# The fields of Liénard and Wiechert and Poynting vector

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:

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

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:

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

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

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

written in a compact form, like:

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

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{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}

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.

[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.

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 $$1/r$$.
• 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? – Sebastiano Mar 24 at 23:05
• Just look at the power of $|\vec{r}-\vec{r}_s|$ in the denominator. Far away from the source, this is just $r$. – G. Smith Mar 24 at 23:12