Vector potential field for a hypothetical magnetic monopole

Question

This is a mathemetical question with perhaps a physical interpretation.

I would like to solve the equation: $$ \vec{\nabla} \times \vec{A} = \frac{1}{r^2} \hat{r} \tag{1}$$

using spherical coordinates $(A_r, A_{\theta}, A_{\phi})$. The form of the rotation $\vec{\nabla} \times$ in spherical coordinates can e.g. be found on wikipedia.
This gives 3 differential equations and I just wonder whether they can be solved?
I was under the impression that there would be plenty examples, but I cannot find any.
The equation could perhaps be considered to be the vector potential of a hypothetical magnetic point-monopole.

• Does anybody know a solution or why it's not possible?

Answer 1

You can find a solution on most of space, but not all of space. To see why, imagine taking the flux integral of $\vec{B} = \hat{r}/r^2$ over the surface of a sphere $S$ of radius $r$. Doing this integral is straightforward, and yields $$ \iint_S \vec{B} \cdot d\vec{a} = 4\pi. $$ Now let's assume that there exists a vector field $\vec{A}$ such that $\vec{\nabla} \times \vec{A} = \vec{B}$. This implies that $$ 4\pi = \iint_S (\vec{\nabla} \times \vec{A}) \cdot d\vec{a}. $$ But according to Stokes' theorem, we can always replace the integral of the curl of a vector field over a surface with a line integral around the surface's boundary $\partial S$: $$ \iint_S (\vec{\nabla} \times \vec{A}) \cdot d\vec{a} = \oint_{\partial S} \vec{A} \cdot d\vec{l}. $$ The problem is, of course, that the surface of a sphere has no boundary! This means that $\partial S$ is empty, the integral vanishes, and we conclude that $$ 4\pi = \iint_S (\vec{\nabla} \times \vec{A}) \cdot d\vec{a}= \oint_{\partial S} \vec{A} \cdot d\vec{l} = 0, $$ which is a contradiction.^{[citation needed]}

If you really want to work with a vector potential for a monopole, then, you have to avoid enclosing it in a spherical surface. One well-known way to do this is to "remove" the negative $z$-axis ($\theta = \pi$) from the space you're considering. In that case, $S$ can never be a full sphere; it can only be a sphere minus a tiny "puncture" at the south pole, and the boundary around that puncture means that we no longer necessarily have $\oint_{\partial S} \vec{A} \cdot d\vec{l} = 0$. In particular, if you then define $$ \vec{A} = \frac{1 - \cos \theta}{r \sin\theta} \hat{\phi} = \frac{\tan \frac{\theta}{2}}{r} \hat{\phi}, $$ then it's not hard to show that $\vec{\nabla} \times \vec{A} = \hat{r}/r^2$. As we can see, though, $|\vec{A}| \to \infty$ as $\theta \to \pi$, meaning that this vector potential can't be extended over all of space.

Answer 2

Describing a magnetic monopole is a problem in Classical Electromagnetism: essentially, if you have a magnetic monopole, then $\nabla \cdot \mathbf{B} \neq 0$, meaning that we can no longer define a vector potential $\mathbf{A}$, since -- if you remember -- it was precisely the fact that the divergence of $\mathbf{B}$ was always zero in Maxwell's Equations that allowed us to write $\mathbf{B = \nabla \times A}$.

Dirac came up with a solution for a monopole (described in an accessible way here), but it came with it's own "problems": unlike an electric monopole which is only singular at a single point ("$r=0$"), the "magnetic monopole" solution turned turned out to be singular along an infinite line. (It turns out there are two ways to see this, either as a "succession" of dipoles (left), or a tightly wound semi-infinite solenoid (right), both stretching off to infinity, as shown in the figure below. See Jackson's Electrodynamics "6.11 On the Question of Magnetic Monopoles" for more details, and a complete description of the image below.)

A system like this cannot be described by one single vector potential over all space, precisely because it is singular in this fashion. However, it turns out that one can define a vector potential that describes nearly everywhere in space (except along the monopole itself).

$$\mathbf{A} = g\left(\frac{1 - \cos{\theta}}{r \sin \theta} \right)\hat{\mathbf{\varphi}}.$$

Such a potential would indeed produce a magnetic field $$\mathbf{B} = \frac{g}{r^2}\hat{\mathbf{r}},$$

but it is singular along the line $\theta = \pi$, which is the "direction" we assume the monopole to be in. Such a singularity is known as a "Dirac String". You can read up more about it here. (The actual argument is a little more involved, it contains two "patches", one for the upper hemisphere and one for the lower one, depending on where the point you're considering is positioned and so on, but it's a little too involved for this discussion. You can find more about it online if you're interested.)

One of the curious results of Dirac's monopole was that he showed that if even one magnetic monopole existed in the universe, it would explain why charge was quantised. Which is cool...

Answer 3

It’s not possible because the divergence of the left side of your equation is identically zero while the divergence of the right side is a Dirac delta function at the origin.

Answer 4

The equation $$\vec{\nabla}\times\vec{A} = \frac{1}{r^2} \hat{r}$$ has no solution. You can prove it as follows. Apply the divergence operator ($\vec{\nabla}\cdot$) to the equation above. You get $$\vec{\nabla}\cdot(\vec{\nabla}\times\vec{A}) = \vec{\nabla}\cdot\frac{1}{r^2}\hat{r}$$

The left side is zero for every $\vec{A}$, because divergence of curl is zero. The right side is a Dirac delta function, as derived for example in "Divergence of $\vec{f} = \frac{1}{r^2} \hat{r}$". Hence we get $$0 = 4\pi\delta(\vec{r}),$$ which obviously is a contradiction.

Answer 5

Apply the divergence operator $\nabla\cdot$ on both sides of OP's eq. (1). The left-hand side vanishes identically, but the right-hand side is proportional to the 3D Dirac delta distribution. So OP's eq. (1) is inconsistent at the origin.