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%TCIDATA{Created=Tue Sep 17 13:59:09 1996}
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\markright{{\sf Radiation by a Localized Source}}
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\begin{document}


\section{9.1\quad Radiation by a Localized Source.}

\subsection{9.1.1}

\quad \thinspace \textbf{a.} The treatment of radiating systems in this
section is limited to sources describable in terms of their electric and
magnetic dipole moments and their electric quadrupole moment. A more
complete theory of multipolar expansions is given in Ch.16.

\smallskip\ 

\textbf{b}. Jackson's formulae are all written for a source density 
\[
\rho (\mathbf{x,}t)=\func{Re}\left[ \rho (\mathbf{x,\,}\omega )e^{-i\omega
t}\right] =\frac 12\left[ \rho (\mathbf{x,\,}\omega )e^{-i\omega t}+\rho
^{*}(\mathbf{x,\,}\omega )e^{i\omega t}\right] 
\]
\begin{equation}
\mathbf{J(x,}t)=\func{Re}\left[ \mathbf{J}(\mathbf{x,\,}\omega )e^{-i\omega
t}\right] =\frac 12\left[ \mathbf{J}(\mathbf{x,\,}\omega )e^{-i\omega t}+%
\mathbf{J}^{*}(\mathbf{x,\,}\omega )e^{i\omega t}\right] .  \label{1}
\end{equation}
We shall follow his notation, although the factor $1/2$ is rather
unfortunate. A general fourier expansion of the sources gives 
\[
\quad \quad \mathbf{J(x,}t)=\int_{-\infty }^{+\infty }\frac{d\omega ^{\prime
}}{2\pi }\mathbf{J}(\mathbf{x,\,}\omega ^{\prime })e^{-i\omega ^{\prime
}t}\quad \quad \text{for a continuous spectrum} 
\]
or 
\[
\mathbf{J}(\mathbf{x,}t)=\sum_{n=-\infty }^\infty \mathbf{J}(\mathbf{x,\,}%
\omega _n)e^{-i\omega _nt}\quad \quad \text{for a discrete spectrum.} 
\]
(For a periodic source, $\omega _n=n\omega _1).$ The last formula suggests
that, if we concentrate attention on a single frequency $\omega _n,$ we
should write (using also $\mathbf{J}(\mathbf{x,\,-}\omega _n)=\mathbf{J}^{*}(%
\mathbf{x,\,}\omega _n)$): 
\[
\mathbf{J}(\mathbf{x,\,}\omega _n)e^{-i\omega _nt}+\mathbf{J}^{*}(\mathbf{%
x,\,}\omega _n)e^{i\omega _nt}\,. 
\]

\smallskip\ 

\textbf{c. }This is the convention of Baym (formula (13.82)), while Jackson
has an extra factor $1/2.$ In other words $\mathbf{J}^{Baym}=\frac 12\mathbf{%
J}^{Jackson},$ and this holds for $\rho ,\mathbf{p,m,}$ etc. The factor of $%
2 $ comes from conventions, is \textsl{not} quantum-mechanical.

\newpage

\subsection{9.1.2}

\quad \thinspace \textbf{a.} With the sinusoidal time dependence in (\ref{1}%
) understood, the solution for the vector potential in the Lorentz gauge
gives 
\begin{equation}
\mathbf{A(x,\,}\omega )=\frac 1c\int \mathbf{J}(\mathbf{x}^{\prime }\mathbf{%
,\,}\omega )\frac{e^{ik\left| \mathbf{x-x}^{\prime }\right| }}{\left| 
\mathbf{x-x}^{\prime }\right| }d^3x^{\prime }  \label{2}
\end{equation}
with $k=\omega /c.$

\smallskip\ 

\textbf{b. }The magnetic field is $\mathbf{B=\nabla \times A}$ and the
electric field outside the source is $\mathbf{E=(}i/k)\mathbf{\nabla \times
B.}$ (This is of course for $\omega \neq 0).$ Proceeding in this way one
does not need the scalar potential $\phi $ at all, except for the static
field (Jackson, p. 394).

\smallskip\ 

\textbf{c.} The multipole expansion is valid for $r>d$, where $d$ is a
typical dimension of the source and $r=\left| \mathbf{x}\right| .$

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It is based on the expansion (Jackson 16.22) of the Green function in
spherical harmonics, valid for $r>r^{\prime },$%
\begin{equation}
\frac{e^{ik\left| \mathbf{x-x}^{\prime }\right| }}{\left| \mathbf{x-x}%
^{\prime }\right| }=4\pi ik\sum_{l=0}^\infty j_l(kr^{\prime
})h_l(kr)\sum_{m=-l}^lY_{lm}^{*}(\theta ^{\prime },\varphi ^{\prime
})Y_{lm}(\theta ,\varphi )  \label{3}
\end{equation}
where $j_l$ and $h_l$ are spherical Bessel and Hankel functions.

\smallskip\ 

\textbf{d. }The expansion (\ref{3}) is not simply an expansion in powers of $%
r^{\prime }/r.$ It reduces to such an expansion in the limit $kr\rightarrow
0 $ (which implies $kr^{\prime }\rightarrow 0).$ We have then 
\begin{equation}
\frac{e^{ik\left| \mathbf{x-x}^{\prime }\right| }}{\left| \mathbf{x-x}%
^{\prime }\right| }\rightarrow \frac 1{\left| \mathbf{x-x}^{\prime }\right|
}=4\pi \sum_{l=0}^\infty \frac{r^{\prime l}}{r^{l+1}}%
\sum_{m=-l}^lY_{lm}^{*}(\theta ^{\prime },\varphi ^{\prime })Y_{lm}(\theta
,\varphi ).  \label{4}
\end{equation}
This is the \textsl{near field} ($r\rightarrow 0)$ or \textsl{static} ($%
\omega \rightarrow 0)$ case.

\newpage

\subsection{9.1.3}

\quad \thinspace \textbf{a.} Another limit when simpler formulae are
obtained is $kr\rightarrow \infty .$ Then all the Hankel functions $%
h_{l}(kr),$ and therefore $\mathbf{A}$ itself, are simply proportional to $%
\exp (ikr)/r.$ This occurs in the \textsl{far field} ( $r\rightarrow \infty )
$ or \textsl{radiation} zone.

\smallskip\ 

\textbf{b. }Although the multipolar expansion can be used for sources of any
size, it is really useful when $kr^{\prime }$ is also small for all $\mathbf{%
x}^{\prime }$ contributing to the integral (\ref{2}), i.e. when $kd\ll 1,$
or $d\ll \lambda .$ The reason is twofold: first there is no near zone ($%
kr\rightarrow 0)$ unless $kd\rightarrow 0$ (because $r>d);$ second a large
number of multipoles contribute to the far field, unless $kd$ is small. If $%
kd\ll 1,$ one can replace $j_l(kr^{\prime })$ by its leading term $%
(kr^{\prime })^l/(2l+1)!!$ and the successive multipoles give contributions
to the radiation field that are proportional to successive powers of $(kd).$

\smallskip\ 

\textbf{c. }It is possible to get a complete picture of the radiation field
and a correct treatment of dipole (and electric quadrupole) terms without
the full machinery based on (\ref{3}) and (\ref{4}). Simply use $\left| 
\mathbf{x-x}^{\prime }\right| \simeq r-\left( \mathbf{x/}r\right) \cdot 
\mathbf{x}^{\prime }$ and find to leading order, with $\mathbf{k=}k(\mathbf{%
x/}r):$%
\begin{equation}
\frac{e^{ik\left| \mathbf{x-x}^{\prime }\right| }}{\left| \mathbf{x-x}%
^{\prime }\right| }=e^{ikr}e^{-i\mathbf{k\cdot x}^{\prime }}\left( \frac 1r+%
\frac{\mathbf{x\cdot x}^{\prime }}{r^3}\right)  \label{5}
\end{equation}

\smallskip\ 

\textbf{d. }In the radiation region we have, inserting the leading order of (%
\ref{5}) in (\ref{2}) 
\begin{equation}
\mathbf{A(x})=\frac{e^{ikr}}{cr}\int \mathbf{J}(\mathbf{x}^{\prime })e^{-i%
\mathbf{k}\cdot \mathbf{x}^{\prime }}d^3x^{\prime }  \label{6}
\end{equation}
The multipolar expansion of the radiation field for $kd\ll 1$ follows from
the power series expansion of $\exp (-i\mathbf{k\cdot x}^{\prime }).$

\end{document}
