# In simple words, why does a Lorenz Gauge does not have any physical effects?

I'm studying vector calculus via Arfken & Weber's "Mathematical Methods for Physicists", and, in page 40, he is deriving the electromagnetic wave equation.

During the demonstration he states that we can fix $$\nabla \cdot \mathbf{A} + \frac{1}{c^2} \frac{\partial \phi}{\partial t} = 0$$ Where $\mathbf{A}$ is the vector magnetic potential, $\phi$ is the non-static electric potential, and $c$ is the speed of light. He states that "this option for fixing the divergence of the vector potential, named Lorenz Gauge, is to unlink the equations of both the potentials. It has no physical effect".

The last phrase is not clear to me. How does fixing the value of a vector field when studying electromagnetic equations will not have any implications, physically speaking?

As for my background, i've never been introduced formally to gauge theory. I know the basics of Group Theory and a little bit of representation theory, if that helps. (I've heard that classical electrodynamics has a $U(1)$ symmetry, does that have anything to do with this?)

I don't believe the assertion of lack of physical effect refers specifically to the Lorenz gauge; rather it refers to the freedom to fix a gauge in in the first place. Any condition that can be fulfilled by a gauge transformation $A_\mu \mapsto A_\mu + \partial_\mu\,\psi$ for some $C^2$ scalar field $\psi$ is fair game: we solve our boundary problem for the 4-potential $A$ and then the physically measurable field is $F=\mathrm{d} A$. This latter $F$ is independent of the gauge because it is invariant under the gauge transformation.
The assertion is simply a statement of gauge symmetry, and that the fixing of the gauge cannot affect the measurable field $F$,
• @VitorCGoergen Sorry, I thought the problem might be that the wording suggested this statement was something specific to the Lorentz gauge (the text's wording is a bit ambiguous). In that case, what it means is that the only physically measurable field is $F$. Therefore, any transformation we can make on $A$ that does not affect $F$ is not physically observable. So we are free to make such changes to $A$. This is gauge symmetry - it is the redundancy in Maxwells equations when they are formulated in terms of $A$ rather than $F$. – Selene Routley Sep 25 '17 at 22:46
• Thanks! That is a lot clearer now. So, when we came the transformation $\mathbf{A} \rightarrow \mathbf{A} + \nabla \phi$, the $\nabla \phi$ is irrelevant to our "reality" because $\mathbf{B} = \nabla \times \mathbf{A} = \nabla \times (\mathbf{A} + \nabla \phi) = \nabla \times \mathbf{A} + \nabla \times \nabla \phi = \nabla \times \mathbf{A}$. Is that correct? I've also verified that $\nabla \phi$ also unchanges Stokes Theorem. For me, it seems like the idea of $\nabla \phi$ is similar to that of a constant of integration in real analysis. Thanks. It was very helpful. – embedded_dev Sep 26 '17 at 11:46