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EDIT: I know that the electric and magnetic fields depend not only on speed but also on acceleration and can both be expressed as the sum of two contributions:

\begin{equation} \overline{E} (\bar{r},t)=\overline{E}_{u} (\bar{r},t)+\overline{E}_{a} (\bar{r},t), \quad \overline{B} (\bar{r},t)=\overline{B}_{u}(\bar{r},t)+\overline{B}_{a}(\bar{r},t) \end{equation}

where $\overline{E}_{u}(\bar{r},t)$ and $\overline{B}_{u}(\bar{r},t)$ only depend on speed, whereas $\overline{E}_{a}(\bar{r},t)$ and $\overline{B}_{a}(\bar{r},t)$ depend on both speed and acceleration and meet the conditions:

\begin{equation} \overline{E}_{a}(\bar{r},t)=\overline{B}_{a}(\bar{r},t)=0\quad\Longleftrightarrow\quad\bar{a}=0 \end{equation}

For the differentiation of $\overline{B} (\bar{r},t)=\overline{B}_{u}(\bar{r},t)+\overline{B}_{a}(\bar{r},t)$ please note that:

\begin{equation} \overline{B}_u (\bar{r},t)=\frac{1}{c}\boldsymbol{[}\hat{\mathbf R}\boldsymbol{]} \times \overline{E}_u (\bar{r},t), \qquad \overline{B}_a (\bar{r},t)=\frac{1}{c}\mathbf{[}\hat{\mathbf R}\mathbf{]} \times \overline{E}_a (\bar{r},t) \end{equation}

written in a compact form, like:

\begin{equation} \overline{B}_{u,a} (\bar{r},t)=\frac{1}{c}\boldsymbol{[}\hat{\mathbf R}\boldsymbol{]} \times \overline{E}_{u,a} (\bar{r},t) \end{equation}

The presence of square brackets is important because the amount(s) should be calculated at the delay time $\tau=t'$. Obviously the magnetic field of a point charge is always perpendicular to the electric field and to the vector from the delayed point.

But $$ \begin{align} \overline E_u(\bar{r},t)&=k_eq\frac{(1-\beta^2)(\bar{r}-R\overline \beta)}{\kappa^3} \tag{1}\\ \overline E_a(\bar{r},t)&=\frac{k_eq}{c^2}\frac{\bar{r}\times \{(\bar{r}-R\overline \beta)\times \overline{A}\}}{\kappa^3} \tag{2} \end{align} $$

where $$\kappa(t')_{\mathbf{vacuum}}\mathrel{\mathop:}=\kappa(\tau)=R(\tau)-\bar{r}(\tau)\cdot\overline{\beta}(\tau), \quad t'=\tau$$ and

where $R(\tau)$ is the distance between the position of the charge $q$ and the point of the observation $P$:

$$R(t')=|\bar{r}(t')|=|\bar{r}'-\bar{r}_{q} (t')|=\sqrt{\left(|\bar x'-\bar {x}_{q} (t')|\right)^2+\left(|\bar y'-\bar {y}_{q} (t')|\right)^2+\left(|\bar z'-\bar {z}_{q} (t')|\right)^2}.$$

The quantity expressed by the $(1)$ is called the generalised Coulombian field and is it not depend by the acceleration. It is sometimes also called the speed range. The amount $(2)$ of the $\overline E$ field is called radiation range or, since it is proportional to $a$, acceleration range. It tends to zero like the inverse of the first power of the $R$ distance and is therefore dominant at great distances.

Now, let be $\overline{E}_{u}$ is the component of electric field that it depends by the velocity $u$ of a charge $q$ that it radiates, $\overline{E}_{a}$ is the component of electric field that it depends by the acceleration. Similary for the $\overline{B}_{u}$ and $\overline{B}_{a}$. After, the Poynting vector $\overline{S}$ can be written as:

\begin{equation} \begin{aligned} \overline{S}&=\frac{1}{\mu_{0}}\overline{E}\times\overline{B}=\frac{1}{\mu_{0}}\left\{(\overline{E}_{u}+\overline{E}_{a})\times(\overline{B}_{u}+\overline{B}_{a})\right\}=\\ &=\frac{1}{\mu_{0}}\left\{ \overline{E}_{u}\times\overline{B}_{u}+\overline{E}_{u}\times\overline{B}_{a}+\overline{E}_{a}\times\overline{B}_{u}+\overline{E}_{a}\times\overline{B}_{a}\right\} \end{aligned} \end{equation}

Explicitly, starting from my formulas, I would like to understand the dependence of the Poynting vector $\overline S$ mathematically or algebraically by $R^2$, $R^3$, which means there's an asymptotic dependence like that:

$$ \overline{S}\asymp\frac{1}{\mu_{0}}\left\{ \frac{\overline \Lambda}{R^{4}}+\frac{\overline \Theta}{R^{3}}+\frac{\overline \epsilon}{R^{2}}\right\}, $$

In other words I would like to know mathematically how I come to understand or prove that:

  1. $\overline{E}_{u}\times\overline{B}_{u}\propto 1/R^4$;

  2. $\overline{E}_{u}\times\overline{B}_{a}+\overline{E}_{a}\times\overline{B}_{u} \propto 1/R^3$;

  3. $\overline{E}_{a}\times\overline{B}_{a}\propto 1/R^2$?

Thank you.

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[NOTE: The OP substantially edited the question while I was composing this answer. He previously had no expressions for the fields, and had not previously mentioned Lienard and Wiechert.]

Once you know the formulas for $\vec{E}_u$, $\vec{E}_a$, $\vec{B}_u$, and $\vec{B}_a$, those dependencies on distance are obvious.

The fields of a point source charge $q$ undergoing arbitrary motion $\vec{r}_s(t)$ can be derived from the Lienard-Wiechert potentials; they are

$$\vec{E}(\vec{r},t)=\frac{1}{4\pi\epsilon_0}\left[\frac{q\,(\hat{n}-\vec{\beta})}{\gamma^2(1-\hat{n}\cdot\vec{\beta})^3|\vec{r}-\vec{r}_s|^2}+\frac{q \,\hat{n}\times((\hat{n}-\vec{\beta})\times\dot{\vec{\beta}}}{c\,(1-\hat{n}\cdot\vec{\beta})^3|\vec{r}-\vec{r}_s|}\right]_{\,t_r}$$

and

$$\vec{B}(\vec{r},t)=\frac{\mu_0}{4\pi}\left[\frac{q\,c\,(\vec{\beta}\times\hat{n})}{\gamma^2(1-\hat{n}\cdot\vec{\beta})^3|\vec{r}-\vec{r}_s|^2}+\frac{q\,\hat{n}\times(\hat{n}\times((\hat{n}-\vec{\beta})\times\dot{\vec{\beta})}}{\,(1-\hat{n}\cdot\vec{\beta})^3|\vec{r}-\vec{r}_s|}\right]_{\,t_r}.$$

Here

$$\vec{\beta}(t)=\frac{\dot{\vec{r_s}}(t)}{c},$$

$$\hat{n}(t)=\frac{\vec{r}-\vec{r}_s(t)}{|\vec{r}-\vec{r}_s(t)|},$$

$$\gamma(t)=\frac{1}{\sqrt{1-|\vec{\beta}|^2}},$$

and

$$t_r=t-\frac{|\vec{r}-\vec{r}_s(t)|}{c}$$

is the so-called "retarded time".

By looking at these fields, you can see that the second term depends on the acceleration of the charge (as $\dot{\vec{\beta}}$), while the first term is independent of the acceleration. Thus the first terms are $\vec{E}_u$ and $\vec{B}_u$, while the second terms are $\vec{E}_a$ and $\vec{B}_a$.

Furthermore, you can see that at large distances the $r$-dependence of the first ("Coulombic") terms is $1/r^2$ and the $r$-dependence of the second ("radiative") terms is $1/r$. This then explains the $r$-dependence of the terms in the Poynting vector. It is the product $\vec{E}_a\times\vec{B}_a$ that carries energy away to infinity; this is because it drops off as $1/r^2$ and thus has a finite flux over a sphere of infinite radius.

ADDENDUM:

The OP asked for a series expansion of the fields in $1/r$. The dependence on $\vec{r}$ is through the $|\vec{r}-\vec{r}_s|$ or $|\vec{r}-\vec{r}_s|^2$ in the denominator. We have

$$\begin{align} \frac{1}{|\vec{r}-\vec{r}_s|^n}&=\frac{1}{[(\vec{r}-\vec{r}_s)\cdot(\vec{r}-\vec{r}_s)]^{n/2}}\\ &=(r^2-2\,\vec{r}_s\cdot\vec{r}+r_s^2)^{-n/2}\\ &=\frac{1}{r^n}\left(1-\frac{2\,\vec{r}_s\cdot\vec{r}}{r^2}+\frac{r_s^2}{r^2}\right)^{-n/2}\\ &=\frac{1}{r^n}\left(1+\frac{n\,\vec{r}_s\cdot\vec{r}}{r^2}+\dots\right)\\ &=\frac{1}{r^n}+O\left(\frac{1}{r^{n+1}}\right) \end{align}.$$

Thus $\vec{E}_u$ and $\vec{B}_u$ fall off as $1/r^2$ while $\vec{E}_a$ and $\vec{B}_a$ fall off as $1/r$.

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  • $\begingroup$ Thank you for your answer. +1. But it may be that at the moment I am not well but I can not understand the dependence on your and my formulas of the inverse of the square of the distance, from $R^2$ and from $R^3$ and $R^4$. What does it take to find this dependency a serial development? $\endgroup$ – Sebastiano Mar 24 at 23:05
  • $\begingroup$ For me, the simpler the answer is made, the more I understand it the better. Then having so many different notations confuses my ideas:-( $\endgroup$ – Sebastiano Mar 24 at 23:06
  • $\begingroup$ Just look at the power of $|\vec{r}-\vec{r}_s|$ in the denominator. Far away from the source, this is just $r$. $\endgroup$ – G. Smith Mar 24 at 23:12
  • $\begingroup$ Thank you very much for your patience. I have edited (:-() again my question. I try to understand (you can rest assured) but after your formulas could you not take for granted many things? I am a high school teacher and with all due respect I am not a university teacher. Could you add the series expansion of the denominator please? Thank you. $\endgroup$ – Sebastiano Mar 25 at 12:56
  • $\begingroup$ I've added the series expansion. $\endgroup$ – G. Smith Mar 25 at 16:21

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