1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – ACuriousMind May 2 '15 at 12:02
  • $\begingroup$ For magnetic monopoles, you might interested in this question $\endgroup$ – ACuriousMind May 2 '15 at 12:03
  • $\begingroup$ 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? $\endgroup$ – SuperCiocia May 2 '15 at 12:05
  • $\begingroup$ Perhaps also have a look at What is the basis of gauge theory? $\endgroup$ – ACuriousMind May 2 '15 at 12:13
  • 2
    $\begingroup$ 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. $\endgroup$ – Qmechanic May 2 '15 at 21:27
2
$\begingroup$

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: $$A_\nu(x)\rightarrow A_\nu(x)+L$$

Now the statement about the Maxwell's equation mentioned, is that we can find a solution to it at any small neighborhood of the manifold, but when we try to patch together solutions in nearby neighborhoods to find a solution that will satisfy the equation everywhere, then this is impossible (for manifolds that have "holes").

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.

So in the example above, if the constant magnetic field along the circle is physical then what creates it? We can think of its source as an electric current flowing through a long wire passing through the center of the circle and being perpendicular to the plane of the circle, or equivalently we can think about it as having placed a magnetic monopole at the center of the circle. This simple example hopefully demonstrates why magnetic monopoles are needed to be introduced to justify global solutions of Maxwell's equaitons on a space with non-trivial topology (ie with holes).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.