The Far Zone

Exactly the opposite is true in the far zone. Here $kr » 1$ and the exponential oscillates rapidly. We can approximate the argument of the exponential as follows:

$$\vert \mbox{\boldmath$x$}- \mbox{\boldmath$x$}' \vert = \sqrt{r^2 + r'^2 - 2 r{\bf n} \cdot \mbox{\boldmath$x$}'}$$

$$= r \left\{ 1 - \frac{2}{r} {\bf n} \cdot \mbox{\boldmath$x$}' + \frac{r'^2}{r^2} \right\}^{1/2}$$

$$= r - {\bf n} \cdot \mbox{\boldmath$x$}' + {\cal O}\left( \frac{1}{r} \right)$$ (11.76)

where we have assumed that $r'_{\max} < d « r$ and used a binomial expansion of the root sum. We neglect higher order terms. Note that this approximation is good independent of k and may be good even in the near zone.

Then

$$\lim_{ (k)r \rightarrow \infty} vA(\mbox{\boldmath$x$}) = \f... ...k\hat{\mbox{\boldmath$n$}} \cdot \mbox{\boldmath$x$}'} d^3x'.$$ (11.77)

In the far zone, the solution behaves like an outgoing spherical wave times an amplitude that depends on integral over the source that depends on angles in an intricate fashion.

At this point I could continue and extract

$$\lim_{ (k)r \rightarrow \infty} vA(\mbox{\boldmath$x$}) = \f... ...hat{\mbox{\boldmath$n$}} \cdot \mbox{\boldmath$x$}')^n d^3x'$$ (11.78)

(if the source is actually small enough to allow expansion of the exponential in a series^11.5). This would give us a cheap introduction into multipoles. But it is so sloppy!

Instead we are going to do it right. We will begin by reviewing the solutions to the homogeneous Helmholtz equation (which should really be discussed before we sweat solving the inhomogeneous equation, don't you think?) and will construct the multipolar expansion for the outgoing and incoming (and stationary) wave Green's function. Using this, it will be a trivial matter to write down a formally exact and convergent solution to the integral equation on all space that we can chop up and approximate as we please. This will provide a much more natural (and accurate) path to multipolar radiation. So let's start.