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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?

Qmechanic
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jensen paull
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1 Answers1

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Mathematically...

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 $\mathrm{curl}(F) \equiv \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}$.)

Physically...

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.

Albatross
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