# Poincaré' lemma and EM potential $A^{\mu}$

My lecturer said that given the sourceless Maxwell's equations $$\partial_{\mu}\, ^ *F^{\mu\nu} = 0$$,

we can find a solution $$F^{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu},$$

that only exists locally.

The fact that there always is a $A_{\nu}$ that can locally give $F^{\mu\nu}$ is a consequence of Poincaré's lemma, and extending it to a global solution would introduce/require magnetic monopoles.

I tried to read up on the Poincore lemma but it's all maths and I was hoping for a physical explanation. Does it make sense that $A_{\nu}$ is local? How do mangetic monopoles enter the picture?

• What do you mean "Does it make sense that $A$ is local?"? In electromagnetism, we define $A$ to be the potential of $F$ such that $F = \mathrm{d}A$. This definition rests on the Poincare lemma, you simply can't define the gauge field from given $F$ without it, so the very definition of $A$ is already local. (This is generally so for gauge theories, btw - the gauge field is always only defined locally on spacetime) – ACuriousMind May 2 '15 at 12:02
• For magnetic monopoles, you might interested in this question – ACuriousMind May 2 '15 at 12:03
• It's the first time that I come across local stuff so I'm not used to it. Why are gauge field defined locally and not globally? – SuperCiocia May 2 '15 at 12:05
• Perhaps also have a look at What is the basis of gauge theory? – ACuriousMind May 2 '15 at 12:13
• Comment to the question (v2): The equation $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ also holds in the presence of electric sources. No need to go to source-free Maxwell equations. – Qmechanic May 2 '15 at 21:27

First of all some clarifications to make sure we are on the same page. The phrase "$A_\nu$ is local" is slightly misleading as it implies that A is only defined in some (small) neighborhood of the total manifold/space. This is not true. A is defined everywhere on the manifold. The correct expression is that "the theory is invariant under a local symmetry" ie that transformations of the form $$A_\nu(x)\rightarrow A_\nu(x)+L(x)$$ do not change the physics. The above are known as local/gauge transformations and they are distinctly different from global transformations in which the function does not depend on the position, ie: $$A_\nu(x)\rightarrow A_\nu(x)+L$$
The basic (physical) reason behind this is that for manifolds with no holes a field of the form $A=$constant is not physical and we can do a gauge transformation of the type described above to bring it to the form $A=0$ everywhere. However, for manifolds with holes (an example would be the circle) a field that is constant everywhere is physical, it cannot be gauged away and has observable consequences. This is another manifestation of the Aharonov–Bohm effect.