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?
2 Answers
In the near field of a Hertzian dipole, $\tilde{\mathbf{E}}\times\tilde{\mathbf{H}}^*$ is (almost) purely imaginary, so the time-averaged Poynting vector \begin{align} \mathbf{S}=\frac{1}{2}\text{Re}\left[\tilde{\mathbf{E}}\times\tilde{\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.
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.
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$\begingroup$ Yes, I know, but if you expand the Poynting vector by writing the electric and magnetic field as a sum of a "velocity" field and "acceleration field", you get four terms, one with a 1/r^4 dependence (neglected in the question so as to not get into too much detail), two with a 1/r^3 dependence and one with a 1/r^2 dependence. My question is if the sum of the other three Poynting vectors (the ones other than the one with 1/r^2 dependence) can have a non-zero power, and if yes, then I wanna know where that energy cones from, if not from the charge. $\endgroup$ Commented Mar 27, 2021 at 13:48
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$\begingroup$ Or, rather, I guess that this non-zero power would have to come from the EM field itself (which has a certain energy density), but then why can't I say that the power from the 1/r^2 dependent Poynting vector also comes from the field, and not from the charge. $\endgroup$ Commented Mar 27, 2021 at 13:59
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$\begingroup$ Yes, sum of the three terms can be non-zero at any point, that is obvious, so there is in general a non-zero energy flux density, it can be zero at some points but not all. When integrated over a sphere in the near field region, we get flux due to those three non-wave-field terms. This flux does not have to be zero, as energy at any interspherical space can decrease of increase in time, depending on the motion of the charge. Only in stationary oscillation, the time average of this flux has to be zero, because energy in the interspherical space is a bounded function. $\endgroup$ Commented Mar 27, 2021 at 15:52
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$\begingroup$ "Where the energy comes from, from charge or the field" it comes from both, and also can come from other bodies. Whatever makes the charge accelerate against electric force that acts on it, does the work that increases energy of EM field. $\endgroup$ Commented Mar 27, 2021 at 15:55