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.


1 Answer 1


[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

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









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.


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 $1/r$.

  • $\begingroup$ 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, 2019 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, 2019 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, 2019 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, 2019 at 12:56
  • $\begingroup$ I've added the series expansion. $\endgroup$
    – G. Smith
    Mar 25, 2019 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.