I have always imagined the magnetic field of wires, as the superposition of infinitely many curl elements.

I, naturally, wanted to see what a function with a single point of curl would look like.

The most obvious scenario is:

$$\nabla × \vec{F} = \delta^3(r)\hat i.$$

However in doing so you run into many flaws

Taking the divergence of both sides:

$$0 = \nabla \cdot [\delta^3(r)\hat i]$$

With right side being non zero.

There is a clear contradiction here, namely the curl of a vector function cannot be a single point of curl.

So my question is, why not? I know mathematically not, I can also see its issue in relation to stokes theorem.

This has the same parallels that amperes law has before maxwell's addition, that the divergence of $j$ must be zero in order to satisfy this relation.

What is the best other alternative for a single point of curl? And/or what other things can be said about this scenario?

  • 1
    $\begingroup$ I'm not sure I understand your question. However, I don't know if this is what you meant, a wire isn't $\delta^3(r)\hat{i}$, it is $\delta(y)\delta(z)\hat{i}$. So there is no contradiction in this case. If this is what you wanted, please tell me so I can post it as an answer. $\endgroup$
    – Habouz
    Jul 16, 2022 at 20:38
  • 1
    $\begingroup$ Yes I am aware an infinitely straight wire follows a 2d dirac function and thus its divergence is zero. But no, I am looking for a three-dimensional dirac function as the curl of a vector field. Or more of an intuition into the "infinite sum of infinitely many curl elements" that make up the wire. As the infinite sum of curl elements picture isn't really valid as per my question. $\endgroup$ Jul 16, 2022 at 20:45
  • 1
    $\begingroup$ I would be interested in what we are actually suming up, since a single point of curl is invalid, but it is still sensible to talk about a function having a defined curl at a point [aslong as its divergence is 0]. the sum of these points making up the vector field is valid, but these thinking about these points on their own?.doesn't seem to be valid $\endgroup$ Jul 16, 2022 at 20:46
  • 1
    $\begingroup$ I can think somewhere that the intuition lies in, for example, taking the curl of biot savart, the solution doesn't satisfy the original equation, unless the sources divergence is zero. The sum of infinitely many of these, leaves the function divergenceless, and thus the sum of these "point" elements satisfies the original equation $\endgroup$ Jul 16, 2022 at 20:49
  • $\begingroup$ Oh, I think I understand, but I don't quite see a contradition. You could take a slice of a curl around a wire and set everything to zero. Divergence would be zero everywhere as the form a loop, and you got a function that only has a curl at the slice of the wire. $\endgroup$
    – Habouz
    Jul 16, 2022 at 20:53

1 Answer 1



When we say that $\mathrm{div}\left(\hat r/r^2\right) = \delta^3(\vec r)$, what we mean is that

  1. $\mathrm{div}\left(\hat r/r^2\right)=0$ for all points where $\hat r/r^2$ is defined, and
  2. For any smooth surface $\Sigma$ enclosing the origin, we have that $$\oint_\Sigma \frac{\hat r}{r^2} \cdot \hat n \ \mathrm dS = 1$$

If a domain $U\subseteq \mathbb R^3$ is contractable, then for any $\vec F$ defined on $U$ we have that $\mathrm{div}(\vec F) = 0 \implies \vec F = \mathrm{curl}(\vec A)$ for some vector field $\vec A$ - this is a version of Poincare's lemma. If we integrate $\mathrm{curl}(\vec A)$ over a smooth surface $\Sigma$, then

$$\int_\Sigma \mathrm{curl}(\vec A)\cdot \hat n \mathrm dS = \oint_{\partial \Sigma} \vec A \cdot \mathrm d\vec r$$ via Stokes' theorem, where $\partial \Sigma$ is the boundary of $\Sigma$. In particular, if $\Sigma$ is a closed surface with no boundary, then $\partial \Sigma = \emptyset$ and so $$\oint_{\partial \Sigma} \vec A \cdot d\vec r = \int_\Sigma \vec F \cdot \hat n \mathrm dS = 0$$

To summarize, if $U\subseteq \mathbb R^3$ is contractable and $\vec F$ is a vector field defined on $U$, then $$\mathrm{div}(\vec F) = 0 \implies \oint_\Sigma \vec F \cdot \hat n \mathrm dS = 0$$ for any smooth, closed surface $\Sigma \subset U$. The reason points (1) and (2) above can coexist is that if we remove a single point from $\mathbb R^3$, it is no longer contractable, and Poincare's lemma no longer applies.

Turning our attention to curl rather than div, we immediately run into a problem. What would $\mathrm{curl}(\vec F) = \delta^3(\mathbf r)\hat i$ (for example) actually mean?

  1. $\mathrm{curl}(\vec F)=0$ for all points where $\vec F$ is defined, and
  2. (???) For any smooth closed curve $C$ enclosing the origin, we have that $\oint_C \vec F \cdot d\vec r = 1$ (???)

The problem is that in 3D, a closed curve (unlike a closed surface) cannot meaningfully be said to enclose a point. After all, if $\mathrm{curl}(\vec F)=0$ everywhere except at the origin, then we could simply apply Stokes' theorem to a surface whose boundary is $C$ and which doesn't intersect the origin, which would immediately give us that $\oint_C \vec F \cdot d\vec r = 0$.

More abstractly, the variation of Poincare's lemma which applies to curls is much stronger than the one which applies to divergences in the following sense: If $\mathrm{div}(\vec F)=0$, we may only conclude that $\vec F = \mathrm{curl}(\vec A)$ if the domain is contractable; however, if $\mathrm{curl}( \vec F)= 0$, then we may conclude that $\vec F = \mathrm{grad}(\varphi)$ for some scalar $\varphi$ if the domain is merely simply-connected, which is a far weaker constraint on the domain.

Case in point, if we remove a single point from $\mathbb R^3$ then it is no longer contractable, but it is simply connected, which means that $\mathrm{curl}(\vec F)=0 \implies \vec F= \mathrm{grad}(\varphi)$, rendering any variation of (4) impossible. The only way to circumvent this is to render $\mathbb R^3$ non-simply-connected - which would require (at minimum) removing an entire line from the space.

(Of course, if you're working in 2D then all of this goes out the window. $\mathbb R^2$ with a point removed is no longer simply connected, and you can have (distribution-valued) "curls" which are defined at a single point, where we define the 2D pseudoscalar $\mathrm{curl}(F) \equiv \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}$.)


The divergence operation measures the local outward flux of a vector field at a point. $\mathrm{div}(\vec F)(\vec r)=\delta^3(\vec r)$ means that the outward flux of $F$ through any surface enclosing $\vec r=0$ is 1. If we substitute the electrostatic field $\vec E$ and consider Maxwell's equations, this simply means that the charge distribution is a point charge (which we might imagine as the limit of a spherically-symmetric source in the limit as it shrinks to zero radius).

In contrast, the curl operation measures the local circulation of a vector field at a point. $\mathrm{curl}(\vec F)(\vec r)=\delta^3(\vec r)$ (presumably) means that the circulation of $\vec F$ around any loop containing $\vec r=0$ is 1, but as stated above it doesn't make sense to say that a loop does or does not contain a point in $\mathbb R^3$. If we substitute the magnetostatic field $\vec B$ and consider Maxwell's equations, such a thing would correspond to a point current, but likewise this would not make much sense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.