\mathrm{somewhere}, \\ \nabla\times\mathbf{F}_{\mathrm{curl}}(\mathbf{x}) &\neq 0\ \quad\mathrm{somewhere}, \\ \nabla\cdot \mathbf{F}_{\mathrm{har}}(\mathbf{x}) & = 0\ \quad\mathrm{everywhere},\ \mathrm{and} \\ \nabla\times \mathbf{F}_{\mathrm{har}}(\mathbf{x}) & = 0\ \quad\mathrm{everywhere}. \end{align} The subscript of $\mathbf{F}_{\mathrm{har}}(\mathbf{x})$ is meant to be short for "harmonic" in analogy to harmonic functions because it can always be expressed in the region of interest as the negative gradient of a harmonic function. You could also call the harmonic term $\mathbf{F}_{\mathrm{ext}}(\mathbf{x})$, since it can also be modeled as being produced by sources external to the region of interest (when that region has finite size).

In this categorization, the reconstruction would work like this: \begin{align} \mathbf{F}_{\mathrm{div}}(\mathbf{x}) & = \frac{1}{4\pi} \int_V \frac{\mathbf{x}-\mathbf{x}'}{\left|\mathbf{x}-\mathbf{x}'\right|^3} \nabla'\cdot \mathbf{F}(\mathbf{x}') \operatorname{d}V' \\ \mathbf{F}_{\mathrm{curl}}(\mathbf{x}) & = \frac{1}{4\pi} \int_V \nabla\times\frac{\nabla'\times \mathbf{F}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} \operatorname{d}V' \\ \mathbf{F}_{\mathrm{har}}(\mathbf{x}) & = \mathbf{F}(\mathbf{x}) - \mathbf{F}_{\mathrm{div}}(\mathbf{x}) - \mathbf{F}_{\mathrm{curl}}(\mathbf{x}). \end{align}

What this exercise makes clear is if you try to take the limit as $V\rightarrow 0$, the region will either have some kind of delta function source that leaves the field ill defined there (e.g. point charge, line charge, or sheet charge) or the integrals that define the divergent and solenoidal components vanish, leaving only a locally harmonic field. Thus, the classification is only useful when done in regions of finite size, and only when the region of interest is the entire possible space.

That is why the Wikipedia formulae quoted in the question do not have an ambiguity in their derivation, they have an unstated assumption that no sources beyond the boundary contribute to the local field. As an example, consider a uniform field in the region of interest. Was it produced by one (or more) infinite charged sheet(s) (divergent source), or infinite current sheet(s) (solenoidal source)? Assuming the region of interest is a cube with faces aligned parallel and perpendicular to the field, the Wikipedia sources would have a combination of charged sheets on the faces the field pierces, and a current sheet circulating around the other four.

Point being, the process is only unambiguous when sources are only allowed in a closed region of interest.

  • $\begingroup$ Sorry, but I really don't see how this answer meets the extremely high bar you've set for proof in the question - or indeed how it answers the OP's core concern at all. $\endgroup$ Commented Dec 29, 2017 at 19:25
  • $\begingroup$ @EmilioPisanty Admittedly, showing a counter-example, if it exists, would be trivial, and part of why I set the bar high was because I thought a counter example could exist. The rest comes down to the fact that I presented an argument that could serve as the outline to a more rigorous proof, and not an actual proof. The argument covers the inherent ambiguity of when the influence of outside sources impinges on the region of interest, and what happens to the classification scheme when one attempt to make a local classification ($V\rightarrow 0$). $\endgroup$ Commented Dec 29, 2017 at 19:38
  • $\begingroup$ Sorry but I just don't see it - I don't see anything here that begins to address the enormous set of possible local decompositions, let alone show that they are unsuitable for even one example. $-1$ for the misleading use of the accept mark (though nothing here is explicitly wrong - it's just not an answer to the question as you posed it). $\endgroup$ Commented Dec 29, 2017 at 19:43
  • $\begingroup$ @EmilioPisanty Provide a more rigorous answer, and I'll accept it. As it is, the main thing I was looking for was showing that the $V\rightarrow 0$ attempt with the equations given falls into ambiguity taken with a real situation where outside sources are allowed. $\endgroup$ Commented Dec 29, 2017 at 19:48
  • $\begingroup$ I can't provide a more rigorous answer, but I am indeed interested in one if it exists; the acceptance makes the posting of such an answer less likely - hence the downvote. Otherwise, I have little more to add here. $\endgroup$ Commented Dec 29, 2017 at 19:52

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