Every source I have ever seen derives the retarded and advanced potentials by finding the Green's functions of the inhomogeneous Lorenz gauge conditions, and I have always thought that any linear combination of retarded or advanced potentials would satisfy the Lorenz conditions, as the PDE is linear.

I am now taking my first graduate course in Electromagnetism, and my professor keeps telling me that only the addition of the advanced and retarded potentials satisfies the gauge condition, because that way they aren't violating time symmetry. This confuses me, since I don't really see how this isn't just some hand-wavy justification, especially since I can just put the integral solutions for the retarded potentials into the Lorenz gauge conditions and show that these satisfy it, at least from a mathematical standpoint.

So can anyone explain to me what my professor is saying here?

Equations of interest:

Homogenous Lorenz Guage Condition: $$\nabla\cdot \mathbf{A}+\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}=0$$

Inhomogenous Lorenz Gauge Condtions: $$\nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}=-\frac{\rho}{\epsilon_0}$$ $$\nabla^2\mathbf{A}-\frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2}=-\mu_0\mathbf{J}$$

  • $\begingroup$ I'd think, like you, either would satisfy the PDE's derived from assuming the Lorentz Gauge. I can see a symmetry problem though if you don't use it. The Lorentz Gauge implies a Lorentz invariant potential 4-vector. This has some bearing also: en.wikipedia.org/wiki/Wheeler%E2%80%93Feynman_absorber_theory $\endgroup$
    – R. Romero
    Commented Mar 14, 2022 at 18:53
  • 2
    $\begingroup$ Why don't you ask your professor to explain in more detail at his or her office hours? $\endgroup$
    – hft
    Commented Mar 14, 2022 at 19:23
  • 1
    $\begingroup$ @R.Romero It is actually 'Lorenz' gauge as in the post. $\endgroup$
    – my2cts
    Commented Mar 14, 2022 at 19:29
  • $\begingroup$ Lorentz and Lorenz ... looks like a law firm, no? $\endgroup$
    – user121330
    Commented Mar 14, 2022 at 19:35
  • 2
    $\begingroup$ Answering the title question. Yes as they are derived from the inhomogenous wave equation which is only valid in the lorenz gauge $\endgroup$ Commented Mar 14, 2022 at 21:54

1 Answer 1


Might have found a counter-example to your professor's statement.

$\nabla \cdot \vec{B}=0 \implies \vec{B}=\nabla \times \vec{A}$.

$\nabla(\nabla \cdot \vec{A})-\nabla^2\vec{A}=\mu_0\vec{J}+\frac{1}{c^2}\frac{\partial}{\partial t}(-\nabla V -\frac{\partial \vec{A}}{\partial t})$

$\implies \nabla(\nabla \cdot \vec{A}+\frac{1}{c^2}\frac{\partial V}{\partial t})-\nabla^2\vec{A}+\frac{1}{c^2}\frac{\partial\vec{A}}{\partial t^2}=\mu_0 \vec{J}$

$\vec{E}=-\nabla V -\frac{\partial \vec{A}}{\partial t}$

$\nabla \cdot \vec{E}=\rho/\epsilon_0=-\nabla^2 V-\frac{\partial \cdot (\nabla \cdot \vec{A})}{\partial t}$

$\implies \rho/\epsilon_0-\frac{1}{c^2}\frac{\partial ^2 V}{\partial t^2}=-\nabla^2 V-\frac{1}{c^2}\frac{\partial ^2 V}{\partial t^2}-\frac{\partial \cdot (\nabla \cdot \vec{A})}{\partial t}$

So rewriting Maxwell's Equations in terms of the potentials, one gets homogenous equations in the absence of sources if $G(A',V)=\nabla \cdot \vec{A}+\frac{1}{c^2}\frac{\partial V}{\partial t}$ is any constant whereas the Lorenz Gauge requires that value be 0.

Potentials $\vec{A'}$ and $V'$ yield the same fields if:

$\vec{A'}=\vec{A}+\nabla f$

$V'=V-\frac{\partial f}{\partial t}.$

So the potentials can be modified

$\nabla \cdot \vec{A'}=\nabla \cdot \vec{A}+\nabla^2 f$

$\frac{1}{c^2}\frac{\partial V'}{\partial t}=\frac{1}{c^2}\frac{\partial V}{\partial t}-\frac{1}{c^2}\frac{\partial^2 f}{\partial t^2}$

So $G(A,V)$ remains constant so long as the l'Alembertian of $f$ is $0$.

So if $G(A,V)$ is a constant, and not necessarily $0$ as in the Lorenz Gauge, Maxwell's Equations imply that the Wave-Equation (D'Alembertian) is zero for both fields as well as the vector and scalar potentials. Further all the PDEs remain unchanged if an $f$ is chosen as above.

Violating the Lorenz Gauge doesn't violate the symmetry of the equations.


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