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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?

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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.

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

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  • $\begingroup$ The Lorenz gauge condition does not seem to be an arbitrary choice for $\nabla \cdot \mathbf{A}$. It seems to enforce the analogy that just as $\mathbf{J}$ corresponds to a flux of $\rho$, $\mathbf{A}$ corresponds to a flux of $\phi$. An arbitrary choice would not lead to the retarded potentials solution that matches with experiments. $\endgroup$
    – Engr. Ravi
    May 30 '20 at 14:56
  • $\begingroup$ @Engr.Ravi I did not say that the Lorenz gauge is an arbitrary choice, it is certainly convenient in many cases. However, all gauges are physically equivalent, and one can use an arbitrary choice and still get equally satisfactory agreement with experimental results. $\endgroup$
    – akhmeteli
    May 30 '20 at 17:16
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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.

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  • $\begingroup$ "This equation establishes a one-to-one (bijective) relation between $A^\nu$ and "j^\nu$. " Why do you say that? You can add any 4-potential with zero d'Alembertian and get the same current. $\endgroup$
    – akhmeteli
    May 31 '20 at 3:41
  • $\begingroup$ I don't understand it. So, for example, a solution with a constant 4-potential (having zero field) is unphysical? $\endgroup$
    – akhmeteli
    Jun 2 '20 at 1:41
  • $\begingroup$ How could it be unphysical? It is 4-potential, not field. It describes the same physical field as the 4-potential without the constant, so the two 4-potentials are equally physical (or equally unphysical, if you wish). Your statement looks very much like just your opinion - I don't see (and have difficulty trying to imagine) any arguments in its favor. $\endgroup$
    – akhmeteli
    Jun 2 '20 at 11:59
  • $\begingroup$ Your argument does not seem to hold water. For example, if the source is just one electron, then, due to charge conservation, it existed since minus eternity, so its light cone is always the entire 3d space. Moreover, when you forbid adding a constant to the potential, you actually forbid to choose an arbitrary energy level as the zero level. This looks ridiculous. Of course, you can choose any boundary condition, but your personal choice does not make all other boundary conditions unphysical. $\endgroup$
    – akhmeteli
    Jun 3 '20 at 4:19
  • $\begingroup$ I don't think this is correct. The gauge transformation includes both adding a gradient to the 4-potential and changing the phase of the wave function. The mass squared, say, in the Klein-Gordon equation equals, roughly speaking, the square of the sum of the 4-momentum and 4-potential, so adding a constant to the electron's potential together with the change of the phase of the wave function does not change its mass. $\endgroup$
    – akhmeteli
    Jun 4 '20 at 1:27
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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).

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