What is the physical meaning of Lorenz gauge condition? [closed]

Question

What is the physical meaning of Lorenz gauge condition? And what part of the solutions we throw?

Answer 1

Although I think this is a good question - the finding of meaning and relationships between physics notions is always worthwhile - "physical meaning" is not a good choice of words here. This is because gauge invariance is a redundancy in the mathematical description of a system; it means that we can partition the solutions of the description into equivalence classes such that all members of a class are a description of exactly the same physics (see this question here). In electromagnetism, the equivalence classes are big enough that one can always find a description for any particular electromagnetic physics that fulfills the Lorenz gauge.

Even so: there is one piece of physics here and that is that in electromagnetism, only the exterior derivative $\mathrm{d} \mathscr{A}$ (as a four-potential) is physically meaningful. This comes to mean exactly the same as the statement:

We can choose the divergence of A (as a three-vector potential) to be any differentiable vector field we like

Accordingly, we are free to choose any statement we like about $\nabla \cdot A$ without changing the physics, which explains why the Lorenz gauge is a statement about $\nabla\cdot A$, just as the Coulomb gauge ($\nabla\cdot A = 0$) is another such statement.

Writen in more componentish form, our statement that only $\mathrm{d} \mathscr{A}$ is physically meaningful is that only $\nabla\times A$ (as a three-vector potential) and $-\partial_t A-\nabla\phi$ are meaningful. There is a (IMO) very clear way to visualize these statements in Fourier space, where the gradient $\phi\mapsto\nabla\phi$, divergence $A\mapsto\nabla A$ and curl $A\mapsto\nabla\times A$ become $\tilde{\phi} \mapsto \tilde{\phi} \,k$, $\tilde{A}\mapsto k\cdot\tilde{A}$ and $\tilde{A}\mapsto k\times\tilde{A}$, respectively. Only $\nabla\times A$ is meaningful, so, only the component of $\tilde{A}$ orthogonal to the ray joining it to the origin is meaningful. That is, we can choose the component $k\cdot \tilde{A}$ (corresponding to the divergence) along $k$ to be anything. We must still leave the value of $-\partial_t A-\nabla\phi$ (the electric field) unchanged, but this statement just says we can change $A$ as long as we can compensate the change in $\partial_t A$ with a gradient: i.e. a radially directed $\tilde{\phi}\,k$ vector field in Fourier space. So we see that we are altogether free to choose the divergence of $A$ to be anything we want, even though the requirement to leave $-\partial_t A-\nabla\phi$ unchanged seems to be a further constraint that might rule out particular changes to $\nabla\cdot A$. We can make an arbitrary choice for $\nabla\cdot A$, and the scalar potential can be adjusted afterwards.

The Lorenz gauge is a little tricky insofar that the above would only seem to work if one chooses the divergence of $A$ then adjusts $\phi$, whereas the Lorenz condition is a statement on both at once. Let's see how this works. We adjust $A$ by adding a (radial-in-Fourier-space) component $\nabla\psi$ to it. So, to keep $-\partial_t A-\nabla\phi$ unchanged, we must take $\partial_t\psi$ away from our electric potential. Suppose that we have a valid solution $\mathscr{A}=(\phi,\,A)$ to Maxwell's equations. $\mathscr{A}^\prime=(\phi^\prime,\,A^\prime) = (\phi - \partial_t\psi,\,A+\nabla\psi)$ is also a solution with the same physics; then (in natural units), we have:

$$\nabla\cdot A^\prime+\partial_t\phi^\prime = \nabla\cdot A+\partial_t\phi + \left(\nabla^2\psi -\partial_t^2\psi\right) $$

and so we can always annul the quantity $\nabla\cdot A^\prime+\partial_t\phi^\prime$ by solving the inhomogeneous Helmholtz equation $\nabla^2\psi -\partial_t^2\psi=\zeta$, where $\zeta$ is the spacetime function defined by the original solution $\zeta=-\nabla\cdot A-\partial_t\phi$. Of course we never actually solve this equation; just the statement that a solution exists under mild conditions guarantees that we can partially fix the gauge through the Lorenz condition.

So let's summarize the meanings that we have found:

The Lorenz gauge exists by dint of the principles that (1) we are free to choose the divergence of $A$ to be anything we want without changing the system's electromagnetic physics and (2) that under very general conditions solutions to the inhomogeneous D'Alembert equation $\nabla^2\psi -\partial_t^2\psi=\zeta$ exist, where $\zeta$ is defined through $\zeta=-\nabla\cdot A-\partial_t\phi$ by any solution $(\rho,\,A)$ of Maxwells equations that we wish to adjust to be a new valid solution defining the same physics fulfilling the Lorenz gauge.

Notice I said partially fix above; by the foregoing discussion we can add any solution to the homogeneous D'Alembert equation $\nabla^2 -\partial_t^2\psi=0$ and still have a solution in the Lorenz gauge. So, strictly speaking, a Lorenz gauge still defines a nontrivial equivalence class of solutions. However, suitable boundary condition assumptions (e.g. the Sommerfeld radiation condition on the rate of attenuation of solutions over large distances) can totally fix the potentials.

Some other meanings of the Lorenz gauge that are worth committing to memory:

1. It is Lorentz covariant (note the "t" in the Dutch Hendrik Lorentz, a different fellow from the Lorenz gauge Danish guy Ludvig Lorenz). So, unlike the convenient Coulomb gauge, it survives arbitrary transformations in special and general relativity;
2. It can be thought of as a continuity equation, although there is no physical fluid involved. If we postulate a (nonphysical) fluid, $\phi$ can be its Lorentz-invariant charge or mass density then $A$ would be its flux; this means that the volume integral over all space of $\phi$ is a constant of any evolution of the electromagnetic field;

Answer 2

The Lorenz gauge condition is nice if you want to have electromagnetism mediated by a massive force carrier.

Of course the mass would have to be very very tiny to not be immediately contradicted by the good experimental fit of the inverse square law. But since experimental results always have nonzero error bars, there will always be a nonzero mass consistent with the data, it will just be a smaller mass when you have smaller error bars.

If you are sticking to Maxwell, another factor is if you want to associate a particular electromagnetic field with a charge and current distribution. Maxwell alone would require boundary conditions before you get a specific electromagnetic field since for any configuration of charge and current you can always add a vacuum Maxwell solution and get another solution. So knowing the charge and current simply doesn't yield unique fields.

But if you want to pick a particular solution (such as Jefimenko's) for Maxwell then using very particular solutions to the wave equation for the potentials is a standard technique. And that basically uses the Lorenz condition.

If you are going to use boundary conditions on a topologically trivial base manifold for a massless carrier, then every gauge should produce the same classical electromagnetic field. But you don't even need potentials if that's all you are doing.

Some people also like that it is Lorentz invariant (though it isn't the only such gauge condition).