What is the physical interpretation of the Lorenz gauge condition?

Question

If we integrate both sides of the Lorenz gauge condition, $\nabla \cdot \mathbf{A} = -\frac{1}{c^2}\frac{\partial \phi}{\partial t}$, over a small volume (free of charges for simplicity), we get:

$$ \int_V \nabla \cdot \mathbf{A} \, dV = -\frac{1}{c^2} \int_V \frac{\partial \phi}{\partial t} \, dV $$

Applying the Gauss divergence theorem: $$ \oint_S \mathbf{A} \cdot d\mathbf{S} = -\frac{1}{c^2}\frac{\partial}{\partial t} \int_V \phi \, dV $$

This says that the rate of decrease of $\phi$ in $V$ is proportional to the flux of $\mathbf{A}$ from the volume $V$.

The retarded potentials solution of Maxwell's equations that result from applying the Lorenz gauge condition seems to support this interpretation:

$$ \phi(t) = \frac{1}{4 \pi \epsilon_0} \int_{V_\rho} \frac{[\rho]}{r} dV_\rho \\ \mathbf{A}(t) = \frac{1}{c^2} \frac{1}{4 \pi \epsilon_0} \int_{V_\rho} \frac{[\rho \mathbf{v}]}{r} dV_\rho $$

The solutions say that two potentials are emitted and propagated from a moving charge. A strong scalar potential $\phi$, and a $\frac{1}{c^2}$ weaker, vector potential $\mathbf{A}$, that is proportional to the velocity, $\mathbf{v}$, of the moving 'emitting' charge.

If all the charges are stationary, then $\phi$ in any region is constant and so is $\int_V \phi \, dV$. If there are moving charges, the $[\rho \mathbf{v}]$ term from the weaker potential, $\mathbf{A}$, corresponds to a flux of $\phi$ that can lead to an increase or decrease of $\int_V \phi \, dV$.

Is this a correct way to interpret the Lorenz gauge condition?

Answer 1

Your interpretation sounds pretty correct. The gauge conditions can be viewed as nothing but electro-magnetic versions of the continuity equation. The Coulomb Gauge is simply a stationary, or very crudely speaking, a more "classical" limit of the Lorentz gauge. You might find further relevant information here.

Answer 2

A gauge condition does not change any physics, so I don't think the Lorenz gauge condition has a physical interpretation.

Answer 3

Yes there is a clear physical meaning to the Lorenz gauge condition. Take a look at the wave equation $$\partial_\mu \partial^\mu A^\nu = -j^\nu / \epsilon_0 ~.$$ This equation establishes a one-to-one (bijective) relation between $A^\nu$ and $j^\nu$. The current conservation law $\partial_\mu j^\mu =0 $ is therefore imaged in the solution space as the Lorenz condition $\partial_\mu A^\mu =0$. See my peer reviewed and published paper at https://arxiv.org/abs/physics/0106078.

What about the other gauge choices? They correspond to the same E and B but for a choice of the current that is not conserved.

Answer 4

May I offer perspective on the practical side of things? Since gauge transformations are ultimately just changes in our description of the system which don't alter the physics, one of our primary purposes for choosing a gauge is ease of calculation.

The Lorenz gauge condition is, importantly, a Lorentz invariant gauge condition since we're contracting the 4-indices of $A_\mu$ and $\partial_\mu$. This is advantageous for a multitude of reasons. One of these reasons is that when we compute the equations of motion for $A^\mu$, we get that our equation of motion in a vacuum is $$ \partial_\mu \partial^\mu A^\nu =0 ,$$ which means that the photon propagator in Fourier space can be simply written (after doing a Fourier transform and solving for our Green's function): $$D_{\mu \nu }(k) = - \frac{i \eta^{\mu \nu}}{k^2}$$. You can attempt to find the propagator in the Coulomb gauge but it's harder.

Imposing the Lorenz gauge at quantisation level on Hilbert space also offers interesting interpretations too, which you can google about (Gupta-Bleuer conditions).