# Variational derivation of *all* covariant Maxwell's equations?

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?

• 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 $F$? Commented Sep 26, 2021 at 17:53

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 $$F\propto \mathrm{d}A$$, $$G\propto\star F$$, $$\mathrm{d}G \propto \star J$$.
• Ok, this action provides the correct equations when $\alpha = -c/8\pi$, $\beta = -c/4\pi$, and $\gamma = 1$. Thank you! Commented Sep 26, 2021 at 21:21