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%TCIDATA{Created=Tue Sep 17 14:26:17 1996}
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%TCIDATA{Language=American English}

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\markright{{\sf Electric Dipole Fields and Radiation}}
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\begin{document}


\section{9.2\quad Electric Dipole Fields and Radiation.}

\subsection{9.2.1}

\quad \thinspace \textbf{a.} The leading term in the multipolar expansion,
for $d\ll \lambda ,$ is found simply by putting $\mathbf{x}^{\prime }=$ $0$
in eq.(\ref{2}), p. 9.1.2. Then 
\begin{equation}
\mathbf{A(x})=\frac{e^{ikr}}{cr}\int \mathbf{J}(\mathbf{x}^{\prime
})d^3x^{\prime }=-ik\mathbf{p}\frac{e^{ikr}}r  \label{1}
\end{equation}
where $\mathbf{p=}\int \mathbf{x}^{\prime }\rho (\mathbf{x}^{\prime
})d^3x^{\prime }$ is the \textsl{electric dipole moment}, as in
electrostatic.

\smallskip\ 

\textbf{b}. The fields are found by using $\mathbf{B=\nabla \times A,}$ $%
\mathbf{E=(}i/k\mathbf{)\nabla \times B}$ and a fair amount of vector
algebra. The steps are outlined in Appendix. The result can be written 
\begin{equation}
\mathbf{B}=k^2(\mathbf{n\times p)}\frac{e^{ikr}}r\left( 1-\frac
1{ikr}\right) \quad \quad \quad \quad  \label{2}
\end{equation}

\begin{equation}
\quad \quad \mathbf{E}=k^2\mathbf{p}_{\perp }\frac{e^{ikr}}r+\mathbf{\nabla }%
\left[ \mathbf{p\cdot \nabla }\left( \frac 1r\right) \right] e^{ikr}(1-ikr)
\label{2'}
\end{equation}
\[
\]
\[
\text{with\quad \quad \quad \quad }\mathbf{p}_{\perp }=(\mathbf{n\times
p)\times n=p-n(p\cdot n),\quad \quad n=\nabla }r=\frac{\mathbf{x}}r 
\]
\[
\text{and\quad \quad \quad \quad }\mathbf{\nabla }\left[ \mathbf{p\cdot
\nabla }\left( \frac 1r\right) \right] =-\mathbf{\nabla }\left( \mathbf{%
p\cdot }\frac{\mathbf{x}}{r^3}\right) =\frac 1{r^3}\left[ 3\mathbf{n(n\cdot
p)}-\mathbf{p}\right] . 
\]

\smallskip\ 

\textbf{c. }In the \textsl{near zone}, $kr\ll 1,$ $\mathbf{E=\nabla }\left[ 
\mathbf{p\cdot \nabla }\left( \dfrac 1r\right) \right] $ as in
electrostatics and $\mathbf{B}$ is smaller by a factor $(kr).$

\smallskip\ 

\textbf{d. }In the \textsl{radiation zone}, $kr\gg 1,$ $\mathbf{E=B\times n}$
and 
\begin{equation}
\mathbf{B}=k^2(\mathbf{n\times p)}\frac{e^{ikr}}r,\quad \quad \quad \mathbf{E%
}=k^2\mathbf{p}_{\perp }\frac{e^{ikr}}r.  \label{3}
\end{equation}

\newpage

\subsection{9.2.2}

\quad \thinspace \textbf{a.} The time-averaged power radiated per unit solid
angle is 
\begin{equation}
\frac{dP}{d\Omega }=\frac c{8\pi }\func{Re}\left[ r^2\mathbf{n\cdot E\times B%
}^{*}\right] =\frac c{8\pi }k^4\left| \mathbf{p}_{\perp }\right| ^2
\label{4}
\end{equation}
using 
\[
\mathbf{p}_{\perp }\times (\mathbf{n\times p}^{*}\mathbf{)\cdot n=p}_{\perp
}\cdot (\mathbf{n\times p}^{*}\mathbf{)\times n=p}_{\perp }\cdot \mathbf{p}%
_{\perp }^{*}. 
\]

\smallskip\ 

\textbf{b. }Note that $\mathbf{p}$ can be complex, as for instance in the
case of a charge in circular motion. Generally, the \textsl{real} dipole
moment $\mathbf{p(}t)$ is fourier-analyzed 
\[
\mathbf{p(}t)=\sum_n\mathbf{p}_ne^{-i\omega _nt} 
\]
and each $\mathbf{p}_n$ contributes separately to (\ref{3}) and (\ref{4}).
Because of Jackson's conventions about factors of 2 (see 9.1.1c), one must
actually insert 2$\mathbf{p}_n$ in the place of $\mathbf{p}$ in formulae (%
\ref{1})-(\ref{4}), for all $\omega _n\neq 0.$

\smallskip\ 

\textbf{c. }The polarization of the emitted radiation is in the direction of 
$\mathbf{p}_{\perp }.$ This is illustrated by the figure for a real $\mathbf{%
p}$ with fixed direction\FRAME{dtbpF}{153.8125pt}{129.6875pt}{0pt}{}{}{%
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magnitude of $\mathbf{p}_{\perp }$ is $p\sin \theta $ and the radiated power
is given by 
\begin{equation}
\frac{dP}{d\Omega }=\frac{c}{8\pi }k^{4}p^{2}\sin ^{2}\theta 
\end{equation}
On a polar plot the angular distribution of emitted radiation is shown below
in cross-section and in a three-dimensional view:

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\smallskip \ 

\textbf{d. }For a general $\mathbf{p}$, one can use a decomposition in
linear components, such as $\mathbf{p}=p_{x}\mathbf{e}_{x}+p_{y}\mathbf{e}%
_{y}+p_{z}\mathbf{e}_{z}.$ Sometimes the decomposition $\mathbf{p}=p_{+}%
\mathbf{e}_{+}+p_{-}\mathbf{e}_{-}+p_{z}\mathbf{e}_{z}$ is more convenient.

\newpage

\subsection{9.2.3}

\quad \thinspace \textbf{a.} If $\mathbf{p}$ is decomposed in an orthogonal
basis, as in the examples of 2.d, each component contributes independently
not only to $\mathbf{E}$ and $\mathbf{B,}$ but also to the radiated power (%
\ref{4}).

\smallskip\ 

\textbf{b. }The total radiated power is 
\[
P=\frac{ck^4}3\left| \mathbf{p}\right| ^2. 
\]
This is shown for a linear component by integration of (\ref{5}) ($%
\int_{-1}^1\sin ^2\theta d(\cos \theta )=4/3).$ A general $\mathbf{p}$ is
broken up first into linear components and the square modulus of each is
integrated separately, with the result 
\[
P=\frac{ck^4}3\left( \left| p_x\right| ^2+\left| p_y\right| ^2+\left|
p_z\right| ^2\right) , 
\]
as stated.

\smallskip\ 

\textbf{c.} The intensity of light emitted with polarization $\mathbf{%
\epsilon }$ is given by the component of $\mathbf{p}_{\perp }$ in the
direction of $\mathbf{\epsilon }$, which is $\mathbf{p}_{\perp }\cdot 
\mathbf{\epsilon }^{*}.$ Since $\mathbf{\epsilon }^{*}\cdot \mathbf{n=}0,$
this is the same as $\mathbf{p}\cdot \mathbf{\epsilon }^{*}.$ Hence the
result 
\[
\frac{dP}{d\Omega }=\frac c{8\pi }k^4\left| \mathbf{p\cdot \epsilon }%
^{*}\right| ^2.\quad \quad \quad \text{(polarization }\mathbf{\epsilon }%
\text{)} 
\]

\smallskip\ 

\textbf{d. }As an example, the center-fed linear antenna with a current
distribution given approximately by $I_0(1-2\left| z\right| /d)\cos \omega t$
gives, by (\ref{1}) 
\[
A_z=\frac{e^{ikr}}{cr}\int_{-d/2}^{d/2}I_0\left( 1-\frac{2\left| z\right| }%
d\right) dz=\frac{e^{ikr}}{cr}\frac{I_0d}2 
\]
(setting $I_0d/2c$ $=-ikp_z$ gives $p_z=iI_0d/2\omega $ as in Jackson).

\smallskip\ 

\textbf{e. }The radiated power varies like $\sin ^2\theta $ and the total
power is $I_0^2(kd)^2/12c.$ Any power loss can be attributed to a linear
damping $\gamma $ or a resistance $R:$ $P=(1/2)RI_0^2,$ so $R=(kd)^2/6c.$
Remember that all this holds only for $kd\ll 1.$

\end{document}
