Why does Coulomb gauge condition $\partial_i A_i =0$ pick exactly one configuration from each gauge equivalence class?

There are infinitely many configurations of a vector field $$A_\mu$$ that describe the same physical situation. This is a result of our gauge freedom $$A_\mu (x_\mu) \to A'_\mu \equiv A_\mu (x_\mu) + \partial_\mu \eta(x_\mu ),$$ where $$\eta (x_\mu)$$ is an arbitrary scalar function.

Therefore, each physical situation can be described by an equivalence class of configurations. All members within a given equivalence class are related by a gauge transformation. Configurations in different equivalence classes describe physically distinct situations and therefore are not related by gauge transformations.

To fix the gauge, we need to pick exactly one member from each such equivalence class. A popular way to accomplish this is by demanding $$$$\partial_i A_i =0 \, .$$$$ Apparently this works because there is only exactly one member in each equivalence class that fulfills this additional condition. How can this be shown and understood?

PS: I asked a very similar question recently, but made a typo in the gauge condition (Lorenz gauge instead of Coulomb gauge). The Lorenz gauge condition, of course, leaves a residual gauge freedom, while the Coulomb gauge is a physical gauge.

The Coulomb gauge actually also leaves residual gauge freedom, just as the Lorentz gauge does. This is another example of the Gribov ambiguity mentioned in my answer to the other question. In general, any gauge-fixing condition defined by a linear partial differential equation will have a Gribov ambiguity that corresponds to the kernel of the differential operator, unless you specify enough boundary conditions to pin down a unique gauge field configuration.

For example, in the case of Coulomb gauge, almost the exact same story holds as with Lorentz gauge: two distinct but physically equivalent gauge fields $$A_\mu$$ and $$A_\mu' = A_\mu + \partial_\mu \eta$$ are both in Coulomb gauge if (a) either one of them is and (b) the transition function $$\eta$$ satisfies the Laplace equation $$\partial_i \partial_i \eta = \nabla^2 \eta \equiv 0$$ for all times, so that $$\eta$$ is a harmonic function. Since there are infinitely many harmonic functions on $$\mathbb{R}^n$$, there are infinitely many gauge fields in Coulomb gauge that correspond to a given electromagnetic field configuration.

Just as before, the solution is to impose suitable boundary conditions to pin down the remaining gauge freedom. In situations where all of the sources are confined to a finite spatial region, the natural boundary condition to impose is that the gauge fields go to zero at spatial infinity. This leads to the usual Biot-Savart-law- and Coulomb's-law-like formulas for the gauge fields in terms of the instantaneous sources with a $$1/r$$ spatial falloff. But in situations where the sources are infinitely extended, there isn't always a unique natural gauge-fixing choice, and you need to just arbitrarily pick one. For example, for an infinitely long uniformly charged wire, you need to just pick an arbitrary reference distance at which the electric potential becomes zero.

• Thanks for your answer. My problem, however, is in understanding why a condition like the Coulomb condition is useful at all. We can translate it into an equation for the gauge function $\eta(x_\mu)$ and thus it helps us to choose one specific gauge function. But why is choosing a gauge function useful in getting rid of the redundancy? Even if we choose one specific gauge function $\eta(x_\mu)$, there are still infinitely many equivalent configurations of $A_\mu$ because we haven't specified which $A_\mu$ we start with.
– jak
Sep 28, 2019 at 10:44
• To be more specific: Let's say we have a solution of the equation of motion $A_\mu$ and another configuration $A'_\mu$ that is related to $A_\mu$ by a gauge transformation. If we now choose one specific gauge function $\eta$, we still have no idea whether we should use $A_\mu +\partial_\mu \eta$ or $A'_\mu +\partial_\mu \eta$ and thus the gauge redundancy is still there. After all, we could equally say that $A'_\mu$ is our original solution of the equation of motion and $A_\mu$ a gauge transformed version of it.
– jak
Sep 28, 2019 at 10:47
• I reformulated my problem in terms of a new question: physics.stackexchange.com/questions/505179/…
– jak
Sep 28, 2019 at 11:18