I was reading the proof of Helmholtz decomposition theorem where I found the relation between the rotational and the irrotational fields are not symmetric. And by that I mean if the divergence of the gradient field is supposed to be zero except where the particle exist aka:

$$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$

Then why is the divergence of the curl always zero?

$$\nabla\cdot\nabla\times \bf A = 0$$

and not: $$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

where $\bf c $ is a vector with speed of light magnitude.

Edit: This question is about asymmetry of electromagnetic potential fields and I have no idea why it was migrated to math forum. Anyways, if mathematicians can see this question my question is:

Suppose there is a vector field $ F=\nabla(1/r)+\nabla \times \bf A $ made out of a scalar potential $1/r$ and a vector potential $\bf A$ where these relations hold: $$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$ and: $$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

So both potential fields have critical points, considering $\bf F$ should have been sufficiently smooth, can we still apply Helmholtz decomposition theorem?

  • $\begingroup$ I'm very confused by your question: why do you think that a relation that holds for the gradient should automatically hold for a curl? Especially when the first relation (divergence of the gradient of $1/r$) is only true for one special function ($1/r$() and the second is a general theorem that can easily be proved mathematically? Also, what is $\mathbf{c}$ in your last equation? $\endgroup$
    – Philip
    Nov 5, 2020 at 7:31
  • $\begingroup$ @Philip I added what I mean by c. I am looking for those special functions A and 1/r that correspond to electromagnetism. $\endgroup$ Nov 5, 2020 at 7:36
  • $\begingroup$ correct me if I'm crazy, but with c having a non-zero magnitude, one could very well say that your last equation is true because $\delta^3(\bf c)$ would be $0$. I mean, what is the point of writing it that way even if it is correct? Taking the delta function of a non-zero constant makes no sense; you know the answer $\endgroup$
    – Jim
    Nov 5, 2020 at 13:34
  • $\begingroup$ @Jim You got it. That's my point. I think that these two may not always equal and we should explicitly consider it. So, I was looking for a physicist to show me why we always neglect that delta function. My guess is that having some critical points for the vector field, A, downgrades it from a physical quantity like the velocity to a mathematical abstraction of it, which ends in gauge theory where we have to implicitly exclude those critical points. I mean I'm not sure if Helmholtz decomposition theorem is in general applicable to electromagnetism. $\endgroup$ Nov 5, 2020 at 15:16

5 Answers 5


That the divergence of a curl is zero, and that the curl of a gradient is zero are exact mathematical identities, which can be easily proven by writing these operations explicitly in terms of components and derivatives.

On the other hand, a Laplacian (divergence of gradient) of a function is not necessarily zero. Equating it to a charge or another source is a matter of specific physical interpretation, which goes beyond pure math.

  • $\begingroup$ Thanks. That's exactly why I asked this question here on physics stack exchange not the math one. What I don't understand is that during 1800s quaternions developed by Hamilton to describe rotations but Helmholtz decomposition theorem lead physicists to abandon quaternions and split fields into rotational and irrotational ones, despite we may have critical points on A field. what I am looking for is the "beyond pure math" part where physicists neglect the critical points and define gauges. $\endgroup$ Nov 5, 2020 at 8:43
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    $\begingroup$ I am not sure that I fully understand the context. Quaternion formulations are still in use, particularly in engineering. This ranges from purely unambiguous angle specification in robotics to some serious Green's function descriptions of wave guides, as done by Schwinger. But this is as far as my knowledge of the subject goes. $\endgroup$
    – Roger V.
    Nov 5, 2020 at 8:48
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    $\begingroup$ @Vadim Quaternions used to be used where vector analysis is used now. For the Maxwell equations, for fluid flow and so on. After Gibbs and Heaviside, their use largely diminished. $\endgroup$ Nov 6, 2020 at 7:30
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    $\begingroup$ @MartinSpinoza Are you saying that you're interested in knowing what happens when the divergence of the curl is allowed to be nonzero? $\endgroup$ Nov 6, 2020 at 11:20
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    $\begingroup$ Hai! I wanted to say one thing I cant on the biology site. I liked your last answer and wanted to upvote. But I couldnt. I was suspended for holding a view. So, just for your information...+1 $\endgroup$ Jul 24, 2021 at 6:36

This is a good question and the answer lies in the misuse of the notation $\nabla$, and it is also why I like to write $\operatorname{\mathbf{grad}}, \operatorname{\mathbf{curl}}, \operatorname{div}$ instead of $\nabla, \nabla \times, \nabla \cdot$, resp. That both $\operatorname{div} \operatorname{\mathbf{curl}} \mathbf{v}=0$ and $\operatorname{\mathbf{curl}}\operatorname{\mathbf{grad}} f=0$ hold for arbitrary $f, \mathbf{v}$ scalar or vector fields, resp., is nothing but the equality of mixed partial derivatives $\frac{\partial^2}{\partial x \partial y}=\frac{\partial^2}{\partial y \partial x} $as it is easily seen by writing them out in components. But this means neither $\nabla \cdot \nabla \equiv 0$ nor $\nabla \times \nabla \equiv \bf{0}$ for such repeated operator is not defined; $\nabla$ is not a vector and it is neither perpendicular to nor parallel with itself in any sense.

There is a formulation of vector analysis using (exterior) differential forms where a differentiation operator is introduced that does act as you would like $\nabla$ behave. It can replace $\operatorname{\mathbf{grad}}, \operatorname{\bf{curl}}, \operatorname{div}$ with a single operator and can also repeatedly act on its argument with the result that you would expect. Differential forms are the natural generalization of vector analysis (including the vector product) for higher than 3 dimensional space, and there are attempts to replace and teach all of vector analysis in that way including the 3D case. You can find much written on this subject here in phys.stack.

  • $\begingroup$ Nabla might not be a vector, but treating it as a vector is a very useful shorthand for manipulations, as shown.g. in Gibbs and Wilson (1901), without resorting to components in a particular coordinate system (and risk that the result will only hold in this system). $\endgroup$ Nov 5, 2020 at 17:43
  • $\begingroup$ @VladimirF I think what is meant here is that vector analysis defines grad, curl and div as rather general limiting forms. Their interpretation in terms of $\nabla$ is a useful shortcut, but this is not what they really mean. $\endgroup$
    – Roger V.
    Nov 6, 2020 at 6:23
  • $\begingroup$ Hmm, I note that you wrote "$\operatorname{\mathbf{curl}}\operatorname{\mathbf{grad}} f=0$ for arbitrary f", and not "$\operatorname{\mathbf{curl}}\operatorname{\mathbf{grad}} =0$" (without the f) like you did with "$\nabla \times \nabla \equiv \bf{0}$". Wouldn't it be equally ok to say "$\nabla \times \nabla f \equiv \bf{0}$ for arbitrary f"? (bar any formatting mistakes I made) $\endgroup$
    – ilkkachu
    Nov 6, 2020 at 11:47
  • $\begingroup$ @ikkachu yes, it would be correct to write $\nabla \times \nabla f = 0$ for all $f$ $\endgroup$
    – hyportnex
    Nov 6, 2020 at 12:40

There are actually cases when $\frac{\partial^2}{\partial x \partial y} \ne\frac{\partial^2}{\partial y \partial x}$ for functions with asymmetric Hessian matrices. Such functions have $\nabla\cdot\nabla\times {\vec A} \ne 0$. Such functions are sometimes mentioned because of an unmotivated assumption that h > 0 in the definition of the derivative

$$ k = \frac{f(x+h) - f(x)}{h}. $$

h can also be < 0.


Just to give some quick intuition as to why this identity is true (doesn't seem like it was mentioned here):

Consider a vector field $F$ and some ball $B$. By Stoke's theorem, the flux of $curl(F)$ through $\partial B$ is 0. By the divergence theorem, the flux of $curl(F)$ through $\partial B$ is equal to the triple integral of $div(curl(F))$ inside $B$.

If $div(curl(F))$ is continuous and non-zero at some point $P$, there exists a small ball around $P$ such that the triple integral of $div(curl(F))$ inside of it is non-zero. Yikes, contradiction; this means that the flux of $curl(F)$ through the boundary of the ball is non-zero.


From the comments you've written, I think that what you're asking is this:

Math tells us that the divergence of a curl is always zero. But suppose we go beyond pure math. What happens when the divergence of the curl is allowed to be nonzero?

And the answer is that it's just not possible. The divergence of the curl is zero, always, everywhere, under all circumstances, in theory and in practice, in the real world and in imaginary worlds.

Asking what happens when the divergence of the curl is nonzero is like asking what happens when two people are both taller than each other. It just can't happen.

  • $\begingroup$ I'm confused by the downvotes on this answer. The asker wanted to know what happens when the divergence of the curl is nonzero. I tried to answer the question clearly and directly by pointing out, in no uncertain terms, that that's totally impossible. I must be misunderstanding something. $\endgroup$ Aug 21, 2021 at 13:08
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    $\begingroup$ The issue in your answer is that OP wants to know why this is impossible. There is a reason - it is not like asking why two people are taller than each other. $\endgroup$
    – Gary Allen
    Mar 18, 2022 at 17:35
  • $\begingroup$ @GaryAllen You may be right. The impression I got from the OP's comments is that they had a misconception that was something along the lines of "in mathematics, the divergence of the curl is always zero, but outside of mathematics, I bet it doesn't have to be zero." I hoped to dispel that misconception with this answer. $\endgroup$ Mar 18, 2022 at 19:00

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