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?
1 Answer
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.
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$\begingroup$ 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. $\endgroup$– DarkeninJun 12, 2020 at 18:09
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$\begingroup$ 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. $\endgroup$ Jun 12, 2020 at 18:15