Meaning of Sommerfeld radiation conditions

EmilyRuck

Hello!
Promising that I will not make other new questions in the next days , I have a doubt about the meaning of a pair of expressions.
Sommerfeld's conditions for an electromagnetic field produced by a finite source bounded by a finite volume are:

[itex]\lim_{r \to +\infty} r|\mathbf{E}| < q\\
\lim_{r \to +\infty} r|\mathbf{H}| < q\\
\lim_{r \to +\infty} r \left[\mathbf{E} - \eta \mathbf{H} \times \mathbf{\hat{u}}_k \right] = 0\\
\lim_{r \to +\infty} r \left[\mathbf{H} - \displaystyle \frac{\mathbf{\hat{u}}_k \times \mathbf{E}}{\eta} \right] = 0[/itex]

where [itex]q[/itex] are finite quantities, [itex]\eta[/itex] is the wave impedance in the considered medium (for example the free space), [itex]\mathbf{\hat{u}}_k[/itex] is the direction of propagation and [itex]r[/itex] is the distance from the source.
The first two state that the fields' module must decrease at least as [itex]1/r[/itex].
The last two state that the fields must be similar to those of a plane wave: they must be mutually orthogonal and also both orhogonal to the direction of propagation. Moreover, the "part" of [itex]\mathbf{E}[/itex] (in the first) and [itex]\mathbf{H}[/itex] (in the second) which does not contribute to that plane wave must decrease at least as [itex]1/r^2[/itex].

Why these last two conditions are called "radiation conditions"? As a matter of fact, in the electric dipole non-radiative field components decrease as [itex]1/r^2[/itex] or [itex]1/r^3[/itex]. But why this is a necessary requirement to build a "radiation"? Couldn't we have a not-radiating component which decreases as [itex]1/r[/itex]? Why?
Thank you, again, for having read.

Emily

mfb

Every component which decreases with 1/r looks like radiation, and therefore it is called radiation.