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%TCIDATA{Created=Wed Sep 18 23:36:36 1996}
%TCIDATA{LastRevised=Wed Sep 25 14:21:30 1996}
%TCIDATA{Language=American English}

\input tcilatex
\pagestyle{myheadings}
\markright{{\sf Magnetic Dipole and Electric Quadrupole}}
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\begin{document}


\section{9.3\quad Magnetic Dipole and Electric Quadrupole.}

\subsection{9.3.1}

\quad \thinspace \textbf{a.} We start again from the exact formula 
\[
\mathbf{A(x})=\frac{1}{c}\int \mathbf{J}(\mathbf{x}^{\prime })\,\frac{%
e^{ik\left| \mathbf{x-x}^{\prime }\right| }}{\left| \mathbf{x-x}^{\prime
}\right| }d^{3}x^{\prime }, 
\]
insert 
\[
\frac{e^{ik\left| \mathbf{x-x}^{\prime }\right| }}{\left| \mathbf{x-x}%
^{\prime }\right| }\simeq e^{ikr}e^{-ik\mathbf{n\cdot x}^{\prime }}\left( 
\frac{1}{r}+\frac{\mathbf{n\cdot x}^{\prime }}{r^{2}}\right) 
\]
from 9.1.3.c, expand 
\[
e^{-ik\mathbf{n\cdot x}^{\prime }}\simeq 1-ik\mathbf{n\cdot x}^{\prime } 
\]
and pick up the terms proportional to $\mathbf{n\cdot x}^{\prime }:$%
\begin{equation}
\mathbf{A(x})=\frac{e^{ikr}}{cr}\left( \frac{1}{r}-ik\right) \int \mathbf{J}(%
\mathbf{x}^{\prime })\mathbf{n\cdot x}^{\prime }d^{3}x^{\prime }.  \label{1}
\end{equation}
We recall that this is valid for $kd\ll 1$ and leads to an expansion in
powers of $d/r$ for $kr<1$ and in powers of $kd$ for large $kr\gg 1.$

\smallskip\ 

\textbf{b}. The integrand in (\ref{1}) involves the tensor $J_\alpha x_\beta
^{\prime }.$ The symmetric and antisymmetric parts of this tensor lead
respectively to the electric quadrupole and magnetic dipole fields.

\smallskip\ 

\textbf{c. }The antisymmetric part, 
\[
(1/2)(J_{\alpha }x_{\beta }^{\prime }-x_{\alpha }^{\prime }J_{\beta })
\]
gives 
\[
\frac{1}{2}\left[ \mathbf{J(n\cdot x}^{\prime })-\mathbf{x}^{\prime }(%
\mathbf{n\cdot J)}\right] ,
\]
which is recognized as $-(1/2)\mathbf{n\times (x}^{\prime }\times \mathbf{J).%
}$ Hence the magnetic dipole vector potential is
\[
\mathbf{A(x})=ik(\mathbf{n\times m)}\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}%
\right) 
\]
where 
\[
\mathbf{m}=\frac{1}{2c}\int \mathbf{x}^{\prime }\times \mathbf{J(x}^{\prime
})d^{3}x^{\prime }
\]
is the magnetic moment.

\newpage

\subsection{9.3.2}

\quad \thinspace \textbf{a.} The task of finding $\mathbf{E}$ and $\mathbf{B}
$ for the \textsl{magnetic dipole} is greatly simplified if one recalls that
only the divergenceless part of the current contributes to the magnetization
(see \textbf{c}.). Hence the charge density $\rho $ vanishes by the
continuity equation, the scalar potential $\phi $ vanishes and $\mathbf{E=-(}%
1/c)\partial \mathbf{A/}\partial t=i(\omega /c)\mathbf{A}=ik\mathbf{A.}$
Thus 
\begin{equation}
\mathbf{E=-}k^{2}(\mathbf{n\times m)}\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}%
\right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad
\label{2}
\end{equation}
\begin{equation}
\,\,\,\,\mathbf{B=\nabla \times A}=k^{2}\mathbf{m}_{\perp }\frac{e^{ikr}}{r}+%
\mathbf{\nabla }\left[ \mathbf{m\cdot \nabla }\left( \frac{1}{r}\right)
\right] e^{ikr}(1-ikr).  \label{2'}
\end{equation}
The calculation leading to $\mathbf{B}$ is exactly the same as the
calculation giving $\mathbf{E}$ for the electric dipole.

\smallskip\ 

\textbf{b. }The analogy between magnetic and electric dipole is another
example of the symmetry of Maxwell's equations \textsl{outside the} \textsl{%
sources} in the electric and magnetic quantities. Once the electric problem
is solved, we need only let $\mathbf{E\rightarrow H,}$ $\mathbf{B\rightarrow
-D,}$ $\varepsilon \rightarrow \mu $ and $\mathbf{p\rightarrow m.}$

\smallskip\ 

c. Jackson shows on page 188 that a magnetization density $\mathbf{M}$ has
the same effect as a current $\mathbf{J}=c\mathbf{\nabla \times M,}$\ which
is clearly divergenceless. If further proof is needed, use 
\[
\frac{1}{2}\varepsilon _{\alpha \beta \gamma }\int \frac{\partial }{\partial
x_{\beta }}(r^{2}J_{\gamma })d^{3}x=\frac{1}{2}\int r^{2}\varepsilon
_{\alpha \beta \gamma }\frac{\partial J_{\gamma }}{\partial \beta }%
d^{3}x+\int \varepsilon _{\alpha \beta \gamma }x_{\beta }J_{\gamma }d^{3}x.
\]
The l.h.s. gives a vanishing integral over the boundary, so that 
\begin{equation}
\mathbf{m=-}\frac{1}{2}\int r^{2}\mathbf{\nabla \times J(x)}d^{3}\mathbf{x.}
\label{3}
\end{equation}
Now $\mathbf{J}$ can be split, $\mathbf{J=J}_{l}\mathbf{+J}_{t}$\ with $%
\mathbf{\nabla \times J}_{l}=0$ and $\mathbf{\nabla \cdot J}_{t}=0$ and
clearly only the divergenceless part $\mathbf{J}_{t}$\ contributes to $%
\mathbf{m}$.

\newpage

\subsection{9.3.3}

\quad \thinspace \textbf{a.} For the symmetric part of $J_\alpha x_\beta
^{\prime }$ we use 
\[
\sum_\gamma \frac \partial {\partial x_\gamma ^{\prime }}(x_\alpha ^{\prime
}x_\beta ^{\prime }J_\alpha )=J_\alpha x_\beta ^{\prime }+J_\beta x_\alpha
^{\prime }+x_\alpha ^{\prime }x_\beta ^{\prime }\sum_\gamma \frac{\partial
J_\gamma }{\partial x_\gamma ^{\prime }}, 
\]
which gives, integrating and discarding the boundary term: 
\[
\frac 12\int \left[ \left( \mathbf{n\cdot x}^{\prime }\right) \mathbf{%
J+\left( \QTR{mathbf}{n\cdot \QTR{mathbf}{J}}\right) x}^{\prime }\right]
d^3x^{\prime }=-\frac 12\int \mathbf{x}^{\prime }\left( \mathbf{n\cdot x}%
^{\prime }\right) \mathbf{\nabla \cdot J(x}^{\prime })d^3x^{\prime }. 
\]

\smallskip\ 

\textbf{b. }Note that only $\mathbf{J}_{l}$ (see previous page) contributes
now. The same was true for the electric dipole part. Using the continuity
equation, $\mathbf{\nabla \cdot J}=i\omega \rho ,$ we find from (\ref{1})
that the symmetric part of  $J_{\alpha }x_{\beta }^{\prime }$ gives rise to
the vector potential 
\begin{equation}
\mathbf{A(x)=-}\frac{k^{2}}{2}\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}\right)
\int \mathbf{x}^{\prime }\left( \mathbf{n\cdot x}^{\prime }\right) \rho 
\mathbf{(x}^{\prime })d^{3}x^{\prime }.  \label{4}
\end{equation}

\smallskip\ 

\textbf{c. }This part of $\mathbf{A}$ is now seen to be given by the tensor $%
x_{\alpha }^{\prime }x_{\beta }^{\prime }\rho \mathbf{(x}^{\prime })$ of the
second moments of the charge distribution. Only the traceless part of this
tensor actually contributes to the fields for $\omega \neq 0.$ This follows
from the fact that $r^{\prime \,2}\delta _{\alpha \beta }\,\rho $ leads to 
\[
\mathbf{A}=\frac{ik}{2}\left( \int r^{\prime \,2}\rho \mathbf{(x}^{\prime
})d^{3}x^{\prime }\right) \mathbf{\nabla }\left( \frac{e^{ikr}}{r}\right) ,
\]
which implies $\mathbf{\nabla \times A=0.}$

\smallskip\ 

\textbf{d.} We therefore define the (electric) \textsl{quadrupole moment
tensor} 
\begin{equation}
Q_{\alpha \beta }=\int \left( 3x_{\alpha }x_{\beta }-r^{2}\delta _{\alpha
\beta }\right) \rho \mathbf{(x})d^{3}x  \label{5}
\end{equation}
so that $\sum_{\alpha }Q_{\alpha \alpha }=0.$ When the tensor is referred to
principal axes, only two numbers (say $Q_{33}$ and $(1/2)(Q_{11}-Q_{22})$)
specify the properties of the system.

\smallskip\ 

\textbf{e. }We also define a vector $\mathbf{Q(n)}$ by 
\begin{equation}
Q_\alpha =\sum_\beta Q_{\alpha \beta }n_\beta .  \label{6}
\end{equation}

\newpage

\subsection{9.3.4}

\quad \thinspace \textbf{a.} The quadrupole fields are obtained from (\ref{4}%
) by $\mathbf{B=\nabla \times A,}$ $\mathbf{E=(}i/k)\mathbf{\nabla \times B.}
$ In the radiation zone, $kr\gg 1,$ we can replace $\mathbf{\nabla \times }$
by $ik\mathbf{n\times },$ because the derivatives of $\exp (ikr)$ give the
leading order. Therefore, with the definition (\ref{6}): 
\begin{equation}
\mathbf{B=-}\frac{ik^{3}}{6}\frac{e^{ikr}}{r}\mathbf{n\times Q(n)\quad \quad
\quad \quad E=-n\times B}  \label{7}
\end{equation}
\begin{equation}
\frac{dP}{d\Omega }=\frac{cr^{2}}{8\pi }\func{Re}\left[ \mathbf{n\cdot
E\times B}^{*}\right] =\frac{ck^{6}}{288\pi }\left| Q_{\perp }\mathbf{(n)}%
\right| ^{2}  \label{8}
\end{equation}
with 
\[
Q_{\perp }\mathbf{(n)}=\left[ \mathbf{n\times Q(n)}\right] \times \mathbf{%
n=Q(n)-n}\left[ \mathbf{Q(n)\cdot n}\right] 
\]
so that 
\[
\left| Q_{\perp }\right| ^{2}=\mathbf{Q}^{*}\mathbf{\cdot Q-}\left| \mathbf{%
n\cdot Q}\right| ^{2}=\sum_{\alpha \beta \gamma }Q_{\alpha \beta
}^{*}Q_{\alpha \gamma }n_{\beta }n_{\gamma }-\sum_{\alpha \beta \gamma
\delta }Q_{\alpha \beta }^{*}Q_{\gamma \delta }n_{\alpha }n_{\beta
}n_{\gamma }n_{\delta }. 
\]

\smallskip\ 

\textbf{b.} To find the total radiated power we need the integrals 
\[
\int n_{\beta }n_{\gamma }d\Omega =\frac{4\pi }{3}\delta _{\beta \gamma } 
\]
\[
\int n_{\alpha }n_{\beta }n_{\gamma }n_{\delta }d\Omega =\frac{4\pi }{15}%
\left[ \delta _{\alpha \beta }\delta _{\gamma \delta }+\delta _{\alpha
\gamma }\delta _{\beta \delta }+\delta _{\alpha \delta }\delta _{\beta
\gamma }\right] . 
\]
It is clear, for instance, that 
\[
\int n_{1}n_{2}d\Omega =\int xy/r^{2}d\Omega 
\]
vanishes because the integrand is odd in $x$. One is then reduced to compute 
\[
\int n_{3}^{2}d\Omega =2\pi \int \cos ^{2}\theta d(\cos \theta )=4\pi
/3,\int n_{3}^{4}d\Omega =2\pi \int \cos ^{4}\theta d(\cos \theta )=4\pi /15 
\]
and 
\[
\int n_{3}^{2}n_{1}^{2}d\Omega =\int \cos ^{2}\theta \sin ^{2}\theta d(\cos
\theta )\int \cos ^{2}\varphi d\varphi =4\pi /15. 
\]

\smallskip\ 

\textbf{c. }We find then, using also $\sum_\alpha Q_{\alpha \alpha }=0,$%
\[
P=\int \frac{dP}{d\Omega }d\Omega =\frac{ck^6}{360}\sum_{\alpha \beta
}\left| Q_{\alpha \beta }\right| ^2. 
\]

\newpage

\subsection{9.3.5}

\quad \thinspace \textbf{a.} In practice it is convenient to refer $%
Q_{\alpha \beta }$ to its principal axes, which are often obvious from the
symmetry of the system. For instance, if the system has an $n$-fold rotation
axis, with $n$ $\geq 3,$ this axis is a principal axis of $Q_{\alpha \beta
}, $ say $x_3.$ Further, in this case $Q_{11}=Q_{22}$ and both equal $%
-Q_{33}/2, $ since $Q_{11}+Q_{22}+Q_{33}$ must vanish. Then $Q_{33}$ is
simply called the quadrupole moment.

\smallskip\ 

\textbf{b. }With a diagonal $Q_{\alpha \beta }$ we find from 4.a 
\[
\left| \mathbf{Q}_{\perp }\mathbf{(n)}\right| ^2=\sum_\alpha \left|
Q_{\alpha \alpha }\right| ^2n_\alpha ^2-\sum_{\alpha \gamma }Q_{\alpha
\alpha }^{*}n_\alpha ^2Q_{\gamma \gamma }n_\gamma ^2. 
\]

\smallskip\ 

\textbf{c. }If $Q_{11}=Q_{22}$ = $-Q_{33}/2$ and $\left| Q_{33}\right| =Q_0$%
\[
\left| Q_{\perp }\mathbf{(n)}\right| ^2=Q_0^2\left( \frac{\sin ^2\theta }%
4+\cos ^2\theta \right) -Q_0^2\left( -\frac{\sin ^2\theta }2+\cos ^2\theta
\right) ^2=\frac 94Q_0^2\sin ^2\theta \cos ^2\theta 
\]
so that 
\[
\frac{dP}{d\Omega }=\frac{ck^6}{128\pi }Q_0^2\sin ^2\theta \cos ^2\theta . 
\]

\smallskip\ 

\textbf{d. }The total radiated power is now 
\[
P=\frac{ck^6Q_0^2}{240}\,. 
\]

\smallskip\ 

\textbf{e. }The radiation pattern is obtained by rotating the ``cloverleaf''
in the picture on this page around the vertical axis (the $z$ axis).
Three-dimensional views are on next page. The intercept in any direction is
proportional to the intensity radiated in that direction.

Here is a picture of $\sin ^{2}\theta \cos ^{2}\theta $ in cross-section%
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\newpage Here is a 3-d view of $\sin ^{2}\theta \cos ^{2}\theta $ from the
side, and a cut of the same

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Here are full views at 18 and 36 degrees from the vertical (the $z$ axis),
and a cut view at 18 degrees:

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\end{document}
