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

  • $\begingroup$ Just find some vector potential for it, then add an arbitrary constant to it. Now you have the original vector potential and another one which are not equal. In fact, instead of a constant you can add any $\nabla m$ with $m$ a continuously differentiable scalar function. $\endgroup$ – Ruslan Dec 4 '14 at 9:24
  • $\begingroup$ @Ruslan That comment applies to any vector potential. I'm sure something more is required here. $\endgroup$ – Rob Jeffries Dec 4 '14 at 9:27
  • $\begingroup$ @RobJeffries that theorem applies to any vector potential, proving it for all potentials includes that of a static monopole. I don't see what more one has to do here if it's not required by the formulation of the problem. $\endgroup$ – Ruslan Dec 4 '14 at 9:30
  • $\begingroup$ @Ruslan I know that the potential is not uniquely defined since the curl of a gradient is 0... If I have well understood this kind of field shouldn't have a vector potential globally defined on all space (without the origin) and I don't know how to see that. $\endgroup$ – Worldsheep Dec 4 '14 at 9:53
  • $\begingroup$ Ah, so by unique you mean global, right? $\endgroup$ – Ruslan Dec 4 '14 at 9:58

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!)

  • $\begingroup$ Yes it pretty much what they do in "Geometry, Topology and Gauge Fields: Foundations". Thank you. $\endgroup$ – Worldsheep Dec 4 '14 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. "


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