#  Electrodynamics of an oscillating particle without cheating PART I : In
vacuo. PART II : Near a dispersive bulk

M. Garcia-Vergara  Departamento de Física, Facultad de Ciencias, Universidad
Nacional Autónoma de México, Ciudad Universitaria, Av. Universidad 3000,
Mexico City, 04510, Mexico  G. Demésy  Aix Marseille Univ, CNRS, Centrale
Marseille, Institut Fresnel, Marseille, France  A Nicolet  Aix Marseille Univ,
CNRS, Centrale Marseille, Institut Fresnel, Marseille, France  F. Zolla

###### Abstract

In this paper, the electromagnetic radiation from an oscillating particle
placed in the vicinity of an object of size comparable to the wavelength is
studied. Although this problem may seem academic at first sight, the details
of the calculations are presented throughout without any detail left under the
carpet. A polyharmonic decomposition of the radiation sources allows the
diffraction problem to be fully characterised while satisfying energy
conservation. Finally, the source expressions obtained are suitable for use in
a numerical code. A 3D illustration using finite elements is provided.

##  1 Introduction

This paper consists in two main parts: In the first one we will address the
problem of the electromagnetic (EM) field generated by an oscillating particle
_in vacuo_ and the second one is devoted to its interaction near to a bulk
made of some dispersive material. Albeit the apparently pure academic nature
of this problem, many applications to this case of configurations can be
found. Starting from the study of antennas made by Hertz and Sommerfeld [1, 2,
3] to the more recently study of quantum nanoemitters and its interaction with
nano spheres [4, 5]. For this kind of phenomena, it is common to use an
approximation to describe the field generated by an oscillating charged
particle, the most common of these approximations for the far field is to use
a _dipole approximation_ as described in [6, 7, 8, 2]. However, it is
important to remark here that the dipole approximation is only valid when
dealing with the far field.

It is then necessary to find a better way to describe the field generated by
an oscillating charge, the immediate approach is to use the _Liénard-Wiechert_
fields, which describe in quite a compact form the fields produced by a moving
charge. The problem with this solution is that is not tractable from the
practical (_i.e._ numerical) point of view as we will show in Section 3.2.1.
Other approaches have been to consider the moving charge as a delta
distribution that can be approximated by a Taylor’s series [9] or a harmonic
expansion [10]. Here we propose a different way for obtaining the harmonic
representation of the fields which explicitly depends on time and space.
Finally, once sources are obtained in harmonic form we use them as an incident
field of the diffraction problem to obtain the diffracted field by a sphere by
using the Finite Elements Method.

##  2 Mathematical description of the problem

The mathematical description of the interaction of an EM field with a time
dispersive bulk is given by the Maxwell equations:

$\displaystyle\mathbf{\nabla}\times\mathbf{E}_{\mathrm{\textit{$\textit{$\xi$}$}}}=-\partial_{t}\mathbf{B}_{\mathrm{\textit{$\xi$}}},$
(1)
$\displaystyle\mathbf{\nabla}\times\mathbf{H}_{\mathrm{\textit{$\xi$}}}=\partial_{t}\mathbf{D}_{\mathrm{\textit{$\xi$}}}+\mathbf{\boldsymbol{\jmath}}_{\mathrm{}},$
(2)
$\displaystyle\mathbf{\nabla}\cdot\mathbf{D}_{\mathrm{\textit{$\xi$}}}=\rho,$
(3) $\displaystyle\mathbf{\nabla}\cdot\mathbf{B}_{\mathrm{\textit{$\xi$}}}=0$
(4)

where $\textit{$\xi$}=\\{I,II\\}$ represents the restriction of the fields,
$I$ corresponds to the total field _outside the bulk_ and $II$ to the field
_inside the bulk_. The constitutive relations are then
$\mathbf{H}_{\mathrm{\textit{$\xi$}}}=\mu_{0}^{-1}\mathbf{B}_{\mathrm{\textit{$\xi$}}}$
(_i.e._ the dispersive bulk has no magnetization) and
$\mathbf{D}_{\mathrm{\textit{$\xi$}}}=\epsilon_{0}(\epsilon_{r,\textit{$\xi$}}*\mathbf{E}_{\mathrm{\textit{$\xi$}}})$,
where

$\epsilon_{r,\textit{$\xi$}}(t)=\begin{cases}2\pi\delta(t)&\textrm{ if
}\xi=I,\\\ 2\pi\delta(t)+\chi(t)&\textrm{ if }\xi=II,\end{cases}$ (5)

and $\chi(t)$ is the causal electric susceptibility with its support within
the bulk. The convolution is the one corresponding to the Fourier Transform
convention in Appendix A. Thus, our new system of equations in terms of the
($\mathbf{E}_{\mathrm{\xi}},\mathbf{B}_{\mathrm{\xi}}$) fields reads:

$\displaystyle\mathbf{\nabla}\times\mathbf{E}_{\mathrm{\textit{$\xi$}}}=-\partial_{t}\mathbf{B}_{\mathrm{\textit{$\xi$}}},$
(6)
$\displaystyle\mathbf{\nabla}\times\mathbf{B}_{\mathrm{\textit{$\xi$}}}=\frac{1}{c^{2}}\partial_{t}(\epsilon_{r,\textit{$\xi$
}}*\mathbf{E}_{\mathrm{\textit{$\xi$}}})+\mu_{0}\mathbf{\boldsymbol{\jmath}}_{\mathrm{}},$
(7)
$\displaystyle\mathbf{\nabla}\cdot(\epsilon_{r,\textit{$\xi$}}*\mathbf{E}_{\mathrm{\textit{$\xi$}}})=\frac{\rho}{\epsilon_{0}},$
(8) $\displaystyle\mathbf{\nabla}\cdot\mathbf{B}_{\mathrm{\textit{$\xi$}}}=0.$
(9)

The sources $\rho$ and $\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}$ represent
the charge density and current corresponding to an oscillating particle and
their explicit expression will be given in Section 4. The fields generated by
these sources _in vacuo_ are:

$\displaystyle\mathbf{\nabla}\times\mathbf{E}_{\mathrm{}}^{0}=-\partial_{t}\mathbf{B}_{\mathrm{}}^{0},$
(10)
$\displaystyle\mathbf{\nabla}\times\mathbf{B}_{\mathrm{}}^{0}=\frac{1}{c^{2}}\partial_{t}\mathbf{E}_{\mathrm{}}^{0}+\mu_{0}\mathbf{\boldsymbol{\jmath}}_{\mathrm{}},$
(11)
$\displaystyle\mathbf{\nabla}\cdot\mathbf{E}_{\mathrm{}}^{0}=\frac{\rho}{\epsilon_{0}},$
(12) $\displaystyle\mathbf{\nabla}\cdot\mathbf{B}_{\mathrm{}}^{0}=0$ (13)

where the constitutive relations
$\mathbf{D}_{\mathrm{}}^{0}=\epsilon_{0}\mathbf{E}_{\mathrm{}}^{0}$ and
$\mathbf{H}_{\mathrm{}}^{0}=\mu_{0}^{-1}\mathbf{B}_{\mathrm{}}^{0}$,
$\xi=\\{I,II\\}$ have been used. Let us remark here that
($\mathbf{E}_{\mathrm{}}^{0}$,$\mathbf{B}_{\mathrm{}}^{0}$) is the incident
field of our diffraction problem and that it is defined evrywhere in space.
The next step is to consider a new set of fields defined as:
$\mathbf{E}_{\mathrm{\textit{$\xi$}}}^{1}:=\mathbf{E}_{\mathrm{\textit{$\xi$}}}-\mathbf{E}_{\mathrm{}}^{0}$,
$\mathbf{B}_{\mathrm{\textit{$\xi$}}}^{1}:=\mathbf{B}_{\mathrm{\textit{$\xi$}}}-\mathbf{B}_{\mathrm{}}^{0}$
which, in the present linear case, satisfy the system of so called diffracted
fields:

$\displaystyle\mathbf{\nabla}\times\mathbf{E}_{\mathrm{\textit{$\xi$}}}^{1}=-\partial_{t}\mathbf{B}_{\mathrm{\textit{$\xi$}}}^{1},$
(14)
$\displaystyle\mathbf{\nabla}\times\mathbf{B}_{\mathrm{\textit{$\xi$}}}^{1}=\frac{1}{c^{2}}\partial_{t}(\epsilon_{r,\textit{$\xi$}}*\mathbf{E}_{\mathrm{\textit{$\xi$}}}^{1})+\mu_{0}\mathbf{\boldsymbol{\jmath}}_{\mathrm{II}}^{0},$
(15)
$\displaystyle\mathbf{\nabla}\cdot(\epsilon_{r,\textit{$\xi$}}*\mathbf{E}_{\mathrm{\textit{$\xi$}}}^{1})=\frac{\rho^{0}_{II}}{\epsilon_{0}},$
(16)
$\displaystyle\mathbf{\nabla}\cdot\mathbf{B}_{\mathrm{\textit{$\xi$}}}^{1}=0,$
(17)

whose sources are defined as:

$\rho^{0}_{II}:=-\epsilon_{0}\mathbf{\nabla}\cdot([\epsilon_{II}-2\pi\delta]*\mathbf{E}_{\mathrm{}}^{0}),$
(18)

and

$\mathbf{\boldsymbol{\jmath}}_{\mathrm{II}}^{0}:=\epsilon_{0}\partial_{t}([\epsilon_{II}-2\pi\delta]*\mathbf{E}_{\mathrm{}}^{0}),$
(19)

and from these definitions it is easy to see that they satisfy charge
conservation. Notice that the support of these new sources is within the bulk
and depends on the electromagnetic field generated by the oscillating
particle. However obtaining a handy expression of such field requires a very
careful crafting as it will be shown in the next section.

##  3 The search for the EM Field generated by an oscillating charge _in
vacuo_

The problem of obtaining the electromagnetic field generated by a charged
oscillating particle is a very important problem _per se_. This section will
start with the very general approach of finding the
($\mathbf{E}_{\mathrm{}}$,$\mathbf{B}_{\mathrm{}}$) fields generated by an
arbitrary pair of charge and current distributions ($\rho$,$\jmath$). Usually
these solutions are given by the so-called _Jefimenko’s equations_. However,
it will be shown how by following a similar method as in Ref. [11] for the far
field approximations, it is possible to express the electric field mostly in
terms of the current density. Next we will discuss some of the efforts that
have been made in order to represent the EM field created by and oscillating
charge.

###  3.1 The EM Field generated by an arbitrary charge distribution

First, we consider the system of equations (10-13) (in the sequel the
superscript $0$ denoting the incident field radiated by the oscillating
particle in freespace has been removed to alleviate notations). By considering
the _Lorentz gauge_ , one can obtain the electric and magnetic fields
($\mathbf{E}_{\mathrm{}}$, $\mathbf{B}_{\mathrm{}}$) using the expressions:

$\displaystyle\mathbf{E}_{\mathrm{}}=-\partial_{t}\mathbf{A}_{\mathrm{}}-\mathbf{\nabla}\phi,$
(20)
$\displaystyle\mathbf{B}_{\mathrm{}}=\mathbf{\nabla}\times\mathbf{A}_{\mathrm{}}$
(21)

where $\phi$ and $\mathbf{A}_{\mathrm{}}$ are the retarded potentials [7, 6]

$\displaystyle\phi$
$\displaystyle=\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\frac{\rho(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(22) $\displaystyle\mathbf{A}_{\mathrm{}}$
$\displaystyle=\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}^{3}}\frac{\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(23)

with the retarded time $t_{r}=t-\frac{R}{c}$ and $R=|\mathbf{R}_{\mathrm{}}|$
with
$\mathbf{R}_{\mathrm{}}=\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}$.
The next step consists in plugging the retarded potentials Eqs. (22,23) into
Eqs. (20,21) and then take the appropriate derivatives. Nevertheless, this
implies to take time derivatives with respect to a function which is in terms
of the retarded time $t_{r}$. In order to avoid this difficulty and carry our
calculations further, we propose to use the Fourier transform in time (See
Annexe A, $\hat{f}$ denoting the Fourier transform of $f$) and then consider
only the spatial derivatives. Starting by transforming the potential $\phi$
one has:

$\displaystyle\hat{\phi}=$
$\displaystyle\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\frac{1}{R}\mathcal{F}_{t\rightarrow\omega}\\{\rho(\mathbf{x}_{\mathrm{}}^{\prime},t-R/c)\\}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
$\displaystyle=$
$\displaystyle\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\frac{1}{R}e^{iRk_{0}}\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(24)

where $k_{0}=\frac{\omega}{c}$. In a similar way the vector potential in the
frequency domains reads:

$\mathbf{\hat{A}}=\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}^{3}}e^{+iRk_{0}}\frac{\mathbf{\hat{\boldsymbol{\jmath}}}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(25)

Before proceeding, the following identities are necessary:

$\displaystyle\mathbf{\nabla}R=\frac{\mathbf{R}_{\mathrm{}}}{R},$ (26)
$\displaystyle\mathbf{\nabla}\frac{1}{R}=-\frac{1}{R^{2}}\mathbf{\nabla}R=-\frac{\mathbf{R}_{\mathrm{}}}{R^{3}},$
(27)
$\displaystyle\mathbf{\nabla}e^{iRk_{0}}=ik_{0}e^{iRk_{0}}\mathbf{\nabla}R=ik_{0}e^{iRk_{0}}\frac{\mathbf{R}_{\mathrm{}}}{R},$
(28)

and combining equations (27) and (28) we get:

$\mathbf{\nabla}\bigg{(}\frac{e^{iRk_{0}}}{R}\bigg{)}=\bigg{(}ik_{0}-\frac{1}{R}\bigg{)}\frac{e^{iRk_{0}}}{R^{2}}\mathbf{R}_{\mathrm{}}.$
(29)

Equipped with these tools, it is quite easy to obtain $\mathbf{B}_{\mathrm{}}$
by simply taking the curl of Eq. (25):

$\displaystyle\mathbf{\hat{B}}=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}^{3}}\mathbf{\nabla}\times\bigg{(}\frac{e^{+iRk_{0}}}{R}\mathbf{\hat{\boldsymbol{\jmath}}}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\bigg{)}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
(30) $\displaystyle=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}^{3}}\mathbf{\nabla}\bigg{(}\frac{e^{iRk_{0}}}{R}\bigg{)}\times\mathbf{\hat{\boldsymbol{\jmath}}}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
(31) $\displaystyle=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}^{3}}\bigg{(}ik_{0}-\frac{1}{R}\bigg{)}\frac{e^{iRk_{0}}}{R^{2}}\mathbf{R}_{\mathrm{}}\times\mathbf{\hat{\boldsymbol{\jmath}}}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
(32)

And by taking the inverse Fourier transform, we arrive to the expression:

$\mathbf{B}_{\mathrm{}}=\mathbf{B}_{\mathrm{int}}+\mathbf{B}_{\mathrm{rad}},$
(33)

where $\mathbf{B}_{\mathrm{int}}$ is the intermediate field and
$\mathbf{B}_{\mathrm{rad}}$ is the radiated field which are given by the
integrals:

$\displaystyle\mathbf{B}_{\mathrm{int}}=\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}^{3}}\frac{\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})\times\mathbf{R}_{\mathrm{}}}{R^{3}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(34) $\displaystyle\mathbf{B}_{\mathrm{rad}}=\frac{\mu_{0}}{4\pi
c}\int_{\mathbb{R}^{3}}\frac{\partial_{t}\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})\times\mathbf{R}_{\mathrm{}}}{R^{2}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(35)

In order to obtain the electric field, it is necessary to consider the Fourier
transform of Eq. (20):

$\mathbf{\hat{E}}=+i\omega\mathbf{\hat{A}}-\mathbf{\nabla}\hat{\phi}.$ (36)

The first term on the right hand side of Eq. (36) is quite easy to calculate,

$i\omega\mathbf{\hat{A}}=\frac{1}{4\pi\epsilon_{0}c}\int_{\mathbb{R}^{3}}ik_{0}\mathbf{\hat{\boldsymbol{\jmath}}}(\omega,\mathbf{x}_{\mathrm{}}^{\prime})\frac{e^{iRk_{0}}}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(37)

The second term is much more tricky and its derivation goes as follows:

$\displaystyle-\mathbf{\nabla}\hat{\phi}=$
$\displaystyle\frac{-1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\mathbf{\nabla}\bigg{(}\frac{e^{iRk_{0}}}{R}\bigg{)}\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
$\displaystyle=$
$\displaystyle\frac{-1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\mathbf{\nabla}\bigg{(}\frac{e^{iRk_{0}}}{R}\bigg{)}\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
$\displaystyle=$
$\displaystyle\frac{-1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}ik_{0}\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\frac{e^{iRk_{0}}}{R^{2}}\mathbf{R}_{\mathrm{}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
$\displaystyle+\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\frac{e^{iRk_{0}}}{R^{3}}\mathbf{R}_{\mathrm{}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
(38)

Up to this point most textbooks _e.g._ Refs. [6, 7, 12] simply take the
inverse Fourier transform of Eq. (38) and give the electric field in terms of
the derivative of the density of charges and obtain the so called _Jefimenko’s
equations_ [12, 7]. Despite the straightforward nature of this derivation, we
are going to carry the calculations further. The interest of doing so will be
shown later. For our purposes, the second integral in Eq. (38) is already in
optimum form, and we will focus our attention on the integral defined within a
bounded volume $\Omega$.

${\mathbf{I}_{\mathrm{}}}_{\Omega}:=\int_{\Omega}ik_{0}\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\frac{e^{iRk_{0}}}{R^{2}}\mathbf{R}_{\mathrm{}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(39)

Notice that the integral we are looking for, is the limit case when
$\Omega\rightarrow\mathbb{R}^{3}$. Next, we will make use of the continuity
equation in the frequency domain:

$i\omega\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)=\mathbf{\nabla}^{\prime}\cdot\mathbf{\hat{\boldsymbol{\jmath}}}$
(40)

and by defining the function:

$F(R):=\frac{e^{iRk_{0}}}{cR^{2}},$ (41)

the integral ${\mathbf{I}_{\mathrm{}}}_{\Omega}$ in Eq. (39) can be written in
a more compact way (omitting the $\mathbf{x}_{\mathrm{}}^{\prime}$, $R$ and
$\omega$ dependencies) as:

${\mathbf{I}_{\mathrm{}}}_{\Omega}:=\int_{\Omega}F\mathbf{\nabla}^{\prime}\cdot\mathbf{\hat{\boldsymbol{\jmath}}}\mathbf{R}_{\mathrm{}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(42)

It is very important to remark here that the divergence is being taken with
respect to the primed coordinates (hence the prime superscript in
$\mathbf{\nabla}^{\prime}$). Due to the fact that we are working with
Cartesian coordinates, it is possible to write

$\mathbf{R}_{\mathrm{}}=\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}=\sum_{\eta=x,y,z}(\eta-\eta^{\prime}){\mathbf{e}_{\mathrm{}}}_{\eta},$
(43)

where
$(\mathbf{e}_{\mathrm{x}},\mathbf{e}_{\mathrm{z}},\mathbf{e}_{\mathrm{z}})$
denote the cartesian unit vectors, and then:

${\mathbf{I}_{\mathrm{}}}_{\Omega}=\sum_{\eta=x,y,z}\int_{\Omega}(\eta-\eta^{\prime})F\mathbf{\nabla}^{\prime}\cdot\mathbf{\hat{\boldsymbol{\jmath}}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}{\mathbf{e}_{\mathrm{}}}_{\text{$\eta$}}=\sum_{\eta=x,y,z}I_{\Omega,\eta}{\mathbf{e}_{\mathrm{}}}_{\text{$\eta$}}.$
(44)

Each one of these $\eta$ integrals can be evaluated by means of the identity
$(\eta-\eta^{\prime})F\mathbf{\nabla}^{\prime}\cdot\mathbf{\hat{\boldsymbol{\jmath}}}=\mathbf{\nabla}^{\prime}\cdot\big{[}(\eta-\eta^{\prime})F\mathbf{\hat{\boldsymbol{\jmath}}}\big{]}-\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{\nabla}^{\prime}\big{[}(\eta-\eta^{\prime})F\big{]}$
and the Green-Ostrogradsky’s theorem [13, 14] as follows:

$I_{\Omega,\eta}=\int_{\partial\Omega}(\eta-\eta^{\prime})F\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{n}_{\mathrm{out}}\big{|}_{\partial\Omega}\,\mathrm{d}S^{\prime}-\int_{\Omega}\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{\nabla}^{\prime}\big{[}(\eta-\eta^{\prime})F\big{]}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
(45)

Taking the limit $\Omega\rightarrow\mathbb{R}^{3}$ and keeping in mind that
the boundary term vanishes as $\frac{1}{R}$ we have:

$I_{\mathbb{R}^{3},\eta}=-\int_{\mathbb{R}^{3}}\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{\nabla}^{\prime}\big{[}(\eta-\eta^{\prime})F\big{]}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(46)

The gradient (with respect to the primed coordinates) can be computed
explicitly:

$\displaystyle\mathbf{\nabla}^{\prime}\big{[}(\eta-\eta^{\prime})F\big{]}=$
$\displaystyle
F\mathbf{\nabla}^{\prime}(\eta-\eta^{\prime})+(\eta-\eta^{\prime})\mathbf{\nabla}^{\prime}F$
(47) $\displaystyle=$
$\displaystyle-F\mathbf{e}_{\mathrm{\text{$\eta$}}}-(\eta-\eta^{\prime})\mathbf{R}_{\mathrm{}}\bigg{(}\frac{ik_{0}}{R}-\frac{2}{R^{2}}\bigg{)}F$
(48)

and then

$\mathbf{I}_{\mathrm{\mathbb{R}^{3},\text{$\eta$}}}=\int_{\mathbb{R}^{3}}F\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{e}_{\mathrm{\text{$\eta$}}}+(\eta-\eta^{\prime})\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}}\bigg{(}\frac{ik_{0}}{R}-\frac{2}{R^{2}}\bigg{)}F\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(49)

Plugging Eq. (49) into Eq. (44) we get:

$\mathbf{I}_{\mathrm{\mathbb{R}^{3}}}=\int_{\mathbb{R}^{3}}\bigg{[}\mathbf{\hat{\boldsymbol{\jmath}}}+\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})\bigg{(}\frac{ik_{0}}{R}-\frac{2}{R^{2}}\bigg{)}\bigg{]}F\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(50)

Before taking the inverse Fourier transform, we will try to express the term
between square brackets (which we will call
$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}$) in a more illuminating way. First,
we rearrange $\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}$ as:

$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}=\mathbf{\hat{\boldsymbol{\jmath}}}-\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})\frac{2}{R^{2}}+ik_{0}\frac{\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})}{R}.$
(51)

Now we consider the vector identity
$(\mathbf{\hat{\boldsymbol{\jmath}}}\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}=\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})-R^{2}\mathbf{\hat{\boldsymbol{\jmath}}}$
[14], and from this we have:

$\displaystyle\frac{\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})}{R}=\frac{(\mathbf{\hat{\boldsymbol{\jmath}}}\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}}{R}+R\mathbf{\hat{\boldsymbol{\jmath}}},$
(52)
$\displaystyle\mathbf{\hat{\boldsymbol{\jmath}}}-2\frac{\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})}{R^{2}}=-\frac{(\mathbf{\hat{\boldsymbol{\jmath}}}\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}}{R^{2}}-\frac{\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})}{R^{2}}.$
(53)

Then $\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}$ can be seen as:

$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}=-\bigg{[}\frac{(\mathbf{\hat{\boldsymbol{\jmath}}}\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}}{R^{2}}+\frac{\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})}{R^{2}}\bigg{]}+ik_{0}\bigg{[}\frac{(\mathbf{\hat{\boldsymbol{\jmath}}}\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}}{R}+R\mathbf{\hat{\boldsymbol{\jmath}}}\bigg{]}.$
(54)

Substituting this result into Eq. (50) and then plugging that new integral
into Eq. (38) we finally arrive to the expression:

$\displaystyle-\mathbf{\nabla}\hat{\phi}=$
$\displaystyle\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\hat{\rho}\frac{e^{iRk_{0}}}{R^{3}}\mathbf{R}_{\mathrm{}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
$\displaystyle-\frac{1}{4\pi\epsilon_{0}c}\int_{\mathbb{R}^{3}}ik_{0}\frac{(\mathbf{\hat{\boldsymbol{\jmath}}}\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}}{R^{3}}e^{iRk_{0}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
$\displaystyle+\frac{1}{4\pi\epsilon_{0}c}\int_{\mathbb{R}^{3}}\bigg{[}\frac{(\mathbf{\hat{\boldsymbol{\jmath}}}\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}}{R^{4}}+\frac{\mathbf{R}_{\mathrm{}}(\mathbf{\hat{\boldsymbol{\jmath}}}\cdot\mathbf{R}_{\mathrm{}})}{R^{4}}\bigg{]}e^{iRk_{0}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
$\displaystyle-\frac{1}{4\pi\epsilon_{0}c}\int_{\mathbb{R}^{3}}ik_{0}\mathbf{\hat{\boldsymbol{\jmath}}}(\omega,\mathbf{x}_{\mathrm{}}^{\prime})\frac{e^{iRk_{0}}}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
(55)

From Eq. (37) we recognize the last integral as $-i\omega\mathbf{\hat{A}}$ and
then, after taking the inverse Fourier transform, we get:

$\mathbf{E}_{\mathrm{}}=\mathbf{E}_{\mathrm{c}}+\mathbf{E}_{\mathrm{int}}+\mathbf{E}_{\mathrm{rad}},$
(56)

where $\mathbf{E}_{\mathrm{c}}$ is the Coulomb field,
$\mathbf{E}_{\mathrm{int}}$ the intermediate field and
$\mathbf{E}_{\mathrm{rad}}$ the radiated field given by the integrals:

$\displaystyle\mathbf{E}_{\mathrm{c}}=\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\frac{\rho(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})}{R^{3}}\mathbf{R}_{\mathrm{}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(57)
$\displaystyle\mathbf{E}_{\mathrm{int}}=\frac{1}{4\pi\epsilon_{0}c}\int_{\mathbb{R}^{3}}\bigg{[}\frac{(\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}}{R^{4}}+\frac{(\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})\cdot\mathbf{R}_{\mathrm{}})}{R^{4}}\mathbf{R}_{\mathrm{}}\bigg{]}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(58)
$\displaystyle\mathbf{E}_{\mathrm{rad}}=\frac{1}{4\pi\epsilon_{0}c^{2}}\int_{\mathbb{R}^{3}}\frac{(\partial_{t}\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})\times\mathbf{R}_{\mathrm{}})\times\mathbf{R}_{\mathrm{}}}{R^{3}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(59)

At this point the reader may be wondering the reason for why we have made all
these extra steps when the Jefimenko’s equations already provide an explicit
expression for the electric field. This is because when dealing with the
Jefimenko’s equations, the magnetic field is terms of the electric current
$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}$ and the electric field is in terms
of the distribution of charge $\rho$ [12, 7]. This point of view albeit
intuitively is very clear, makes it difficult to compare the terms
corresponding to the radiation field. By making the manipulations described
above, we have ensured the fact that the intermediate and radiated electric
and magnetic fields are all expressed in terms of the electric current solely.
The utility of this approach will be shown later.

###  3.2 The different approximations to the oscillating source problem

####  3.2.1 The Liénard-Wiechert’s Field approach

The academic problem of describing the $\mathbf{E}_{\mathrm{}}$ and
$\mathbf{B}_{\mathrm{}}$ fields generated by a charged particle that moves
along a given trajectory $\mathbf{u}_{\mathrm{}}(t)$, which are called the
_Liénard-Wiechert_ fields, has been studied in many books [6, 7, 10, 15, 16,
11]. The basic idea is to consider a charge density $\rho$ and an electric
current $\mathbf{j}_{\mathrm{}}$ given by:

$\displaystyle\rho(\mathbf{x}_{\mathrm{}},t)=q\delta(\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t)),$
(60)
$\displaystyle\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=q\delta(\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t))\mathbf{v}_{\mathrm{}}(t),$
(61)

where $\delta$ is a Delta distribution and
$\mathbf{v}_{\mathrm{}}(t)=\frac{d\mathbf{u}_{\mathrm{}}}{dt}$. And from here
there are many ways to tackle the problem of obtaining the
$\mathbf{E}_{\mathrm{}}$ and $\mathbf{B}_{\mathrm{}}$ generated fields:
Jackson [6] and Landau [10] consider an elegant formalism using quadrivector
approach. Panofsky [11] and Heald [15] use the so called Liénard-Wiechert
potentials which can be obtained by direct substitution on equations (22-23)
and then carrying all the necessary derivatives. The deduction of the fields
$\mathbf{E}_{\mathrm{}}$ and $\mathbf{B}_{\mathrm{}}$ following this procedure
can be seen in Ref. [7]. For this section we have decided not to follow any of
these approaches, but rather to proceed by direct substitution of the sources
(60) and (61) into equations (34-35) and (57-59), that is:

$\displaystyle\mathbf{B}_{\mathrm{int}}=\frac{qc\mu_{0}}{4\pi}\int_{\mathbb{R}^{3}}\delta(\mathbf{x}_{\mathrm{}}^{\prime}-\mathbf{u}_{\mathrm{}})\frac{\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}}{R^{2}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(62)
$\displaystyle\mathbf{B}_{\mathrm{rad}}=\frac{qc\mu_{0}}{4\pi}\partial_{t}\int_{\mathbb{R}^{3}}\delta(\mathbf{x}_{\mathrm{}}^{\prime}-\mathbf{u}_{\mathrm{}})\frac{\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}}{cR}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(63)
$\displaystyle\mathbf{E}_{\mathrm{c}}=\frac{q}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\frac{\delta(\mathbf{x}_{\mathrm{}}^{\prime}-\mathbf{u}_{\mathrm{}})}{R^{2}}\mathbf{n}_{\mathrm{}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}$
(64)
$\displaystyle\mathbf{E}_{\mathrm{int}}=\frac{q}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\frac{\delta(\mathbf{x}_{\mathrm{}}^{\prime}-\mathbf{u}_{\mathrm{}})}{R^{2}}\big{[}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})\times\mathbf{n}_{\mathrm{}}+(\boldsymbol{\beta}\cdot\mathbf{n}_{\mathrm{}})\mathbf{n}_{\mathrm{}}\big{]}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(65)
$\displaystyle\mathbf{E}_{\mathrm{rad}}=\frac{q}{4\pi\epsilon_{0}}\partial_{t}\int_{\mathbb{R}^{3}}\frac{\delta(\mathbf{x}_{\mathrm{}}^{\prime}-\mathbf{u}_{\mathrm{}})}{cR}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})\times\mathbf{n}_{\mathrm{}}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(66)

where we have introduced the ususal following short hand conventions:

$\boldsymbol{\beta}:=\frac{\mathbf{v}_{\mathrm{}}}{c},\qquad\mathbf{n}_{\mathrm{}}:=\frac{\mathbf{R}_{\mathrm{}}}{R}.$
(67)

The procedure that is shown in Annexe 2, follows the ideas expressed by Heald
and Marion in Ref. [16], the main difference is that while Heald and Marion
consider the Jefimenko’s equations, we are going to use equations (62-66). We
have decided to include the full deduction of the Liénard-Wiechert fields
because as far as we have seen this result is quoted but the steps towards its
obtention are not shown. Griffiths just states that the deduction is _very
difficult_ and Heald and Marion say that it is necessary to perform _heroic
algebra_. Therefore, we believe that it is important to show, as best as we
can, how to obtain one of the main results in classical electrodynamics.

Once the fields $\mathbf{E}_{\mathrm{}}$ and $\mathbf{B}_{\mathrm{}}$ are
given by Eq. (240) and Eq. (234), namely

$\mathbf{B}_{\mathrm{}}=\frac{qc\mu_{0}}{4\pi}\bigg{[}\frac{\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}}{K^{2}Rc^{2}}+\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{c^{2}K^{3}R}+\frac{(1-\beta^{2})(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{K^{3}R^{2}}\bigg{]}_{t_{r}}$

and

$\mathbf{E}_{\mathrm{}}=\frac{q}{4\pi\epsilon_{0}}\bigg{[}\frac{(1-\beta^{2})(\mathbf{n}_{\mathrm{}}-\boldsymbol{\beta})}{K^{3}R^{2}}+\frac{\mathbf{a}_{\mathrm{}}\times(\mathbf{n}_{\mathrm{}}-\boldsymbol{\beta})\times\mathbf{n}_{\mathrm{}}}{c^{2}K^{3}R}\bigg{]}_{t_{r}},$

with
$\mathbf{a}_{\mathrm{}}=Kc\boldsymbol{\accentset{\mbox{\large.}}{\beta}}$, it
would be easy to think that the fields produced by an oscillating particle
could be retrieved by considering the specific trajectory:

$\mathbf{u}_{\mathrm{}}(t_{r})=a\cos(\omega_{0}t_{r})\mathbf{e}_{\mathrm{z}},$
(68)

where $a$ and $\omega_{0}$ are the oscillation amplitude and frequency
respectively. Nevertheless, as pointed by Spohn in [15], the _Liénard
Wiechert_ fields are less explicit than they appear to be. This is due to the
fact that Eq. (240) and Eq. (234) depend on the retarded time which is still
itself a solution of a (in general non trivial) transcendental equation,
namely:

$t_{r}=t-\frac{|\mathbf{x}_{\mathrm{}}-a\cos(\omega_{0}t_{r})\mathbf{e}_{\mathrm{z}}|}{c}.$
(69)

Let us notice here that if the particle is at constant speed with a straight
trajectory, the _Liénard Wiechert_ fields can be almost straightforwardly [7].
However, the solution of this problem, when dealing with an oscillating
charge, in this case the retarded time is a function of the present time and
the position ($t_{r}:=t_{r}(t,\mathbf{x}_{\mathrm{}})$).

####  3.2.2 The Landau’s Spectral resolution approach

In _The classical theory of fields_ , Landau _et al._ [10] consider that the
fields produced by moving charges can be expanded as a superpostion of
monochromatic waves. Assuming that $\rho(\mathbf{x}_{\mathrm{}},t)$ and
$\mathbf{\jmath}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$ have a Fourier
integral representation, we can write

$\displaystyle\rho(\mathbf{x}_{\mathrm{}},t)$
$\displaystyle=\int_{\omega\in\mathbb{R}}\hat{\rho}(\mathbf{x}_{\mathrm{}},\omega)e^{-i\omega
t}\,\mathrm{d}\omega,$ (70)
$\displaystyle\mathbf{\jmath}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$
$\displaystyle=\int_{\omega\in\mathbb{R}}\mathbf{\hat{\jmath}}(\mathbf{x}_{\mathrm{}},\omega)e^{-i\omega
t}\,\mathrm{d}\omega.$ (71)

According to Landau: _It is clear that each Fourier component of
$\rho(\mathbf{x}_{\mathrm{}},t)$ and
$\mathbf{\jmath}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$ is responsible for the
creation of the corresponding monochromatic component of the field._ Thus, it
is natural to consider the following Fourier integral representations of the
potentials $\phi(\mathbf{x}_{\mathrm{}},t)$ and
$\mathbf{A}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$:

$\displaystyle\phi(\mathbf{x}_{\mathrm{}},t)$
$\displaystyle=\int_{\omega\in\mathbb{R}}\hat{\phi}(\mathbf{x}_{\mathrm{}},\omega)e^{-i\omega
t}\,\mathrm{d}\omega,$ (72)
$\displaystyle\mathbf{A}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$
$\displaystyle=\int_{\omega\in\mathbb{R}}\mathbf{\hat{A}}(\mathbf{x}_{\mathrm{}},\omega)e^{-i\omega
t}\,\mathrm{d}\omega.$ (73)

For the sequel, we will only work with $\phi(\mathbf{x}_{\mathrm{}},t)$,
because all the discussion applies also to
$\mathbf{A}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$. and substituting Eq. (70)
into Eq. (22) we get

$\displaystyle\phi(\mathbf{x}_{\mathrm{}},t)$
$\displaystyle=\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\int_{\omega\in\mathbb{R}}\hat{\rho}(\mathbf{x}_{\mathrm{}},\omega)e^{-i\omega
t_{r}}\,\mathrm{d}\omega\frac{1}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(74) $\displaystyle=\int_{\omega\in\mathbb{R}}e^{-i\omega
t}\bigg{(}\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\frac{e^{i\omega
R/c}}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}\bigg{)}\,\mathrm{d}\omega.$
(75)

Comparing this last equality with Eq. (72), one gets that

$\hat{\phi}(\mathbf{x}_{\mathrm{}},\omega)=\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}^{3}}\hat{\rho}(\mathbf{x}_{\mathrm{}}^{\prime},\omega)\frac{e^{i\omega
R/c}}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$ (76)

Remembering that $\hat{\rho}(\mathbf{x}_{\mathrm{}},\omega)$ is the Fourier
transform of $\rho(\mathbf{x}_{\mathrm{}},\tau)$ and after some manipulations
$\hat{\phi}(\mathbf{x}_{\mathrm{}},\omega)$ can be written as per

$\hat{\phi}(\mathbf{x}_{\mathrm{}},\omega)=\frac{1}{4\pi\epsilon_{0}}\frac{1}{2\pi}\int_{\tau\in\mathbb{R}}\int_{\mathbb{R}^{3}}\rho(\mathbf{x}_{\mathrm{}}^{\prime},\tau)\frac{e^{i\omega(t+|\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}|/c)}}{|\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}|}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}\,\mathrm{d}\tau.$
(77)

Now, we consider the singular charge distribution as in Eq. (60) to obtain

$\hat{\phi}(\mathbf{x}_{\mathrm{}},\omega)=\frac{q}{4\pi\epsilon_{0}}\frac{1}{2\pi}\int_{\tau\in\mathbb{R}}\frac{e^{i\omega(\tau+|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(\tau)|/c)}}{|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(\tau)|}\,\mathrm{d}\tau.$
(78)

Upon substitution of this expression into Eq. (72) we arrive to

$\phi(\mathbf{x}_{\mathrm{}},t)=\frac{q}{4\pi\epsilon_{0}}\frac{1}{2\pi}\int_{\omega\in\mathbb{R}}\int_{\tau\in\mathbb{R}}\frac{e^{i\omega(\tau+|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(\tau)|/c-t)}}{|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(\tau)|}\,\mathrm{d}\tau\,\mathrm{d}\omega,$
(79)

and similarly for $\mathbf{A}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$

$\mathbf{A}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=\frac{q\mu_{0}}{4\pi}\frac{1}{2\pi}\int_{\omega\in\mathbb{R}}\int_{\tau\in\mathbb{R}}\mathbf{v}_{\mathrm{}}(\tau)\frac{e^{i\omega(\tau+|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(\tau)|/c-t)}}{|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(\tau)|}\,\mathrm{d}\tau\,\mathrm{d}\omega.$
(80)

From the above expressions, it’s evident that the right-hand side of the
equation explicitly depends on the present time. This feature avoids issues
related to retarded time, which is a notable problem in the Liénard-Wiechert
fields. However, difficulties emerge when attempting to explicitly compute
these integrals. This arises due to the fact that the particle’s trajectory,
denoted by $\mathbf{u}$, is, in principle, unrestricted in its choice of any
argument $\tau$. Moreover, the exponential function in equations (79) and (80)
necessitates a sweep over all the $(\omega,\tau)$ values in $\mathbb{R}^{2}$.

####  3.2.3 The Raimond’s Taylor series expansion approach

In [9] J.M. Raimond proposes another way to represent the charge density
$\rho(\mathbf{x}_{\mathrm{}},t)=q\delta(\mathbf{x}_{\mathrm{}}-a\cos(\omega_{0}t)\mathbf{e}_{\mathrm{z}})$
by using a Taylor series expansión of the Dirac delta in the sense of
distributions:

$\delta(\mathbf{x}_{\mathrm{}}-g(t))=\delta(\mathbf{x}_{\mathrm{}})-g(t)\mathbf{e}_{\mathrm{z}}\cdot\nabla\delta(\mathbf{x}_{\mathrm{}})+\frac{g^{2}(t)}{2}\mathbf{e}_{\mathrm{z}}\cdot\nabla[\mathbf{e}_{\mathrm{z}}\cdot\nabla\delta(\mathbf{x}_{\mathrm{}})]+\dots,$
(81)

where $g(t)=a\cos(\omega_{0}t)$. In this way one can rewrite the charge
density:

$\rho(\mathbf{x}_{\mathrm{}},t)=\sum_{n=0}^{\infty}\rho_{n}(\mathbf{x}_{\mathrm{}},t)=\sum_{n=0}^{\infty}(g(t))^{n}\varrho_{n}(\mathbf{x}_{\mathrm{}}),$
(82)

where the $\rho_{n}(\mathbf{x}_{\mathrm{}},t)$ is the $n$-th charge density
and $\varrho_{n}(\mathbf{x}_{\mathrm{}})$ is given by

$\varrho_{n}(\mathbf{x}_{\mathrm{}})=\frac{q[-\mathbf{e}_{\mathrm{z}}\cdot\nabla]^{n}\delta(\mathbf{x}_{\mathrm{}})}{n!}=\frac{q(-1)^{n}}{n!}\frac{\partial^{n}\delta(\mathbf{x}_{\mathrm{}})}{\partial
z^{n}}.$ (83)

The reader may identify $\varrho_{n}(\mathbf{x}_{\mathrm{}})$ as the singular
charge distribution of a $2^{n}$-pole (_i.e._ $n=0$ monopole, $n=1$ dipole,
$n=2$ quadrupole, etc.) [17, 18]. Now, by considering the charge conservation
we can write:

$\displaystyle\frac{\partial\rho}{\partial t}=$
$\displaystyle\sum_{n=0}^{\infty}n(g(t))^{n-1}g^{\prime}(t)\frac{q(-1)^{n}}{n!}\frac{\partial^{n}\delta(\mathbf{x}_{\mathrm{}})}{\partial
z^{n}}$ $\displaystyle=$ $\displaystyle 0-\frac{\partial}{\partial
z}\sum_{n=1}^{\infty}(g(t))^{n-1}g^{\prime}(t)\frac{q(-1)^{n-1}}{(n-1)!}\frac{\partial^{n-1}\delta(\mathbf{x}_{\mathrm{}})}{\partial
z^{n-1}}$ $\displaystyle=$ $\displaystyle 0-\frac{\partial}{\partial
z}\sum_{n=1}^{\infty}(g(t))^{n-1}g^{\prime}(t)\varrho_{n-1}(\mathbf{x}_{\mathrm{}})$
$\displaystyle=$ $\displaystyle 0-\frac{\partial}{\partial
z}\sum_{n=1}^{\infty}g^{\prime}(t)\rho_{n-1}(\mathbf{x}_{\mathrm{}},t).$ (84)

From this last expression, we can see that the current distribution
$\mathbf{\jmath}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$ can be written as:

$\mathbf{\jmath}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=\sum_{n=0}^{\infty}\mathbf{\jmath}_{\mathrm{n}}(\mathbf{x}_{\mathrm{}},t),$
(85)

where
$\mathbf{\jmath}_{\mathrm{0}}(\mathbf{x}_{\mathrm{}},t)=\mathbf{0}_{\mathrm{}}$
and
$\mathbf{\jmath}_{\mathrm{n}}(\mathbf{x}_{\mathrm{}},t)=g^{\prime}(t)\rho_{n-1}(\mathbf{x}_{\mathrm{}})\mathbf{e}_{\mathrm{z}}$.
From the above expressions it is easy to see that there is a conservation of
charge between the $n$-th current density
$\mathbf{\jmath}_{\mathrm{n}}(\mathbf{x}_{\mathrm{}},t)$ and the $n$-th charge
density $\rho_{n}(\mathbf{x}_{\mathrm{}},t)$ for $n\geq 0$. Thus one can
consider the charge distribution pairs
$(\rho_{n}(\mathbf{x}_{\mathrm{}},t),\mathbf{\jmath}_{\mathrm{n}}(\mathbf{x}_{\mathrm{}},t))$
in order to obtain the fields generated by an oscillating charge. In this case
the retarded potentials are:

$\displaystyle\phi(\mathbf{x}_{\mathrm{}},t)$
$\displaystyle=\frac{q}{4\pi\epsilon_{0}}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\expectationvalue{\frac{\partial^{n}\delta(\mathbf{x}_{\mathrm{}}^{\prime})}{\partial
z^{\prime
n}},\frac{(g(t-c^{-1}||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||))^{n}}{||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||}},$
$\displaystyle=\frac{q}{4\pi\epsilon_{0}}\sum_{n=0}^{\infty}\frac{1}{n!}\frac{\partial^{n}}{\partial
z^{\prime n}}\bigg{[}\frac{(a\cos(\omega
t-k||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||))^{n}}{||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||}\bigg{]}_{\mathbf{x}_{\mathrm{}}^{\prime}=\mathbf{0}_{\mathrm{}}}.$
(86)

In the same fashion, one can write the vectorial magnetic potential as:

$\displaystyle\mathbf{A}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$
$\displaystyle=\frac{q\mu_{0}}{4\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n-1)!}\expectationvalue{\frac{\partial^{n-1}\delta(\mathbf{x}_{\mathrm{}}^{\prime})}{\partial
z^{\prime(n-1)}},\frac{g^{\prime}(t-c^{-1}||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||)(g(t-c^{-1}||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||))^{n}}{||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||}}\mathbf{e}_{\mathrm{z}}$
$\displaystyle=\frac{q\mu_{0}}{4\pi}\sum_{n=1}^{\infty}\frac{-\omega
a^{2}}{(n-1)!}\frac{\partial^{n-1}}{\partial
z^{\prime(n-1)}}\bigg{[}\frac{\sin(\omega
t-k||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||)(\cos(\omega
t-k||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||))^{n}}{||\mathbf{x}_{\mathrm{}}-\mathbf{x}_{\mathrm{}}^{\prime}||}\bigg{]}_{\mathbf{x}_{\mathrm{}}^{\prime}=\mathbf{0}_{\mathrm{}}}\mathbf{e}_{\mathrm{z}}.$
(87)

The potentials in equations (3.2.3) and (3.2.3) are also expressed in terms of
the present time $t$. However, the multiple derivatives that must be computed
render the expressions impractical. One could argue that for a large value of
$||\mathbf{x}||$, only a few terms are necessary to obtain a good
approximation, as demonstrated in [9] by retaining up to the dipole term.
However, this approach would essentially involve a far-field approximation
once again [6]. In the following section, we will present a more practical and
elegant method for representing an oscillating charge.

##  4 Harmonic decomposition of the sources

As we saw in the previous section, the _Liénard-Wiechert_ fields are not the
best way to obtain the fields produced by an oscillating particle. The main
problem is that the source terms depend on the trajectory that describes the
charge. For this reason it will be convenient to find a way to decompose the
source terms in a polyharmonic way. This idea has been previously considered
by Landau [10] and, as evident from equations (3.2.3) and (3.2.3), also by
Raimond [9], albeit indirectly. However, as demonstrated in the previous
section, the expressions derived from Landau’s and Raimond’s ideas are
challenging to implement in practice. Thus, a new approach for describing the
sources is necessary.

In this section we propose another way inspired in quantum mechanics, which
can be summarized as: _The superposition of waves spread in a certain domain
can be seen as a particle_. Physically, this means that a very localized
source can be seen as the interference of a certain kind of waves.
Mathematically speaking, we are looking to find a sequence such that, we can
have convergence in the sense of distributions to a Dirac delta [19, 20, 21].
It is worth noting that a similar concept is employed in references [22, 23]
for the case of a charged particle in uniform motion.

###  4.1 Two Fourier expansions for the sources

Let us start our analysis by having the charge density $\rho$ and its current
density $\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}$ simply given by:

$\rho(\mathbf{x}_{\mathrm{}},t)=q\delta(\mathbf{x}_{\mathrm{\bot}})\otimes\varrho(z,t),\qquad\mathbf{x}_{\mathrm{\bot}}=(x,y)$
(88)

with $\varrho(z,t):=\delta(z-a\cos(\omega_{0}t))$ and

$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=q\delta(\mathbf{x}_{\mathrm{\bot}})\otimes\jmath(z,t)\mathbf{e}_{\mathrm{z}},$
(89)

with $\jmath(z,t):=-a\omega_{0}\sin(\omega_{0}t)\varrho(z,t)$ out of charge
conservation. Notice that $\varrho$ and $\jmath$ are not multiplied by the
charge. It then turns out that the charge density is not harmonic despite the
harmonic motion of the particle. In other words, no complex function
$\underline{\varrho}(z)$ can be found in such a way that
$\varrho(z,t)=\real\mathrm{e}\\{\underline{\varrho}(z)e^{i\omega t}\\}$. This
is quite important to remark because in references as [6, 17] this is the
starting point when representing and oscillating dipole. Nevertheless it is
apropos to notice that the distribution $\varrho(z,\cdot)$ is a
$\frac{2\pi}{\omega_{0}}$-periodic distribution. Thus, $\varrho$ as well as
$\jmath(z,t)$ can be expanded as a Fourier series (See Annex B) :

$\varrho(z,t)=\sum_{l\in\mathbb{Z}}\varrho_{l}^{F}(z,t),\quad\jmath(z,t)=\sum_{l\in\mathbb{Z}}\jmath_{l}^{F}(z,t),$
(90)

with

$\varrho_{l}^{F}(z,t)=w(z)T_{l}\bigg{(}\frac{z}{a}\bigg{)}e^{+il\omega_{0}t}$
(91)

and

$\boldsymbol{\jmath}_{l}^{F}(z,t)=-a\omega_{0}\sin(\omega_{0}t)w(z)T_{l}\bigg{(}\frac{z}{a}\bigg{)}e^{+il\omega_{0}t},$
(92)

where $T_{l}$ are the Chebyshev polynomials of the first kind [24] and $w(z)$
is the weight function

$w(z):=\frac{1}{\pi\sqrt{a^{2}-z^{2}}}\chi_{[-a,a]}(z)\,$ (93)

with $\chi_{[-a,a]}(z)$ a characteristic function.

###  4.2 Continuity equation for the harmonic components of the sources

In the previous paragraph, an expansion for $\rho$ and
$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}$ are linked by the so-called charge
conservation, namely:
$\mathbf{\nabla}\cdot\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}+\accentset{\mbox{\large.}}{\rho}=0$.
Notice that for moving point particles this equation has to be understood in
the sense of distributions [13, 19, 20, 21]. What about the different
components $\varrho_{l}^{F}$ and $\jmath_{l}^{F}$? In other words, is there
any transference of energy, between the different waves oscillating with the
different frequencies at stake $\omega_{0}$, $2\omega_{0}$, etc$\dots$ ? To
answer this question, we have to care much more about the notation referring
this $l$. While $\varrho_{l}^{F}$ oscillates with frequency $l\omega_{0}$, the
scalar function $\jmath_{l}^{F}$ is a mix of two oscillations with different
frequencies namely $(l-1)\omega_{0}$ and $(l+1)\omega_{0}$ due to the presence
of $\sin(\omega_{0}t)$ term in Eq. (92). Making use of the complex
representation of $\sin(\omega_{0}t)$ it follows:

$\jmath(z,t)=\sum_{l\in\mathbb{Z}}\boldsymbol{\jmath}_{l}^{F}(z,t)=\sum_{l\in\mathbb{Z}}\frac{a\omega_{0}}{2i}w(z)T_{l}(\frac{z}{a})[e^{i(l-1)\omega_{0}t}-e^{i(l+1)\omega_{0}t}].$
(94)

As we can see, this representation is not in a convenient form, instead we
would like to see each term in the series oscillating at frequency
$l\omega_{0}$ (where $l$ is a dummy index), namely:

$\jmath(z,t)=\sum_{l\in\mathbb{Z}}\jmath_{l}^{T}(z)e^{+il\omega_{0}t}.$ (95)

This can be easily achieved after renaming indices ($l+1\rightarrow l$ and
$l-1\rightarrow l$) for the corresponding terms $e^{i(l-1)\omega_{0}t}$ and
$e^{i(l+1)\omega_{0}t}$ which allows to obtain:

$\boldsymbol{\jmath}_{l}^{T}(z):=\frac{a\omega_{0}}{2i}[\textit{$\xi$}_{l+1}(z)-\textit{$\xi$}_{l-1}(z)],$
(96)

where $\textit{$\xi$}_{l}(z)=w(z)T_{l}(\frac{z}{a})$. Analogously for
$\rho(z,t)$ we get:

$\rho(z,t)=\sum_{l\in\mathbb{Z}}\varrho_{l}^{F}(z,t)=\sum_{l\in\mathbb{Z}}\varrho_{l}^{T}(z)e^{+il\omega_{0}t}=\sum_{l\in\mathbb{Z}}\textit{$\xi$}_{l}(z)e^{+il\omega_{0}t}.$
(97)

The arcane meaning of the superscripts $T$ and $F$ is therefore clear: $T$
(resp. $F$) means true (resp. false) in the sense that each spatial
coefficient corresponds to only one $l\in\mathbb{Z}$. Correspondingly the
conservation of charge can be now formulated for each multiple of the
frequency $\omega_{0}$. By the definition of $\rho(\mathbf{x}_{\mathrm{}},t)$
and $\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$ The
conservation of charge implies:

$\partial_{t}\varrho(z,t)+\partial_{z}\jmath(z,t)=0,$ (98)

and plugging the harmonic expansions of $\varrho(z,t)$ and $\jmath(z,t)$ one
gets:

$\sum_{l\in\mathbb{Z}}\bigg{[}il\omega_{0}\textit{$\xi$}_{l}(z)+\frac{a\omega_{0}}{2i}\partial_{z}(\textit{$\xi$}_{l+1}(z)-\textit{$\xi$}_{l-1}(z))\bigg{]}e^{+il\omega_{0}t}=0.$
(99)

Given the fact that $e^{+il\omega_{0}t}$ with $l\in\mathbb{Z}$ is a basis [19,
20] for our polyharmonic decomposition, all the terms between square brackets
are equal to zero. Thus the following identity is obtained:

$l\textit{$\xi$}_{l}(z)=\frac{a}{2}\partial_{z}(\textit{$\xi$}_{l+1}(z)-\textit{$\xi$}_{l-1}(z)).$
(100)

And the conservation of charge for each $l$-th term in the expressions Eq.
(95) and Eq. (97) follows. Upon demonstrating that there is no transfer among
the distinct harmonic components of the charge density
$\rho(\mathbf{x}_{\mathrm{}},t)$ and the current density
$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$, we may
proceed to employ these Fourier expansions to derive the electric and magnetic
induction fields for an oscillating charged particle.This will be show in the
following section.

##  5 The building of E and B via polyharmonic computations

The main consequence of the harmonic decomposition of the sources and the
conservation of charge term by term is that the electric and magnetic
induction fields, respectively $\mathbf{E}_{\mathrm{}}$ and
$\mathbf{B}_{\mathrm{}}$, can be seen as a superposition of fields
$\mathbf{E}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}},t)$ and
$\mathbf{B}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}},t)$, respectively. That is,
fields generated by the harmonic densities of charge
$\varrho_{l}^{T}(z)e^{+il\omega_{0}t}$ and current
$\jmath_{l}^{T}(z)e^{+il\omega_{0}t}$. Each one of these fields oscillates in
terms of multiples of the fundamental frequency $\omega_{0}$. This section is
devoted to this issue starting to work with equations (34-35) and (57-59),
considering
$\mathbf{R}_{\mathrm{}}=\mathbf{x}_{\mathrm{}}-\mathbf{x^{\prime}}_{\mathrm{}}$,
$R=|\mathbf{R}_{\mathrm{}}|$ and $t_{r}=t-\frac{R}{c}$ the retarded time. It
is important to remark here that in this case the retarded time is not in
terms of a transcendental equations but rather explicitly given in terms of
the present time $t$.

###  5.1 The Fourier expansion of the Fields E and B

After We start by applying the delta distribution
$\delta(\mathbf{x}_{\mathrm{\bot}}^{\prime})$ from equations Eq. (88) and Eq.
(89) into the expressions (34-35) and (57-59), which implies we get that
$\mathbf{E}_{\mathrm{}}$ and $\mathbf{B}_{\mathrm{}}$ are given by:

$\displaystyle\mathbf{E}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=$
$\displaystyle\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}}\frac{q}{c\tilde{R}^{3}}\bigg{[}\frac{\jmath(z^{\prime},\tilde{t}_{r})}{\tilde{R}}+\frac{\partial_{t}\jmath(z^{\prime},\tilde{t}_{r})}{c}\bigg{]}(\mathbf{e}_{\mathrm{z}}\times\mathbf{\tilde{R}}_{\mathrm{}})\times\mathbf{\tilde{R}}_{\mathrm{}}dz^{\prime}$
$\displaystyle+\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}}\frac{q}{\tilde{R}^{3}}\bigg{[}\varrho(z^{\prime},\tilde{t}_{r})+\jmath(z^{\prime},\tilde{t}_{r})\frac{\mathbf{e}_{\mathrm{z}}\cdot\mathbf{\tilde{R}}_{\mathrm{}}}{\tilde{R}c}\bigg{]}\mathbf{\tilde{R}}_{\mathrm{}}dz^{\prime},$
(101) $\displaystyle\mathbf{B}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}}\frac{q}{\tilde{R}^{2}}\bigg{[}\frac{\jmath(z^{\prime},\tilde{t}_{r})}{\tilde{R}}+\frac{\partial_{t}\jmath(z^{\prime},\tilde{t}_{r})}{c}\bigg{]}\mathbf{e}_{\mathrm{z}}\times\mathbf{\tilde{R}}_{\mathrm{}}dz^{\prime},$
(102)

where
$\mathbf{\tilde{R}}_{\mathrm{}}=\mathbf{x}_{\mathrm{}}-z^{\prime}\mathbf{e}_{\mathrm{z}}$,
$\tilde{R}=|\mathbf{\tilde{R}}_{\mathrm{}}|$ and
$\tilde{t}_{r}=t-\frac{\tilde{R}}{c}$. The next step is to define the
functions:

$\displaystyle Q(z^{\prime},\mathbf{\tilde{R}}_{\mathrm{}},\tilde{t}_{r})$
$\displaystyle:=\varrho(z^{\prime},\tilde{t}_{r})+\boldsymbol{\jmath}(z^{\prime},\tilde{t}_{r})\frac{\mathbf{e}_{\mathrm{z}}\cdot\mathbf{\tilde{R}}_{\mathrm{}}}{\tilde{R}c},$
(103) $\displaystyle
K(z^{\prime},\mathbf{\tilde{R}}_{\mathrm{}},\tilde{t}_{r})$
$\displaystyle:=\frac{\jmath(z^{\prime},\tilde{t}_{r})}{\tilde{R}}+\frac{\partial_{t}\jmath(z^{\prime},\tilde{t}_{r})}{c}.$
(104)

And then the electric and magnetic fields can be written in a more compact way
as:

$\displaystyle\mathbf{E}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=$
$\displaystyle\frac{1}{4\pi\epsilon_{0}}\int_{\mathbb{R}}\frac{q}{c\tilde{R}^{3}}\,\mathbf{F}_{\mathbf{E}}(\mathbf{x},\mathbf{x}^{\prime})dz^{\prime},$
(105)

with

$\mathbf{F}_{\mathbf{E}}(\mathbf{x},\mathbf{x}^{\prime}):=\bigg{(}cQ(z^{\prime},\tilde{R},\tilde{t}_{r})\mathbf{\tilde{R}}_{\mathrm{}}+K(z^{\prime},\tilde{R},\tilde{t}_{r})(\mathbf{e}_{\mathrm{z}}\times\mathbf{\tilde{R}}_{\mathrm{}})\times\mathbf{\tilde{R}}_{\mathrm{}}\bigg{)}$
(106)

and

$\displaystyle\mathbf{B}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}}\frac{q}{\tilde{R}^{2}}K(z^{\prime},\tilde{R},\tilde{t}_{r})\mathbf{e}_{\mathrm{z}}\times\mathbf{\tilde{R}}_{\mathrm{}}dz^{\prime}.$
(107)

From the definitions of $\varrho(z,t)$ in 205 and $\boldsymbol{\jmath}(z,t)$
in 95 we have that equations (103) and (104) read:

$\displaystyle Q(z^{\prime},\mathbf{\tilde{R}}_{\mathrm{}},\tilde{t}_{r})$
$\displaystyle=\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}t}Q_{l}(z^{\prime},\mathbf{\tilde{R}}_{\mathrm{}}),$
(108) $\displaystyle
K(z^{\prime},\mathbf{\tilde{R}}_{\mathrm{}},\tilde{t}_{r})$
$\displaystyle=\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}t}K_{l}(z^{\prime},\mathbf{\tilde{R}}_{\mathrm{}}),$
(109)

where

$\displaystyle
Q_{l}(z^{\prime},\mathbf{\tilde{R}}_{\mathrm{}}):=\bigg{[}\varrho_{l}^{T}(z^{\prime})+\jmath_{l}^{T}(z^{\prime})\frac{\mathbf{e}_{\mathrm{z}}\cdot\mathbf{\tilde{R}}_{\mathrm{}}}{\tilde{R}c}\bigg{]}e^{-ilk_{0}\tilde{R}},\qquad
k_{0}=\frac{\omega_{0}}{c}$ (110) $\displaystyle
K_{l}(z^{\prime},\mathbf{\tilde{R}}_{\mathrm{}}):=\bigg{[}\frac{1}{\tilde{R}}+\frac{il\omega_{0}}{c}\bigg{]}\jmath_{l}^{T}(z^{\prime})e^{-ilk_{0}\tilde{R}}.$
(111)

Notice that in these expressions there a phase shift $e^{-ilk_{0}\tilde{R}}$
due to the retarded time $t_{r}$. Therefore
$\mathbf{E}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$ and
$\mathbf{B}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)$ can be seen as a
superposition of elementary harmonic terms, _i.e._ :

$\displaystyle\mathbf{E}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=\sum_{l\in\mathbb{Z}}\mathbf{E}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}},t)=\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}t}\mathbf{E}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}}),$
(112)
$\displaystyle\mathbf{B}_{\mathrm{}}(\mathbf{x}_{\mathrm{}},t)=\sum_{l\in\mathbb{Z}}\mathbf{B}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}},t)=\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}t}\mathbf{B}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}}),$
(113)

with spatially dependent coefficients given by:

$\displaystyle\mathbf{E}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}})=\frac{1}{4\pi\epsilon_{0}}\int_{[-a,a]}\frac{q}{c\tilde{R}^{3}}\bigg{(}cQ_{l}(z^{\prime},\tilde{R})\mathbf{\tilde{R}}_{\mathrm{}}+K_{l}(z^{\prime},\tilde{R})(\mathbf{e}_{\mathrm{z}}\times\mathbf{\tilde{R}}_{\mathrm{}})\times\mathbf{\tilde{R}}_{\mathrm{}}\bigg{)}dz^{\prime},$
(114)
$\displaystyle\mathbf{B}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}})=\frac{\mu_{0}}{4\pi}\int_{[-a,a]}\frac{q}{\tilde{R}^{2}}K_{l}(z^{\prime},\tilde{R})\mathbf{e}_{\mathrm{z}}\times\mathbf{\tilde{R}}_{\mathrm{}}dz^{\prime}.$
(115)

where we have used the fact that the support of $\varrho_{l}^{T}(z^{\prime})$
and $\jmath_{l}^{T}(z^{\prime})$ is within the interval $[-a,a]$. These
coefficients can be computed numerically, and thus the building of the
$\mathbf{E}_{\mathrm{}}$ and $\mathbf{B}_{\mathrm{}}$ fields is complete.

###  5.2 A geometrical description of the fields

Despite the complicated appearance of equations (114-115) it is posible to
extract some _a priori_ information about the geometric behaviour of the
($\mathbf{E}_{\mathrm{}}$,$\mathbf{B}_{\mathrm{}}$) fields. Starting by using
the vector identity
$(\mathbf{e}_{\mathrm{z}}\times\mathbf{\tilde{R}}_{\mathrm{}})\times\mathbf{\tilde{R}}_{\mathrm{}}=(\mathbf{e}_{\mathrm{z}}\cdot\mathbf{\tilde{R}}_{\mathrm{}})\mathbf{\tilde{R}}_{\mathrm{}}-\tilde{R}^{2}\mathbf{e}_{\mathrm{z}}$
we can rewrite the spatial coefficients as per:

$\displaystyle\mathbf{E}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}})=$
$\displaystyle\displaystyle\frac{1}{4\pi\epsilon_{0}}\bigg{\\{}\int_{\mathbb{R}}\frac{q}{c\tilde{R}^{3}}\Phi_{l}^{sph}(z^{\prime},\mathbf{x}_{\mathrm{}})\,dz^{\prime}\mathbf{x}_{\mathrm{}}$
$\displaystyle-\int_{\mathbb{R}}\frac{q}{c\tilde{R}^{3}}\Phi_{l}^{flat}(z^{\prime},\mathbf{x}_{\mathrm{}})\,dz^{\prime}\mathbf{e}_{\mathrm{z}}\bigg{\\}},$
(116)

where $\mathbf{E}_{\mathrm{}}^{sph}$ and $\mathbf{E}_{\mathrm{}}^{flat}$ are
defined by

$\Phi_{l}^{flat}(z^{\prime},\mathbf{x}_{\mathrm{}}):=-z^{\prime}\big{(}cQ_{l}(z^{\prime},\tilde{R})+(z-z^{\prime})K_{l}(z^{\prime},\tilde{R})\big{)}+K_{l}(z^{\prime},\tilde{R})\tilde{R}^{2}$
(117)

and

$\Phi_{l}^{sph}(z^{\prime},\mathbf{x}_{\mathrm{}}):=cQ_{l}(z^{\prime},\tilde{R})+(z-z^{\prime})K_{l}(z^{\prime},\tilde{R})$
(118) $\displaystyle\mathbf{B}_{\mathrm{l}}(\mathbf{x}_{\mathrm{}})=$
$\displaystyle\frac{\mu_{0}}{4\pi}\int_{\mathbb{R}}\frac{q}{\tilde{R}^{2}}K_{l}(z^{\prime},\tilde{R})\mathbf{e}_{\mathrm{z}}\times\mathbf{x}_{\mathrm{\perp}}dz^{\prime},$
(119)

From this representation it is easy to see that the first integral term of
$\mathbf{E}_{\mathrm{}}$, which will be called $\mathbf{E}_{\mathrm{}}^{sph}$,
is a field with spherical symmetry. However due to the action of the second
integral term, in the sequel $\mathbf{E}_{\mathrm{}}^{flat}$, the total field
is flattened out in the direction perpendicular to the motion. On the other
hand, the field lines of $\mathbf{B}_{\mathrm{}}$ circle around the $z$-axis
and, as expected, are perpendicular to the field lines of the electric field.
Figure 1 and ( resp.3) show imaginary (resp. real) part of the the harmonic
field components $\mathbf{E}_{\mathrm{l}}$ (resp. $\mathbf{B}_{\mathrm{l}}$)
for $l=1,2,3,4$. Whereas figure 5 represents the real part of the Poynting
vector
$\mathbf{S}_{\mathrm{l}}=\frac{1}{2\mu_{0}}\mathbf{E}_{\mathrm{l}}\times\mathbf{B}_{\mathrm{l}}$.
Albeit the electric and magnetic fields seem to be more or less the same, the
figures that show the fields at the canonical planes $y=0$ and $x=0$ reveal a
quite different behavior, that is: the electric field shows the expected
geometrical behavior (this is more evident for fig. 1(d)) and the magnetic
field circles around the $z$-axis (see for instance fig. 3(a) ). Finally, the
projection of the Poynting vector field on the canonical planes is shown in
figure 6.

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 1: Harmonic components of the electric field generated by an
oscillating particle, $\mathbf{E}_{\mathrm{l}}$ for l = $1,2,3,4$

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 2: Harmonic components of the electric field generated by an
oscillating particle, $\mathbf{E}_{\mathrm{l}}$ for l = $1,2,3,4$ (cut in
canonical planes view)

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 3: Harmonic components of the real part of magnetic induction field
generated by an oscillating particle, $\mathbf{B}_{\mathrm{l}}$ for l =
$1,2,3,4$ (cut in canonical planes view)

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 4: Harmonic components of the real part of magnetic induction field
generated by an oscillating particle, $\mathbf{B}_{\mathrm{l}}$ for l =
$1,2,3,4$ (cut in canonical planes view)

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 5: Harmonic components of the real part of Poynting vector field
generated by an oscillating particle, $\mathbf{S}_{\mathrm{l}}$ for l =
$1,2,3,4$ (cut in canonical planes view)

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 6: Harmonic components of the real part of Poynting vector field
generated by an oscillating particle, $\mathbf{S}_{\mathrm{l}}$ for l =
$1,2,3,4$ (cut in canonical planes view)

##  6 Far Field radiated power

As it was mentioned before, the main approach towards the study of the EM
field generated by an oscillating charge requires to consider the _far field
approximation_. In this section we will demonstrate how, by using the
polyharmonic expression from previous sections, it is possible to retrive the
same results as reported in the literature [6].

###  6.1 The polyharmonic representation of the EM far field approximations

As we know, the radiated electric and magnetic induction fields are given by:

$\displaystyle\mathbf{B}_{\mathrm{rad}}=q\frac{\mu_{0}}{4\pi
c}\int_{\mathbb{R}^{3}}\frac{\partial_{t}\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})\times\mathbf{n}_{\mathrm{}}}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime},$
(120)
$\displaystyle\mathbf{E}_{\mathrm{rad}}=q\frac{1}{4\pi\epsilon_{0}c^{2}}\int_{\mathbb{R}^{3}}\frac{(\partial_{t}\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r})\times\mathbf{n}_{\mathrm{}})\times\mathbf{n}_{\mathrm{}}}{R}\,\mathrm{d}\mathbf{x}_{\mathrm{}}^{\prime}.$
(121)

where $\mathbf{n}_{\mathrm{}}=\frac{\mathbf{R}_{\mathrm{}}}{R}$. Using the
fact that the electric current is defined as
$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},t_{r}):=q\delta(\mathbf{x}_{\mathrm{\bot}})\otimes\boldsymbol{\jmath}(z,t_{r})\mathbf{e}_{\mathrm{z}}$,
the fields read:

$\displaystyle\mathbf{B}_{\mathrm{rad}}=q\frac{\mu_{0}}{4\pi
c}\int_{\mathbb{R}}\frac{\partial_{t}\boldsymbol{\jmath}(\mathbf{x}_{\mathrm{}}^{\prime},\tilde{t}_{r})\mathbf{e}_{\mathrm{z}}\times\mathbf{\tilde{n}}_{\mathrm{}}}{\tilde{R}}dz^{\prime},$
(122)
$\displaystyle\mathbf{E}_{\mathrm{rad}}=q\frac{1}{4\pi\epsilon_{0}c^{2}}\int_{\mathbb{R}}\frac{(\partial_{t}\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}(\mathbf{x}_{\mathrm{}}^{\prime},\tilde{t}_{r})\times\mathbf{\tilde{n}}_{\mathrm{}})\times\mathbf{\tilde{n}}_{\mathrm{}}}{\tilde{R}}dz^{\prime}.$
(123)

and as we did before
$\mathbf{\tilde{R}}_{\mathrm{}}=\mathbf{x}_{\mathrm{}}-z^{\prime}\mathbf{e}_{\mathrm{z}}$,
$\mathbf{\tilde{n}}_{\mathrm{}}=\frac{\mathbf{\tilde{R}}_{\mathrm{}}}{\tilde{R}}$,
and $\tilde{t}_{r}=t-\frac{\tilde{R}}{c}$. Now, in order to compute the fields
radiated _at infinitum_ we make the following approximations for when
$|\mathbf{x}_{\mathrm{}}^{\prime}|\ll|\mathbf{x}_{\mathrm{}}|$:

$\displaystyle\mathbf{\tilde{n}}_{\mathrm{}}\simeq\frac{\mathbf{x}_{\mathrm{}}}{|\mathbf{x}_{\mathrm{}}|}=\boldsymbol{\nu},$
(124)
$\displaystyle\frac{1}{\tilde{R}}\simeq\frac{1}{|\mathbf{x}_{\mathrm{}}|},$
(125) $\displaystyle\tilde{t}_{r}\simeq
t-\frac{|\mathbf{x}_{\mathrm{}}|}{c}+\bigg{(}\frac{z}{|\mathbf{x}_{\mathrm{}}|}\bigg{)}\bigg{(}\frac{z^{\prime}}{c}\bigg{)}=\tau_{r}.$
(126)

By plugging these approximations into equations (122) and (123), one arrives
to the expressions:

$\displaystyle\mathbf{B}_{\mathrm{rad}}\simeq\mathbf{B}_{\mathrm{\infty}}=$
$\displaystyle q\frac{\mu_{0}}{4\pi
c|\mathbf{x}_{\mathrm{}}|}\int_{\mathbb{R}}\partial_{t}\boldsymbol{\jmath}(z^{\prime},\tau_{r})dz^{\prime}\mathbf{e}_{\mathrm{z}}\times\boldsymbol{\nu},$
(127)
$\displaystyle\mathbf{E}_{\mathrm{rad}}\simeq\mathbf{E}_{\mathrm{\infty}}=$
$\displaystyle
q\frac{1}{4\pi\epsilon_{0}c^{2}|\mathbf{x}_{\mathrm{}}|}\int_{\mathbb{R}}\partial_{t}\boldsymbol{\jmath}(z^{\prime},\tau_{r})dz^{\prime}(\mathbf{e}_{\mathrm{z}}\times\boldsymbol{\nu})\times\boldsymbol{\nu},$
$\displaystyle=$ $\displaystyle
c\mathbf{B}_{\mathrm{\infty}}\times\boldsymbol{\nu}.$ (128)

From this last equation it is clear that we can focus our attention just in
$\mathbf{B}_{\mathrm{\infty}}$. Therefore, the next step is to consider its
polyharmonic representation. Remembering the definition of
$\boldsymbol{\jmath}(z^{\prime},\tau_{r})$ in equation (95) and $\tau_{r}$ in
(126) we have that

$\mathbf{B}_{\mathrm{\infty}}=\sum_{l\in\mathbb{Z}}e^{il\omega_{0}t}\mathbf{B}_{\mathrm{\infty}}^{l},$
(129)

with the spatial coefficient $\mathbf{B}_{\mathrm{\infty}}^{l}$ defined as:

$\mathbf{B}_{\mathrm{\infty}}^{l}:=il\omega_{0}\frac{q\mu_{0}}{4\pi
c|\mathbf{x}_{\mathrm{}}|}e^{-ilk_{0}|\mathbf{x}_{\mathrm{}}|}\int_{\mathbb{R}}\boldsymbol{\jmath}_{l}^{T}(z^{\prime})\exp[ilk_{0}\frac{z}{|\mathbf{x}_{\mathrm{}}|}z^{\prime}\bigg{]}dz^{\prime}(\mathbf{e}_{\mathrm{z}}\times\boldsymbol{\nu}).$
(130)

Here it is important to emphasize that $\mathbf{B}_{\mathrm{\infty}}^{0}=0$.
Thus the results obtained in this section are for $l\neq 0$. Notice that the
integral term resembles to a spatial Fourier transform that goes from
$z^{\prime}\rightarrow\eta_{l}$ with the new variable

$\eta_{l}:=lk_{0}\frac{z}{|\mathbf{x}_{\mathrm{}}|}.$ (131)

Thus

$\mathbf{B}_{\mathrm{\infty}}^{l}=il\omega_{0}\frac{q\mu_{0}}{4\pi
c|\mathbf{x}_{\mathrm{}}|}e^{-ilk_{0}|\mathbf{x}_{\mathrm{}}|}2\pi\mathcal{F}_{z^{\prime}\rightarrow\eta_{l}}\big{\\{}\boldsymbol{\jmath}_{l}^{T}(z^{\prime})\big{\\}}(\mathbf{e}_{\mathrm{z}}\times\boldsymbol{\nu}).$
(132)

This Fourier transform can be easily computed by using the definition in
Eq.(96):

$2\pi\mathcal{F}_{z^{\prime}\rightarrow\eta_{l}}\big{\\{}\boldsymbol{\jmath}_{l}^{T}(z^{\prime})\big{\\}}=2\pi\bigg{[}\mathcal{F}_{z^{\prime}\rightarrow\eta_{l}}\big{\\{}\textit{$\xi$}_{l+1}(z^{\prime})-\mathcal{F}_{z^{\prime}\rightarrow\eta_{l}}\big{\\{}\textit{$\xi$}_{l-1}(z^{\prime})\big{\\}}\bigg{]}\frac{a\omega_{0}}{2i}$
(133)

Fortunately, the Fourier transform of $\textit{$\xi$}_{l}$ is given in
equation (206) and after using identity (242) one gets:

$2\pi\mathcal{F}_{z^{\prime}\rightarrow\eta_{l}}\big{\\{}\boldsymbol{\jmath}_{l}^{T}(z^{\prime})\big{\\}}=ai^{l}l\omega_{0}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a},$
(134)

which implies that:

$\displaystyle\mathbf{B}_{\mathrm{\infty}}^{l}$
$\displaystyle=i^{l+1}\frac{q\mu_{0}}{4\pi
c}(l\omega_{0})^{2}\frac{a}{|\mathbf{x}_{\mathrm{}}|}e^{-ilk_{0}|\mathbf{x}_{\mathrm{}}|}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a}(\mathbf{e}_{\mathrm{z}}\times\boldsymbol{\nu}),$
(135)
$\displaystyle=i^{l+1}\frac{\mu_{0}}{4\pi}c(lk_{0})^{2}qa\frac{e^{-ilk_{0}|\mathbf{x}_{\mathrm{}}|}}{|\mathbf{x}_{\mathrm{}}|}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a}(\mathbf{e}_{\mathrm{z}}\times\boldsymbol{\nu}),$
(136)
$\displaystyle=\frac{\mu_{0}}{4\pi}c(lk_{0})^{2}\frac{e^{-ilk_{0}|\mathbf{x}_{\mathrm{}}|}}{|\mathbf{x}_{\mathrm{}}|}(\boldsymbol{\nu}\times\mathbf{p}_{\mathrm{z}})\bigg{(}i^{l-1}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a}\bigg{)},$
(137)

where we have defined the dipole moment vector
$\mathbf{p}_{\mathrm{z}}=qa\mathbf{e}_{\mathrm{z}}$. Notice that this last
expression is very similar to the one in [6] for the _magnetic dipole field_
in the radiation zone. Calling

$A_{l}=i^{l-1}\frac{\mu_{0}}{4\pi}c(lk_{0})^{2}e^{-ilk_{0}|\mathbf{x}_{\mathrm{}}|},$
(138)

we get:

$\mathbf{B}_{\mathrm{\infty}}^{l}=A_{l}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a}\frac{(\boldsymbol{\nu}\times\mathbf{p}_{\mathrm{z}})}{|\mathbf{x}_{\mathrm{}}|},$
(139)

and its square norm is given by:

$\displaystyle|\mathbf{B}_{\mathrm{\infty}}^{l}|^{2}=$
$\displaystyle\frac{|A_{l}|^{2}}{|\mathbf{x}_{\mathrm{}}|^{2}}\bigg{(}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a}\bigg{)}^{2}\bigg{[}1-\bigg{(}\frac{z}{|\mathbf{x}_{\mathrm{}}|}\bigg{)}^{2}\bigg{]}|\mathbf{p}_{\mathrm{z}}|^{2},$
(140) $\displaystyle=$
$\displaystyle\frac{|A_{l}|^{2}}{|\mathbf{x}_{\mathrm{}}|^{2}}\bigg{(}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a}\bigg{)}^{2}\bigg{[}1-\bigg{(}\frac{a\eta_{l}}{alk_{0}}\bigg{)}^{2}\bigg{]}|\mathbf{p}_{\mathrm{z}}|^{2},$
(141)

where it has been used the fact that:
$\eta_{l}=lk_{0}\frac{z}{|\mathbf{x}_{\mathrm{}}|}$. This last result will be
used in the study of the radiated power.

###  6.2 Radiated power

The expression for the far field radiated power for an oscillating will be
derived. First for the relativistic case and later for the non relativistic
one.

####  6.2.1 Relativistic case

We know from the definition of the Poynting vector and (6.1) that:

$\mathbf{S}_{\mathrm{\infty}}=\mathbf{E}_{\mathrm{\infty}}\times\mathbf{H}_{\mathrm{\infty}}=\frac{c}{\mu_{0}}(\mathbf{B}_{\mathrm{\infty}}\times\boldsymbol{\nu})\times\mathbf{B}_{\mathrm{\infty}}=\frac{c}{\mu_{0}}|\mathbf{B}_{\mathrm{\infty}}|^{2}\boldsymbol{\nu}.$
(142)

Remembering the Fourier expansion of $\mathbf{B}_{\mathrm{\infty}}$ in Eq.
(129) we have that:

$|\mathbf{B}_{\mathrm{\infty}}|^{2}=\sum_{\begin{subarray}{c}l\in\mathbb{Z}\\\
l\neq 0\end{subarray}}\sum_{\begin{subarray}{c}m\in\mathbb{Z}\\\ m\neq
0\end{subarray}}e^{i(l-m)\omega_{0}t}\mathbf{B}_{\mathrm{\infty}}^{l}\overline{\mathbf{B}_{\mathrm{}}}^{m}_{\infty}.$
(143)

Now, we consider the time-averaged Poynting vector, given the fact that
$\mathbf{B}_{\mathrm{\infty}}$ is $\frac{2\pi}{\omega_{0}}$-periodic we
integrate Eq. (142) from $-\frac{\pi}{\omega_{0}}$ to $\frac{\pi}{\omega_{0}}$
to have:

$\innerproduct{\mathbf{S}_{\mathrm{\infty}}}{\mathbf{S}_{\mathrm{\infty}}}=\frac{c}{\mu_{0}}\innerproduct{|\mathbf{B}_{\mathrm{\infty}}|^{2}}{|\mathbf{B}_{\mathrm{\infty}}|^{2}}\boldsymbol{\nu},$
(144)

where the quantities between brackets are time averaged. By the orthogonality
of the complex exponential we have:

$\innerproduct{|\mathbf{B}_{\mathrm{\infty}}|^{2}}{|\mathbf{B}_{\mathrm{\infty}}|^{2}}=\sum_{\begin{subarray}{c}l\in\mathbb{Z}\\\
l\neq
0\end{subarray}}\frac{|A_{l}|^{2}}{|\mathbf{x}_{\mathrm{}}|^{2}}\bigg{(}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a}\bigg{)}^{2}\bigg{[}1-\bigg{(}\frac{a\eta_{l}}{alk_{0}}\bigg{)}^{2}\bigg{]}|\mathbf{p}_{\mathrm{z}}|^{2}.$
(145)

For obtaining the power it is customary to calculate the flux of the averaged
Poynting vector through a spherical surface of radius $R_{0}$ (it could be any
surface encompassing the EM sources, but the spherical surface is the one that
allows the simplest calculations). That is:

$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}\innerproduct{\mathbf{S}_{\mathrm{\infty}}}{\mathbf{S}_{\mathrm{\infty}}}R_{0}^{2}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi$
$\displaystyle=\frac{c}{\mu_{0}}\int_{0}^{2\pi}\int_{0}^{\pi}\innerproduct{|\mathbf{B}_{\mathrm{\infty}}|^{2}}{|\mathbf{B}_{\mathrm{\infty}}|^{2}}R_{0}^{2}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi,$
(146) $\displaystyle=\sum_{\begin{subarray}{c}l\in\mathbb{Z}\\\ l\neq
0\end{subarray}}\frac{2\pi
c}{\mu_{0}}|A_{l}|^{2}\int_{0}^{\pi}\bigg{(}\frac{\mathrm{J}_{l}(\eta_{l}a)}{\eta_{l}a}\bigg{)}^{2}\bigg{[}1-\bigg{(}\frac{a\eta_{l}}{alk_{0}}\bigg{)}^{2}\bigg{]}|\mathbf{p}_{\mathrm{z}}|^{2}\sin\theta\,\mathrm{d}\theta.$
(147)

Next, we perform the change of variable:

$u=a\eta_{l}=alk_{0}\frac{z}{R_{0}}=alk_{0}\cos\theta$ (148)

then the differential $du$ reads

$\mathrm{d}u=-alk_{0}\sin\theta\,\mathrm{d}\theta.$ (149)

After these calculations, the flux of the averaged Poynting vector reads:

$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}\innerproduct{\mathbf{S}_{\mathrm{\infty}}}{\mathbf{S}_{\mathrm{\infty}}}R_{0}^{2}\sin\theta
d\theta d\phi=$ (150) $\displaystyle\sum_{\begin{subarray}{c}l\in\mathbb{Z}\\\
l\neq 0\end{subarray}}\frac{2\pi
c}{\mu_{0}alk_{0}}|A_{l}|^{2}\int_{-alk_{0}}^{alk_{0}}\bigg{(}\frac{\mathrm{J}_{l}(u)}{u}\bigg{)}^{2}\bigg{[}1-\bigg{(}\frac{u}{alk_{0}}\bigg{)}^{2}\bigg{]}|\mathbf{p}_{\mathrm{z}}|^{2}du.$

Calling:

$D_{l}:=\frac{c(2\pi)^{2}}{\omega_{0}\mu_{0}alk_{0}}|A_{l}|^{2}=\frac{2\pi
c}{\mu_{0}alk_{0}}\bigg{(}\frac{\mu_{0}}{4\pi}c(lk_{0})^{2}\bigg{)}^{2}=\frac{Z_{0}c^{2}(lk_{0})^{4}}{8\pi(alk_{0})},$
(151)

and because $D_{-l}=-D_{l}$ one has that the averaged energy flux can be seen
as:

$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}\innerproduct{\mathbf{S}_{\mathrm{\infty}}}{\mathbf{S}_{\mathrm{\infty}}}R_{0}^{2}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi$
$\displaystyle=\sum_{l=1}^{\infty}2D_{l}\int_{-alk_{0}}^{alk_{0}}\bigg{(}\frac{\mathrm{J}_{l}(u)}{u}\bigg{)}^{2}\bigg{[}1-\bigg{(}\frac{u}{alk_{0}}\bigg{)}^{2}\bigg{]}\,\mathrm{d}u|\mathbf{p}_{\mathrm{z}}|^{2},$
(152) $\displaystyle=\sum_{l=1}^{\infty}P_{l}$ (153)

with $P_{l}$ the $l$-th contribution given by:

$\displaystyle P_{l}$
$\displaystyle=4D_{l}\int_{0}^{alk_{0}}\bigg{(}\frac{\mathrm{J}_{l}(u)}{u}\bigg{)}^{2}\bigg{[}1-\bigg{(}\frac{u}{alk_{0}}\bigg{)}^{2}\bigg{]}\,\mathrm{d}u|\mathbf{p}_{\mathrm{z}}|^{2},\qquad
l\geq 1,$ (154)
$\displaystyle=4D_{l}\bigg{[}\int_{0}^{alk_{0}}\frac{\mathrm{J}_{l}^{2}(u)}{u^{2}}\,\mathrm{d}u-\frac{1}{(alk_{0})^{2}}\int_{0}^{alk_{0}}\mathrm{J}_{l}^{2}(u)\,\mathrm{d}u\bigg{]}|\mathbf{p}_{\mathrm{z}}|^{2}$
(155)

where by means of the identities obtained in Annexe D one can see that the
integrals are given by:

$\displaystyle\mathcal{I}_{l}^{S}(alk_{0})=\int_{0}^{alk_{0}}\frac{\mathrm{J}_{l}^{2}(u)}{u^{2}}\,\mathrm{d}u=\frac{alk_{0}}{2l}\bigg{[}\frac{\mathrm{J}_{l}\mathrm{J}^{\prime}_{l+1}-\mathrm{J}_{l+1}\mathrm{J}^{\prime}_{l}}{1+2l}+\frac{\mathrm{J}_{l}\mathrm{J}^{\prime}_{l-1}-\mathrm{J}_{l-1}\mathrm{J}^{\prime}_{l}}{1-2l}\bigg{]}_{alk_{0}}\quad
l\geq 1,$ (156)
$\displaystyle\mathcal{I}_{l}^{R}(alk_{0})=\int_{0}^{alk_{0}}\mathrm{J}_{l}^{2}(u)\,\mathrm{d}u=\bigg{[}\mathcal{I}_{l-2}^{R}-\frac{2(l-1)}{alk_{0}}\mathrm{J}_{l-1}^{2}-2(l-1)\mathcal{I}_{l-1}^{S}\bigg{]}_{alk_{0}}\quad
l>1,$ (157)
$\displaystyle\mathcal{I}_{1}^{R}(ak_{0})=\int_{0}^{ak_{0}}\mathrm{J}_{1}^{2}(u)\,\mathrm{d}u=\big{[}\mathcal{I}_{0}^{R}-ak_{0}(\mathrm{J}_{1}^{2}-\mathrm{J}_{0}^{2})\big{]}_{ak_{0}}\quad
l=1,$ (158)

where the function $\mathcal{I}^{R}_{0}$ is defined trough an integral (see
261). By employing equations (156)-(158) it is possible to compute in a semi
analytical manner the radiated power flux. Let us In particular, if we focus
our attention to the case when $l=1$:

$P_{1}=4D_{1}\int_{0}^{ak_{0}}\bigg{(}\frac{\mathrm{J}_{1}(u)}{u}\bigg{)}^{2}\bigg{[}1-\bigg{(}\frac{u}{ak_{0}}\bigg{)}^{2}\bigg{]}\,\mathrm{d}u|\mathbf{p}_{\mathrm{z}}|^{2},$
(159)

or in a more explicit fashion

$P_{1}=4D_{l}\bigg{\\{}\frac{ak_{0}}{2}\bigg{[}\frac{\mathrm{J}_{1}\mathrm{J}_{2}^{\prime}-\mathrm{J}_{2}\mathrm{J}_{1}^{\prime}}{3}+\mathrm{J}_{1}\mathrm{J}_{0}^{\prime}-\mathrm{J}_{0}\mathrm{J}_{1}^{\prime}\bigg{]}_{ak_{0}}+\\\
\frac{\mathcal{I}_{0}^{R}(ak_{0})}{ak_{0}}-\frac{[\mathrm{J}_{1}^{2}(ak_{0})+\mathrm{J}_{0}^{2}(ak_{0})]}{(ak_{0})^{2}}\bigg{\\}}|\mathbf{p}_{\mathrm{z}}|^{2}.$
(160)

It is important to notice that in this case we have made no restrictions
regarding the velocity of the charged particle, albeit it can not be faster
than $c$ (this will be studied in a more deeper way in subsection 6.3). Thus,
the expression derived here are valid even for charges whith speed near $c$.

####  6.2.2 Non-relativistic case

We are then in the case of $alk_{0}\ll 1$. For $l=1$, we have $0<u<ak_{0}\ll
1$. In that case, the behavior of $\mathrm{J}_{1}$ near the origin is well
known namely $\mathrm{J}_{1}(u)\sim\frac{u}{2}$ and as a result:

$\frac{\mathrm{J}_{1}^{2}(u)}{u^{2}}\sim\frac{1}{4}$ (161)

and we obtain

$P_{1}\sim
D_{1}\,|\mathbf{p}_{\mathrm{z}}|^{2}\int_{0}^{ak_{0}}\left(1-\frac{u^{2}}{(ak_{0})^{2}}\right)\,\,\mathrm{d}u$
(162)

i.e.

$P_{1}\sim
D_{1}\frac{2}{3}ak_{0}|\mathbf{p}_{\mathrm{z}}|^{2}=\frac{Z_{0}c^{2}(k_{0})^{4}}{8\pi(ak_{0})}\frac{2}{3}ak_{0}|\mathbf{p}_{\mathrm{z}}|^{2}=\frac{Z_{0}c^{2}(k_{0})^{4}}{12\pi}|\mathbf{p}_{\mathrm{z}}|^{2}.$
(163)

The reader may recognize this last equality as the expression for the total
power radiated in [6]. Finally, figure 7 shows a comparison between the power
$P_{l}$ (for $l=1,2,3$) obtained analytically by means of 154 when using the
far field approximation and the power numerically computed by considering the
Poynting vector flux across the surface of a pill box that encloses the
trajectory of the charged particle. The Poynting vector is obtained via
equations (114-115) and the numerical integrations is performed by using the
solver GetDP [25]. As we can see, both approaches fit perfectly.

Figure 7: Comparison of the powers $P_{l}$ ($l=1,2,3$) obtained analytically
and numerically.

###  6.3 The particle cannot be supraluminal

It is common sense that for the $l-th$ component of the magnetic induction
field in 127, namely $\mathbf{B}_{\infty}^{l}$, the $l$-th averaged Poynting
vector
$\innerproduct{\boldsymbol{S}_{\infty}^{l}}{\boldsymbol{S}_{\infty}^{l}}:=\frac{c}{\mu_{0}}|\mathbf{B}_{\mathrm{\infty}}^{l}|^{2}$
(see 6.1) as well as for the power $P_{l}$ (given by equations (150-154)) in
the value converges with growing $l$. A divergence would result in the total
value of infinity. Due to the connection between $\mathbf{B}_{\infty}^{l}$,
$\innerproduct{\boldsymbol{S}_{\infty}^{l}}{\boldsymbol{S}_{\infty}^{l}}$ and
$\mathcal{P}_{l}$ it follows, that if the magnetic field diverges, the two
others diverge, too.
The $l^{th}$ magnetic field depends on $l$ only in

$\mathbf{B}_{\infty}^{l}\sim l\cdot\mathrm{J}_{l}(l\cdot ak_{0}\cos\theta)$
(164)

For large $l$, $\mathrm{J}_{l}(l\beta)$ can be written as follows [26]:

$\mathrm{J}_{l}(l\leavevmode\nobreak\
ak_{0}\cos\theta)\sim\frac{e^{l\,(\tanh(a_{0})-a_{0})}}{\sqrt{\frac{1}{2}\pi\,l\,\tanh(a_{0})}}\
\ ,\ \text{with
}a_{0}=\mathrm{arccosh}\left(\frac{1}{ak_{0}\cos\theta}\right)$ (165)

With this, $\mathbf{B}_{\infty}^{l}$ goes with

$\mathbf{B}_{\infty}^{l}\sim\sqrt{\frac{2l}{\pi\sqrt{1-(ak_{0}\cos\theta)^{2}}}}e^{l(\sqrt{1-(ak_{0}\cos\theta)^{2}}-\mathrm{arcosh}(\frac{1}{ak_{0}\cos\theta}))}$
(166)

This only converges with $l\rightarrow\infty$ if the argument of the
exponential function is negative. That leads in the form of Eq. 165 to the
following inequalities:

$\displaystyle a_{0}>\mathrm{tanh}(a_{0})$ $\displaystyle\Leftrightarrow
a_{0}>0$ $\displaystyle\leftrightarrow 0<ak_{0}\cos\theta<1$
$\displaystyle\Leftrightarrow ak_{0}<1$ $\displaystyle\Leftrightarrow
v:=a\omega_{0}<c$ (167)

With $v=a\omega_{0}$ the maximum velocity of our point charge in his
sinusoidal movement, the condition for convergence is matched in the
physically sense of Albert Einstein’s postulate that nothing is faster than
light [7, 6]. It is important to remark here the fact that this result holds
irrespective of whether the particle has got a mass.

##  7 Obtaining the diffracted field

After all this work, we have that the fields $\mathbf{E}_{\mathrm{}}^{0}$ and
$\mathbf{B}_{\mathrm{}}^{0}$ of the system of equations (10-13) can be seen as
a superposition of waves in the form of Eq. (114) and Eq. (115). Even more,
the analysis fo the radiated energy from previous section, show that from our
expressions of $\mathbf{E}_{\mathrm{}}^{0}$ and $\mathbf{B}_{\mathrm{}}^{0}$
one can retrieve the classical results when the non relativistic far field
approximations are considered. The second part of this chapter work (which is
going to be considerably shorter that the first one) can be tackled in a
straight forward fashion. First, the sources in Eq. (18) can be easily
retrieved by noticing that:

$\displaystyle[\epsilon_{r,II}-2\pi\delta]*\mathbf{E}_{\mathrm{}}^{0}=$
$\displaystyle\frac{1}{2\pi}\int_{\tau\in\mathbb{R}}[\epsilon_{r,II}(\tau)-2\pi\delta]\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}(t-\tau)}\mathbf{E}_{\mathrm{l}}^{0}(\mathbf{x}_{\mathrm{}})d\tau,$
$\displaystyle=$
$\displaystyle\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}t}\mathbf{E}_{\mathrm{l}}^{0}(\mathbf{x}_{\mathrm{}})\frac{1}{2\pi}\int_{\tau\in\mathbb{R}}[\epsilon_{r,II}(\tau)-2\pi\delta]e^{-il\omega_{0}\tau}d\tau,$
$\displaystyle=$
$\displaystyle\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}t}\mathbf{E}_{\mathrm{l}}^{0}(\mathbf{x}_{\mathrm{}})[\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})-1].$
(168)

Then:

$\displaystyle\rho^{0}_{II}=$
$\displaystyle-\epsilon_{0}\sum_{l\in\mathbb{Z}}[\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})-1]e^{+il\omega_{0}t}\mathbf{\nabla}\cdot\mathbf{E}_{\mathrm{l}}^{0}(\mathbf{x}_{\mathrm{}}),$
(169) $\displaystyle\mathbf{\boldsymbol{\jmath}}_{\mathrm{II}}^{0}=$
$\displaystyle\epsilon_{0}\sum_{l\in\mathbb{Z}}il\omega_{0}[\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})-1]e^{+il\omega_{0}t}\mathbf{E}_{\mathrm{l}}^{0}(\mathbf{x}_{\mathrm{}}).$
(170)

It is then natural to propose as solutions of the system (10-13):

$\displaystyle\mathbf{E}_{\mathrm{\textit{$\xi$}}}(\mathbf{x}_{\mathrm{}},t)=$
$\displaystyle\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}t}\mathbf{E}_{\mathrm{\textit{$\xi$},l}}(\mathbf{x}_{\mathrm{}}),$
(171)
$\displaystyle\mathbf{B}_{\mathrm{\textit{$\xi$}}}(\mathbf{x}_{\mathrm{}},t)=$
$\displaystyle\sum_{l\in\mathbb{Z}}e^{+il\omega_{0}t}\mathbf{B}_{\mathrm{\textit{$\xi$},l}}(\mathbf{x}_{\mathrm{}}),$
(172)

Plugging these solutions into equations (14-17) (and recalling that
$\mathbf{E}_{\mathrm{\textit{$\xi$},l}}^{1}:=\mathbf{E}_{\mathrm{\textit{$\xi$},l}}-\mathbf{E}_{\mathrm{}}^{0}$),
we arrive to the following system which must be satisfied for each
$l\in\mathbb{Z}$.

$\displaystyle\mathbf{\nabla}\times\mathbf{E}_{\mathrm{\textit{$\xi$},l}}^{1}=-il\omega_{0}\mathbf{B}_{\mathrm{\textit{$\xi$},l}}^{1},$
(173)
$\displaystyle\mathbf{\nabla}\times\mathbf{B}_{\mathrm{\textit{$\xi$},l}}^{1}=\frac{il\omega_{0}}{c^{2}}\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})\mathbf{E}_{\mathrm{\textit{$\xi$},l}}^{1}+\frac{il\omega_{0}}{c^{2}}[\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})-1]\mathbf{E}_{\mathrm{l}}^{0}(\mathbf{x}_{\mathrm{}}),$
(174)
$\displaystyle\mathbf{\nabla}\cdot\mathbf{E}_{\mathrm{\textit{$\xi$},l}}^{1}=-\frac{[\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})-1]}{\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})}\mathbf{\nabla}\cdot\mathbf{E}_{\mathrm{l}}^{0}(\mathbf{x}_{\mathrm{}}),$
(175)
$\displaystyle\mathbf{\nabla}\cdot\mathbf{B}_{\mathrm{\textit{$\xi$},l}}^{1}=0$
(176)

Taking the curl on Faraday’s Law we get:

$\mathbf{\nabla}\times\mathbf{\nabla}\times\mathbf{E}_{\mathrm{\textit{$\xi$},l}}^{1}-\bigg{(}\frac{l\omega_{0}}{c}\bigg{)}^{2}\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})\mathbf{E}_{\mathrm{\textit{$\xi$},l}}^{1}=\bigg{(}\frac{l\omega_{0}}{c}\bigg{)}^{2}[\overline{\hat{\epsilon}}_{r,II}(l\omega_{0})-1]\mathbf{E}_{\mathrm{l}}^{0},$
(177)

where the ring hand side of this expression is a source term. Moreover
equation can be solved for instance, by using Finite Elements Method as it is
explained in Refs. [27, 28].

##  8 Numerical illustration

In order to illustrate our discussion, we present the following numerical
results for the case of an oscillating charged particle close to a sphere
(albeit this method can be applied for more complicated geometries). For this
example it is considered that the particle oscillates at a frequency
$\omega_{0}=\frac{2\pi c}{\lambda_{0}}$ with $\lambda_{0}=900$ nm. The
amplitude of oscillation is $2a$ with $a=\frac{\lambda_{0}}{7}$. The radius of
the sphere is $\frac{a}{2}$ and its center is separated from the $z$-axis by a
distance of $a$, in addition the permittivity of the sphere is 9+$i$. Finally,
all geometries and conformal meshes have been obtained using the Gmsh software
[29] and all the finite element formulations in this article are implemented
thanks to the flexibility of the finite element software GetDP [25] The
incident field was hard-coded as well in GetDP). The mesh size (see Fig. 8) is
set to $\lambda_{0}/20$, which is very fine at $\omega_{0}$ and resonably
coarse at $\omega_{4}=4\omega_{0}\leftrightarrow\lambda_{4}=\lambda_{0}/4$.
The 3D scattering problem uses high order Webb hierarchical edge elements [30,
31, 32] with 20 unknowns per tetrahedron (2 unknowns per edge, 2 unknowns per
face). The direct problem described in Eq. (177) is solved using the direct
solver MUMPS [33] interfaced in GetDP. Next, we present our results as
follows:

  * •

Figures 10 and 11 show the $\mathbf{E}_{\mathrm{l}}^{0}$ and
$\mathbf{B}_{\mathrm{l}}^{0}$ fields respectively, for $l=1,2,3,4$. The
computation of these fields has been made by means of equations (114) and
(115). In order to compute these integrals numerically, the following change
of variables is used: $z^{\prime}=a\sin(\theta)$, this allows to eliminate the
term $\frac{1}{\sqrt{a^{2}-z^{\prime 2}}}$ that comes from the definition of
$w(z^{\prime})$ (see 197). Once the change of variable is made the formulae
are coded into GetDP using a simple trapezoidal rule with 300 integration
points. The only drawback we have encountered is that
$\mathbf{E}_{\mathrm{l}}^{0}$ might be singular between $(0,0,-a)$ and
$(0,0,+a)$. This is a problem that needs to be tackled in the future. The
illustration of their Poynting vector
$\mathbf{S}_{\mathrm{l}}^{0}=\frac{1}{2\mu_{0}}\mathbf{E}_{\mathrm{l}}^{0}\times\overline{\mathbf{B}_{\mathrm{l}}}^{0}$
is shown in figure 12.

  * •

Figure 13 shows the harmonic components of diffracted field
$\mathbf{E}_{\mathrm{l}}^{1}$ for $l=1,2,3,4$. Notice that this field was
obtained as a numerical solution of 177 by using FEM. Moreover, due to the
fact that the support of source term on the right hand side of 177 is within
the sphere, the possible singularity of $\mathbf{E}_{\mathrm{l}}^{0}$ does not
affect our numerical results. PML’s were used to truncate the surrounding free
space.

  * •

Figure 14 shows the total electric field
($\mathbf{E}_{\mathrm{l}}=\mathbf{E}_{\mathrm{l}}^{1}+\mathbf{E}_{\mathrm{}}^{0}$)
and its interaction with the sphere, whereas the total Poynting vector has
been computed as
$\mathbf{S}_{\mathrm{l}}=\frac{1}{2\mu_{0}}\mathbf{E}_{\mathrm{l}}\times\overline{\mathbf{B}_{\mathrm{l}}}$
and can be seen in Figure 15 for $l=1,2,3,4$. It is important to see in the
case of the Poynting vector how this one is kind of _pulled_ by the sphere.
This is due to the passivity of the material (remember that the permittivity
of the sphere is 9+$i$).

Finally, all our results have been corroborated by considering an energy
balance that measures the total energy flux that crosses a pill-box that
surrounds the sphere. This is shown in Figure 9 .

(a) Mesh showing the different regions of integration

(b) Zoom enclosing the sphere and the region where the particle oscillates.

Figure 8: Mesh showing the regions where the different fields are obtained via
FEM.

(a) Integration box that shows the

(b) Zoom enclosing the sphere

Figure 9: Pill-box surface enclosing the sphere. The tube below the sphere in
Figure 9(a) is just there to refine the mesh around the trajectory of the
particle (see figure 8(b)) and to represent the total field.

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 10: Harmonic components of the imaginary part of the incident electric
field $\mathbf{E}_{\mathrm{l}}^{0}$ for l = $1,2,3,4$

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 11: Harmonic components of the real part of the incident magnetic
induction field $\mathbf{B}_{\mathrm{l}}^{0}$ for l = $1,2,3,4$

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 12: Harmonic components of the real part of the incident Poynting
vector field $\mathbf{S}_{\mathrm{l}}^{0}$ for l = $1,2,3,4$

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 13: Harmonic components of the real part of the diffracted electric
field $\mathbf{E}_{\mathrm{l}}^{1}$ for l = $1,2,3,4$

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 14: Harmonic components of the real part of the total electric field
$\mathbf{E}_{\mathrm{l}}$ for l = $1,2,3,4$

(a) $\omega_{0}$

(b) $2\omega_{0}$

(c) $3\omega_{0}$

(d) $4\omega_{0}$

Figure 15: Harmonic components of the real part of the total Poynting vector
field $\mathbf{S}_{\mathrm{l}}$ for l = $1,2,3,4$

##  9 Conclusion

In this chapter work the problem of an oscillating particle near to a nano-
sphere has been studied by using the diffracted field formalism. The first
problem to tackle was the description of the electromagnetic field generated
by an oscillating particle _in vacuo_. Albeit this problem can be solved in
principle by the Liénard-Wiechert fields, the resulting equations are not
useful in practice. In order to overcome this difficulty we propose a harmonic
decomposition of the sources $\rho$ and
$\mathbf{\boldsymbol{\jmath}}_{\mathrm{}}$ which will allow us to find the
harmonic representation of the electromagnetic fields. Next, we make an
analysis of the radiated fields for the relativistic and non relativistic
cases. Finally, by using the finite Elements Method, we solve the full problem
that involves the interaction between the oscillating charge and the
dispersive nano sphere. The solutions obtained have been validated by an
energy balance.

##  Appendix A Annexe 0

In this appendix we set the Fourier transform convention (in the classical
sense) for this work as:

$\displaystyle\hat{f}(\omega):=\frac{1}{2\pi}\int_{t\in\mathbb{R}}f(t)e^{+i\omega
t}dt,$ (178) $\displaystyle
f(t):=\int_{\omega\in\mathbb{R}}\hat{f}(\omega)e^{-i\omega
t}\,\mathrm{d}\omega.$ (179)

If we want to extend these definitions to a wider family of mathematical
objects (_e.g._ $\delta$’s, unit steps, ramps, sines, cosines), it is
necessary to consider the definition of the Fourier transform in the sense of
distributions as per [34, 35, 36]:

$\innerproduct{\mathcal{F}_{t\rightarrow\omega}\\{f(t)\\}(\omega),\varphi(\omega)}{\mathcal{F}_{t\rightarrow\omega}\\{f(t)\\}(\omega),\varphi(\omega)}_{\omega\in\mathbb{R}}:=\innerproduct{f(t),\hat{\varphi}(t)}{f(t),\hat{\varphi}(t)}_{t\in\mathbb{R}},$
(180)

accordingly the inverse Fourier transform is defined as:

$\innerproduct{\mathcal{F}^{-1}_{\omega\rightarrow
t}\\{\hat{f}(\omega)\\}(t),\hat{\varphi}(t)}{\mathcal{F}^{-1}_{\omega\rightarrow
t}\\{\hat{f}(\omega)\\}(t),\hat{\varphi}(t)}_{t\in\mathbb{R}}:=\innerproduct{\hat{f}(\omega),\varphi(\omega)}{\hat{f}(\omega),\varphi(\omega)}_{\omega\in\mathbb{R}},$
(181)

where the brackets denote integration in the whole real line when dealing with
regular distributions and $\phi(\omega)$ is a gaussian test function as it is
explained in [34].

##  Appendix B Annexe 1

In this Annexe we show to obtain the poly harmonic representation of the
sources.

For this, the distribution $\varrho(z,t)$ has to be considered as a
$\delta$-distribution of a function $f(t):=z-a\cos(\omega_{0}t)$. Using the
expansion given in [24, 6], we obtain:

$\varrho(z,t)=\delta(f(t))=\sum_{l\in\mathbb{Z}}\frac{1}{\dot{f}(t_{l})}\delta(t-t_{l}),$
(182)

where $t_{l}$ are the zeros of the function $f(t)$, _i.e._

$\cos(\omega_{0}t)=\frac{z}{a}.$ (183)

Given the fact that the absolute value of the cosine is bounded by 1, there
are only two solutions within the range $[-a,a]$. Then, by considering the
principal branch of the arc cosine function and denoting
$t_{0}(z):=\frac{1}{\omega_{0}}\mathrm{arccos}(\frac{z}{a})$, these zeros are

$\displaystyle t_{l}^{+}(z)$ $\displaystyle=t_{0}(z)+2l\pi/\omega_{0},$ (184)
$\displaystyle t_{l}^{-}(z)$ $\displaystyle=-t_{0}(z)+2l\pi/\omega_{0}.$ (185)

In addition, it turns out that for every $t_{l}$ ($t^{+}_{l}$ or $t^{-}_{l}$)
$|\dot{f}(t)|=\omega_{0}\sqrt{a^{2}-z^{2}}$ and the $\delta$-distribution
reads:

$\varrho(z,t)=\delta(f(t))=\frac{1}{\omega_{0}\sqrt{a^{2}-z^{2}}}(W_{+}(t)+W_{-}(t))\chi_{[-a,a]}(z),$
(186)

where $W_{\pm}(t,z)=\sum_{l\in\mathbb{Z}}\delta(t-t^{\pm}(z))$. These last two
series of distributions (Dirac’s combs) are expandable in a Fourier series as
follows [35]:

$W_{\pm}(t,z)=\sum_{l\in\mathbb{Z}}\delta\bigg{(}t\mp t_{0}(z)-\frac{2\pi
l}{\omega_{0}}\bigg{)}=\sum_{m\in\mathbb{Z}}\Phi^{\pm}_{m}e^{i\frac{m\pi}{T_{0}}t},$
(187)

with $T_{0}=\frac{\pi}{\omega_{0}}$. Making the change of variables $t=\tau\pm
t_{0}(z)$ one gets:

$\sum_{l\in\mathbb{Z}}\delta(\tau-2lT_{0})=\sum_{m\in\mathbb{Z}}\Phi^{\pm}_{m}e^{i\frac{m\pi}{T_{0}}(\tau\pm
t_{0}(z))}.$ (188)

Multiplying this equation by $e^{-i\frac{n\pi}{T_{0}}(\tau\pm t_{0}(z))}$,
taking the integral from $-T_{0}$ to $T_{0}$ and performing the change of
variables $\sigma=\tau-2lT_{0}$, equation (188) can be rewritten as:

$\sum_{l\in\mathbb{Z}}\int_{-(2l-1)T_{0}}^{-(2l+1)T_{0}}\delta(\sigma)e^{-im\frac{\pi\sigma}{T_{0}}}e^{\mp
im\frac{t_{0}(z)}{T_{0}}}d\sigma=2T_{0}\Phi^{\pm}_{m}.$ (189)

or in a more illuminating way

$e^{\mp
im\frac{t_{0}(z)}{T_{0}}}\int_{-\infty}^{+\infty}\delta(\sigma)e^{-im\frac{\pi\sigma}{T_{0}}}d\sigma=2T_{0}\Phi^{\pm}_{m},$
(190)

which implies

$\Phi^{\pm}_{m}=\frac{1}{2T_{0}}e^{\mp
im\frac{t_{0}(z)}{T_{0}}}=\frac{\omega_{0}}{2\pi}e^{\mp
im\omega_{0}t_{0}(z)}.$ (191)

Then

$W_{\pm}(t,z)=\frac{\omega_{0}}{2\pi}\sum_{m\in\mathbb{Z}}e^{\mp
im\omega_{0}t_{0}(z)}e^{im\omega_{0}t},$ (192)

and naturally (recovering the $l$ index)

$\displaystyle W_{+}(t,z)+W_{-}(t,z)$
$\displaystyle=\frac{\omega_{0}}{2\pi}\sum_{l\in\mathbb{Z}}\bigg{[}e^{il\omega_{0}t_{0}(z)}+e^{-il\omega_{0}t_{0}(z)}\bigg{]}e^{im\omega_{0}t},$
(193)
$\displaystyle=\frac{\omega_{0}}{\pi}\sum_{l\in\mathbb{Z}}\cos(l\omega_{0}t_{0}(z))e^{il\omega_{0}t}.$
(194)

Remembering the definition of $t_{0}$ one gets:

$\cos(l\omega_{0}t_{0})=\cos\bigg(l\arccos\bigg(\frac{z}{a}\bigg{missing})\bigg{missing})=T_{l}\bigg{(}\frac{z}{a}\bigg{)},$
(195)

where $T_{l}$ are the Chebyshev polynomials of first kind. Therefore

$W_{+}(t,z)+W_{-}(t,z)=\sum_{l\in\mathbb{Z}}T_{l}\bigg{(}\frac{z}{a}\bigg{)}e^{+il\omega_{0}t}$
(196)

Plugging this equation into Eq. (186) and defining the function

$w(z):=\frac{1}{\pi\sqrt{a^{2}-z^{2}}}\chi_{[-a,a]}(z),$ (197)

the function $\varrho$ reads

$\varrho(z,t)=\sum_{l\mathbb{Z}}w(z)T_{l}\bigg{(}\frac{z}{a}\bigg{)}e^{+il\omega_{0}t}.$
(198)

As a consequence of this, the function $\boldsymbol{\jmath}$ is given by

$\boldsymbol{\jmath}(z,t)=-a\omega_{0}\sin(\omega_{0}t)\sum_{l\mathbb{Z}}w(z)T_{l}\bigg{(}\frac{z}{a}\bigg{)}e^{+il\omega_{0}t}.$
(199)

In short, $\varrho$ and $\boldsymbol{\jmath}$ are in the following form

$\varrho(z,t)=\sum_{l\in\mathbb{Z}}\varrho_{l}^{F}(z,t),\quad\boldsymbol{\jmath}(z,t)=\sum_{l\in\mathbb{Z}}\boldsymbol{\jmath}_{l}^{F}(z,t),$
(200)

with

$\varrho_{l}^{F}(z,t)=w(z)T_{l}\bigg{(}\frac{z}{a}\bigg{)}e^{+il\omega_{0}t},$
(201)
$\boldsymbol{\jmath}_{l}^{F}(z,t)=-a\omega_{0}\sin(\omega_{0}t)w(z)T_{l}\bigg{(}\frac{z}{a}\bigg{)}e^{+il\omega_{0}t}.$
(202)

On the other hand, the Fourier transform
$\varrho(z,t):=\delta(z-a\cos(\omega_{0}t))$ can be derived by considering the
convention:

$\displaystyle\hat{\varrho}(k,t)$ $\displaystyle=$
$\displaystyle\mathcal{F}_{z\rightarrow k}\\{\delta(z-a\cos(\omega_{0}t))\\}$
(203) $\displaystyle=$
$\displaystyle\frac{1}{2\pi}\int_{z\in\mathbb{R}}\delta(z-a\cos(\omega_{0}t))e^{+izk}dz\,.$
(204)

After a suitable change of variable ($\sigma=z-a\cos(\omega_{0}t)$) we obtain
that:

$\hat{\varrho}(k,t)=\frac{1}{2\pi}e^{+ia\cos(\omega_{0}t)k}=\frac{1}{2\pi}\sum_{l\in\mathbb{Z}}i^{l}\mathrm{J}_{l}(ka)e^{+il\omega_{0}t},$
(205)

where the last equality comes from the generating function of the Bessel’s
functions [37, 38, 36]. Taking the Fourier transform of Eq. (198) and keeping
in mind that these two Fourier series are equal term by term we arrive to this
beautiful and unexpected expression:

$\mathcal{F}_{z\rightarrow
k}\bigg{\\{}w(z)T_{l}\bigg{(}\frac{z}{a}\bigg{)}\bigg{\\}}=\frac{i^{l}}{2\pi}\mathrm{J}_{l}(ka)\,,$
(206)

which will be used later.

##  Appendix C Annexe 2

This Annexe is devoted to the deduction of the _Liénard-Wiechert_ fields.

In order to fix ideas, we are going to deal just with
$\mathbf{B}_{\mathrm{int}}$ in Eq. (62). At first sight it would be tempting
to just evaluate everything at
$\mathbf{x}_{\mathrm{}}^{\prime}=\mathbf{u}_{\mathrm{}}$. However, it is
important to remember that $\mathbf{u}_{\mathrm{}}$, and
$\mathbf{n}_{\mathrm{}}$ are functions of the retarded time $t_{r}$ which is
defined as:

$t_{r}=t-\frac{|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|}{c}\,.$
(207)

A change of strategy is then necessary, instead of asking in Eq. (62) which
$\mathbf{x}_{\mathrm{}}^{\prime}$ makes
$\mathbf{x}_{\mathrm{}}^{\prime}-\mathbf{u}_{\mathrm{}}=0$ for each $t_{r}$,
we ask which $t_{r}$ satisfies the transcendental equation (207) for each $t$,
the present time, and the given trajectory $\mathbf{u}_{\mathrm{}}$ [16].
Mathematically, the space integral in 62 is equivalent to:

$\mathbf{B}_{\mathrm{int}}=\frac{c\mu_{0}}{4\pi}\int_{\mathbb{R}}\delta\bigg{(}t_{r}-t+\frac{|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|}{c}\bigg{)}\frac{\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}}{R^{2}}dt_{r}.$
(208)

The next step to evaluate this integral, is to perform a change of variable
with respect to the argument of the delta distribution:

$\tau=t_{r}-t+\frac{|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|}{c}.$
(209)

Then, the differential $d\tau$ is given by:

$d\tau=dt_{r}\bigg{[}1+\frac{1}{c}\frac{d}{dt_{r}}|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|\bigg{]}.$
(210)

The derivative of $|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|$
with respect to the retarded time, can be easily computed by remembering that:

$\frac{d}{dt_{r}}|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|^{2}=\frac{d}{dt_{r}}(\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r}))\cdot(\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})),$
(211)

and taking the derivative from both sides we have:

$2|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|\frac{d}{dt_{r}}|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|=-2\mathbf{v}_{\mathrm{}}(t_{r})\cdot(\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})),$
(212)

which after some elementary manipulations gives:

$\frac{d}{dt_{r}}|\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r})|=-\mathbf{v}_{\mathrm{}}(t_{r})\cdot\mathbf{n}_{\mathrm{}}(t_{r}),$
(213)

and by defining
$K:=1-\boldsymbol{\beta}(t_{r})\cdot\mathbf{n}_{\mathrm{}}(t_{r})$ we finally
obtain:

$d\tau=dt_{r}\big{[}1-\boldsymbol{\beta}(t_{r})\cdot\mathbf{n}_{\mathrm{}}(t_{r})\big{]}=dt_{r}K.$
(214)

Thus, the intermediate magnetic induction field is:

$\mathbf{B}_{\mathrm{int}}=\frac{c\mu_{0}}{4\pi}\int_{\mathbb{R}}\delta(\tau)\frac{\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}}{KR^{2}}d\tau=\frac{c\mu_{0}}{4\pi}\frac{\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}}{KR^{2}}\bigg{|}_{\tau=0},$
(215)

where due to the fact that $\tau=0$ we have the retarded time defined
implicitly as in Eq. (208). _Mutatis mutandis_ , this procedure can be
repeated for the other fields in equations (63-66). Then, the total electric
and magnetic induction fields read:

$\mathbf{B}_{\mathrm{}}=\frac{qc\mu_{0}}{4\pi}\bigg{[}\bigg{(}\frac{\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}}{KR^{2}}\bigg{)}\bigg{|}_{t_{r}}+\partial_{t}\bigg{(}\frac{\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}}{cKR}\bigg{)}\bigg{|}_{t_{r}}\bigg{]},$
(216) $\displaystyle\mathbf{E}_{\mathrm{}}=$
$\displaystyle\frac{q}{4\pi\epsilon_{0}}\bigg{[}\bigg{(}\frac{(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{KR^{2}}\times\mathbf{n}_{\mathrm{}}\bigg{)}\bigg{|}_{t_{r}}+\partial_{t}\bigg{(}\frac{(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{cKR}\times\mathbf{n}_{\mathrm{}}\bigg{)}\bigg{|}_{t_{r}}\bigg{]}$
$\displaystyle+\frac{q}{4\pi\epsilon_{0}}\bigg{(}\mathbf{n}_{\mathrm{}}\frac{1+\mathbf{n}_{\mathrm{}}\cdot\boldsymbol{\beta}}{KR^{2}}\bigg{)}\bigg{|}_{t_{r}}.$
(217)

Now, it is necessary to compute the derivatives with respect to the present
time $t$. Assuming the convention that the doted quantities denote partial
derivation with respect to time $t$, the following identities will be useful:

$\displaystyle\mathbf{\dot{R}}_{\mathrm{}}$
$\displaystyle=\partial_{t}(\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(tr))=-\frac{\partial
t_{r}}{\partial t}\mathbf{v}_{\mathrm{}}(t_{r})=-c\frac{\partial
t_{r}}{\partial t}\boldsymbol{\beta}(t_{r})$ (218)
$\displaystyle\accentset{\mbox{\large.}}{R}$
$\displaystyle=\partial_{t}(\sqrt[]{\mathbf{R}_{\mathrm{}}\cdot\mathbf{R}_{\mathrm{}}})=\frac{1}{R}\mathbf{R}_{\mathrm{}}\cdot\mathbf{\accentset{\mbox{\large.}}{R}}_{\mathrm{}}=-c(\mathbf{n}_{\mathrm{}}\cdot\boldsymbol{\beta})\frac{\partial
t_{r}}{\partial t}$ (219)
$\displaystyle=c\partial_{t}(t-t_{r})=c\bigg{(}1-\frac{\partial
t_{r}}{\partial t}\bigg{)}$ (220)

from equations (219) and (220) one can solve for:

$\frac{\partial t_{r}}{\partial
t}=\frac{1}{1-\mathbf{n}_{\mathrm{}}\cdot\boldsymbol{\beta}}=\frac{1}{K},$
(221)

and upon substitution of Eq. (221) into equations (219) and (220) one gets:

$\displaystyle\mathbf{\accentset{\mbox{\large.}}{R}}_{\mathrm{}}$
$\displaystyle=\partial_{t}(\mathbf{x}_{\mathrm{}}-\mathbf{u}_{\mathrm{}}(t_{r}))=-\frac{\partial
t_{r}}{\partial
t}\mathbf{v}_{\mathrm{}}(t_{r})=-\frac{c}{K}\boldsymbol{\beta}$ (222)
$\displaystyle\dot{R}$
$\displaystyle=\partial_{t}(\sqrt[]{\mathbf{R}_{\mathrm{}}\cdot\mathbf{R}_{\mathrm{}}})=\frac{1}{R}\mathbf{R}_{\mathrm{}}\cdot\mathbf{\accentset{\mbox{\large.}}{R}}_{\mathrm{}}=-\frac{c}{K}(\mathbf{n}_{\mathrm{}}\cdot\boldsymbol{\beta})$
(223) $\displaystyle=c\partial_{t}(t-t_{r})=c\bigg{(}1-\frac{1}{K}\bigg{)}$
(224)

From the expression $R\mathbf{n}_{\mathrm{}}=\mathbf{R}_{\mathrm{}}$ we can
derive with respect to $t$ and after some manipulations we obtain:

$\displaystyle\mathbf{\accentset{\mbox{\large.}}{n}}_{\mathrm{}}$
$\displaystyle=$
$\displaystyle\frac{1}{R}\big{[}\mathbf{\accentset{\mbox{\large.}}{R}}_{\mathrm{}}-\dot{R}\mathbf{n}_{\mathrm{}}\big{]}$
(225) $\displaystyle=$
$\displaystyle\frac{1}{R}\bigg{[}-\frac{c}{K}\boldsymbol{\beta}-c\bigg{(}1-\frac{1}{K}\bigg{)}\mathbf{n}_{\mathrm{}}\bigg{]}$
$\displaystyle=$
$\displaystyle-\frac{c}{KR}\big{[}(K-1)\mathbf{n}_{\mathrm{}}+\boldsymbol{\beta}\big{]}.$

In addition, we have:

$\boldsymbol{\accentset{\mbox{\large.}}{\beta}}=\frac{\mathbf{a}_{\mathrm{}}}{Kc}$
(226)

and

$\displaystyle\accentset{\mbox{\large.}}{K}=$
$\displaystyle-\partial_{t}(\boldsymbol{\beta}\cdot\mathbf{n}_{\mathrm{}})=-\boldsymbol{\accentset{\mbox{\large.}}{\beta}}\cdot\mathbf{n}_{\mathrm{}}-\boldsymbol{\beta}\cdot\mathbf{\accentset{\mbox{\large.}}{n}}_{\mathrm{}}$
$\displaystyle=$
$\displaystyle-\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}}{Kc}+\frac{c}{KR}\big{[}(K-1)\boldsymbol{\beta}\cdot\mathbf{n}_{\mathrm{}}+\beta^{2}\big{]}$
$\displaystyle=$
$\displaystyle-\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}}{Kc}+\frac{c}{KR}\big{[}\beta^{2}-(K-1)^{2}\big{]}$
(227)

Using these identities, we will first deal with $\mathbf{B}_{\mathrm{}}$ in
Eq. (216):

$\mathbf{B}_{\mathrm{}}=\frac{qc\mu_{0}}{4\pi}\bigg{[}\frac{\mathbf{M}_{\mathrm{}}}{R}+\frac{\mathbf{\accentset{\mbox{\large.}}{M}}_{\mathrm{}}}{c}\bigg{]},$
(228)

where we have defined:

$\mathbf{M}_{\mathrm{}}:=\frac{\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}}{KR}.$
(229)

It is then necessary to calculate the derivative of $\mathbf{M}_{\mathrm{}}$:

$\mathbf{\accentset{\mbox{\large.}}{M}}_{\mathrm{}}=\frac{1}{KR}\big{[}\partial_{t}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})-\mathbf{M}_{\mathrm{}}\partial_{t}(KR)\big{]}.$
(230)

The first derivative in the right hand side of Eq. (230) is:

$\displaystyle\partial_{t}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})=$
$\displaystyle\boldsymbol{\accentset{\mbox{\large.}}{\beta}}\times\mathbf{n}_{\mathrm{}}+\boldsymbol{\beta}\times\mathbf{\accentset{\mbox{\large.}}{n}}_{\mathrm{}},$
$\displaystyle=$
$\displaystyle\frac{\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}}{Kc}-\frac{c}{KR}\big{[}(K-1)\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}+\boldsymbol{\beta}\times\boldsymbol{\beta}\big{]},$
$\displaystyle=$
$\displaystyle\frac{\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}}{Kc}-c(K-1)\mathbf{M}_{\mathrm{}},$
(231)

and the second one is given by:

$\displaystyle\partial_{t}(KR)=$
$\displaystyle\accentset{\mbox{\large.}}{K}R+\accentset{\mbox{\large.}}{R}K,$
$\displaystyle=$
$\displaystyle-\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}}{Kc}R+\frac{c}{K}\big{[}\beta^{2}-(K-1)^{2}\big{]}+c(K-1),$
$\displaystyle-\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}}{Kc}R+\frac{c}{K}\big{[}\beta^{2}-1]+c.$
(232)

Plugging Eq. (C) and Eq. (C) into Eq. (230) we obtain:

$\mathbf{\accentset{\mbox{\large.}}{M}}_{\mathrm{}}=\bigg{[}\frac{\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}}{K^{2}Rc}-c\frac{\mathbf{M}_{\mathrm{}}}{R}+\mathbf{M}_{\mathrm{}}\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}}{cK^{2}}+\frac{c}{K^{2}R}\big{[}1-\beta^{2}]\mathbf{M}_{\mathrm{}}\bigg{]},$
(233)

and substituting Eq. (233) into Eq. (228) we obtain the Liénard-Wiechert
magnetic induction field:

$\mathbf{B}_{\mathrm{}}=\frac{qc\mu_{0}}{4\pi}\bigg{[}\frac{\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}}{K^{2}Rc^{2}}+\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{c^{2}K^{3}R}+\frac{(1-\beta^{2})(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{K^{3}R^{2}}\bigg{]}_{t_{r}}.$
(234)

For the case of the electric field, we start again by expressing Eq. (C) in
terms of Eq. (229):

$\mathbf{E}_{\mathrm{}}=\frac{q}{4\pi\epsilon_{0}}\bigg{[}\bigg{(}\frac{\mathbf{M}_{\mathrm{}}}{R}+\frac{\mathbf{\accentset{\mbox{\large.}}{M}}_{\mathrm{}}}{c}\bigg{)}\times\mathbf{n}_{\mathrm{}}+\frac{\mathbf{M}_{\mathrm{}}}{c}\times\mathbf{\accentset{\mbox{\large.}}{n}}_{\mathrm{}}+\frac{2-K}{KR^{2}}\mathbf{n}_{\mathrm{}}\bigg{]}_{t_{r}}.$
(235)

Then, we consider the second term in the right hand side of 235:

$\displaystyle\frac{\mathbf{M}_{\mathrm{}}}{c}\times\mathbf{\accentset{\mbox{\large.}}{n}}_{\mathrm{}}$
$\displaystyle=\frac{-\big{[}(K-1)(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})\times\mathbf{n}_{\mathrm{}}+(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})\times\boldsymbol{\beta}\big{]}}{K^{2}R^{2}},$
$\displaystyle=\frac{-\big{[}(K-1)(\mathbf{n}_{\mathrm{}}(\boldsymbol{\beta}\cdot\mathbf{n}_{\mathrm{}})-\boldsymbol{\beta})-(\boldsymbol{\beta}(\boldsymbol{\beta}\cdot\mathbf{n}_{\mathrm{}})-\beta^{2}\mathbf{n}_{\mathrm{}})\big{]}}{K^{2}R^{2}},$

and using $1-K=\boldsymbol{\beta}\cdot\mathbf{n}_{\mathrm{}}$

$\displaystyle\frac{\mathbf{M}_{\mathrm{}}}{c}\times\mathbf{\accentset{\mbox{\large.}}{n}}_{\mathrm{}}$
$\displaystyle=\frac{-\big{[}(K-1)(\mathbf{n}_{\mathrm{}}(1-K)-\boldsymbol{\beta})-(\boldsymbol{\beta}(1-K)-\beta^{2}\mathbf{n}_{\mathrm{}})\big{]}}{K^{2}R^{2}},$
$\displaystyle=\frac{-\mathbf{n}_{\mathrm{}}}{K^{2}R^{2}}(\beta^{2}-(K-1)^{2}),$
$\displaystyle=-\frac{2-K}{KR^{2}}\mathbf{n}_{\mathrm{}}+\frac{\mathbf{n}_{\mathrm{}}}{K^{2}R^{2}}(1-\beta^{2}).$
(236)

Notice that the first term in Eq. (C) is going to cancel with the third term
in Eq. (235). Then, we can focus our attention in the first term of Eq. (235):

$\displaystyle\bigg{(}\frac{\mathbf{M}_{\mathrm{}}}{R}+\frac{\mathbf{\accentset{\mbox{\large.}}{M}}_{\mathrm{}}}{c}\bigg{)}\times\mathbf{n}_{\mathrm{}}$
$\displaystyle=$
$\displaystyle\bigg{[}\frac{\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}}{K^{2}Rc^{2}}+\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{c^{2}K^{3}R}+\frac{(1-\beta^{2})(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{K^{3}R^{2}}\bigg{]}\times\mathbf{n}_{\mathrm{}}$
(237)

Let us work with the first two terms of Eq. (C):

$\displaystyle\bigg{[}\frac{\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}}{K^{2}Rc^{2}}+\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{c^{2}K^{3}R}\bigg{]}\times\mathbf{n}_{\mathrm{}}=$
$\displaystyle\frac{\big{[}K(\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}})+\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})\big{]}}{c^{2}K^{3}R}\times\mathbf{n}_{\mathrm{}},$
$\displaystyle=$
$\displaystyle\frac{\big{[}(1-\boldsymbol{\beta}\cdot\mathbf{n}_{\mathrm{}})(\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}})+\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})\big{]}}{c^{2}K^{3}R}\times\mathbf{n}_{\mathrm{}}.$

Now, by means of the vector identity
$\mathbf{A}_{\mathrm{}}\cdot(\mathbf{B}_{\mathrm{}}\times\mathbf{C}_{\mathrm{}})\mathbf{D}_{\mathrm{}}=(\mathbf{A}_{\mathrm{}}\cdot\mathbf{D}_{\mathrm{}})(\mathbf{B}_{\mathrm{}}\times\mathbf{C}_{\mathrm{}})+(\mathbf{B}_{\mathrm{}}\cdot\mathbf{D}_{\mathrm{}})(\mathbf{C}_{\mathrm{}}\times\mathbf{A}_{\mathrm{}})+(\mathbf{C}_{\mathrm{}}\cdot\mathbf{D}_{\mathrm{}})(\mathbf{A}_{\mathrm{}}\times\mathbf{D}_{\mathrm{}})$
[14] and letting $\mathbf{A}_{\mathrm{}}=\mathbf{a}_{\mathrm{}}$,
$\mathbf{B}_{\mathrm{}}=\boldsymbol{\beta}$ and
$\mathbf{C}_{\mathrm{}}=\mathbf{D}_{\mathrm{}}=\mathbf{n}_{\mathrm{}}$ we
have:

$\displaystyle\bigg{[}\frac{\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}}{K^{2}Rc^{2}}+\frac{\mathbf{a}_{\mathrm{}}\cdot\mathbf{n}_{\mathrm{}}(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{c^{2}K^{3}R}\bigg{]}\times\mathbf{n}_{\mathrm{}}=$
$\displaystyle\frac{[\mathbf{a}_{\mathrm{}}\times\mathbf{n}_{\mathrm{}}-\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}}+\mathbf{a}_{\mathrm{}}\cdot(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})\mathbf{n}_{\mathrm{}}]}{c^{2}K^{3}R}\times\mathbf{n}_{\mathrm{}},$
$\displaystyle=$
$\displaystyle\frac{\mathbf{a}_{\mathrm{}}\times(\mathbf{n}_{\mathrm{}}-\boldsymbol{\beta})\times\mathbf{n}_{\mathrm{}}}{c^{2}K^{3}R}.$
(238)

On the other hand, the third term on C can be seen as:

$\displaystyle\frac{(1-\beta^{2})(\boldsymbol{\beta}\times\mathbf{n}_{\mathrm{}})}{K^{3}R^{2}}\times\mathbf{n}_{\mathrm{}}$
$\displaystyle=\frac{(1-\beta^{2})(\mathbf{n}_{\mathrm{}}(1-K)-\boldsymbol{\beta})}{K^{3}R^{2}},$
$\displaystyle=\frac{(1-\beta^{2})(\mathbf{n}_{\mathrm{}}-\boldsymbol{\beta})}{K^{3}R^{2}}-\frac{(1-\beta^{2})\mathbf{n}_{\mathrm{}}}{K^{2}R^{2}}$
(239)

Putting Eq. (C), and Eq. (C) into Eq. (C) and then plugging this result and C
into 235, we finally obtain the electric Liénard-Wiechert field:

$\mathbf{E}_{\mathrm{}}=\frac{q}{4\pi\epsilon_{0}}\bigg{[}\frac{(1-\beta^{2})(\mathbf{n}_{\mathrm{}}-\boldsymbol{\beta})}{K^{3}R^{2}}+\frac{\mathbf{a}_{\mathrm{}}\times(\mathbf{n}_{\mathrm{}}-\boldsymbol{\beta})\times\mathbf{n}_{\mathrm{}}}{c^{2}K^{3}R}\bigg{]}_{t_{r}}.$
(240)

We can see that, along expression Eq. (234), these are the same results as
obtained by Griffiths [7] and Heald and Marion [16].

##  Appendix D Annexe 4

Here the goal is to express $P_{l}$ with the minimum of _ad hoc_ special
functions. For this reason, we are going to derive the integrals used in
Section 6. As we know, the Bessel functions of first kind are non diverging
solutions at the origin of the differential equation [39, 37, 36]:

$r^{2}\frac{d^{2}}{dr^{2}}J_{l}(r)+r\frac{d}{dr}J_{l}(r)+J_{l}(r)\bigg{(}r^{2}-l^{2}\bigg{)}=0$
(241)

which satisfy the identities [36, 24, 26]:

$\displaystyle J_{l-1}(r)+J_{l+1}(r)=\frac{2l}{r}J_{l}(r),$ (242)
$\displaystyle J_{l-1}(r)-J_{l+1}(r)=2\frac{d}{dr}J_{l}(r),$ (243)

Equipped with these tools, our first goal is to compute the integral

$\mathcal{I}_{l}^{S}(A)=\int_{0}^{A}\frac{J_{l}^{2}}{r^{2}}dr.$ (244)

In order to do that, consider Eq. (241) and divide it over $r$, after some
elementary manipulations this equation reads:

$\frac{d}{dr}\bigg{[}r\frac{dJ_{l}}{dr}\bigg{]}+J_{l}\bigg{(}r-\frac{l^{2}}{r}\bigg{)}=0,$
(245)

and analogously for $J_{k}$

$\frac{d}{dr}\bigg{[}r\frac{dJ_{k}}{dr}\bigg{]}+J_{k}\bigg{(}r-\frac{k^{2}}{r}\bigg{)}=0.$
(246)

Multiplying Eq. (245) by $J_{k}$, Eq. (246) by $J_{l}$ and taking its
difference one gets:

$J_{k}\frac{d}{dr}\bigg{[}r\frac{dJ_{l}}{dr}\bigg{]}-J_{l}\frac{d}{dr}\bigg{[}r\frac{dJ_{k}}{dr}\bigg{]}+(k^{2}-l^{2})\frac{1}{r}J_{k}J_{l}=0.$
(247)

Next, we take the integral from 0 to $A$ to obtain:

$J_{k}r\frac{dJ_{l}}{dr}\bigg{|}_{0}^{A}-J_{l}r\frac{dJ_{k}}{dr}\bigg{|}_{0}^{A}=(l^{2}-k^{2})\int_{0}^{A}\frac{1}{r}J_{k}J_{l}dr,$
(248)

and then:

$\int_{0}^{A}\frac{1}{r}J_{k}J_{l}dr=\frac{A}{(l^{2}-k^{2})}\bigg{[}J_{k}\frac{dJ_{l}}{dr}-J_{l}\frac{dJ_{k}}{dr}\bigg{]}_{A}.$
(249)

Now, by means of the identity in (242), it is easy to see Eq. (244) as:

$\int_{0}^{A}\frac{1}{r^{2}}J_{l}^{2}dr=\frac{1}{2l}\bigg{[}\int_{0}^{A}\frac{1}{r}J_{l-1}J_{l}dr+\int_{0}^{A}\frac{1}{r}J_{l+1}J_{l}dr\bigg{]}.$
(250)

These last two integrals in the right hand side of Eq. (250) can be easily
obtained via (249) and finally

$\boxed{\mathcal{I}_{l}^{S}(A)=\frac{A}{2l}\bigg{[}\frac{J_{l}J^{\prime}_{l+1}-J_{l+1}J^{\prime}_{l}}{1+2l}+\frac{J_{l}J^{\prime}_{l-1}-J_{l-1}J^{\prime}_{l}}{1-2l}\bigg{]}_{A}.}$
(251)

The second goal, is to compute the integral

$\mathcal{I}_{l}^{R}(A)=\int_{0}^{A}J_{l}^{2}dr.$ (252)

Despite its harmless appearance, this integral is in general not easy to
integrate and there is no analytical expression in the consulted references
[39, 37, 36, 24, 26]. Thus, we propose here a semi analytical approach that
give us a way to obtain this integral in a recursive manner.

Starting by assuming $l>1$, taking the product of equations (242) and (243)
and integrating this resulting equation from 0 to $A$ one gets:

$\int_{0}^{A}J_{l-1}^{2}dr-\int_{0}^{A}J_{l+1}^{2}dr=4l\int_{0}^{A}\frac{1}{r}J_{l}\frac{dJ_{l}}{dr}dr,$
(253)

and after performing integration by parts in the right hand side of this
equation one arrives to the expression:

$\boxed{\mathcal{I}_{l+1}^{R}(A)=\mathcal{I}_{l-1}^{R}(A)-\frac{2l}{A}J_{l}^{2}(A)-2l\mathcal{I}_{l}^{S}(A).}$
(254)

Notice that the last integral in the right hand side of Eq. (254) is given by
(251). This establishes a two step recurrence relation between the integrals
$I_{l+1}^{R}(A)$ and $I_{l-1}^{R}(A)$ as defined in Eq. (252). Next, we are
going to get an expression for $I_{1}^{R}(A)$ by considering Eq. (245) with
$l=0$

$\frac{d}{dr}\bigg{[}r\frac{d}{dr}J_{0}\bigg{]}+J_{0}r=0.$ (255)

Remembering that $\frac{d}{dr}J_{0}=-J_{1}$ we get:

$-\frac{d}{dr}\bigg{[}rJ_{1}\bigg{]}+J_{0}r=0,$ (256)

and multiplying by $\frac{dJ_{0}}{dr}=-J_{1}$

$J_{1}\frac{d}{dr}\bigg{[}rJ_{1}\bigg{]}+\frac{dJ_{0}}{dr}J_{0}r=0$ (257)

which implies

$J_{1}^{2}+\frac{1}{2}\frac{d}{dr}[J_{1}^{2}+J_{0}^{2}]r=0$ (258)

and after performing integration by parts

$\int_{0}^{A}J_{1}^{2}dr-\int_{0}^{A}J_{0}^{2}dr+A[J_{1}^{2}+J_{0}^{2}]_{A}=0.$
(259)

This expression can be arranged in a more illuminating way as per:

$\boxed{\mathcal{I}_{1}^{R}(A)=\mathcal{I}_{0}^{R}(A)-A[J_{1}^{2}-J_{0}^{2}]_{A}}.$
(260)

Therefore Eq. (252) depends at the end only of

$\boxed{\mathcal{I}_{0}^{R}(A)=\int_{0}^{A}J_{0}^{2}dr,}$ (261)

which can be evaluated numerically in a very precise way by the Periodisation
method described for instance in [40].

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