# Why is the electromagnetic duality an S-duality?

One of the examples that Wikipedia gives of S-duality is the EM duality. Namely that \begin{align} \mathbf{E} &\rightarrow\mathbf{B} \\ \mathbf{B} &\rightarrow -\frac{1}{c^2}\mathbf{E} \quad \text{ or }\quad (E, B) \rightarrow (E \cos \theta - B \sin \theta, B \cos \theta + E \sin \theta) \end{align} or if you prefer $$F^{\mu \nu} \mapsto{ }^{\star} F^{\mu \nu} \quad{ }^{\star} F^{\mu \nu} \mapsto-F^{\mu \nu}.$$ As far as I understand, an S-duality maps an strongly interactive theory to a weakly interactive one, i.e. generally, $$g\mapsto 1/g$$. Where can I see this?

I am assuming one would see this at the Hamiltonian/action level but I am a bit lost on how this would work out...

You can see it in lots of ways. Here is one of them: Consider the first-order Lagrangian for the Maxwell theory $$L=-\frac{1}{4g^2}F^{\mu\nu}F_{\mu\nu}+\partial_{\mu}\widehat{A}_{\nu}(\star F)^{\mu\nu}.$$ The fields $$F^{\mu\nu}$$,$$\widehat{A}_\mu$$ are assumed to be independent and $$F^{\mu\nu}$$ is not the field strength of a 1-form at this level. Varying this Lagrangian w.r.t. $$\widehat{A}_\mu$$ will give you the Bianchi identity on $$F$$, namely $$\partial_{[\mu}F_{\nu\rho]}=0\,,$$ which can be locally solved as $$F_{\mu\nu}=\partial_{[\mu}A_{\nu]}$$ for another 1-form $$A_\mu$$. Plugging this solution back into $$L$$ will give you the standard second order Maxwell Lagrangian for the $$A$$ field.

Alternatively, one can vary $$L$$ w.r.t. $$F$$ and obtain the so-called duality relation (up to an irrelevant overall factor that can be absorbed into the field $$\widehat{A}$$) $$F_{\mu\nu}\propto 2g^2\star(\partial_{[\mu}\widehat{A}_{\nu]}):=g^2(\star\widehat{F})_{\mu\nu}.$$ Using this relation, you can see that the first order Lagrangian $$L$$ becomes the second order Maxwell theory (this time for the dual field $$\widehat{A}$$) but with the reverse coupling constant, i.e. $$L\propto -\frac{g^2}{4}\widehat{F}^{\mu\nu}\widehat{F}_{\mu\nu}.$$ Thus, you can see that the coupling gets inverted. This justifies why EM duality is a strong-weak one.

• Thank you for your answer! Could you perhaps explain a bit more explicitly where the duality comes in to play? namely, is $\hat{F}^{\mu \nu} =^{\star} F^{\mu \nu}$ or something analogous? Feb 3, 2021 at 15:06
• Dear @FriendlyLagrangian, unfortunately there was a tedious typo in my first reply and it is related to your comment. I edited my answer and now everything should be clear.
– Ozz
Feb 3, 2021 at 15:27
• So if I understood you correctly, $F^{\mu \nu} \mapsto *F^{\mu \nu}$ and $*F^{\mu \nu} \mapsto -F^{\mu \nu}$ is a symmetry (I agree). Further more, we can define an object $\widehat{F}_{\mu \nu} \propto -\frac{1}{g^2}*F_{\mu \nu}$ and its dual (using $*^2=-1$). With $\widehat{F}_{\mu \nu}$ we arrive at a second order Maxwell theory but with inverse coupling constant $g$. What is the claim? $\widehat{F}^{\mu \nu} \neq * F^{\mu \nu}$ Feb 3, 2021 at 17:29
• Dear @FriendlyLagrangian, yes that is exactly the point. Another way to think about this: Let me switch to standard differential notation for brevity and suppose that we have the duality relation $F=g^2\star\widehat{F}$. If you act on both sides with the differential $d$, you will get $0=d\star d\widehat{F}$ (the Maxwell equation of motion for $\widehat{A}$), simply because $F$ satisfies the Bianchi identity $dF=0$. Now, let's act with $d\star$ on the duality relation. The result should convince you that, under EM duality, one has an ''exchange'' of Bianchi identities with equations of motion.
– Ozz
Feb 3, 2021 at 18:47
• Regarding the citation, my answer relies on the first-order approach. This is a very basic approach and I cannot really cite one particular reference. I guess you can find the same approach (even for free theories involving different gauge fields in arbitrary number of spacetime dimensions) in any textbook or relevant article :)
– Ozz
Feb 3, 2021 at 18:52