 Solution

The Maxwell field has the gauge symmetry $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$ for an arbitrary smooth scalar function $\alpha$. Commuting the two derivatives shows that $F_{\mu \nu }$ is unchanged, so the Maxwell Lagrangian is invariant. Vary the action, use antisymmetry of $F$ and integrate by parts with variations vanishing on the boundary:The Euler-Lagrange field equations are therefore ${\partial }_{\mu }{F}^{\mu \nu }=0$. Expanding $F$ givesIn Lorenz gauge the second term vanishes, leavingThe PDF calls this “Lorentz gauge”; the standard name is Lorenz gauge. A residual gauge symmetry preserves this condition when $\square \alpha =0$.