Let's say I have an electromagnetics problem in a spatially varying medium. After I impose Maxwell's equations, the Lorenz gauge choice, boundary conditions, and the Sommerfeld radiation condition, I still have more unknowns than equations and the solution for (say) the magnetic vector potential is not uniquely determined by the above. This does actually happen when you formulate the equations in plane-stratified media in the plane-wave / Fourier domain. The way to proceed that I have seen is to choose that one of the cartesian components of the vector potential is zero, i.e. impose one of the following as an additional condition to ensure uniqueness of the potentials: $A_x=0$, $A_y=0$ or $A_z=0$. We could of course make up an infinite number of other conditions that leave the fields invariant.

My question is, can I call the above arbitrary choice about the vector potential a "gauge choice"? The reason for imposing it seems to be identical to the reasons we normally impose the Lorenz or Coulomb gauge, namely that the field equations don't dictate anything about certain potential quantities, the choice makes solving the equations uniquely possible, and the physical $\mathbf{E},\mathbf{H}$ fields are invariant to the extra condition on the potentials.


3 Answers 3


Setting $A_z = 0$ is usually called 'axial gauge'. (The choice of the $z$-direction is arbitrary, of course.) This is a perfectly valid gauge condition, and used a lot in QCD.

However, I don't think it makes sense in general to simultaneously impose Lorenz gauge and axial gauge. This is too many constraints; you'll kill off physical degrees of freedom. However, in specific situations, it may be that your boundary conditions are actually imposing these constraints, so accidentally imposing them a second time with the gauge condition causes no harm. It's hard to say more without knowing exactly what you're doing.

Is it possible that you are missing the secondary constraint implied by the gauge choice?

  • $\begingroup$ I read on Wikipedia (en.wikipedia.org/wiki/Gauge_fixing#Lorenz_gauge) that the Lorenz gauge is an "incomplete" gauge, meaning that it doesn't actually fix all the physical degrees of freedom. Something to the effect of adding a solution to the wave equation to the potentials leaves the fields invariant. I was hoping someone would shed light on exactly this...I can provide more detail about my multilayered medium problem if desired. $\endgroup$
    – rajb245
    Jul 26, 2013 at 13:58
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    $\begingroup$ It would be good to spell out your computation, so that we don't have to guess how you end up with too few constraints. $\endgroup$
    – user1504
    Jul 29, 2013 at 18:26
  • $\begingroup$ @rajb245 : It's called residual gauge. They are the class of (residual) gauge transformations that leave your gauge choice invariant. Sort of a gauge freedom within a gauge. In the Lorenz gauge you fix $\partial_\mu A^\mu = 0$. So if $A_mu\rightarrow A_mu = \partial_\mu f$ is a gauge transformation then your residual gauge must satisfy: $\partial_\mu^2 f = 0$. These transformations leave the Lorenz gauge intact. All $f$'s which satisfy this equation (+ boundary conditions) describe your residual gauge. $\endgroup$
    – Olaf
    Jul 30, 2013 at 13:29
  • $\begingroup$ Thanks for this. I'll ask a follow-on question describing how / why I have too few constraints. $\endgroup$
    – rajb245
    Aug 2, 2013 at 14:22

Per your comment to user1504, you are correct: the Lorenz gauge contains considerable arbitrariness.

To wit, suppose potentials $\Phi, \boldsymbol{A}$ satisfy:

$$ \nabla^2 \Phi - \frac{1}{c^2} \frac{\partial^2 \Phi}{\partial t^2} = - 4 \pi \rho $$

$$ \nabla^2 \boldsymbol{A} - \frac{1}{c^2} \frac{\partial^2 \boldsymbol{A}}{\partial t^2} = - \frac{4 \pi}{c} \boldsymbol{J} $$

$$ \boldsymbol{\nabla \cdot A} + \frac{1}{c} \frac{\partial \Phi}{\partial t} = 0 \quad \text{ (Lorenz condition)}$$

Then, if $ \Lambda $ is a scalar that satisfies the wave equation: $$ \nabla^2 \Lambda - \frac{1}{c^2} \frac{\partial ^2 \Lambda}{\partial t^2} = 0 $$ the restricted gauge transformation (restricted because of the condition on $\Lambda$): $$ \boldsymbol{A} \rightarrow \boldsymbol{A} + \boldsymbol{\nabla} \Lambda \quad, \quad \Phi \rightarrow \Phi - \frac{1}{c} \frac{\partial \Lambda}{\partial t}$$

preserves the Lorenz condition.

Caveat! This analysis applies in vacuum. Extension to a homogeneous, isotropic medium looks straightforward, but for a spatially varying background like you posit, I don't know...

  • $\begingroup$ It is the Lorenz, not Lorentz, gauge. It's not the same person as the Lorentz transformations. $\endgroup$
    – Muphrid
    Jul 30, 2013 at 3:36
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    $\begingroup$ Thank you @Muphrid, I stand corrected. Jackson uses "Lorentz" in his 2nd edition, but notes in his review paper that Lorenz used it much earlier. $\endgroup$
    – Art Brown
    Jul 30, 2013 at 3:48
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    $\begingroup$ And the wiki article on the Lorenz gauge points out it's Ludvig Lorenz, not Hendrik Lorentz. This is also what I was taught in my graduate EM class. $\endgroup$
    – Muphrid
    Jul 30, 2013 at 13:29

You can only make the gauge choice $A_z=0$ if for any given potential you can set it 0 as the result of a Gauge transformation, a transformation which doesn't change the Lagrangian (or rather the action - or Maxwell's equations in your question, which are the equations of motion of some Lagrangian). This is, yes, a more complicated way of saying we can change the potentials as long as the fields don't change. The symmetries of the Lagrangian form a Lie group, which for EM is U(1). The addition of the divergence of a scalar adds a boundary term to the action which vanishes for physical potentials.

For electromagnetism, it ends up that by adding scalar divergences we can kill off two degrees of freedom and be left with two (I'm not really sure how to prove this from knowing the Lie group, if someone could expand here that would be great). These are transverse in the radiation gauge, but we can kill them off any way we like. In matter, I'm pretty sure you can't just kill off the longitudinal DOF. There will still be only two DOF but a different mix of $\Phi,A_x,A_y$ and $A_z$, i.e. two constraints on the 4 DOF.


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