Lorenz gauge condition

The Lorenz gauge condition isor ${\partial }_{\mu }{A}^{\mu }=0$ in covariant notation.

For a transverse monochromatic electromagnetic wave with angular frequency $\omega \ne 0$, ${\partial }_t^2\mathbf{E}=-{\omega }^2\mathbf{E}$. The displayed electric potential and magnetic vector potential therefore give $-\nabla \Phi -{\partial }_t\mathbf{A}=\mathbf{E}$. Transversality gives $\nabla \cdot \mathbf{A}=0$, so the Lorenz gauge holds. For $\mathbf{E}=E_0(\cos \theta ,\sin \theta ,0)$ and $\theta =\omega t-Kz$, taking a curl yields $\mathbf{B}=(K/\omega )\hat{\mathbf{z}}\times \mathbf{E}$. The vacuum Maxwell equations require $K^2={\omega }^2/c^2$. The field has circular polarization, with $|\mathbf{B}|=|\mathbf{E}|/c$ and perpendicular electric and magnetic directions.

With convention ${\nabla }^{\mu }{F}_{\mu \nu }=-{\mu }_0{J}_{\nu }$, write the electromagnetic field tensor as $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$. Commuting the contracted covariant derivatives gives ${\nabla }^{\mu }{\nabla }_{\nu }{A}_{\mu }={\nabla }_{\nu }({\nabla }^{\mu }{A}_{\mu })+R^{\alpha }{}_{\nu }{A}_{\alpha }$. The Lorenz gauge condition removes the gradient-of-divergence term, leaving the displayed wave equation for a covector field. A locally inertial coordinate system does not eliminate the Ricci tensor term.

An Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$ changes $\partial \cdot A$ by $\square \lambda$. Consequently gauge fixing by a quadratic function of $\partial \cdot A$ retains transformations satisfying $\square \lambda =0$, subject to the same boundary conditions. Eliminating those residual transformations is part of specifying an inverse with a particular Green function prescription.