Why does near-field attenuate at $\frac{1}{r^6}$?

Question

So far-field makes intuitive sense to me, it attenuates at $\frac{1}{r^2}$ much like gravity. This is just common sense since the area for the surface of a sphere relative to its radius has a $r^2$ relationship.

However the near-field attenuates at a rate of $\frac{1}{r^6}$ instead. Is there some equally intuitive explanation for this? How is this derived?

The above numbers can be confirmed from the following Wikipedia article. To quote the article:

According to Maxwell's equation for a radiating wire, the power density of far-field transmissions attenuates or rolls off at a rate proportional to the inverse of the range to the second power ($\frac{1}{r^2}$) or −20 dB per decade. This slow attenuation over distance allows far-field transmissions to communicate effectively over a long range. The properties that make long range communication possible are a disadvantage for short range communication systems.

NFMI systems are designed to contain transmission energy within the localized magnetic field. This magnetic field energy resonates around the communication system, but does not radiate into free space. This type of transmission is referred to as "near-field." The power density of near-field transmissions is extremely restrictive and attenuates or rolls off at a rate proportional to the inverse of the range to the sixth power ($\frac{1}{r^6}$) or −60 dB per decade.

Answer 1

This answer overlaps with the answer by Roger Vadim, which quotes from this Wikipedia article, but I had to read his answer more than once before I understood, and my clarification outgrew the comment box.

If you have a function which diverges at the origin and vanishes at infinity, a Taylor-expansion-ish thing to do is to expand in powers of $1/r$:

$$ f(\omega,t,r) = \frac{a_1(\omega,t)}r + \frac{a_2(\omega,t)}{r^2} + \cdots $$

So at large distances you only care about $a_1$, but there may be some intermediate distance where $a_2$ starts to “win,” while at even closer distances the “winner” becomes $a_3$, then maybe $a_4$, and so on.

For antennas, this expansion is done for the field amplitudes. The power density is proportional to the square of the amplitude, and is therefore expanded in even powers of $r$. The statement that “the near field power varies like $r^{-6}$” is approximately equivalent to “we can usefully describe this antenna keeping only the first three terms in the multipole expansion of the field.”

Answer 2

This is not necessarily the case. For example, Wikipedia article Near and Far Field states:

The separation of the electric and magnetic fields into components is mathematical, rather than clearly physical, and is based on the relative rates at which the amplitude of different terms of the electric and magnetic field equations diminish as distance from the radiating element increases. The amplitudes of the far-field components fall off as $1/r$, the radiative near-field amplitudes fall off as $1/r^{2}$, and the reactive near-field amplitudes fall off as $1/r^{3}$. Definitions of the regions attempt to characterize locations where the activity of the associated field components are the strongest. Mathematically, the distinction between field components is very clear, but the demarcation of the spatial field regions is subjective. All of the field components overlap everywhere, so for example, there are always substantial far-field and radiative near-field components in the closest-in near-field reactive region.

Note that here they are talking about a dipole antenna, i.e., about cylindrical symmetry, which is why it decreases as $1/r$ rather than $1/r^2$, although one could give the same argument as in the OP.

Overall one can rigorously define only the far field, as the slowest decaying component of the field. The rest is a matter of convention.