I have a question concerning transverse (solenoidal) current in the Coulomb gauge. This current is the one that enables the radiation, since we have a wave equation for the vector potential:
$\nabla^2 \vec A - \mu_0\epsilon_0\frac{\partial^2\vec A}{\partial t^2} = -\mu_0\vec{J}_{\perp}$
Where the transverse current can be written as
$J_\perp = \nabla \times \nabla \times\int\frac{\vec J(\vec x')\textrm{d}^3x'}{4\pi|\vec x - \vec x'|}$
This transverse current extends to all space. We can solve the wave equation using Green's function.
$\vec A(\vec x, t) = \frac{\mu_0}{4\pi}\int\frac{\vec J_\perp(\vec x', t')}{|\vec x - \vec x'|} \delta(|\vec x- \vec x'|/c-(t-t'))\textrm{d}^3x'\textrm{d}t'$
The point is that, when we want to keep only terms that will result in fields falling like $1/r$, neglecting $1/r^2$ and so on, it seems that we don't have to consider the complicated expression for the transverse current. It seems to be enough in some cases to keep the projection of the current perpendicular to the observation vector $\vec x$ and derivate the vector potential with respect to $t$, though this is not obvious, as $\vec J_\perp$ covers all the space.
I checked two different cases -an instantaneous dipole at the origin and an instantaneously accelerating point charge, and the field that goes as $1/r$ is created, as expected, by currents transverse to the line of sight. However, I do not have a rigorous explanation to ignore the non-perpendicular currents in the integral for $\vec A$. I've asked my boss, checked some books and searched in the Internet, but I couldn't find any hint or proof. Do you know of any place? Thank you very much.
