Existence of the Coulomb gauge In reading about the Coulomb gauge, my mind seems to have painted itself into a corner. For, lets assume that Maxwells equations for the physics of the problem are solved by the magnetic vector potential A´. We know that the magnetic vector potential (1) A = A´$-\nabla f$  will also be a solution, and we can use this gauge freedom to get (2) $\nabla\cdot$A$ =0$. Plugging (1) into (2), we see that we must have (3) $\nabla\cdot$A´$=\nabla^2f $. Or $\nabla\cdot$(A´)$=\nabla \cdot (\nabla f $). Lets say we find an $f$ that satisfies (3). Certainly, we dont want to have it satisfying (4) $\nabla f =$ A´? Because then we would have, by (1) and (4), A = A´$-\nabla f = $$\nabla f-\nabla f = 0$. But what condition would $\nabla f $ satisfy?
 A: Not really sure what your saying, but I'll try my best.
$\nabla \cdot \vec{B} = 0$
Which allows us to write
(1) $\nabla × \vec{A} = \vec {B}$
Consider the transformation
$\vec{A}->\vec{A'}$
Where,
(2) $\vec{A'} = \vec{A} + \nabla f$
Applying the transformation and taking the curl
$\nabla × \vec{A'} $
$\nabla × (\vec{A} + \nabla f) $
$\nabla × \vec{A} $
Which we know equals $\vec{B}$
Thus a gauge transformation that leaves the field invariant is
$\vec{A'} = \vec{A} + \nabla f$
Suppose further, that I perform a gauge transformation as described above, and I would ALSO like this new potential to satisfy the condition that:
(3) $\nabla \cdot \vec{A'} = 0$
How would I find the correct function f that conscribes the required condition above?
Well, we know what the form $\vec{A'}$ must take to keep the $\vec{B}$ field invariant.
Substituting (2) into (3) gives:
$\nabla \cdot (\vec{A} + \nabla f) = 0$
$\nabla \cdot \vec{A} + \nabla^2 f = 0$
$\nabla^2 f = -\nabla \cdot \vec{A}$
All that I have done so far is prove that there exists some function f, that leaves the field invariant, whilst also following the coulomb gauge. Meaning in the potential formulation of maxwells equations, I can use the coloums gauge.
A: Be careful, because $\nabla·\mathbf{A'} = \nabla·(\nabla f)$ doesn't necessarilly mean that $\mathbf{A'} = \nabla f$ as it can also mean that $\mathbf{A'} = \nabla f \ +\ \nabla \times \mathbf{F}$, being $\mathbf{F}$ an arbitrary vector since:
$$\nabla · \mathbf{A'} = \nabla · (\nabla f+\nabla \times\mathbf{F})$$
$$\nabla · \mathbf{A'} = \nabla · (\nabla f)+\nabla · (\nabla \times\mathbf{F})$$
$$\nabla · \mathbf{A'} = \nabla · (\nabla f)\ \forall \ \mathbf{F} \ \epsilon \ \mathbf{R}^3$$
as a property of vector calculus. So, as
$$\mathbf{A'} = \nabla f \ +\ \nabla \times \mathbf{F}$$
to find an $f$ that satisfies the Coulomb gauge you have that
$$ \nabla f  =\mathbf{A'} \ - \ \nabla \times \mathbf{F}, \ \forall \ \mathbf{F} \ \epsilon \ \mathbf{R}^3$$
where $\mathbf{A} \rightarrow \mathbf{A'}-\nabla f$
