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I'm familiar with Electric-Magnetic duality, where in the absence of source fields one can exchange the $F_{\mu \nu}$ field with the dual field: $\tilde{F}_{\mu \nu}={\epsilon}_{\mu \nu \alpha \beta} {F^{\alpha \beta}} $ and the Maxwell equations remain the same.

I was wondering if it's possible to extend such discrete symmetry to a continuous one where one can principally derive a Noether current for the symmetry.

Basically I'd start with transforming the field to a new one:

$${F^{'} } _{\mu \nu} = {A}_{\mu \nu \alpha \beta} {F^{\alpha \beta}}$$

Where ${A}_{\mu \nu \alpha \beta}$ is a generalized parameter dependent (an internal unknown symmetry parameters) pseudo-tensor which reduces to the Levi-Civita pseudo-tensor in some points of the parameter space.

To fix the number of parameters one might impose some normalization conditions on the field based on energy conservation, though I'm not really sure if it necessarily fixes them completely. But can perhaps reduce their number.

Qmechanic
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Bastam Tajik
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  • Related: https://physics.stackexchange.com/q/223991/2451 – Qmechanic Jul 15 '22 at 12:42
  • @Qmechanic I read your answer to the post you attached. Interesting. But I wonder if there's any way to show that the number of symmetry parameters can be reduced to one that you've denoted as $\alpha$ or there might be a greater symmetry group. – Bastam Tajik Jul 15 '22 at 13:06

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Wald's books General Relativity and Advanced Classical Electromagnetism discuss a continuous duality, but I'm not sure whether it leads to an interesting Noether current. Namely, the transformation $$F_{\mu\nu} \to \cos\alpha F_{\mu\nu} + \sin\alpha \tilde{F}_{\mu\nu}$$ generalizes the duality transformation you exhibited. Wald calls it a "duality rotation" and it is mentioned on Chap. 4, Ex. 2 of the GR book. Also mentions it on Eq. (5.10) of the E&M book. Neither of them have deep discussions nor get even close to the Noether theorem, but I think it could be a starting point if you want to fill in the details.

Níckolas Alves
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  • Could not believe if such an idea in not already applied by the father's of the field, say, Montonen or Olive. Really provocative answer. Thanks. – Bastam Tajik Jul 15 '22 at 11:06
  • It's not clear to me whether Noether's theorem can actually be applied to this symmetry, since it's not a symmetry that transforms the fundamental fields of the Lagrangian $A_\mu$. (Which isn't to say that it's impossible, just that I don't know whether Noether's theorem can be applied to such a case.) – Michael Seifert Jul 15 '22 at 11:41
  • @MichaelSeifert What if one takes $F_{\mu \nu}$ as the fundamental field in this particular case and applies the Noether theorem to this field? – Bastam Tajik Jul 15 '22 at 12:41
  • @BastamTajik: If you take $F_{\mu \nu}$ as the fundamental equation in the Lagrangian $\mathcal{L} = \frac14 F_{\mu \nu} F^{\mu \nu}$, then the equations of motion are $F_{\mu \nu} = 0$ over all space and time. That would imply that any quantity constructed out of $F_{\mu \nu}$ (including the Noether current, whatever it is) will be conserved under the equations of motion, because any such quantity will also be zero. ... – Michael Seifert Jul 15 '22 at 12:52
  • The other option would be to construct the Noether current treating $F_{\mu \nu}$ as a fundamental field and then see if that current is conserved under the "real" Maxwell evolution equations $\partial_\nu F^{\mu \nu} = 0$. However, the proof of Noether's theorem relies on the symmetries between the variations of the Lagrangian with respect to the fundamental field (i.e., the Lagrangian) and the variation of the fundamental fields under the symmetry. Using two different fields for these variations means that an important step in the proof may no longer hold. – Michael Seifert Jul 15 '22 at 12:57
  • @MichaelSeifert I liked Qmechanic's answer to this post: physics.stackexchange.com/q/223991/2451 But I'm still wondering if there's any formal way to determine the exact number of symmetry parameters. They assume it's 1 since it works. But I'm afraid maybe there's a bigger symmetry group. I'm even thinking of adding fermionic currents in a symmetric fashion, both electric and magnetic currents so that the symmetry doesn't fail by addition of matter, and maybe even a spontaneous symmetry breaking of such symmetry, to justify the non-existence of magnetic current in nature. – Bastam Tajik Jul 15 '22 at 13:53
  • It's quite inspired by the global axial symmetry breaking in QED by coupling to Higgs, after the phase transition. – Bastam Tajik Jul 15 '22 at 13:55
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There's an interesting approach to the continuous electric-magnetic duality symmetry in 'Spacetime algebra as a powerful tool for electromagnetism' by Justin Dressel, Konstantin Y. Bliokh, and Franco Nori: https://arxiv.org/abs/1411.5002

They use the Geometric Algebra approach (Real Clifford algebras, interpreted geometrically) to unify the electric and magnetic components into a single 'complex' bivector field, where the pseudoscalar of the Geometric Algebra for spacetime is used as the imaginary unit. This gives the complex numbers a geometric interpretation, which fixes the reference frame-dependence problems of the Riemann-Silberstein complex vector interpretation.

The corresponding Noether current is described in section 8.8 as the 'helicity pseudocurrent'. Essentially, it results in conservation of helicity in the vaccuum case.

Nullius in Verba
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