Just a quick sanity check here:

I'm preparing a tutorial for a class on classical electrodynamics and I wanted to show an example of a gauge condition which leads to a contradiction, so I simply thought of $$(A^\mu)' = A^\mu - \partial^\mu \chi \stackrel{!}{=} 0\quad \forall \mu \in \{0,1,2,3\}\qquad (1)$$ For the 0th component this gives me the Weyl gauge ($\phi = 0$), which is allowed since I will always find a $\chi$ satisfying $$\phi = \frac1c \partial_t\chi.$$ I do, however, have a contradiction in the spatial components, since $$\vec{A} = \vec{\nabla}\chi$$ is only true for $\vec{A}$ with $\vec{\nabla}\times\vec{A}=0$ (on a simply-connected region), so if a given vector potential has non-vanishing rotation, I won't be able to find a $\chi$ so that the transformed field vanishes in all components.

Does anyone have some other examples?

(1) - As mentioned by QMechanic: Here are actually four gauge-fixing conditions disguised as one; this is unusual since common gauges (e.g. $\vec{\nabla}\cdot\vec{A}=0$) generally eliminate only one degree of freedom.


Well $F_{\mu\nu}$ is gauge invariant, so you can't gauge it away. The other reason is to remember that the photon has two polarizations (this is really one and the same, as you can show two gauge redundancies), so you can't set more than two components of $A_\mu$ = 0. In short, if you start with the Lorentz gauge, $\partial_\mu A^\mu = 0$, you can further find a solution to $\frac{\partial^2\phi}{\partial t^2} = \nabla^2 \phi$ in order to gauge the scalar field away.

| cite | improve this answer | |
  • $\begingroup$ Oh right - the thing about $F_{\mu\nu}$ is kind of obvious, I'll use that :) Thanks! $\endgroup$ – Wojciech Morawiec Nov 28 '13 at 11:03
  • $\begingroup$ Incidently, 1+1 dimensional electrondynamics is non-dynamical. $\endgroup$ – lionelbrits Nov 28 '13 at 11:29

OP's proposal $A_{\mu}=0$ are four$^1$ gauge-fixing conditions (in 4D), not just one.

Usually in EM we only specify a single gauge-fixing condition $\chi=0$, such as, e.g.

  • temporal gauge: $A_0=0$;

  • axial gauge: $\vec{n} \cdot \vec{A}=0$, where $\vec{n}$ is some constant 3-vector;

  • Coulomb gauge: $\vec{\nabla} \cdot \vec{A}=0$;

  • Lorenz gauge: $d_{\mu}A^{\mu}=0$;

etc. This is because we only have one gauge symmetry $\delta A_{\mu}=\partial_{\mu}\Lambda$ in EM.

Geometrically, a gauge-fixing condition is supposed to intersect each gauge-orbit precisely once. See also Gribov ambiguity.

A bit more general (e.g. in non-Abelian Yang-Mills theory), if we have $m$ gauge symmetries, we would need $m$ gauge-fixing conditions.

For a more general gauge theory (the generators of) the gauge symmetries are not necessarily independent. In that case we get (a new level of) gauge symmetry among the original gauge symmetries. We say that such theory has reducible gauge symmetry, see e.g. Ref. 1. The corresponding gauge-fixing procedure for reducible gauge symmetry becomes quite technical, and is most systematically developed via the Batalin-Vilkovisky formalism.


  1. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.


$^1$ Strictly speaking, each condition $\chi(x)=0$ in field theory (FT) is implemented in each space-time point $x^{\mu}$. Since space-time naively has $\infty^4$ many points, $\chi(x)=0$ is then $1 \times \infty^4$ conditions. In that spirit, OP's proposal $A_{\mu}(x)=0$ then is $4 \times \infty^4$ conditions. Often in FT one can get away in counting $\chi=0$ as just one condition (and ignoring the $\infty^4$ from the spacetime volume factor.) We will assume that here.

| cite | improve this answer | |
  • $\begingroup$ Thanks for the comment on my notation, I clarified that in the post. It is actually a very interesting insight that the usual gauge-fixing conditions eliminate only one degree of freedom, where I want to eliminate all four of them - do you have any examples of incorrect (i.e. leading to contradictory or not generally valid statements) gauge-fixing conditions that eliminate only one degree of freedom? Regarding your list: Could you clarify what you mean by $\vec{n}$? $\endgroup$ – Wojciech Morawiec Nov 28 '13 at 17:41
  • $\begingroup$ $\vec{n}$ is just some constant 3-vector, e.g. $\vec{n}=(0,0,1)$. $\endgroup$ – Qmechanic Nov 28 '13 at 17:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.