# Transverse current equivalence in Coulomb gauge

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.

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When you say you checked some books, have you checked Jackson's Classical Electrodynamics? :) –  Lagerbaer Jun 8 '11 at 23:36
Indeed :D In Jackson's book I found a problem concerning my doubt (6.20 in the 3rd edition). After many calculations one finds that only perpendicular currents affect to pure radiation. Jackson doesn't comment this, and it surprises me because if you carry out the calculations this result is not at all obvious to get from the start (at least for me). He doesn't say anywhere I looked that this property can be generalised, and that's what I want to know. –  Daniel Jun 9 '11 at 8:32
All my problems come from the fact that I found in an article that the transverse current could be replaced at large distances by a perpendicular current, and that could be shown. The result they get is, moreover, correct. However, I asked one of the authors where could I see the proof, and he had no idea :S –  Daniel Jun 9 '11 at 8:47

## 1 Answer

Ok, I'll answer my own question, since I've just found a solution thanks to Mr. Jackson. We can find in this article: http://arxiv.org/pdf/physics/0204034v2, that the equation (3.16) implies that every field with dependence $1/r$ is created by a current that is transverse to the direction of observation.

$\vec{A}_{Coulomb-radiation} = \frac{\mu}{4\pi}\int \frac{\textrm{d}^2 x'}{r} [\vec{J}(\vec x',t')-\hat r(\hat r \cdot \vec{J}(\vec x',t')]_{retarded}$

And precisely the current present in the integral is the perpendicular component to the line of sight.

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