# Why does Lorenz gauge condition $\partial_\mu A^\mu =0$ pick exactly one configuration from each gauge equivalence class?

For a vector field $$A_\mu$$, there are infinitely many configurations 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 $$\begin{equation} \partial_\mu A^\mu =0 \, . \end{equation}$$ 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?

The Lorenz gauge condition does not fix the gauge completely.

Let $$A^\mu$$ be a field satisfying the Lorenz gauge condition $$\partial_\mu A^\mu = 0$$. Given a scalar function $$f$$, let $$B^\mu = A^\mu + \partial^\mu f$$. $$B^\mu$$ can also satisfy the Lorenz gauge condition if

$$\partial_\mu B^\mu = \partial_\mu\partial^\mu f = 0,$$ i.e. if $$f$$ is the Minkowski space equivalent of a harmonic function. Therefore it is not true that the condition picks exactly one function per gauge-equivalence class. In each class, there is a whole (non-trivial) vector space of functions satisfying the Lorenz gauge.

• +1, although $\partial_\mu \partial^\mu f = 0$ isn't the usual definition of harmonic function, but instead $\partial_i \partial^i f = 0$, where $\partial_i \partial^i$ represents the (definite-signature) Laplacian operator. – tparker Sep 27 '19 at 15:46
• thanks @tparker, I edited the answer (feel free to improve the wording) – fqq Sep 27 '19 at 17:07
• @tparker Not really. $\partial_\mu\partial^\mu f=0$ is the usual definition of a harmonic function. A harmonic function is one such that $\Delta=0$, for whatever $\Delta$ you are working with. In particular, for a metric $\eta_{\mu\nu}$, $\Delta=\partial_\mu \partial^\mu$. The metric being positive definite is not a standard requirement in the definition of harmonic. – AccidentalFourierTransform Sep 28 '19 at 12:14
• @AccidentalFourierTransform Wikipedia, Mathworld, the Encyclopedia of Math, Encyclopedia Britannica, Axler Bourdon and Ramey's "Harmonic Function Theory", Gamelin's "Complex Analysis", and everyone whom I've ever heard use the term all define a "harmonic function" as obeying Laplace's equation with the Euclidean metric. I think that generalizing it to include any function annihilated by the Laplace-Beltrami operator is nonstandard usage. – tparker Sep 28 '19 at 14:18
• @AccidentalFourierTransform Some sources give an "equivalent definition" of a harmonic function that at every point the function attains its average value over any hypersphere centered at that point, which implicitly assumes Euclidean space (or at least a Riemannian manifold). – tparker Sep 28 '19 at 14:23

There isn't exactly true: the condition you state doesn't uniquely fixes the vector potential, in fact you have what's called residual gauge freedom, which means that with the condition $$\partial_\mu A^\mu = 0$$ you haven't completely fixed the gauge. Let's prove this.

Suppose you have $$A^\mu$$ such that $$\partial_\mu A^\mu=0$$ then you have infinite Vector potentials which satisfy that relationship and are related by the following gauge transformation: $$A^\mu \rightarrow {A'}^\mu = A^\mu + \partial^\mu \theta$$. Where $$\theta$$ is an armonic function, that is it satisfies $$\Box \theta = \partial_\mu \partial ^\mu \theta=0$$

Therefore you can see that the potential $$A'$$ satisfies your gauge fixing condition.

As the other answers point out, Lorentz gauge is actually just a partial gauge fixing that leaves residual degrees of freedom. The remaining gauge freedom is what is known as a Gribov ambiguity. To fully specify a gauge, you need to additionally specify enough boundary conditions to fix a particular solution to the wave equation for your transition function $$\eta$$. For example, if all the sources vanish sufficiently far back in the past, then the usual causal formula for the gauge field in terms of retarded potentials sourced by sources on the past light cone comes from the additional gauge-fixing condition that the gauge field also vanish sufficiently far back in the past.

The wave equation, $$\partial_\mu \partial^\mu A^\nu = - j^\nu/ \epsilon_0$$ for the potential implies a unique , bijective relation between source and potential. The source term is restricted to be conserved. The image of this is a relation between the field components, the Lorenz condition. The Gribov ambiguity is dealt with by requiring that the potential of a source element is zero outside its light cone. I published this in a peer reviewed journal and the paper can also be found at https://arxiv.org/abs/physics/0106078.