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In a following reference [David M. Pozar 2012, 4ed, Microwave Engineering, p 661], it is told that the far field distance of a relatively large antenna is given from the formula \begin{equation} R_{ff}=\frac{2D^2}{\lambda} \end{equation} where D is a maximum dimension of an antenna, $\lambda$ is a wavelength. And it is told that:"This result is derived from the condition that the actual spherical wave front radiated by the antenna departs less than $\pi/8 =22.5^\circ$ from a true plane wave front over the maximum extent of the antenna". Does this mean that we have to compare the "far-field" component of electric field with a "near-field" component and at phase $\pi/8$ they are equal or something else? I think that I miss the point.

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  • $\begingroup$ "far field" and "near field" describe places in which the fields macroscopically look different; a point in space either is in "near field", or in "far field" $\endgroup$ Commented Oct 23, 2021 at 22:04
  • $\begingroup$ @Marcus Müller, I want to say that we can calculate the EM wave with the Green function formalism from a given current, thus we have an exact solution. Next, as far as I know, we take into account different terms when talking about far field and near field. $\endgroup$ Commented Oct 23, 2021 at 22:07
  • $\begingroup$ And there are terms that will be more important in near field and other in far field $\endgroup$ Commented Oct 23, 2021 at 22:08
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    $\begingroup$ If you go far enough away from any finite source the resulting wavefront is asymptotically plane. By Rayleigh's formula the far-field is defined so that the actual wavefront having phase deviation relative to an ideal plane one is not to exceed $\pi/8$, which is a reasonable approximation to a plane wavefront. $\endgroup$
    – hyportnex
    Commented Oct 23, 2021 at 22:18
  • $\begingroup$ It is unlikely that you can compute the full field (near-field and far-field regime) in a single expression without approximation. If anything the regimes are physically different. The criterion given here is (in some sense) arbitrary and depends on how accurate you want your fields to be. $\endgroup$ Commented Oct 23, 2021 at 23:50

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The condition on the far field in the free space is obtained from wave equations. First of all, the equation on the retarded potentials is

\begin{equation} \label{eq:dAlembert} \Box{\vec{A}=-\mu_0 \vec{j}} \end{equation}

where the temporal part of of potential is $e^{j \omega t}$, so we can get the following equation assuming $k = \omega /c$

\begin{equation} \label{eq:Helmholtz} \Delta\vec{A}+ k^{2}\vec{A} = -\mu_0\vec{j} \end{equation}

Next we solve this equation assuming Green function formalism.

\begin{equation} G(\vec{r}) = \frac {-e^{\pm ikr}}{4\pi r} \end{equation}

After a long calculation and using Laurenz jauge, we can achieve precise equations on electric and magnetic field. For exemple, for the electric field we have

\begin{equation} \vec{H} = -\frac{1} {4 \pi} \int \ e^{ik|\vec{r}-\vec{r}\,'|} \frac {1} {|\vec{r}-\vec{r}\,'|^2} (-\frac{\vec{r}-\vec{r}\,'} {|\vec{r}-\vec{r}\,'|} + ik (\vec{r}-\vec{r}\,') ) \times \vec{j}(\vec{r}\,') \, d^3\vec{r}\,' \end{equation} We are interested in the term $ e^{ik|\vec{r}-\vec{r}\,'|} $, where the Taylor series of the following function is \begin{equation} \label{Eq:Abs_r-r'} |\vec{r}-\vec{r}\,'| = \sqrt{(\vec{r}-\vec{r}\,')\cdot(\vec{r}-\vec{r}\,')}\approx r (1-\vec{e}_r \vec{r}\,'/r +\frac{{\vec{r}\,'}^2} {2 r^2}) \end{equation}

In order to decrease the last term in the previous equation we have the far field criteria:

\begin{equation} \frac{{r\,'}^2} {2 r^2} < \lambda/16 \end{equation} or equally, $$phase = k\frac{{r\,'}^2} {2 r^2} < \frac{\pi}{8}$$.

It is well explained in [p83, A.B. Smolders, H.J. Visser, U. Johannsen, Modern Antennas and Microwave Circuits, September 2020, Eindhoven University of Technology].

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