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The standard way to write the Maxwell equations (say in vacuum in absence of charges) as Euler-Lagrange (EL) equations is to take the first pair of the Maxwell equaitons and to deduce from it existence of electromagnetic potential. Substituting it into the second pair of Maxwell equations one gets second order equations for the potential. They can be presented as Euler-Lagrange equations for the potential.

I am wondering if there is a way to present the Maxwell equations as an EL-equation in terms of electromagnetic fields only rather than potential.

I think I can prove that this is impossible if one requires in addition that the Lagrangian density is invariant under the Poincare group.

Qmechanic
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MKO
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  • Related: https://physics.stackexchange.com/q/441172/2451 , https://physics.stackexchange.com/q/55291/2451 and links therein. – Qmechanic Dec 25 '20 at 19:36
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    @Qmechanic: I tihnk the answers there do not answer my question. – MKO Dec 25 '20 at 19:40
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    The questions and answers Qmechanic refers to give you an action/Lagrangian in terms of the stress tensor (which contains the fields) and also give you the Maxwell equations solely in terms of the stress tensor. So yes, they do answer your question. You dont need the potential, it's just a corrolary of the EL equation $dF=0$ together with Poincare's lemma. – NDewolf Dec 26 '20 at 10:03
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    @NDewolf : I am not agree. While the Lagrangian density does not use the potential, to get EL equations coinciding with Maxwell equations, one has to use that $F_{\mu,\nu}$ is expressed via potential and then use integration by parts. – MKO Dec 26 '20 at 10:45
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    I see, you also want to express the intermediate steps without potentials. I'm not sure a gauge theory approach is useful then, since this built around the potiential/gauge connection. However, is there a good reason why you'd want to do this? – NDewolf Dec 26 '20 at 12:14
  • x-posted on math.OF: https://mathoverflow.net/q/379750/106114 – AccidentalFourierTransform Dec 26 '20 at 21:53
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    The question says "in terms of electromagnetic fields only", which suggests that additional fields (auxiliary fields) are not allowed. If auxiliary fields were allowed, then the question would have already been answered in https://physics.stackexchange.com/a/343082. The question also seems to require that all of Maxwell's equations must be contained in the Euler-Lagrange equations alone, with no additional constraints. (I tried posting an answer that respects those rules, but the answer was flawed for other reasons, so I deleted it.) – Chiral Anomaly Dec 27 '20 at 01:37
  • This is an interesting question. I would have asked this question myself, one motivation being that: how does choice of field in a Lagrangian can affect the interaction picture in QFT. Usually in classical theory, the equation of motion is more fundamental, and given this e.o.m, one can cook up a Lagrangian whose E-L eqn will produce the same dynamics. In QFT, the action is considered more fundamental, both in canonical quantization or path integral formulation, so maybe a good choice of field can simplify scattering amplitude calculation (?) – KP99 Jul 21 '21 at 05:17
  • Related : (1) Deriving Lagrangian density for electromagnetic field. (2) Derivation of Maxwell's equations from field tensor lagrangian. – Frobenius Jul 21 '21 at 07:32

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In the case of time-harmonic fields, it can be formulated in terms of fields without employing the auxiliary potentials. The Lagrangian in this case is $$L=\int_V \frac{1}{2}i\omega\mu H^2-\frac{1}{2}i\omega\epsilon E^2-\mathbf{E}\cdot \nabla \times\mathbf{H}+\frac{1}{2}\sigma E^2+\mathbf{J}\cdot \mathbf{E} \; d^3r.$$ The Maxwell's equations in time-harmonic case are $$\nabla \times \mathbf{E}=i\omega\mu\mathbf{H},$$ $$\nabla\times \mathbf{H}=-i\omega\epsilon\mathbf{E}+\mathbf{J},$$ $$\nabla\cdot\epsilon\mathbf{E}=\rho,$$ $$\nabla\cdot\mu\mathbf{H}=0.$$ The continuity equation then becomes $$\nabla\cdot\mathbf{J}=i\omega\rho.$$ For the general case, as far as I know, no such Lagrangian has been found so far.

Kksen
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