# Magnetic monopole and vector potential

Does anyone know how to prove (in a simple way if possible) that it is impossible to define a single-valued globally defined magnetic vector potential $\vec{A}$ on the manifold $M=\mathbb{R}^3\backslash\{0\}$ for a static magnetic monopole placed at the origin?

Hypothesis: $\vec{B} = g \vec{r}/(4\pi r^3)$

• Both actually, Wu-Yang proved that you can define 2 potentials in different regions of space for such a field which are linked to each other by a gauge transoformation. But out of this context I definitely meant global. But, still maybe I'm miss-understanding something. Dec 4, 2014 at 10:24
• So far I can's see any reason not to be globally defined... Dec 4, 2014 at 10:49
• In fact, here it's said: "Because the divergence of B is equal to zero almost everywhere, except for the locus of the magnetic monopole at r = 0, one can locally define the vector potential such that the curl of the vector potential A equals the magnetic field B. However, the vector potential cannot be defined globally precisely because the divergence of the magnetic field is proportional to the Dirac delta function at the origin". Dec 4, 2014 at 11:07
• Try reading the book 'Geometry, Topology and Gauge Fields: Foundations', the author (Naber) have this discussion right on the introductory chapter and points how the impossibility to define a proper vector potential on ${\mathbb R}^3-{0}$ is linked with it's topology (the second homotopy group is non-trivial), and also how classical dirac monopoles rise from this. Obs: Your hypothesis have non-zero flux integral, just like the equivalent solution for the electric point charge field Dec 4, 2014 at 11:38
• Thank you, I'll have a look! That's exactly why I need this... To understand why Dirac needed to introduce the string! Dec 4, 2014 at 11:48

You seek a 1-form $A$ on $\mathbb{R} - \{0\}$ such that $\mathrm{d}A = B$. On all of $\mathbb{R} - \{0\}$, $\mathrm{d}B = 0$, so this could exist. But, since you have magnetic flux, you require that the integral of $B$ over any 2-sphere around the origin should be $g$. Therefore, by Stokes' theorem,

$$g = \int_{S^2}B = \int_{S^2} \mathrm{d}A = \int_{\partial S^2} A = \int_{\emptyset} A = 0$$

which is a contradiction for $g \neq 0$. Therefore, such an $A$ cannot exist.

This is a more formal rephrasing of Holographer's answer.

The magnetic flux through any closed surface enclosing the origin is just $g$ (the magnetic charge enclosed). If the magnetic field comes from a vector potential $\vec{B}=\nabla\times\vec{A}$, this surface integral by Stokes' theorem is an integral of $\vec{A}$ around the boundary of the surface. But the surface is closed, so has no boundary, so the answer must vanish. This is a contradiction if $g$ is nonzero, so no such $\vec{A}$ exists.

(This has a more sophisticated interpretation in the language of differential forms and de Rham cohomology, but that's not really necessary here!)

• Yes it pretty much what they do in "Geometry, Topology and Gauge Fields: Foundations". Thank you. Dec 4, 2014 at 14:05

user23873 answered my question in the comments. I quote: " Try reading the book 'Geometry, Topology and Gauge Fields: Foundations', the author (Naber) have this discussion right on the introductory chapter and points how the impossibility to define a proper vector potential on ℝ3−0 is linked with it's topology (the second homotopy group is non-trivial), and also how classical dirac monopoles rise from this. Obs: Your hypothesis have non-zero flux integral, just like the equivalent solution for the electric point charge field. "