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In Is it possible to elevate the electric-magnetic duality discrete symmetry to a continuous one? the OP asks about how to derive the Noether current associated with the electromagnetic duality transformation. Michael Seifert points out that the Lagrangian involves the potentials $A_\mu$ and the duality transformation is not a transformation of $A_\mu$, which prevents us from being able to apply Noether's theorem.

Clearly, the duality transformation is only physical in free space (i.e. no sources) since otherwise it introduces unphysical magnetic charges and currents. In free space, how can we prove that the electromagnetic duality transformation $$\vec E \to \vec E \, \cos \theta + \vec B \, \sin \theta; \quad \vec B \to \vec B \, \cos \theta - \vec E \, \sin \theta$$ can't be expressed as a continuous, local transformation of the potentials $\varphi$ and $\vec A$?

Qmechanic
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Brian Bi
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It depends what you mean by 'impossible'. If you give the potential a 'complex' component to represent the magnetic currents and charges, and use $\tfrac{1}{2}\left< (\nabla A)(\nabla A^*)\right>_0$ instead of $\tfrac{1}{2}\left< (\nabla A)^2\right>_0$, then the modified Lagrangian is continuous dual-symmetric and you can apply Noether's theorem. (The magnetic component can subsequently be disposed of with electroweak symmetry breaking.) I'd not expect it to be possible to get dual-symmetry if you stick to the traditional non-dual-symmetric Lagrangian.

See sections 8.1 and 8.2 of the Dressel, Bliokh, and Nori paper referenced in the previous discussion, section 8.9 for the symmetry breaking argument, and section 8.8 for the application of Noether's theorem. https://arxiv.org/abs/1411.5002

Nullius in Verba
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