# Why can we pick the divergence of the vector potential? [duplicate]

I'm aware that the vector and scalar potential in E&M can be modified using a function $$\lambda(t)$$ in the following way:

$$\mathbf{A}' = \mathbf{A} + \nabla\lambda,\;\; \textrm{ and } \;\;\Phi' = \Phi - \frac{\partial\lambda}{\partial t}.$$

However, I'm confused when Griffiths claims that we can merely choose $$\nabla\cdot\mathbf{A} = 0$$ as in the Coulomb gauge, or $$\nabla\cdot\mathbf{A} = -\mu_0\epsilon_0\dfrac{\partial \Phi}{\partial t}$$ as in the Lorenz gauge. How is it that we know that such choices for $$\nabla\cdot\mathbf{A}$$ are valid? What are we allowed to choose for $$\nabla\cdot\mathbf{A}$$ and why?

Edit: This question Do we fix divergence of the vector potential $A$, because $\nabla \cdot \nabla \psi \ne 0$? doesn't explain why there's always a solution to $$\nabla^2 \lambda = -\nabla \cdot \mathbf{A}$$.

## marked as duplicate by Kyle Kanos, Dvij Mankad, tparker, Rob Jeffries, ZeroTheHeroMar 2 at 3:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – Kyle Kanos Feb 26 at 13:22
• @KyleKanos Maybe, but if it is a duplicate, I'm having trouble making sense of the answer. – PiKindOfGuy Feb 26 at 13:28
• Then you need to explicitly state in the question (an edit, not just a comment here) what it is about the duplicate you don't understand and/or how your question differs from the other. – Kyle Kanos Feb 26 at 13:31

## 1 Answer

If $$\phi$$ is the electric potential correspnding to an electric field $${\bf E}$$, then $$\nabla^2 \phi = -{\bf \nabla} \cdot {\bf E} = -\rho/\epsilon_0$$. We know that this can be turned around: given any charge density $$\rho$$, we can find a corresponding $$\phi$$ using the usual formulas.

You're asking why there's always a solution to $$\nabla^2 \lambda = -{\bf \nabla} \cdot {\bf A}$$ for any $${\bf A}$$. By analogy with the previous equation, we can define $$\chi := {\bf \nabla} \cdot {\bf A}$$. Then no matter what form $$\chi$$ takes, we can solve $$\nabla^2 \lambda = -\chi$$ by integrating the usual Coulomb $$1/r$$ potential times $$\chi$$ as if $$\chi$$ were a charge density.