# Equivalence between Lorenz gauge and continuity equation

I want to show that the Lorenz gauge condition$$\nabla\cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial\Phi}{\partial t}~~=~~0 \,,$$where $\mathbf{A}$ and $\Phi$ are the vector and scalar potential of the electromagnetic field, is equivalent to the continuity equation$$\nabla \cdot \mathbf{J}+\frac{\partial \rho}{\partial t}~~=~~0 \,,$$ where $\mathbf{J}$ is the electric current and $\rho$ the charge density, using the general expression of the potential using retarded Green functions\begin{alignat}{7} \Phi & ~~=~~ & \frac{1}{4\pi\varepsilon_0} & \int \mathrm{d}^3x' \frac{\rho\left(\mathbf{x'},t-\frac{1}{c}|\mathbf{x}-\mathbf{x'}|\right)}{|\mathbf{x}-\mathbf{x'}|} \\ \mathbf{A} & ~~=~~ & \frac{\mu_0}{4\pi} & \int \mathrm{d}^3x' \frac{\mathbf{J}\left(\mathbf{x'},t-\frac{1}{c}|\mathbf{x}-\mathbf{x'}|\right)}{\left|\mathbf{x}-\mathbf{x'}\right|} \end{alignat}

My first instinct is to simply plug the expression of the potential in the Lorenz gauge, which yields\begin{alignat}{7} \nabla\cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial\Phi}{\partial t} & ~~=~~ && \frac{\mu_0}{4\pi}\int \mathrm{d}^3x' \nabla_{\mathbf{x}}\cdot\frac{\mathbf{J}\left(\mathbf{x'},t-\frac{1}{c}|\mathbf{x}-\mathbf{x'}|\right)}{\left|\mathbf{x}-\mathbf{x'}\right|} \\ && ~~+~~ & \frac{\mu_0}{4\pi}\int \mathrm{d}^3x' \frac{\partial}{\partial t}\frac{\rho\left(\mathbf{x'},t-\frac{1}{c}\left|\mathbf{x}-\mathbf{x'}\right|\right)}{|\mathbf{x}-\mathbf{x'}|} \\ \\ &~~=~~&&\frac{\mu_0}{4\pi} % \left( \begin{array}{rl} & \displaystyle{\int{\mathrm{d}^3x' \frac{\nabla_{\mathbf{x}}\cdot\mathbf{J}\left(\mathbf{x'},t-\frac{1}{c}|\mathbf{x}-\mathbf{x'}|\right)}{\left|\mathbf{x}-\mathbf{x'}\right|}}} \\ - & \displaystyle{\int{\mathrm{d}^3x' \mathbf{J}\left(\mathbf{x'},t-\frac{1}{c}\left|\mathbf{x}-\mathbf{x'}\right|\right)\cdot\frac{\mathbf{x}-\mathbf{x'}}{\left|\mathbf{x}-\mathbf{x'}\right|^3}}} \\ + & \displaystyle{\int{\mathrm{d}^3x' \frac{\partial}{\partial t}\frac{\rho\left(\mathbf{x'},t-\frac{1}{c}\left|\mathbf{x}-\mathbf{x'}\right|\right)}{|\mathbf{x}-\mathbf{x'}|}}} \end{array} \right)_{\Large{,}} \end{alignat} using $$\nabla \cdot \psi \mathbf{A} ~~=~~ \psi \nabla\cdot \mathbf{A} + \mathbf{A}\cdot\nabla\psi \,.$$

Now, the first and last term in the last expression are the continuity equation, but that middle term ruins everything. I don't see why it should be zero, and if it shouldn't, where I'm wrong.

• What am I missing here. They can't be equivalent. The continuity equation must hold, but the Lorenz condition doesn't have to. – garyp Oct 14 '17 at 15:56
• I can see your point. This is what an exercise series is asking me to do. Those are the expression of the potentials in the case of homogeneous boundary conditions in empty space, so I guess that in this simple case the equivalence holds? – user2723984 Oct 14 '17 at 16:14
• Possible duplicate of Ensuring Lorenz Gauge condition in Green Function solution – AccidentalFourierTransform May 20 '18 at 2:53
• Hi @Nat, I've noticed you've been editing ~~=~~ into equations lately. Any particular reason? It doesn't look good, and it is not a recommended practice. – AccidentalFourierTransform May 20 '18 at 3:11
• @AccidentalFourierTransform Mostly just to spread 'em out. I try to keep expressions more spacially local to those that they interact with first on parsing, so low-order-of-operations-priority operands like = tend to get more spacing around them. I think that most folks tend to use \quad or \qquad, but ~~ just seems a bit cleaner and more adjustable to me. If it looks off, is the concern that there's not enough spacing or too much? – Nat May 20 '18 at 3:14