# Uniqeness of Coulomb gauge

Say I have some magnetic vector potential $$A$$ which is not in Coulomb gauge, meaning $$\nabla \cdot A \neq 0$$. I can set it to Coulomb gauge by adding some scalar potential function $$\nabla \phi$$ (and that's okay because the rotor - the magnetic force, will stay the same). My questions is whether there is only one unique potential function we can add to $$A$$ to make it in Coulomb's gauge. It doesn't seem right to me that there is only one way to do it, is that so? If I'm wrong, why?

You can find the answer just by computing the gauge transformation. You want to look for a escalar field $$\phi$$ whose gradient, added to the vector potential $$\vec{A}$$ makes a new vector potential, say $$\vec{A}'$$ which satisfies the Coulomb gauge: $$\nabla \cdot \vec{A}'=0.$$ Now, plug in the expression of $$\vec{A}'$$ $$\nabla \cdot \vec{A}'=\nabla \cdot \left( \vec{A}+\nabla\phi\right)=\nabla\cdot \vec{A}+\nabla\cdot (\nabla\phi),$$ for the divergence operator is linear. Now, you can identify the laplacian ($$\nabla^2\equiv\Delta$$) of $$\phi$$ in the right-hand side of the equation. According to the Coulomb gauge, this quantity equals zero, then $$\nabla \cdot \vec{A}'=\nabla \cdot \vec{A}+\Delta\phi=0 \Rightarrow \Delta\phi=-\nabla\cdot \vec{A}$$ So as you expected, the choice of $$\phi$$ isn't unique, and will depend on the form of the magnetic vector potential. You can look further on the resolution of this problem as it is a very well-known problem called Laplace-Poisson's problem. You can also verify that the result I posted is true, for example try to do problem 12.3 of Griffith's Introduction to Electrodynamics.
• Thank you, that's the exact process I have done. Upon reaching the $\Delta \phi = -\nabla \cdot A$ part It looks like I can make many choices. I can for example try to find $\phi (z)$ or $\phi (\theta)$ that will both fulfill this equation. Are they both correct? Can you expand on the "will depend on the form of the magnetic vector potential" part? I'm familiar with Laplace-Poisson's equations but I don't see any boundary conditions I need to fulfill. Jun 12, 2020 at 18:09
• By saying that I'm assuming that you know the expression of the vector potential, so you actually have a Poisson's equation to solve. BCs are also influenced by this. In general, I couldn't give you any precise answer. In principle, any choice of the field $\phi$ is valid if it fulfills the previous condition. Jun 12, 2020 at 18:15