Near and far field EM radiation

Question

When we talk about radiation, we often talk about the near and far field radiation. For the near field, the Poynting vector falls off like $1/r^3$ and the far field Poynting vector falls off like $1/r^2$. Once the distance gets big, the far field term dominates, hence the name. Now, it is the far field term that is associated with the EM radiation emitted by a particle. Since the area of a sphere increases like $r^2$, the far field power emitted by the particle (the integral of the Poynting vector over the area of the sphere) is constant. My question is if the near field also has such a power (integral of Poynting over the area of a sphere) associated with it, or if the various Poynting vectors nicely cancel out as to make the total power zero (if that's the case, then I'd love a proof that it's $0$). I know that the far field dominates over large distances, but if the near field also has such a power associated with it, then the total power must change with distance, which would seem to break energy conservation. And if the near field also has a power and it somehow doesn't break energy conservation, then why is it only the far field power that is associated with the energy emitted by the particle?

Ján Lalinský · Answer

At any point of space, there is just one EM field and there is just one associated Poynting vector
$$
mathbf S = mathbf E times mathbf B/mu_0.
$$
At large enough distances, the total field can be approximated by the "far field", i.e. by a component that decays as $1/r$. That's why one can use the far field there instead of total field - they are almost the same.
The vector field $mathbf S$ obeys local conservation law, there is no energy conservation breaking anywhere.
It does not matter that near field decays with distance differently than the far field; because near field isn't a tranversal wave but has strong static component, total field Poynting vector isn't directed radially everywhere, some of the flux goes "around" and "back to the radiator". So the total energy flux that goes away through a small imagined sphere (where near field is strong) is much smaller than the simplistic idea about radial Poynting vector of near field would suggest.

ZeroTheHero · Answer

In the near field of a Hertzian dipole, $tilde{mathbf{E}}timestilde{mathbf{H}}^*$ is (almost) purely imaginary, so the time-averaged Poynting vector
begin{align}
mathbf{S}=frac{1}{2}text{Re}left[tilde{mathbf{E}}timestilde{mathbf{H}}^*right]=0, .
end{align}
In fact the electric field phasor in this case is basically the same as that of a static electric dipole.
This also holds true for finite antennas, although the expressions are more complicated.
In the near-field region the field is basically described by Biot-Savart and Coulomb’s law.

Near and far field EM radiation

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