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If I suppose there exists a 4-"vector potential" $A\in\Omega^1(U)$ such that the Faraday 2-form satisfies $F = dA$ (which is equivalent to assuming the homogeneous Maxwell's equations $dF=0$ are satisfied, provided the domain $U$ I'm working in is nice enough) and start from the action $$ S[A] = -\frac{c}{8\pi}\int_U F\wedge\star F + \int_U A\wedge\star J, $$ by varying $A \to A_\varepsilon := A + \varepsilon a$ with $a\in\Omega^1_0(U)$ and requiring $$0=\frac{\delta S}{\delta A}[A] := \frac{d}{d\varepsilon}\Bigg|_{\varepsilon=0} S[A_\varepsilon] \qquad \forall a \in \Omega^1_0(U)$$ I can easily obtain the inhomogeneous equations $\delta F = \dfrac{4\pi}{c} J$, where $\delta = \star d \star$ is the formal adjoint of $d$ (4-divergence) under the $L^2$-like inner product of forms $(\alpha,\beta) := \int_U \alpha \wedge \star \beta$.

Is there a way to complete the derivation without assuming $dF=0$ is satisfied? Possibly to derive the homogeneous equations variationally as well?

My thought was to assume $F$ is such that there is $A \in \Omega^1(U)$ such that $F = \hat F+dA$ for a fixed 2-form $\hat F$ instead, which can always be achieved, and carry on from there. Defining the action as above, I get the variational principle becomes $$ 0 = \frac{d}{d\varepsilon}\Bigg|_{\varepsilon=0} S[A_\varepsilon] = -\frac{c}{8\pi} \left[\frac{d}{d\varepsilon}\Bigg|_{\varepsilon=0} (\hat F,\hat F) + 2(\hat F,da) +2(dA,da) \right] + (a,J) $$ so that $$0 = \left(-\frac{c}{4\pi} \delta(\hat F + dA), a\right) $$ and thus $\delta \hat F = F + dA$ as wanted, provided $A$ is chosen with Lorenz gauge ($\delta A= 0$; remember $-\square_c A = (d\delta + \delta d)A$). This should be sufficient to show $dF = 0$ as soon as $d\hat F = 0$ as well. Can I avoid assuming $\hat F$ is already closed somehow?

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    $\begingroup$ It's not clear to me what you're asking. In the the action $S[A]$ you've written, $A$ is the dynamical variable and $F$ is defined as $F:=dA$. Are you looking for some other action $\widetilde{S}[F]$ which encodes both of Maxwell's equations for $F$? $\endgroup$
    – Kajelad
    Commented Sep 26, 2021 at 17:53

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I think you mean something like $$S[F,G,A] = \alpha \int F \wedge \star F + \beta \int G \wedge (F - \mathrm{d}A) + \gamma \int A \wedge \star J$$ with $F$, $G$, $A$ independent? Then taking the variation gives $F\propto \mathrm{d}A$, $G\propto\star F$, $\mathrm{d}G \propto \star J$.

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  • $\begingroup$ Ok, this action provides the correct equations when $\alpha = -c/8\pi$, $\beta = -c/4\pi$, and $\gamma = 1$. Thank you! $\endgroup$
    – giobrach
    Commented Sep 26, 2021 at 21:21

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