In Maxwell's equations, since the $\boldsymbol{B}$ field is divergence free, we can find the vector potential $\boldsymbol{A}$ such that $\boldsymbol{B}=\nabla\times\boldsymbol{A}$. However, this identity holds only when the domain $\Omega$ is topologically trivial -- if there is a 'hole' in the domain, then the de Rham complex is not exact and so it may be that $\boldsymbol{B}=\nabla\times\boldsymbol{A}+\boldsymbol{f}$, for some nonzero divergence and curl-free vector $f$, instead of just $\boldsymbol{B}=\nabla\times\boldsymbol{A}$.

Most of the time Maxwell's equations are used with non-trivial domains, especially in the industrial community. Do we have potential formulations on the equations on these domains or do we have to use the the field formulation?

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    $\begingroup$ If it is about computation, there will be a lot of use of the potentials, because keeping track of E and B takes six components, whereas phi and A is four. Massive savings in computational time and storage. We have no choice but to use potentials when dealing with quantum theory. But yes, the issues you mention plagues everything. $\endgroup$ May 23, 2023 at 4:15
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    $\begingroup$ Same; I think almost everybody would find some way to make the domain topologically trivial. Even in the Aharonov-Bohm problem, we would just include the solenoid with magnetic field, and thereby obtain topologically trivial space yet again, even if later on in the interpretations they would pretend that the solenoid is impenetrable. $\endgroup$ May 23, 2023 at 4:39
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    $\begingroup$ re hyportnex: It is true that div$\mathbf{B}=0$ holds everywhere, but it does not mean $\mathbf{B}$=curl$\mathbf{A}$ for some $\mathbf{A}$ unless the domain is trivial. $\endgroup$
    – Eric
    May 23, 2023 at 5:30
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    $\begingroup$ But a potential $(\phi,A)$ always exists locally. Just divide your domain up into trivial patches and then sew the potential together with cocycles. This breaks down your problem into a topological part and a harmonic part, both of which are surely easy to put on a computer for more involved charge configurations/domains? $\endgroup$ May 23, 2023 at 6:49
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    $\begingroup$ what determines if we can write $\mathbf B = curl \mathbf A$ is not whether we can ignore the field inside some region and still calculate the field with reasonable accuracy outside of it but whether the 3D space on which $\mathbf B $ is defined is singly connected or not. $\mathbf B $ permeates everything everywhere all the time and that will ensure the existence of the vector potential. $\endgroup$
    – hyportnex
    May 23, 2023 at 20:54

1 Answer 1


Maxwell’s equation: $$ \nabla\cdot B=0 $$ is assumed to be valid in the entire space which is topologically trivial. You’ll therefore not have issues in practical cases and can define globally a single valued potential vector.

Things start to change when you add magnetic charges which you can equivalently view as changing the topology of the space by excluding the charged region and imposing topological properties on the fields by Gauss’ law. For the domain, the topological property is not simple connectedness, the latter would be relevant to find the potential of an irrotational vector field. It is rather the second Betti number.

You can still use potential vectors, however it will not be a globally defined as a single valued function. Each branch must differ by a gauge transformation. In practice, you divide your space into regions with $b_2=0$ (typically convex regions) where you can define an $A$ and link the different formulas in overlapping regions via gauge transformations.

The typical example is that of a unit magnetic monopole. Using the formalism of 1-forms, in spherical coordinates, you have: $$ B=\frac{\sin\theta}{4\pi}d\theta d\phi $$ which can be viewed as the exterior derivative of: $$ A=\frac{1-\cos\theta}{4\pi}d\phi $$ Equivalently, in terms of vector fields: $$ B=\frac{1}{4\pi r^2}e_r\\ A=\frac{\tan(\theta/2)}{4\pi r}e_\phi $$ The vector potential is defined everywhere except on the half line $\theta=\pi$. You can cover this by producing another vector potential: $$ A=-\frac{1+\cos\theta}{4\pi}d\phi\\ A=-\frac{\cot(\theta/2)}{4\pi r}e_\phi $$ They both differ by a gauge transformation for $0<\theta<\pi$: $$ A=\frac{1}{2\pi}d\phi\\ A=-\frac{1}{2\pi\sin\theta r}e_\phi $$

You can similarly work up all the usual problems of electrostatics using vector potentials by this method.

Hope this helps.


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