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.

  • 3
    $\begingroup$ What am I missing here. They can't be equivalent. The continuity equation must hold, but the Lorenz condition doesn't have to. $\endgroup$ – garyp Oct 14 '17 at 15:56
  • $\begingroup$ 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? $\endgroup$ – user2723984 Oct 14 '17 at 16:14
  • $\begingroup$ Possible duplicate of Ensuring Lorenz Gauge condition in Green Function solution $\endgroup$ – AccidentalFourierTransform May 20 '18 at 2:53
  • $\begingroup$ 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. $\endgroup$ – AccidentalFourierTransform May 20 '18 at 3:11
  • $\begingroup$ @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? $\endgroup$ – Nat May 20 '18 at 3:14

You're trying to prove the right thing with the wrong assumptions.

The continuity equation and the Lorentz gauge equation describe what is happening at the source's place and time, while the potentials given by the retarded Green's integrals describe what an observer measures far away from the source's place-time. So these two couple of equations can not be directly substituited one into the other, for are not directly related.

For an elegant proof that the Lorentz condition is a direct consequence of te equation of continuity, take a look into the section 14-5 (The Hertz Potential) of the book: "Classical Electricity and Magnetism" by Wolfgang Panofsky and Melba Phillips.

  • $\begingroup$ Nowadays it is called the Lorenz condition, not the Lorentz condition. $\endgroup$ – Ján Lalinský Aug 3 '18 at 11:02
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    $\begingroup$ I could not find any such proof in the section 14-5. What they are saying is that when potentials are assumed to be given by the retarded solutions of the wave equations, together with the equation of continuity this implies the Lorenz condition. But the wave equation and the retarded solutions are usually derived while assuming the Lorenz condition, so that's a circular proof. $\endgroup$ – Ján Lalinský Aug 3 '18 at 11:05

In Griffiths' textbook "Introduction to Electrodynamics" (third edition), Problem 10.8 is "Confirm that the retarded potentials satisfy the Loren(t)z gauge condition". The solution is available on-line (search "introduction to electrodynamics solution manual david griffiths"). A comparison of Griffiths' solution with yours revealed that the first and last terms of your last expression were not the continuity equation, as Julio Moros pointed out in his answer.


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