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Replace $\phi$ by $\phi +ϵ\eta$, with $\eta$ vanishing on the boundary, and differentiate the action at zero. Integrate the derivative of $\eta$ by parts:The coefficient of $\eta$ is the functional derivative $\delta S/\delta \phi$.

For the electromagnetic action, $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$. Antisymmetry combines these terms and integration by parts givesThus the inhomogeneous Maxwell equations are ${\partial }_a{F}^{ab}={\mu }_0{J}^b$. The definition of $F$ also gives the homogeneous equations ${\partial }_{[a}{F}_{bc]}=0$, equivalently ${\partial }_a{\tilde{F}}^{ab}=0$. These are the two four-vector Maxwell systems.

Unrestricted variation of the alternative component-gradient action instead gives $\square A^b={\mu }_0{J}^b$. This is equivalent to the first Maxwell system only when supplemented by the Lorenz gauge ${\partial }_a{A}^a=0$, becauseThe alternative action is not by itself gauge-invariant; the additional condition must be supplied consistently, not silently omitted.

Expand the tensor square: $F^{ab}{F}_{ab}=2[{\partial }^b{A}^a{\partial }_b{A}_a-{\partial }^b{A}^a{\partial }_a{A}_b]$. HenceIntegrating this cross term by parts twice expresses its integral as a boundary term plus $\int ({\partial }_a{A}^a{)}^2{d}^{4}x$. Under Lorenz gauge it is therefore purely a boundary term. With fixed or vanishing boundary contributions, the two actions have the same variational equations once the additional gauge condition is included. This is the Lorenz-gauge reduction of the electromagnetic action.