# A question on gauge fixing

As I understand it, a physical theory that has a gauge symmetry is simply one that has redundant degrees of freedom in its description, and as such, is invariant under a continuous group of (in general) local transformations, so-called gauge transformations.

With this in mind, consider electromagnetism as a prototypical example. This has a $U(1)$ gauge symmetry, such that the theory is invariant under transformations of the vector four-potential, $A^{\mu}$ $$A^{\mu}\rightarrow A'^{\mu}=A^{\mu}+\partial^{\mu}\Lambda(x)$$ where $\Lambda(x)$ is some local function of space-time coordinates. Is it then correct to say that the theory is described by an equivalence class of vector four-potentials, $A^{\mu}$ such that $$A'^{\mu}\cong A^{\mu}\iff A'^{\mu}=A^{\mu}+\partial^{\mu}\Lambda(x)$$ Given this, does "choosing a gauge " simply amount to choosing a particular four-potential $A^{\mu}$ from this equivalence class?

Furthermore, does "fixing a gauge" simply amount to specifying some constraint on the choice of $A^{\mu}$ such that it "picks out" a single four-potential $A^{\mu}$ from this equivalence class?

For example, is it correct to say that choosing the Lorenz gauge $\partial_{\mu}A^{\mu}=0$ partially removes gauge freedom, since it restricts ones choice of four-potential $A^{\mu}$ such that it satisfies $\partial_{\mu}A^{\mu}=0$, however, it doesn't fully fix ones choice of $A^{\mu}$, i.e. it doesn't fully "fix the gauge", since there remains a subspace of gauge transformations that preserve this constraint, corresponding to gauge functions $\psi$ that satisfy the wave equation $\partial_{t}^{2}\psi=c^{2}\nabla^{2}\psi$?!

Yes, all those things are correct.

The equivalence class of potentials that are related by gauge transformation is called a gauge orbit, since it is an orbit for the action of the group of gauge transformations on the space of potentials.

Choosing/fixing a gauge means picking out particular representants $$A$$ from each gauge orbit according to a rule encoded by $$F[A] = 0$$ for some functional $$F$$, i.e. you select those potentials which fulfill the equation. A partial gauge fixing is indeed given by things like $$F[A] = \partial_\mu A^\mu$$, for which $$F[A] = 0$$ has more than one solution in a given orbit. These solutions are related by gauge transformations with harmonic parameter functions, and these transformations are called residual gauge symmetry.

In general, it is not possible to fix a gauge that choose only one representant from every orbit. This is known as the Gribov problem.

• I was about to ask a question similar to OP's one. But your answer clarified thing. Just one question. How can I know that a Gauge fixing condition is a suitable one? For example: $\partial_\mu A^\mu=0$, the Lorentz gauge. How do I know this sort of gauge fixing is allowed, given my gauge symmetry $A_\mu \rightarrow A_\mu+\partial_\mu \alpha$? – Luthien Mar 10 '17 at 3:05
• @Luthien To "know" a gauge fixing is allowed, you must show that, starting from arbitrary $A$, you can find an $\alpha$ such that $A_\mu +\partial_\mu \alpha$ fulfills the gauge condition. I.e. you take $F[A+\partial_\mu \alpha] = 0$, treat $A$ as given and show that a solution for $\alpha$ exists. – ACuriousMind Mar 10 '17 at 12:53

"does "fixing a gauge" simply amount to specifying some constraint on the choice of $A^\mu$ such that it "picks out" a single four-potential $A^\mu$ from this equivalence class?" Generally speaking, not quite. For example, if you consider, say, interacting Maxwell and Dirac fields, you have to change the phase of the Dirac field when you change the gauge for the Maxwell field to provide gauge invariance.

• How in practice do you change the phase of the Dirac field? For example, you decide to work with the Coulomb gauge for EMF. What do you do with the Dirac equation solutions? – Vladimir Kalitvianski Dec 7 '16 at 16:13
• @VladimirKalitvianski: See, e.g., hep.phy.cam.ac.uk/theory/webber/GFT/gft_handout4_06.pdf, pp.2-3. I am sure you can sort out the specifics for the Coulomb gauge. – akhmeteli Dec 7 '16 at 16:25
• What I know is that people just solve the Dirac equation whatever gauge fixing they do. – Vladimir Kalitvianski Dec 7 '16 at 17:37
• @VladimirKalitvianski: That is correct. But then the solutions that they get for the Dirac field (the 4-spinor wave function) are adapted to the gauge they chose for the 4-potential of the electromagnetic field, not for any other gauge. Note that, say, the Lagrangian of interacting Dirac and Maxwell fields is only gauge invariant if you simultaneously transform both fields. – akhmeteli Dec 7 '16 at 17:47
• But in practice we do not transform explicitly the Dirac field. The Dirac solutions are different for different gauges and they connected with the gauge transformation for sure, but we never transform them ourselves. – Vladimir Kalitvianski Dec 7 '16 at 17:52