What is a non-wave zone for an EM waves? I'm currently reading about the Zel'dovich amplification effect.Long story short: an axially-symmetric rotating body inisde a resonator cavity can amplify certain modes of an incident EM radiation, at the expense of loose its rotation energy.
The condition to reach this amplification is effect is that:
$$
\Omega \ell > \omega
$$
where $\omega$ is the frequency of the EM radiation, $\ell$ its orbital angular momentum and $\Omega$ the angular velocity of the body.
However both his original articles (http://www.jetpletters.ac.ru/cgi-bin/articles/download.cgi/1604/article_24607.pdf, http://jetp.ac.ru/cgi-bin/dn/e_035_06_1085.pdf, ) cite about the fact that the body is entirely in the non-wave zone, described as tge place where the fields of the wave decrease like high powers of $r^\ell$. I don't have any idea of what that region is.
I've tried considering that, calling $r_0$ the body radius we have:
$$
\begin{cases}
v_{body, z} = \Omega r_0 = \beta c \\
\beta < 1\\
\end{cases}
\longrightarrow r_0 < \frac{c}{\Omega} < \frac{n \ell}{\omega} = \frac{n}{k}\,.
$$
Then I've tried to put this inequality in the wave equation. To do so I've considered that the EM wave has a general form:
$$
A \exp(i \vec{k}\cdot \vec{x} - i \omega t + i \ell \phi)
$$
And then using the plane wave expansion in spherical armonics:
$$
e^{i\vec{k}\cdot\vec{r}} = \sum_{\ell = 0}^{\infty} (2\ell + 1) i^\ell j_{\ell}(kr) P_\ell(\cos\theta)) 
$$
and then consider the fact that the bessel function (that may describe the amplitude of my incoming wave) have a dependence of:
$$j_l(kr)= \sum_{m=0}^\infty \frac{\sqrt{\pi} (-1)^m}{m! \Gamma(m + l + 3/2)} (\frac{k r}{2})^{\ell + 2m}$$
Then the amplitude goes at least as
$$
A_{tot} \propto (k r)^\ell 
$$
or higher powers. But placing inside the inequality obtained before I obtain that:
$$
A_{tot} \propto (kr)^\ell < (\ell)^\ell
$$
and here I see no decreasing, since $\ell$ is an integer.
Why does the article state that this region is where the field decrease? I am missing something?
 A: Just have a look at Near and Far field. Roughly speaking, when one computes the radiation field associated to an accelerating charge one has to consider that the field reaching the test point (point of measurement) was produced in the past, since it takes a while for the field to travel from the charge to the test point. That implies one has to solve Maxwell's equation for a retarded position. (You can find the details in any chapter on radiation in standard textbooks such as Griffiths). So the situation is more complicated and even more so when the body has a non-negligible size.
Probably the point is better illustrated by the following arguments. Consider the electric field generated by a particle which is being accelerated.
The field will have an expression that can be decomposed in different terms according to how strongly they fall with distance. One will have for example under certain assumptions for this case of a point charge particle following some trajectory, $\vec{x}(t)$:
$$\vec{E}(\vec{r},t) \propto \frac{|r'|}{(\vec{r}'\cdot \vec{u})}\left[ (c^2 -v^2) \vec{u} + \vec{r}'\times(\vec{u}\times\vec{a})\right]$$
where $\vec{r}'=\vec{r}-\vec{x}(t_r)$, $|\vec{r}'|=c(t-t_r)$ and $\vec{u} = c \vec{r}'/|\vec{r}'| - \vec{v}$. Where $t_r$ stands for retarded time and is defined implicitly.
As you can see the field it self reduces to the static case if the velocity and acceleration are taken to zero. Now you can see that the first term falls off as inverse square of the distance and the second one as inverse power of distance. These are the origins of the near and far field contributions.
To make contact with your approach, notice that you are assuming the vector potential will already have the plane wave form. A general solution accounting for the complexity of the situation would need to be an expansion in plane waves, or an expansion in spherical harmonics with a radial function. You will end up with a multi-pole expansion in such a case and the correct fall-off behavior.
