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in my antennas book (I do not put its reference because it is not in english) it is written that, at high distance from the source, the electromagnetic field radiated by it is a spherical wave.

This can be shown by analyzing the expression of the EM field at high distance generated by some specific antennas. For instance, for a half-wave dipole (which is not the short herzian dipole) we get (Reference here, page 17):

enter image description here

enter image description here

It is a spherical wave because its dependence on r is given by the term:

\$\frac{e^{-jkr}}{r}\$

I do not understand how can be this physically possible. A spherical wave is ideally generated only from a point source. I easily understand how even the wave generated by a small source (for example a short Herzian dipole) can be, at high distance, approximated as spherical. But in this case, in which the linear current source is not short, I don't understand why the wave cannot be, more plausibly (from a physical point of view), cylindrical. Furthermore, if it is true that at a great distance any wave is spherical, in which situations do cylindrical waves exist? I can't imagine how a cylinder becomes a sphere at a great distance.

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    \$\begingroup\$ If you dropped a brick (rectangular shaped heavy object) into the middle of a calm large pond, what will the waves look like at 1 m and 10 m and 100 m? \$\endgroup\$
    – Andy aka
    May 2, 2020 at 16:00

3 Answers 3

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There's a couple of parts to this. Within the near field, roughly less than couple of wavelengths away from the radiating element, there is no E&M wave as such. In the near field region, the electric and magnetic fields can exist independently of each other.

Once you get more than a handful of wavelengths away from the element, the traditional E&M field takes over (not being precise in my language here), and takes on the classical spherical wavefront.

As you get further and further away from the element, the spherical wavefront starts to flatten as a function of D (distance from the element). Think of the wavefront as being the circumference of a circle.

In the extreme far field, many many wavelengths away, the wavefront is to all intents and purposes flat (planar), as analogsystemsrf said.

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  • \$\begingroup\$ So, in case of a linear source, is the situation like: "undefined" wave (near the source), cylindrical wave (a bit far), spherical wave (very very far), and plane wave (very very far and with a local analysis)? \$\endgroup\$
    – Kinka-Byo
    May 2, 2020 at 15:54
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At high distance, all waves become planar, or near planar.

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  • \$\begingroup\$ I know it is only a local approximation of the wavefront. In fact E field contains the term exp(-jkr)/r which is a spherical wave \$\endgroup\$
    – Kinka-Byo
    May 2, 2020 at 15:51
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Consider two points a meter apart on the surface of a sphere centered on the center of a large dipole. For a small radius sphere, the difference in the angle from the sphere surface to any part of the antenna will be large, thus a difference in field strengths, thus the two points will show an EM field gradient, thus the wave front will not be flat between those two points, thus the wave front can't be spherical.

For a two points a meter apart on a large enough radius sphere, any difference in relative angle to any part of the antenna will go to zero, thus there won't be an EM gradient on the surface of the sphere, thus it will be the wave front.

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