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I have read at many places that the pure Maxwell theory (without any matter) is self-dual. This is the general form for Maxwell Lagrangian density: $$\mathcal{L} = - \frac{1}{4} F_{\mu\nu} F^{\mu\nu},$$ where $F_{\mu\nu} \equiv \partial_\mu A_\nu - \partial_\nu A_\mu.$

There are two ways to see the self-duality that I know of. Either we can write it in terms of electric and magnetic fields, $$ \mathcal{L} = \frac{1}{2}\left(\mathbf{E}^2 - \mathbf{B}^2\right);$$ under the duality transformation $\mathbf{E}\rightarrow\mathbf{B}$ and $\mathbf{B} \rightarrow-\mathbf{E}$, the form of the Lagrangian remains the same (up to a negative sign), and so the equations of motion also remain the same. Alternatively, we can also write (in the language of differential forms): $$\mathcal{L} = -\frac{1}{2} F \wedge *F,$$ where $*F$ is the Hodge dual of $F$. This is invariant under the duality transformation $ F \rightarrow *F \ ,\ *F \rightarrow **F = -F $ which again lands us with the same form of Lagrangian (with a negative sign).

My actual problem concerns the duality in presence of monopoles. But I think my confusion can be answered at the level of pure Maxwell theory itself. As described in this answer, we can introduce an extra gauge field (call it $A^m$, and call the electric gauge field $A^e$) that couples with magnetic monopoles. I am assuming that under the self-duality, the $A^m$ and $A^e$ fields will be swapped (up to a negative sign). But then I don't see how the kinetic term will remain invariant. So perhaps I can word my question like this:

1) How can I express the Maxwell Lagrangian in terms of $A^e$ and $A^m$, so that the self-duality is obvious?

Let me also phrase this question in slightly different way:

2) Where is the $A^m$ field hidden in Maxwell Lagrangian? (Is it hidden as a constraint, for example?)

My understanding might be completely incorrect about this. If you have any comment or reference about it, that will be appreciated.

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The $\newcommand{\m}{\mathrm{m}}\newcommand{\e}{\mathrm{e}}$$A^\m$ that you are postulating is related to $A^\e$ by $$ F^\m :\overset{(*)}{=} \mathrm{d}A^\m\overset{!}{=} \star\, \mathrm{d}A^\e \overset{(*)}{=}:\star\,F^\e. \tag{1}\label{1}$$ Note, however, that while $A^\m$ and $A^\e$ are tenuously related to one another, they are so, non-locally. Namely you have to solve a differential equation to get one from the other. I.e. $A^\m(x)$ depends on the values of $A^\e(y)$ everywhere, and not only at $y=x$.

If you decide to write your action, in the absence of sources, in terms of $A^\m$, it is simply $$S[A^\m] = \frac12\int F^\m\wedge\star F^\m = \frac12\int \star F^\e\wedge F^\e = \frac12\int F^\e\wedge \star F^\e = S[A^\e],$$ so self-duality is guaranteed. Now to add a coupling, $A^\e$ couples to electric charges as $A^\e\wedge\star J^\e$ and $A^\m$ couples to magnetic monopoles by $A^\m\wedge\star J^\m$.

If you are perverse enough to give up local actions to have self-duality manifest in the presence of sources, you can certainly write $$ S = \int \Big(\frac14 F^\e\wedge\star F^\e + \frac14 F^\m\wedge\star F^\m + A^\e\wedge\star J^\e + A^\m \wedge\star J^\m\Big).$$ This action is manifestly invariant under $A^\e\leftrightarrow A^\m$, provided $J^\e=J^\m$$(\dagger)$. Keep in mind, however, that due to equation \eqref{1} you cannot treat $A^\e$ and $A^\m$ as independent fields. In particular you cannot, for example, use them to derive equations of motion.


$(*)$In the absence of topological obstructions. In the presence of those, $F = \mathrm{d}A+N$, where $N$ is a representative of a non-trivial cohomology class capturing integer fluxes.

$(\dagger)$You can fix the appropriate minus signs to have the duality work correctly yourself :)

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There is a continuous version of the electric-magnetic duality symmetry discussed in 'Spacetime algebra as a powerful tool for electromagnetism' by Justin Dressel, Konstantin Y. Bliokh, and Franco Nori: https://arxiv.org/abs/1411.5002

This paper uses a Clifford algebra to represent the electromagnetic field, which can be split into 'real' and 'imaginary' vector parts $F=E+IB$ of a Riemann-Silberstein-like 'complex' representation in a particular reference frame. However, unlike the traditional Riemann-Silberstein approach, the $I$ here is the Clifford pseudoscalar, which has a geometric interpretation, and so transforms correctly under Lorentz transformations. Since $I^2=-1$, it acts a lot like the imaginary unit, but isn't quite the same thing, which is why I keep putting 'complex' in scare quotes.

The equations are invariant under the continuous $U(1)$ transformation $z\rightarrow ze^{I\theta}$, which for the special case of $\theta=\pm\pi/2$ gives the duality transformation $E\rightarrow \pm B$, $B\rightarrow \mp E$ swapping (or more precisely, rotating) electric and magnetic fields into each other.

As discussed in section 8.1 of the paper, the traditional electromagnetic Lagrangian when straightforwardly translated into this form ($\mathcal{L}_{trad}=\frac{1}{2}\left< (\nabla z)^2 \right>_0$ for a 'complex' vector potential $z=a_e+a_mI$ giving a 'complex' field $F=\nabla z=F_e+F_mI$) turns out not to be invariant under the duality transformation. But the similar (and mathematically much nicer) form $\mathcal{L}=\frac{1}{2}\left< (\nabla z)(\nabla z^*) \right>_0$ is. This gives a dual-symmetric Lagrangian that can be expanded in terms of separate vector potentials and fields for electric and magnetic charges as:

$$\mathcal{L}=\frac{1}{2}\left< (\nabla z)(\nabla z^*) \right>_0=\frac{1}{2}(|E_e|^2-|B_e|^2)+\frac{1}{2}(|E_m|^2-|B_m|^2)$$

$$=\frac{1}{4}\sum_{\mu\nu}[(F_e)_{\mu\nu}(F_e)^{\mu\nu}+(F_m)_{\mu\nu}(F_m)^{\mu\nu}]$$

It does look a little arbitrary just sticking two copies of the Lagrangian for the electric and magnetic charges together like that. Of course, things look a lot simpler and more natural if you keep the electric and magnetic parts together as a single 'complex' structure rather than splitting it up into components.

The paper uses the Geometric Algebra approach, so the notation and terminology are not what a lot of physicists are used to, and it does take an initial effort to get into it. Because of that, I am finding it hard to give a self-contained, compact summary here. But I find the elegance of the result makes it worth the effort, though.

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Section 3 in https://arxiv.org/abs/2101.02350 shows a Lagrangian that treats the electric potential $A$ on equal footing with the magnetic potential $B$. The Lagrangian is in eq. (3.2) and the self-duality relation has been derived in eq. (3.25). In Lagrangian (3.2) all the fields except $F=dA$ and $G=dB$ are auxiliary fields which can be integrated out.

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  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Miyase
    Commented Mar 30, 2023 at 15:21

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