4
$\begingroup$

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 $\nabla^2 \lambda = -\nabla \cdot \mathbf{A}$.

$\endgroup$

marked as duplicate by Kyle Kanos, Dvij Mankad, tparker, Rob Jeffries, ZeroTheHero Mar 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.

2
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.