How is $\nabla \cdot A = 0$ solved?

Asked Aug 18 '22 at 00:34

Active Aug 18 '22 at 00:34

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I'm struggling to understand some aspects of a very basic gauge, the coulomb gauge.

So I am asking simpler questions.

If $\nabla \cdot A = 0$ ?

So $\frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} = 0 $

By Poincare's lemma, there should exist a G such that

$A = \nabla \times G$ (I think)

I believe there is a thing called the Helmholtz decomposition theorem that is supposed to help solve this, but to be honest, I would prefer to keep theorems and lemmas out of this.

Are these two facts enough to determine A? If so how does this proceed in the simplest possible way at a basic calculus level?

asked Aug 18 '22 at 00:34

Kevin Njokom

1

This is not enough, you need to specify the result of $\nabla \times \vec A$, with both the divergence and curl of $\vec A$ you could solve for it. – Diógenes Figueroa Aug 18 '22 at 01:04
Ah yes. Let me set $\nabla \times \vec{A}$ that to $\vec{B}$ and try. – Kevin Njokom Aug 18 '22 at 13:41
Then you can use Helmholtz Theorem to solve it yes. Is this the electromagnetic vector potential you are talking about? – Diógenes Figueroa Aug 18 '22 at 13:43
yes it is indeed the magnetic vector potential – Kevin Njokom Aug 18 '22 at 21:02

How is $\nabla \cdot A = 0$ solved?

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