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Taking common equation for the electric field of a Hertzian dipole from Wikipedia:

$${\displaystyle {\begin{aligned}E_{\theta }=i{\frac {\zeta _{0}I\delta \ell }{4\pi }}\left({\frac {k}{r}}-{\frac {i}{r^{2}}}-{\frac {1}{kr^{3}}}\right)e^{-ikr}\sin {\left(\theta \right)}\\[2pt]E_{r}={\frac {\zeta _{0}I\delta \ell }{2\pi }}\left({\frac {1}{r^{2}}}-{\frac {i}{kr^{3}}}\right)e^{-ikr}\cos {\left(\theta \right)}\end{aligned}}}$$

It seems to me that the inverse cubed term is the electrostatic field of the dipole, and please confirm my suspicion that it's because the Hertzian dipole has such a small length that the term reduces to $\frac{1}{kr^3}$ instead of the usual inverse polynomial term where the dipole is more like inverse square close to it and inverse cube further away. The inverse linear term is of course the radiation field although I'm also not sure how to mathematically derive it as 1/r but it seems to make intuitive sense in that the magnetic field in a wave generates extra electric field (that what is available in the electrostaric field at a given distance) and vice versa.

My main question is what is the inverse square term. The only explanation I can find on it is in a book called Classical Electricity and Magnetism from Lawrence Berkeley national laboratory.

the second term varies as the inverse square of the distance and is called the transition field. The transition field will not, contribute to the radiated energy but it does contribute to the energy storage during the oscillation

It is a vague description. What actually is it, why does it intuitively exist and why does it roll off as inverse square?

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The $\frac{1}{r^2}$ term represents the reactive, i.e., capacitive/inductive oscillating field. All conductors have a certain amount of capacitance and inductance and their energy fluctuates back and forth between being electric and magnetic. The $1/r^2$ roll-off is just as you would expect from being that kind of fluctuation of the field intensities from an electric/magnetic pole. When you drive an antenna from a transmission line this capacitive/inductive term shows up in the input impedance of the line as a reactive termination in addition to the $1/r$ radiating part representing the "resistive" term.

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  • $\begingroup$ That is, quite demonstrably, bonkers explanation. Fuck that. We gotta work with what we have available, I guess, but this is a garbage answer. $\endgroup$
    – Chad Brandon
    Oct 20, 2022 at 14:41
  • $\begingroup$ @user346760 @ hyportnex surely the "capacitive" field is the electrostatic field, which decreases with inverse cube – how are you differentiating those two? Interestingly Wikipedia says "By contrast, near-field E and B strength decrease more rapidly with distance: the radiative field decreases by the inverse-distance squared, the reactive field by an inverse-cube law" so it implies that it's radiation that's inverse squared near the dipole, contrary to what the book says. It is called the "radiative" field on Wikipedia $\endgroup$
    – Lewis Kelsey
    Oct 20, 2022 at 17:10
  • $\begingroup$ go to archive.org/details/dli.ernet.15390/page/93/mode/… where you also find $H_{\phi}$. You can see that the $1/r^2$ term is in quadrature to that of the same in $E_{\theta}$. In other words that represents a reactive field, whose time averaged energy is zero. And that is because the energy it represents oscillates between being electric and magnetic, back and forth, just like in an LC resonator. The $1/r^3$ term is just the electrostatic field of the electric dipole and is also not radiated and it has not magnetic counterpart in the $H$ field. $\endgroup$
    – hyportnex
    Oct 20, 2022 at 21:06
  • $\begingroup$ @hyportnex I understand now. The $1/r^2$ electric field arises from the induction of the $1/r^2$ oscillating magnetic field $\endgroup$
    – Lewis Kelsey
    Oct 21, 2022 at 3:23
  • $\begingroup$ note too that the "radiative field decreases by the inverse-distance squared" means that both $E$ and $H$ decrease as $1/r$ so their product the Poynting vector $E \times H$ decreases as $1/r^2$ $\endgroup$
    – hyportnex
    Oct 21, 2022 at 10:41

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