\documentstyle[12pt,psfig]{article}
\message{
 	Copyright, 1993, all rights reserved, Charlie Ebner (Dept.of Physics, 
 The Ohio State University, Columbus OH 43210) and Mark Jarrell (Dept.of 
 Physics, The University of Cincinnati, Cincinnati, OH 45221-0011).  This 
 material may not be reproduced for profit, modified or published in any 
 form (this includes electronic redistribution) without the prior written 
 permission of the authors listed above.	
}
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\title{Chapter Nine\\Radiation}
\author{Heinrich Rudolf Hertz\\(1857 - 1894)}
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\tableofcontents

\vspace{0.3in}

\section{Introduction}

An electromagnetic wave, or electromagnetic {\bf {radiation}}, has as its 
sources electric accelerated charges in motion. We have learned a great deal 
about waves 
but have not given much thought to the connection between the waves and the
sources that produce them. That oversight will be rectified in this chapter.

The {\bf{scattering}} of electromagnetic waves is produced by
bombarding some object (the scatterer) with an electromagnetic wave.
Under the influence of the fields in that wave, charges in the scatterer
will be set into some sort of coherent motion\footnote{ The response to
a harmonic excitation is of the same frequency, and thus coherent} and 
these moving charges will produce radiation, called the scattered wave. 
Hence scattering phenomena are closely related to radiation phenomena.

{\bf{Diffraction}} of electromagnetic waves is similar. One starts
with a wave incident on an opaque screen with holes, or aperture, in it.
Charges in the screen, especially around the apertures, are set in motion
and produce radiation which in this case is called the diffracted wave.

Thus radiation, scattering, and diffraction are closely related. We shall
start our investigation by considering the radiation produced by some 
specified localized distribution of charges and currents in harmonic motion.
We delay until Chap.~14, the discussion of non-harmonic sources.

\centerline{\psfig{figure=DIFFvSCAT.eps,height=5.0in}}

\section{Radiation by a localized source}
Suppose that we are given some charge and current densities $\rhxt$ and
$\Jxt$\footnote{In this chapter, we assume $\ep=\mu=1$}. 
These produce potentials which, in the Lorentz gauge (Chap. 6), can be found
immediately using the retarded Green's function $G\upl(\x,t;\xp,t')$ which
we shall write simply as $\gxt$:
\beqa
\Axt=\frac1c\inxtp\gxt\J(\xp,t')\\
\Phxt=\inxtp\gxt\rh(\xp,t').
\eeqa
The Green's function itself is given by
\beq
\gxt=\frac{\de(t-t'-\xxpa/c)}\xxpa.
\eeq
Because all of the equations we shall use to compute fields are linear in
the fields themselves, we may conveniently treat just one Fourier component
(frequency) of the field at a time. To this end we write
\beq
\Jxt=\frac12\intmp d\om \Jxo\emot
\eeq
where
\beq
\J(\x,-\om)=\J^*(\x,\om)
\eeq
is required in order that $\Jxt$ be real; \eq{5} is known as a ``reality
condition.'' We may equally well, and more conveniently, replace
\eqs{4}{5} by
\beq
\Jxt=\Re\intzp d\om\Jxo\emot.
\eeq
We will do this and will in general not bother to write $\Re$ in front of
every complex expression whose real part must be taken. We will just have
to remember that the real part is the physically meaningful quantity.
Similarly, we introduce the Fourier transform of the charge density,
\beq
\rhxt=\intzp d\om\rhxo\emot.
\eeq

In the following we shall suppose that the sources have just a single
frequency component,
\beqa
\Jxop=\Jx\de(\om-\om'),\hsph\om'>0\\
\rhxop=\rhx\de(\om-\om'),\hsph\om'>0.
\eeqa
Thus, assuming $\om>0$,
\beqa
\Jxt=\Jx\emot\andh\\
\rhxt=\rhx\emot.
\eeqa
From the continuity condition on the sources,
\beq
\pde\rhxt t+\div\Jxt=0,
\eeq
we find that $\rhx$ and $\Jx$ are related by
\beq
\rhx=-i\frac{\div\Jx}\om.
\eeq

Using \eq{10} in \eq{1} and employing \eq{3} for the Green's function, we
find, upon completing the integration over the time, that
\beq
\Axt=\frac1c\inivp\frac{\Jxp}\xxpa\ekxxpa\emot
\eeq
where, as usual, $k\equiv\om/c$. Define $\Ax$ by
\beq
\Axt=\Ax\emot;
\eeq
comparison with \eq{14} gives
\beq
\Ax=\frac1c\inivp\Jxp\frac\ekxxpa\xxpa.
\eeq
From here the recipe is to find $\Bxt$ from the curl of $\Axt$; then the
electric field is found\footnote{Notice that we never have to evaluate the
scalar potential.} from $\curl\Bxt=c^{-1}\partial\Ext/\partial t$, which
holds in regions where the current density vanishes. These fields have the
forms
\beqa
\Bxt=\Bx\emot\hsph\Ext=\Ex\emot
\eeqa
where
\beqa
\Bx=\curl\Ax\hsph\Ex=\frac ik\curl\Bx.
\eeqa

We have reduced everything to a set of straightforward calculations -
integrals and derivatives. Doing them exactly can be tedious, so we should
spend some time thinking about whether there are any {\bf simplifying
approximations} that may have general validity or at least validity in some
cases of interest. There are approximations based on expansions in powers
of some small parameter. We can see what may be possible by realizing that
there are {\em{three relevant lengths}} in any radiating system.  Provided the 
origin of coordinates is taken to be somewhere within the source in the integrals above, these are the size of the source, $r'=|\xp|$; the distance 
of the observer from the source, represented by $r
=|\x|$; and the wavelength of the radiation, $\la=2\pi/k$. The magnitude
$r'$ is never larger than $d$, the size of the source. Focusing on just the
relative size of $\la$ and $r$, we identify the three traditional regimes below.

\centerline{\psfig{figure=fig2.ps,height=3.2in,width=6.0in}}

Here we have specified also that $d$ be much smaller than the other two
lengths. That simplifies the discussion of the three regimes and so is a
convenience but it is not always met in practice nor is it always
necessary. In particular, the fields far away from the source (in the
radiation zone) have characteristic forms independent of the relative size
of $\la$ and $d$ provided $r$ is large enough. Also, man-made sources such
as antennas (and antenna arrays) are often intentionally constructed 
to have $\la\sim d$ and even
$\la<<d$ in which case the inequalities above are not always
satisfied. On the other hand, natural radiating systems, such as atoms and
nuclei, typically do satisfy the condition $d<<\la$ and $d<<r$ for any $r$ at
which it is practical to detect the radiation.

\subsection{The Near Zone}

Consider first the {\bf{near zone}}. Here $d<<\la$ and $r<<\la$, so there is a
simple expansion of the exponential factor,
\beq
\ekxxpa=1+ik\xxpa+...
\eeq
which leads to
\beq
\Axt=\frac1c\inivp\frac\Jxp\xxpa[1+ik\xxpa+...]\emot.
\eeq
The leading term in this expansion is
\beq
\Axt=\frac1c\inivp\frac\Jxp\xxpa\emot.
\eeq
Aside from the harmonic time dependence, this is just the vector potential of a
static current distribution $\Jx$, and that is the origin of the name
``static'' zone; the magnetic induction here has a spatial dependence which is
the same as what one would find for a static current distribution with the
spatial dependence of the actual oscillating current distribution. We find this
result for the simple reason that in the near zone the exponential factor can
be approximated as unity.

\subsection{The Radiation or Far Zone}

In the {\bf{radiation or far zone}} ($r\gg\la\gg d$), the story is completely different because 
in this regime the behavior of the exponential dominates the integral. We can
most easily see what will happen if we first expand the displacement $\xxpa$ in
powers of $r'/r$ ($d/r$):
\beqa
\xxpa=[\xxp\cdot\xxp]^{1/2}=(r^2-2\x\cdot\xp+r'^2)^{1/2}\nonumber\\
=r\leb1-\frac{2\x\cdot\xp}{r^2}+\lep\frac{r'}r\rip^2\rib^{1/2}=r\leb1-2
\frac{\nn\cdot\xp}r+\lep\frac{r'}r\rip^2\rib^{1/2}
\eeqa
where $\nn=\x/r$ is a unit vector in the direction of $\x$. Carrying the
expansion to second order in $r'/r$, we have
\beq
\xxpa=r\leb1-\frac{\nn\cdot\xp}r+\frac12\lep\frac{r'}r\rip^2-\frac12
\lep\frac{\nn\cdot\xp}r\rip^2\rib.
\eeq
Similarly,
\beq
\frac1{\xxpa}=\frac1r\lec1+\frac{\nn\cdot\xp}r+\frac12\lep\frac{r'}r\rip^2
\leb3\lep\frac{\nn\cdot\xp}{r'}\rip^2-1\rib\ric.
\eeq
Using the first of these expansions we can write
\beqa
\ekxxpa=\ekr\emknxp e^{i\frac{kr}2\leb\lep\frac{r'}r\rip^2-\lep\frac{\nn\cdot
\xp}r\rip^2\rib}\nonumber\\=\ekr\emknxp e^{i\frac{kr'^2}{2r}\leb1-\frac{
(\nn\cdot\xp)^2}{r'^2}\rib}.
\eeqa
Given $r>>r'$ and $kr'^2/r<<1$, it is clear that this exponential function
can be approximated by just the first two factors; the third represents a
change of phase by an amount small compared to a radian.
Further, in the far zone it is sufficient to approximate
\beq
\frac1\xxpa=\frac1r.
\eeq

Putting these pieces together we have
\beq
\Axt=\frac\ekrot{cr}\inivp\Jxp\emknxp \;\; (r\gg d\;{\rm and}\;r\gg d^2/\la)\,.
\eeq
This expression is always valid for $r$ ``large enough'' which means
$r>>r'$ and $r>>kr'^2$. The relative size of $\la$ and $d$ is unimportant.

The behavior of $\Axt$ on $r$ and $t$ is explicitly given by the factor in
front of the integral; the integral depends on the direction of $\x$ but
not on its magnitude. Hence in the far zone the vector potential always
takes the form
\beq
\Axt=\frac\ekrot r\ftp
\eeq
where
\beq
\ftp\equiv\frac1c\inivp\Jxp\emknxp;
\eeq
$\th$ and $\ph$ specify, in polar coordinates, the direction of $\x$ (or
$\nn$).


Knowing the form of the potential so precisely makes it easy to see what
must be the form of the fields $\E$ and $\B$ in the far zone. First,
\beq
\Bxt=\curl\Axt=\curl\lep\frac\ekrot r\ftp\rip\approx ik\frac\ekrot r[\nn\times\ftp
]
\eeq
where we have discarded terms of relative order $\la/r$ or $d/r$.
Further, from \eqs{18}{30}, we can find \Ext; to the same order as $\B$, it
is
\beq
\Ext=-ik\frac\ekrot r[\nn\times(\nn\times\ftp)]=\Bxt\times\nn.
\eeq

There are two essential features of these equations.  
\begin{itemize}
\item First, both $\E$ and $\B$ in the radiation zone is that
the field strengths are proportional to $r^{-1}$; this is very different
from the case for static fields which fall off at least as fast as $r^{-2}$
(consider the static zone).
\item Second, the radiation fields are transverse, meaning they are
perpendicular to $\x$ or $\nn$; they are also perpendicular to each other.
\end{itemize}

The Poynting vector in the far zone also has a simple basic form:
\beqa
<\S>=\frac c{8\pi}(\E\times\B^*)=\frac c{8\pi}\lep\frac{-ik}r\rip^2[\nn
\times(\nn\times\ftp)]\times[\nn\times\f^*(\th,\ph)]\nonumber\\
=-\frac{ck^2}{8\pi r^2}\lec[\nn(\nn\cdot\ftp)-\ftp]\times[\nn\times\f^*(\th
,\ph)]\ric\nonumber\\
=-\frac{ck^2}{8\pi r^2}\lec(\nn\cdot\ftp)[\nn(\nn\cdot\f^*(\th,\ph))-\f^*(
\th,\ph)]-\nn|\ftp|^2+\f^*(\th,\ph)(\nn\cdot\ftp)\ric\nonumber\\
=\frac{ck^2}{8\pi r^2}\nn[|\ftp|^2-|\nn\cdot\ftp|^2].
\eeqa
This is presumably the time-averaged energy current density. Because it
points radially outward\footnote{In the near or intermediate zones, there
are non-zero components in other directions as well.}, it also gives
directly the angular distribution of radiated power:
\beq
\der{{\cal P}}\Om=\frac{ck^2}{8\pi}[|\ftp|^2-|\nn\cdot\ftp|^2].
\eeq
If we integrate over the solid angle, we find the total power radiated:
\beq
{\cal P}=\inac<\S>\cdot\nn=\frac{ck^2}{8\pi}
\int d\Om\,[|\ftp|^2-|\nn\cdot\ftp|^2].
\eeq
Notice that the radiated power is, appropriately, independent of $r$.

\section{Multipole Expansion of the Radiation Field}

	Thus far, we have only assumed that $r\gg\lambda,\;d$.  Now we will
consider the $d/\lambda$.  Consider two limits:  
\begin{itemize}
\item If $d/\lambda\ll 1$, then all elements of the source will
essentially be in phase, and thus an observer at $r$ cannot learn
about the structure of the source from the emitted radiation.
In this limit, we need only consider the first finite moment
in $d/\lambda$ (if the series is convergent).
\item If $d/\lambda\agt 1$, then the elements of the source will not radiate 
in phase, and an observer at $r$ may learn about the details of the structure 
of the source by analyzing the interference of the radiation pattern (i.e. 
Bragg diffraction).  In this case, to be discussed in sec.~V, we need to 
retain more terms in the series. 
\end{itemize}

Let us now go back and attempt to evaluate $\ftp$. If $kd<<1$, or $2\pi d/
\la<<1$, then it is not unreasonable to proceed with the evaluation by
expanding the exponential function $\emknxp$,
\beq
\ftp=\frac1c\inivp\Jxp\leb1-ik(\nn\cdot\xp)-\frac12k^2(\nn\cdot\xp)^2+...
\rib.
\eeq

\subsection{Electric Dipole}
The first term in the expansion is just the volume integral of $\Jxp$; one
can write it as
\beq
\frac1c\inivp\Jxp=-\frac1c\inivp[\divp\Jxp]\xp
\eeq
which one can show by doing integration by parts
starting from the right-hand side of this equation. Now employ
\eq{13} to have
\beq
\frac1c\inivp\Jxp=-ik\inivp\xp\rhxp.
\eeq
The right-hand side can be recognized as the electric dipole moment of the
amplitude $\rhx$ of the harmonically varying charge distribution. Let us
define the electric dipole moment $\edm$ in the usual way,
\beq
\edm\equiv\inivp\xp\rhxp.
\eeq

The electric dipole contribution $\f_{ed}(\th,\ph)$ to $\ftp$ is thus
\beq
\f_{ed}(\th,\ph)=-ik\edm;
\eeq
it is in fact independent of $\th$ and $\ph$. The corresponding
contribution to the vector potential, $\A_{ed}(\x,t)$, is
\beq
\A_{ed}(\x,t)=-ik\edm\frac{\ekrot}r.
\eeq

We have used only $d<<\la$ and $d<<r$; no assumption about the relative
size of $\la$ and $r$ has been made. It is somewhat tedious, but nevertheless
instructive, to evaluate the fields without making any assumptions so that
we can see their form in the near, intermediate, and far zones. First, the
magnetic induction is
\beqa
\B_{ed}(\x)&=&\curl\A_{ed}(\x)=-ik\grad\lep\frac\ekr r\rip\times\edm
\nonumber\\&=&-ik\lep ik-\frac1r\rip\frac\ekr r(\nn\times\edm)=
k^2\lep1-\frac1{ikr}\rip\frac\ekr r(\nn\times\edm).
\eeqa
Further, the electric field is
\beqa
\E_{ed}(\x)&=&\frac ik(\curl\B_{ed}(\x))\nonumber\\
&=&ik\grad\leb\frac\ekr r\lep1-\frac1{ikr}\rip\rib\times(\nn\times\edm)+ik
\frac\ekr r\lep1-\frac1{ikr}\rip\curl(\nn\times\edm)\nonumber\\
&=&ik\lec\lep\frac{ik}r-\frac1{r^2}\rip\lep1-\frac1{ikr}\rip+\frac1{ikr^3}\ric
\ekr\nn\times(\nn\times\edm)\nonumber\\
&&+ik\frac\ekr r\lep1-\frac1{ikr}\rip\leb(\edm\cdot\grad)\lep\frac\x r\rip-
\edm\div\lep\frac\x r\rip\rib\nonumber\\
&=&\frac{k^2}r\leb-1+\frac2{ikr}+\frac2{k^2r^2}\rib\ekr\nn\times(\nn\times\edm)
\nonumber\\&&+\frac{ik}r\ekr\lep1-\frac1{ikr}\rip\lep\frac\edm r-\frac{
(\edm\cdot\nn)\nn}r-\frac{3\edm}r+\frac\edm r\rip\nonumber\\
&=&\lec-\frac{k^2}r\nn\times(\nn\times\edm)+\frac1{r^3}(1-ikr)[3\nn(\edm\cdot
\nn)-\edm]\ric\ekr.
\eeqa
The electric field is divided up in this fashion to bring out, first, the
form in the radiation zone which is the first term and, second, the form in
the near zone $r<<\la$ which is the second term. The spatial dependence of
the latter is the same as the field of a static dipole,
$[3\nn(\edm\cdot\nn)-\edm]/r^3$, but do not forget that it oscillates with
angular frequency $\om$. In the intermediate zone where all contributions
are comparable, the field is complex indeed.\footnote{And this is only the
electric dipole part of the field which is, along with the magnetic dipole
part of the field, by far the simplest.}

In the radiation zone, where the fields become quite simple, they are
\beq
\B_{ed}(\x)=\frac{k^2}r\ekr(\nn\times\edm)\andh\E_{ed}=-\frac{k^2}r\ekr
\nn\times(\nn\times\edm).
\eeq
The same conclusion may be reached much more simply from \eqss{30}{31}{39}.
As remarked earlier, the fields in the far zone are transverse to $\nn$. If
we let $\edm$ define the polar axis, then $\B_{ed}$ is azimuthal, i.e., in
the direction of $\phu$. This is a special feature of electric dipole
radiation. Further, $\E_{ed}(\x)$ is in the direction of $\thu$. 

\centerline{\psfig{figure=fig2a.ps,height=1.88in,width=8.0in}}

From\eqs{32}{39} we find that the time-averaged Poynting vector in the 
radiation zone is
\beq
<\S>=\frac{ck^4}{8\pi r^2}[|\edm|^2-|\nn\cdot\edm|^2]\nn=\frac{ck^4}{8\pi
r^2}|\edm|^2\sin^2\th\,\nn,
\eeq
where $\th$ is the usual polar angle, i.e., the angle between the dipole
moment and the direction at which the radiation is observed. The radiated
power per unit solid angle is
\beq
\der{{\cal P}}\Om=r^2<\S>\cdot\nn=\frac{ck^4}{8\pi}|\edm|^2\sin^2\th,
\eeq
and the total power radiated is
\beq
{\cal P}=\int d\Om\,\der{{\cal P}}\Om=\frac{ck^4|\edm|^2}{8\pi}\int d\ph d
\th\,\sin\th\sin^2\th=\frac{ck^4|\edm|^2}3.
\eeq

\pagebreak
\subsubsection{Example: Linear Center-Fed Antenna}
Consider the short, linear, ``center-fed'' antenna shown below.

\centerline{\psfig{figure=fig3.ps,height=4.0in,width=5.0in}}

\noindent For such an antenna, the current density can be crudely
approximated by
\beq
\Jxt=\ept I_0\de(x)\de(y)\lep1-2\frac{|z|}d\rip\emot
\eeq for $|z|<d/2$; for $|z|>d/2$, it is zero. Given this current density,
we can evaluate the divergence and so find the charge density,
\beq
\div\Jx=-\frac{2I_0}d\de(x)\de(y)\frac z{|z|}=-i\om\rhx
\eeq
or
\beq
\rhx=\frac{2iI_0}{\om d}\de(x)\de(y)\frac z{|z|},\hsph |z|<d/2.
\eeq
Hence,
\beq
\edm=\iniv\x\rhx=\frac{iI_0d}{2ck}\ept.
\eeq
Now we have only to plug \eq{50} into \eqs{45}{46} to find
\beq
\der{{\cal P}}\Om=\frac{I_0^2}{32\pi c}(kd)^2\sin^2\th\andh{\cal P}=
\frac{I_0^2k^2d^2}{12c}.
\eeq
The calculation in this example may be expected to provide a good approximation
to the total radiated power provided $kd<<1$ so that the electric dipole term
dominates, unless it vanishes. There are many instances where this happens.

\subsection{Magnetic Dipole}
When this happens it is necessary to look at higher-order terms 
in the expansion of
the phase factor $\ekxxpa$. Let's look now at the next one. Start from
the exact expression for $\Ax$,
\beqa
\Ax&=&\frac1c\inivp\Jxp\frac\ekxxpa\xxpa\nonumber\\
&=&\A_{ed}(\x)+\frac\ekr{cr}\inivp\Jxp\lep\frac{\nn\cdot\xp}r-ik(\nn\cdot\xp)
\rip[1+{\cal O}(d/r,d/\la)].
\eeqa
where the second term in the () comes from the exponential, and the first
comes from the corresponding denominator.  The integral we wish to examine is
\beqa
\frac1c\inivp\Jxp(\nn\cdot\xp)\hsph\hsph\hsph\hsph\hsph\hsph\nonumber\\
=\frac1{2c}\inivp\lec[\Jxp(\nn\cdot\xp)+\xp(\nn\cdot\Jxp)]+[\Jxp(\nn\cdot\xp)-
\xp(\nn\cdot\Jxp)]\ric\nonumber\\
=\frac1{2c}\inivp[\Jxp(\nn\cdot\xp)+\xp(\nn\cdot\Jxp)]
+\frac1{2c}\inivp\nn\times(\Jxp\times\xp)
\eeqa
What is the point of breaking the integral into two pieces, symmetric and
antisymmetric under interchange of $\xp$ and $\Jxp$? There are
several related points. One is that in the near zone the second term on the
right-hand side produces a magnetic induction which has the form of the
induction of a static magnetic dipole while the first term produces an
electric field which has the form of the field of a static electric
quadrupole. Hence the radiation from the former is called magnetic dipole
radiation while that from the latter is known as electric quadrupole
radiation. Also, the magnetic dipole part produces a purely transverse
electric field in all zones while the electric quadrupole part gives a
purely transverse magnetic induction. Recall that for electric dipole
radiation, $\Bxt$ is purely transverse in all zones as well.

Let us look at the magnetic dipole fields first. From \eqs{52}{53} we have
\beqa
\A_{md}(\x) & = & \frac\ekr r\lep\frac1r-ik\rip\nn\times\inivp\frac1{2c}(\Jxp
\times\xp)\nonumber\\ & \equiv & -ik\frac\ekr r\lep1-\frac1{ikr}\rip
(\nn\times\mdm)
\eeqa
where
\beq
\mdm\equiv\frac1{2c}\inivp[\xp\times\Jxp]
\eeq
is the magnetic dipole moment, familiar from our study of magnetostatics.

Rather than plow ahead with with the evaluation of the curl to find $\B$,
etc., let us recall the electric dipole results
\beq
\A_{ed}(\x)=-ik\frac\ekr r\edm
\eeq
\beq
\B_{ed}(\x)=k^2\frac\ekr r\lep1-\frac1{ikr}\rip(\nn\times\edm).
\eeq
We see that $\B_{ed}$ is the same in functional form as $\A_{md}$;
consequently, $\E_{ed}$, which is the curl of $\B_{ed}$, must be the same
in form as the curl of $\A_{md}$, or $\B_{md}$. Hence we can immediately
write, using \eq{43},
\beq
\B_{md}(\x)=k^2\frac\ekr r\nn\times(\nn\times\mdm)+\frac\ekr{r^3}(1-ikr)
[3\nn(\nn\cdot\mdm)-\mdm].
\eeq
Notice that in the near zone $\B_{md}(\x)$ is the same as that of a static
dipole.

As for the corresponding electric field, we could work through the tedious
derivatives of the magnetic induction, but it happens that this is one
time when it is much easier to evaluate the the field from the potentials.
We know that
\beq
\Ext=-\grad\Phxt-\frac1c\pde\Axt t.
\eeq
Also, in the Lorentz gauge
\beq
\div\Axt+\frac1c\pde\Phxt t=0\orh\div\Ax-ik\Phx=0.
\eeq
Using this relation for $\Phx$, we find that
\beq
\Ex=-\frac ik\grad(\div\Ax)+ik\Ax
\eeq
for any electric field which is harmonic in time. From our expression for $
\A_{md}(\x)$, one can easily see that $\div\A_{md}=0$\footnote{An equivalent
statement is that a magnetic dipole is always charge neutral,
so that $\Phi=0$.} and so $\E_{md}$ is
simply proportional to the vector potential,
\beq
\E_{md}(\x)=-k^2\frac\ekr r\lep1-\frac1{ikr}\rip(\nn\times\mdm).
\eeq

\subsection{Comparison of Dipoles}
To summarize: 
\begin{itemize}
\item The electric dipole and magnetic dipole fields are the
same with $\E$ and $\B$ interchanged, $\E_{md}\lra-\B_{ed}$ and $\B_{md}\lra
\E_{ed}$ when $\edm\lra\mdm$. 
\item In the near zone $\E_{ed}$ and $\B_{md}$ have
the form of static dipole fields, while in all zones, $\B_{ed}$ and $\E_{md}$
are purely azimuthal in direction. 
\item The Poynting vector in the far zone has the
same form for both electric dipole and magnetic dipole fields; in the latter
case it is
\beq
<\S>=r^2\nn\frac{ck^4}{8\pi}|\mdm|^2\sin^2\th,
\eeq
leading to results for the power distribution and total power which are
the same as for electric dipole radiation, \eqs{45}{46}, with $\mdm$ in
place of $\edm$. Notice particularly that if one measures $d{\cal P}/d
\Om$, and finds the $\sin^2\th$ pattern, then one knows that the radiation is
indeed\footnote{Higher-order multipoles produce radiation
with distinctive angular distributions which are never proportional to
$\sin^2\th$.} dipole radiation; however, it is impossible to distinguish
electric dipole radiation from magnetic dipole radiation without examining
its polarization.
\end{itemize}

\centerline{\psfig{figure=fig4.ps,height=4.0in,width=8.5in}}


	These formulas {\em suggest} that for a given set of moving
charges, one should get as much power
out of an oscillating magnetic dipole as an oscillating electric dipole.
This is not so, since the magnetic dipole moment is
\beq
\mdm\equiv\frac1{2c}\inivp[\xp\times\Jxp]
\eeq
thus
\beq
|\mdm|\sim \frac{v}{c} Q d
\eeq
where v is the speed of the moving charge, $Q$ is the magnitude of this 
charge, and $d$ is the size of the current loop.  The size of the
corresponding electric dipole moment is roughly $Qd$.  Thus, 
\beq
|\mdm |\sim\frac{v}{c}|\edm |
\eeq
We see that
\beq
{\cal P}_{md}\approx \lep\frac{v}{c}\rip^2{\cal P}_{ed}\,.
\eeq
Thus, in a system with both an electric and magnetic dipole moment,
the latter is usually a relativistic correction to the former.

\subsection{Electric Quadrupole}
Let's go now to the other contribution to the vector potential in \eq{53}.
This is called the electric quadrupole term; knowing that, we naturally expect
to find that it produces an electric field in the near zone which has the
characteristic form of a static electric quadrupole field. The vector potential
is
\beqa
A_{eq}(\x)&=&\frac\ekr r\lep\frac1r-ik\rip\frac1{2c}\inivp[\Jxp(\nn\cdot\xp)+
\xp(\nn\cdot\Jxp)]\nonumber\\
&=&-\frac\ekr r\lep\frac1r-ik\rip\frac1{2c}\inivp(\nn\cdot\xp)(\divp\Jxp)\xp.
\eeqa
To demonstrate this algebraic step, consider the $i^{th}$ component of the
final expression:
\beqa
\frac1{2c}\inivp(\nn\cdot\xp)\lep\sum_j\pde{J_j}{x'_j}\rip x'_i&=&-\frac1{2c}
\sum_j\inivp\pde{}{x'_j}[x'_i(\nn\cdot\xp)]J_j\nonumber\\
&=&-\frac1{2c}\inivp\sum_j[\de_{ij}(\nn\cdot\xp)+x'_in_j]J_j\nonumber\\
&=&-\frac1{2c}\inivp[J_i(\nn\cdot\xp)+x'_i(\nn\cdot\J)]
\eeqa
which matches the $i^{th}$ component of the original expression in \eq{68}.
Making use of \eq{13} for $\div\Jx$ and also using $\om=ck$, we find, from
\eq{68}, that
\beq
\A_{eq}(\x)=-\frac{k^2}2\frac\ekr r\lep1-\frac1{ikr}\rip\inivp\xp(\nn\cdot
\xp)\rhxp.
\eeq

There are nine components to the integral since the factor of $\nn$ in the
integrand can be used to project out three numbers by, for example, letting
$\nn$ be each of the Cartesian basis vectors. That is, the basic integral
over the source, $\rhx$, which appears here is a dyadic, $\iniv\x\x\rhx$.
It is symmetric and so has at most six independent components. Notice also
that $\A_{eq}$ depends on the direction of $\nn$, which was not the case
for either $\A_{ed}$ or $\A_{md}$; the evaluation of the fields is further
complicated as a consequence.

The electric quadrupole vector potential can be written in terms of the
components $Q_{ij}$ of the electric quadrupole moment tensor which we defined
long ago. Recall that
\beq
Q_{ij}\equiv\iniv(3x_ix_j-r^2\de_{ij})\rhx.
\eeq
Take combinations of these to construct the components $Q_i$ of a
vector $\Q$:
\beq
Q_i(\nn)\equiv\sum_{j=1}^3Q_{ij}n_j.
\eeq
From these definitions one can show quite easily that
\beq
\frac13\nn\times\Q(\nn)\equiv\nn\times\lep\inivp\xp(\nn\cdot\xp)\rhxp\rip.
\eeq
To see this, note that the $i^{th}$ component of $\Q/3$ is
\beq
\frac13 Q_i=\inivp[x_i'(\sum_jn_jx'_j)\rhxp-r'^2n_i\rhxp/3].
\eeq
The first term on the right-hand side of this relation produces the
$i^{th}$ component of the integral in \eq{73}; the second term is some
$i$-independent quantity multiplied by $n_i$; the three terms in $\Q$ of
this form give something which is directly proportional to
$\nn$ and so they do not contribute to $\nn\times\Q$.

Thus have we established the validity of \eq{73}. But what good is it? It
tells us that we can write $\nn\times\A_{eq}$ in terms of $\nn\times\Q$,
but does not allow us to write $\A_{eq}$ itself in terms of $\Q$. However,
if we restrict our attention to the radiation zone, then $\nn\times\A$ is all
we will need, because in this zone, $\B=ik(\nn\times\A)$ and $\E=-\nn\times\B$.
Thus, in the far zone,
\beq
\B_{eq}(\\x)=-\frac{ik^3}6\frac\ekr r[\nn\times\Q(\nn)]
\eeq
and
\beq
\E_{eq}(\x)=\frac{ik^3}6\frac\ekr r\nn\times[\nn\times\Q(\nn)].
\eeq
The time-averaged power radiated is
\beqa
\der{{\cal P}}\Om=r^2<\S\cdot\nn>&=&\frac c{8\pi}r^2[\Ex\times\B^*(\x)]\cdot\nn
=\frac c{8\pi}r^2[\B^*(\x)\times\nn]\cdot\Ex\nonumber\\
&=&\frac c{8\pi}\frac{k^6}{36}|\nn\times[\nn\times\Q(\nn)]|^2=
\frac{ck^6}{288\pi}|\nn\times\Q(\nn)|^2.
\eeqa
The right-hand side does not have any single form as a function of $\th$
and $\ph$ that we can extract because $\Q(\nn)$ depends in an unknown way
(in general) on these angles. We can proceed to a general result only up to a
point:
\beqa
|\nn\times\Q|^2=&(\nn\times\Q)\cdot(\nn\times\Q^*)=[(\nn\times\Q)\times
\nn]\cdot\Q^*\nonumber\\=&-\Q^*\cdot[\nn(\Q\cdot\nn)-\Q]=|\Q|^2-|\nn\cdot\Q|^2;
\eeqa
this is a brilliant derivation of the statement that $\sin^2\th=1-\cos^2\th$.
Further,
\beq
\Q\cdot\Q^*-(\nn\cdot\Q)(\nn\cdot\Q^*)=\sum_{ijk}Q_{ij}n_jQ_{ik}^*n_k-
\sum_{ijkl}n_iQ_{ij}n_jn_kQ_{kl}^*n_l.
\eeq
The $n_i$'s are direction cosines and so obey the identities
\beq
\int d\Om\,n_in_j=\frac{4\pi}3\de_{ij}
\eeq
and
\beq
\int d\Om\,n_in_jn_kn_l=\frac{4\pi}{15}(\de_{ij}\de_{kl}+\de_{ik}\de_{jl}
+\de_{il}\de_{jk}).
\eeq
These enable us at least to get a simple result for the total radiated
power:
\beqa
{\cal P}&=&\int d\Om\,\der{{\cal P}}\Om\nonumber\\
&=&\frac {ck^6}{288\pi}\lec\frac{4\pi}3\sum_{ij}|Q_{ij}|^2-\frac{4\pi}{15}
\sum_{ik}[Q_{ii}Q^*_{kk}+|Q_{ik}|^2+Q_{ij}Q^*_{ji}]\ric
\nonumber\\&=&\frac{ck^6}{360}\sum_{ij}|Q_{ij}|^2
\eeqa
where we have used the facts that $Q_{ij}=Q_{ji}$ and $\sum_iQ_{ii}=0$.

By choosing appropriate axes (the principal axes of the quadrupole moment
matrix or tensor) one can always put the matrix of quadrupole moments into
diagonal form. Further, only two of the diagonal elements can be chosen
independently because the trace of the matrix must vanish. Hence any
quadrupole moment matrix is a linear combination of two basic ones.

\subsubsection{Example: Oscillating Charged Spheroid}
A commonly occurring example is an oscillating spheroidal charge distribution leading to
\beq
Q_{33}=Q_0\andh Q_{22}=Q_{11}=-Q_0/2.
\eeq
Then
\beq
Q_i=\sum_j Q_{ij}n_j=Q_{ii}n_i
\eeq
or
\beq
\Q=Q_0\leb\cos\th\ept-\frac12\sin\th(\cos\ph\,\epu+\sin\ph\,\epd)\rib.
\eeq
From this we have

\beqa
|\Q|^2&=&Q_0^2\lep\cos^2\th+\frac14\sin^2\th\rip,\nonumber\\
\nn\cdot\Q&=&Q_0\lep\cos^2\th-\frac12\sin^2\th\rip,\nonumber\\
|\nn\cdot\Q|^2&=&Q_0^2\lep\cos^4\th-\cos^2\th\sin^2\th+\frac14\sin^4\th\rip,
\eeqa
and so
\beqa
|\Q|^2-|\nn\cdot\Q|^2&=&Q_0^2\leb\cos^2\th+\frac14\sin^2\th-\cos^4\th+\sin^2
\th\cos^2\th-\frac14\sin^4\th\rib\nonumber\\
&=&Q_0^2\leb\cos^2\th\sin^2\th+\frac14\cos^2\th\sin^2\th+\cos^2\th\sin^2\th\rib
\nonumber\\&=&\frac94Q_0^2\cos^2\th\sin^2\th.
\eeqa
Hence,
\beq
\der{{\cal P}}\Om=\frac{ck^6}{128\pi}Q_0^2\sin^2\th\cos^2\th.
\eeq
This is a typical, but not uniquely so, electric quadrupole radiation
distribution.

\centerline{\psfig{figure=quad.ps,height=4.0in,width=6.0in}}

Higher-order multipole radiation (including magnetic quadrupole radiation)
is found by expanding the factor $\emknxp$ in \eq{27} to higher order in
powers of $d/\la$. One can do this in a complete and systematic fashion
after first developing some appropriate mathematical machinery by
generalizing the spherical harmonics to vector fields,\footnote{Thereby
producing the so-called {\em vector spherical harmonics}. This is the subject
matter of Chapter 16.} the purpose being to construct an orthonormal set of
basis functions for the electromagnetic fields.

\subsection{Large Radiating Systems}
Before abandoning the topic of simple radiating systems, let us look at one
example of an antenna which is not small compared the wavelength of the
emitted radiation. Our example is an array of antennas, each of which is
itself small compared to $\la$ and each of which will be treated as a point
dipole. We take the current density of this array to be
\beq
\Jx=I_0a\sum_j\de(\x-\x_j)e^{i\ph_j}\ept.
\eeq
One can easily see that this is an array of point antennas located at
positions $\x_j$; they all have the same current but are not necessarily
in phase, the phase of the $j^{th}$ antenna being given by $\ph_j$ (the
additional phase factor $\emot$, common to all antennas, has been omitted, as
usual).

\centerline{\psfig{figure=antfarm.ps,height=1.0in,width=8.5in}}

No matter how large the array, we can specify that $\x$ is large enough
that the observation point is in the far zone in which case the vector
potential can be taken as
\beq
\Ax=\frac\ekr r\ftp
\eeq
where
\beqa
\ftp=\ept\frac{I_0a}c\inivp\sum_j\de(\xp-\x_j)\emknxp e^{i\ph_j}\nonumber\\
=\ept\frac{I_0a}c\sum_je^{i[\ph_j-k(\nn\cdot\x_j)]}.
\eeqa
Referring back to \eq{33} we find that the distribution of radiated power
is
\beq
\der{{\cal P}}\Om=\frac{k^2I_0^2a^2}{8\pi c}\sin^2\th|w|^2
\eeq
where
\beq
w\equiv\sum_je^{i(\ph_j-k\nn\cdot\x_j)}.
\eeq
The factor of $\sin^2\th$ arises because each element of this array is
treated in the dipole approximation. The factor $|w|^2$ contains all of
the information about the relative phases of the amplitudes from the
different elements, i.e., about the interference of the waves from the
different elements. 

\subsubsection{Example: Linear Array of Dipoles}

As a special and explicit example, suppose that
\beq
\ph_j=j\ph_0\andh\x_j=aj\ept,
\eeq
meaning that the elements are equally spaced in a line along the z-axis and
that they have relative phases that increase linearly along the array. 

\centerline{\psfig{figure=array1.ps,height=1.5in,width=8.5in}}

\noindent Then
\beq
w=\sum_je^{i(\ph_0-ka\cos\th)j}\equiv\sum_jx^j
\eeq
where
\beq
x=e^{i(\ph_0-ka\cos\th)}.
\eeq
The sum is easy to evaluate if we know where $j$ begins and ends:
\beq
\sum_{j=j_1}^{j_2}x^j=x^{j_1}\frac{1-x^{j_2-j_1+1}}{1-x},
\eeq
so
\beq
|w|^2=\lel\frac{1-e^{i\al(\th)(j_2-j_1+1)}}{1-e^{i\al(\th)}}\ril^2
\eeq
where $\al(\th)\equiv\ph_0-ka\cos\th$.
Let $j_1=-n$ and $j_2=n$, corresponding to an array of $2n+1$ elements
centered at the origin. Then
\beq
|w|^2=\lel\frac{1-e^{i\al(2n+1)}}{1-e^{i\al}}\ril^2=\frac{1-\cos[(2n+1)\al]
}{1-\cos\al}.
\eeq
This is a function which is in general of order unity and which oscillates
as a function of $\th$. It has, however, a large peak of size $(2n+1)^2$
when $\al(\th)$ is an integral multiple of $2\pi$. The peak occurs at that
angle $\th_0$ where\footnote{More generally, $\al(\th_0)=2m\pi$ where $m$ is an
integer; because we control $\ph_0$, we can make it small enough that the
peak corresponds to the particular case $m=0$.} $\al(\th_0)=0$ or
$\cos\th_0=\ph_0/ka$. If we choose $\ph_0=0$, the peak is at $\th_0=\pi/2$;
further, if $ka<2\pi$ or $a<\la$, there is no other such peak. The width of
the peak can be determined from the fact that the factor $w$ goes to zero
when $(2n+1)\al(\th)=2\pi$. Assuming that $n>>1$, one finds that the
corresponding angle $\th$ differs from $\th_0$ by $\et$ where
\beq
\et=\frac{2\pi}{(2n+1)ka[1-(\ph_0/ka)^2]}\sim(2n+1)^{-1}.
\eeq
Hence the antenna becomes increasingly directional with increasing $n$.
The accompanying figure shows the power distribution in units of
$k^2I_0^2a^2/8\pi c$ for 5 elements with $ka=\pi$ and $\ph_0=0$.

\centerline{\psfig{figure=array.ps,height=3.0in,width=4.5in}}

\section{Multipole expansion of sources in waveguides}
We saw in Chapter~8 that a field in a waveguide could be expanded in the
normal modes of the waveguide with an integral over the sources, consisting
of a current distribution and apertures, determining the coefficients in
the expansion. If the sources are small in size compared to distances over
which fields in the normal modes vary, then one can do the integrals in
an approximate fashion by making a multipole expansion of the sources.

\subsection{Electric Dipole}
Consider first the part of the amplitude which is produced by some explicit
current distribution. It is
\beq
A_\la\upm=-\frac{2\pi Z_\la}c\inivp\Jxp\cdot\E_\la^{(\mp)}.
\eeq
The field in the mode is given by
\beq
\E_\la\upm(\xp)=[\E_\la(x',y')\pm\ept E_{z\la}(x',y')]e^{\pm ik_\la z'}.
\eeq
The origin of the coordinate $\xp$ is at some appropriately chosen point
which is probably not near the center of the source distribution. Let us
use a different coordinate $\x$ having as origin a point near the center of the
source. Also, let's let the electric field in the mode be called $\Ex$ as a
matter of convenience. Then we have to do an integral of the form
\beq
\iniv\Jx\cdot\Ex=\iniv\Jx\cdot\leb\E(0)+\sum_{i.j}\left.\pde{E_i}{x_j}\ril_0
\epi x_j+...\rib
\eeq
where we are assuming that the field varies little over the size of the
source. In this integral the first term can be converted to an integral
over the charge density,
\beq
\iniv\Jx=-i\om\iniv\x\rhx
\eeq
{\bf provided} the integration by parts can be done without picking up a
contribution from the surface of the integration volume. That is not
automatic here because the surface includes some points on the wall of the
waveguide including those points where the current is fed into the guide,
so one has to exercise some care in applying this formula. Assuming that it
is okay, we see that the leading term in the expansion of the integrand
produces to a term in the coefficient $A_\la\upm$ which is proportional to
$\edm\cdot\E(0)$.

The next term in the expansion can, as we have seen, be divided into symmetric
and antisymmetric parts:
\beqa
\sum_{i,j}\left.\pde{E_i}{x_j}\ril_0\iniv J_i(\x)x_j
&=&\frac14\sum_{i,j}\left.\lep\pde{E_i}{x_j}-\pde{E_j}{x_i}\rip\ril_0\iniv
[J_ix_j-J_jx_i]\nonumber\\
&&+\frac12\left.\pde{E_i}{x_j}\ril_0\iniv[J_ix_j+J_jx_i].
\eeqa
The first term on the right-had side is set up in such a way as to display
explicitly a component of $\curl\E$, which is $i(\om/c)\B$, and the same
component of the magnetic dipole moment. Hence this term is proportional to
$\mdm\cdot\B(0)$. The remaining term can be handled in the way that we
treated the electric quadrupole part of the vector potential earlier; it
becomes
\beq
-\frac{i\om}2\sum_{i,j}\left.\pde{E_i}{x_j}\ril_0\iniv x_ix_j\rhx
=-\frac{i\om}6\sum_{i,j}Q_{ij}\left.\pde{E_i}{x_j}\ril_0
\eeq
{\bf provided} one can throw away the contributions that come from the
surface when the integration by parts is done. The final step is achieved
by making use of the fact that $\div\E=0$.

In the next order, not shown in \eq{103}, the antisymmetric part provides
the magnetic quadrupole contribution. Without delving into the algebra of
the derivation, we state that the result is of the same form as \eq{106} but
with the {\em magnetic quadrupole moment tensor} $Q_{ij}^M$ in place of the
electric quadrupole moment tensor, $\B$ in place of $\E$, and an overall
relative $(-)$. The components of the magnetic quadrupole moment tensor are
defined in the same way as those of the electric quadrupole moment tensor
except that in place of $\rhx$ there is
\beq
\rh^M(\x)=-\frac1{2c}\div[\x\times\Jx].
\eeq

The final result for the amplitude $A_\la\upm$ with all indices in place is
\beq
A_\la\upm=i\frac{2\pi\om}c\lec\edm\cdot\E_\la^{(\mp)}(0)-\mdm\cdot\B_\la^{(
\mp)}\right.+\left.\frac16\sum_{i,j}\leb Q_{ij}\left.\pde{E_{i\la}
^{(\mp)}}{x_j}\ril_0-Q_{ij}^M\left.\pde{B_{i\la}^{(\mp)}}{x_j}\ril_0\rib+...
\ric.
\eeq

We can see immediately some interesting features of this result. For
example, if one wants to produce TM modes, which have $z$ components of
$\E$ but not of $\B$, then this is most efficiently done by designing the
source to have a large $p_z$. At the same time, hardly any TE mode will be
generated if there is only a $z$ component of $\edm$ because the TE mode
has no $E_z$ to couple to $\edm$. Hence this expression gives one a good
idea how to design a source to produce, or not produce, modes of a given
kind.

Now let's look at the same expansion if the source of radiation is an
aperture rather than an explicit current distribution. We derived in
chapter~8 that in this case
\beqa
A_\la\upm&=&-\frac{Z_\la}2\int_{S_a}d^2x'\,\nn\cdot\leb\E(\xp)\times\H_\la
\ump(\xp)\rib\nonumber\\&=&
-\frac{Z_\la}2\int_{S_a}d^2x'\,[\nn\times\E(\xp)]\cdot\H_\la\ump(\xp)
\eeqa
where the integral is over the aperture $S_a$, $\E(\xp)$ is the electric field
actually present at point $\xp$, and $\nn$ is the inward directed normal at
the aperture. Notice that only the tangential component of the electric field
contributes to this integral. Assuming that the aperture is small compared
to distances over which the fields in the normal modes vary, we can expand
the latter and find, after changing the origin to a point in the aperture,
\beqa
A_\la\upm=-\frac{Z_\la}2\int_{S_a}d^2x\,(\nn\times\E_{tan})\cdot
\leb\H_\la\ump(0)+\sum_{i,j}\epi\left.\pde{H_{i \la}\upm}{x_j}\ril_0x_j+...
\rib
\nonumber\\=-\frac{Z_\la}2\lec\frac{\B_\la\ump(0)}\mu\int_{S_a}d^2x\,(\nn
\times\E_{tan})+\sum_{i.j}\int_{S_a}d^2x\,(\nn\times\E_{tan})_i\left.
\pde{H_{i\la}\ump}{x_j}\ril_0x_j+...\ric.
\eeqa
The first term here has already an appropriate form. As for the second one,
note that we can break up the integrand into even and odd pieces,
\beqa
(\nn\times\E_{tan})_i\pde{H_{i\la}}{x_j}x_j
&=&\frac12[(\nn\times\E_{tan})_ix_j-(\nn\times\E_{tan})_jx_i]\left.
\pde{H_{i\la}\ump}{x_j}\ril_0\nonumber\\&&+\frac12[(\nn\times\E_{tan})_ix_j
+(\nn\times\E_{tan})_jx_i]\left.\pde{H_{i\la}\ump}{x_j}\ril_0.
\eeqa
Take just the antisymmetric part of this expression and complete the
integral over $\x$. The contribution to the amplitude, aside from a factor of $
-Z_\la/2$, is
\beqa
\frac14\sum_{i,j}\int_{S_a}d^2x[(\nn\times\E_{tan})_ix_j-(\nn\times
\E_{tan})_jx_i]\left.\lep\pde{H_{i\la}\ump}{x_j}-\pde{H_{j\la}\ump}{x_i}
\rip\ril_0\nonumber\\=\frac14\sum_{i,j,k}\int_{S_a}d^2x\,[(\nn\times
\E_{tan})_ix_j-(\nn\times\E_{tan})_jx_i]\ep_{kij}\lep-\frac{i\om\ep}c
E_{k\la}\ump(0)\rip\nonumber\\=-\frac{i\om\ep}{4c}\int_{S_a}d^2x\,\E_\la\ump
(0)\cdot[2\x\times(\nn\times\E_{tan})]\nonumber\\
=-\frac{i\om\ep}{2c}\int_{S_a}d^2x\E_\la\ump(0)\cdot[\x\times(\nn\times
\E_{tan})]=-\frac{i\om\ep}{2c}\int_{S_a}d^2x\,\E_\la\ump(0)\cdot\nn(\x
\cdot\E_{tan}).
\eeqa
In this string of algebra we have made use of the fact that $\curl\H=-i(\om
/c)\ep\E$ and that $\E_{tan}$ is orthogonal to $\E_\la\ump$ in the aperture
because of the boundary conditions on the latter field.

Putting this piece into the expression for $A_\la\upm$ along with the
leading one, we find
\beq
A_\la\upm=-\frac{Z_\la}2\lec\frac1\mu\B_\la\ump(0)\cdot\int_{S_a}
d^2x\,\nn\times\E_{tan}(\x)-\frac{i\om\ep}{2c}\E_\la\ump(0)\cdot\nn
\int_{S_a}d^2x\x\cdot\E_{tan}(\x)\ric
\eeq
Defining
\beq
\edm_{eff}\equiv\frac\ep{4\pi}\nn\int_{S_a}d^2x\,\x\cdot\E_{tan}(\x)
\eeq
and
\beq
\mdm_{eff}\equiv\frac{c}{2\pi i\mu\om}\int_{S_a}d^2x\,\nn\times\E_{tan}(\x
\eeq
we can write
\beq
A_\la\upm=\frac{i\pi\om Z_\la}c[\edm_{eff}\cdot\E_\la\ump(0)-\mdm_{eff}
\cdot\B_\la\ump(0)+...]
\eeq

Thus we find that in the small wavelength limit, apertures are equivalent
to dipole sources. The effective dipole moments are found by solving for,
or using some simple approximation for, the fields in the aperture. See
Jackson for a description of the particular case of circular apertures.

\pagebreak
\section{Scattering of Radiation}

	So far, we have discussed the radiation produced by a harmonically
moving source without discussing the origin of the source's motion.
In this section we will address the fields which set the source in motion.

	Consider the case where the motion is excited by some incident 
radiation.  The incident radiation is absorbed by the source, which
then begins to oscillate coherently, and thus generates new radiation.
This process is generally called scattering.

\centerline{\psfig{figure=fig5.ps,height=2.5in,width=8.5in}}

\begin{figure}[htb]
\centerline{\psfig{figure=scatter.ps,height=3.0in}}
\caption[]{\em{Scattering of waves or particles with wavelength of
roughly the same size as the lattice repeat distance allows us to
learn about the lattice structure.  We will assume that each electron
acts as a dipole scatterer.}}
\label{scatter}
\end{figure}

\subsection{Scattering of Polarized Light from an Electron}
For simplicity, let's consider a plane electromagnetic wave incident
upon a single electron of charge $-e$.  We will assume that the
incident field has the form
\beq
\E_{in}(\x,t)=\epu E_{0} \ekxot
\eeq
Since the equations which govern the motion of the electron are
linear, we expect the electron to respond by oscillating along
$\epu$ at the same frequency (assuming that the magnetic force on the
electron is small as long as the electron's motion is nonrelativistic).
This will produce a time varying electric dipole moment.

	We write the electron position as
\beq
\x(t)=\x_0 +\delta\x(t)
\eeq
If $E_0$ is small enough, then $\de\x$ will be small compared to the
wavelength $\la$ of the incident radiation.  

\centerline{\psfig{figure=fig5a.ps,height=1.75in,width=8.5in}}

\noindent This approximation is also
consistent with the non-relativistic assumption.  When this is case, we
can write
\beq
\ekxot \approx {e^{i(\k\cdot\x_0-\om t)}}
\eeq
so that the field of the incident radiation does not change
over the distance traveled by the oscillating electron.
If we assume a harmonic form for the electronic motion
$\de\x(t)=\de\x\emot$, insert this into Newton's law\footnote{
$\F=m\a\to -e\E=-e\epu E_0 {e^{i\k\cdot\x_0}}= m\delta\ddot\x(t)=-m\omega^2\delta\x$}
and solve, we get
\beq
\de\x=\epu\lep\frac{eE_0 {e^{i\k\cdot\x_0}}}{m\om^2}\rip
\eeq
The dipole moment of the oscillating electron will then be $\p(t)=\p\emot$,
with 
\beq
\p=-e\de\x=\epu\lep\frac{-e^2E_0 {e^{i\k\cdot\x_0}}}{m\om^2}\rip
\eeq


	Now we will solve for the scattered radiation.  Let the
geometry of the scattering be as shown below.

\centerline{\psfig{figure=fig6.ps,height=2.5in,width=8.5in}}

\noindent From our electric dipole formulas, the angular distribution
of the scattered power \eq{45} is

\beq
\der{{\cal P}_{scat}}\Om=r^2<\S>\cdot\nn'=
\frac{ck^4}{8\pi}[|\edm|^2-|\nn'\cdot\edm|^2]
\eeq
or
\beq
\der{{\cal P}_{scat}}\Om=\frac{e^4E_0^2}{8\pi m^2 c^3}
[1-(\nn'\cdot\epu)^2]
\eeq
Note that gives zero radiated power for $\nn'$ along $\epu$.  I.e.
there is no power radiated in the direction along which the dipole
oscillates.

	We may determine the differential cross section for scattering
by dividing the radiated power per  solid angle obtained above
by the incident power per unit area (the incident Poynting
vector).
\beq
\S_{in}=\frac{c}{8\pi} \lep \E\times\B^*\rip = \frac{c}{8\pi} E_0^2
\eeq
thus
\beq
\der{\sigma}{\Om}=\frac{1}{|\S_{in}|}\der{{\cal P}_{scat}}\Om=
\frac{e^4}{m^2 c^4} \lep 1-(\nn'\cdot\epu)^2 \rip
\eeq
or
\beq
\der{\sigma}{\Om}=r_0^2 \lep 1-(\nn'\cdot\epu)^2 \rip
\eeq
where $r_0=e^2/mc^2=2.8\, \times \,10^{13}\,{\rm{cm}}$ is the classical scattering
radius of an electron.  This is the formula for Thomson
scattering of incident light polarized along $\epu$.

\subsection{Scattering of Unpolarized Light from an Electron}

Now suppose that the incident light is unpolarized.  Then the
cross section is the average of the cross sections for the
two possible polarizations $\epu$ and $\epd$ of the incident
wave (why?).
\begin{eqnarray}
\lep\der{\sigma}{\Om}\rip_{unpol}&=&\frac{1}{2}\sum_{i=1}^2r_0^2 \lep 1-(\nn'\cdot\epi)^2 \rip\\
   	&=&r_0^2-\frac{1}{2}\sum_{i=1}^2r_0^2 \lep (\nn'\cdot\epi)^2 \rip.
\end{eqnarray}
From the fact that ($\epu$, $\epd$, and $\nn$) form an orthonormal triad
of unit vectors, one may show that
\beq
\lep\der{\sigma}{\Om}\rip_{unpol}=
\frac{1}{2}r_0^2 \lep 1+(\nn'\cdot\nn)^2 \rip
\eeq
This may be rewritten as
\beq
\lep\der{\sigma}{\Om}\rip_{unpol}=
\frac{1}{2}r_0^2 \lep 1+\cos^2(\th) \rip
\eeq
where $\th$ is the scattering angle of the unpolarized radiation.
From this we can see that the scattering in predominantly forward
($\th=0$), and backscattering ($\th=\pi$).

	The total cross sections may be obtained by integrating
the differential cross sections.  Thus
\beq
\sigma_{pol}=\frac{8\pi}{3} r_0^2
\eeq
and 
\beq
\sigma_{unpol}=\frac{1}{2}\lep \sigma_{pol}(\epu)+\sigma_{pol}(\epd)\rip
=\sigma_{pol}
\eeq

\subsection{Elastic Scattering From a Molecule}
Let us now consider the elastic scattering from a molecule\footnote{
Here we shall use molecule to identify any small collection of
charges} in which
many electrons are generally present.  In  calculating the field produced
by the molecular electrons when light is incident on them, we will
again assume that we are making our observation very far away
from the molecule, so that

\centerline{\psfig{figure=fig7.ps,height=2.5in,width=8.5in}}

We will {\em not} assume however, that $\la \gg d$, since we want
to use the scattering to learn something about the structure of the
molecule.  This is only possible if our resolution (limited by $\la$)
is smaller than $d$.  Thus we expect our results to include interference 
effects from different charges in the molecule
\begin{figure}[htb]
\centerline{\psfig{figure=formfac.ps,height=2.5in}}
\caption[]{\em{
Rays scattered from different elements of the basis, and from different
places on the atom, interfere giving the scattered intensity additional
structure described by the form factor $S$ and the atomic form factor
$f$, respectively.
}}
\label{formfac}
\end{figure}
	
	From page 9, the vector potential produced by the i'th electron
in the molecule is
\beq
\A_i(\x)=\frac{\ekr}{cr}\inivp \J_i(\xp)\emknxp.
\eeq
This expression includes just the current of the i'th electron, and
assumes $x \gg \la$ and $x\gg d$, as well as a harmonic time 
dependence.  To find the current $\J_i(\xp)$ for the i'th electron,
we note that the classical current density of an electron is
\beq 
 \J_i(\xp,t)=-e \v_i(t) \de(\xp-\x_i)
\eeq
where $(\x_i,\v_i)$ are the location and velocity of the i'th electron.
If the electron is exposed to an incident electromagnetic plane wave,
we find that
\begin{eqnarray}
\x_i(t) & = & \x_{i,0}+\de\x_i(t) \\
\de\x_i(t) & = & \epu\lep \frac{e E_0}{m\om^2} e^{i\k\cdot\x_i}\emot\rip
\end{eqnarray}
From this we can find the electron velocity, and hence the current
\beq
\J_i(\xp)=\epu \lep \frac{i e^2 E_0}{m\om} e^{i\k\cdot\x_i} \de(\xp-\x_i)\rip .
\eeq
In this we have taken out the harmonic time term.  Thus the vector
potential of the scattered field generated by the i'th electron in the
molecule is 
\beq
\A_i(\x)=\epu \frac{ie^2 E_0}{m\om} \frac{\ekr}{cr} e^{i(\k-\kp)\cdot\x_i}
\eeq
where $\kp=k\nn'$ is the scattered wave vector and $\k=k\nn$ is the 
incident wave vector.  We define the vector $\q=\k-\kp$ which is
($1/\hbar$ times) the momentum transfer from the photon to the electron, 
then the total vector potential of the scattered field is 
\beq
\A(\x)=\sum_i \A_i(\x)=\epu \frac{ie^2 E_0}{m\om} \frac{\ekr}{cr} 
\sum_i e^{i\q\cdot\x_i}
\eeq

	This differs from the vector potential due to a single scattering
electron only through the factor $\sum_i e^{i\q\cdot\x_i}$.  Since this
factor is independent of the observation point $\x$, the resulting
$\B$ and $\E$ fields are likewise those of a single scattering
multiplied by a factor of $\sum_i e^{i\q\cdot\x_i}$.  Thus we may
immediately write down the differential cross sections.
For polarized light
\beq
\lep \der{\sigma}{\Om}\rip_{pol}=\leb r_0^2 \lep 1-(\nn'\cdot\epu)^2\rip\rib
\leb \left| \sum_i e^{i\q\cdot\x_i}\right|^2 \rib
\eeq
where the first term in brackets is the single-electron result, and the second
term is the structure factor.  For unpolarized light
\beq
\lep \der{\sigma}{\Om}\rip_{unpol}=
\leb \frac{1}{2} r_0^2 \lep 1+(\nn'\cdot\nn)^2\rip\rib
\leb \left| \sum_i e^{i\q\cdot\x_i}\right|^2 \rib
\eeq


We see that from a measurement of the differential cross section, we
may learn about the structure of the object from which we are scattering
light.  Let's examine the structure factor in more detail.
\begin{eqnarray}
\sum_i e^{i\q\cdot\x_i} & = & \inv \sum_i e^{i \q\cdot\x_i} \de(\x-\x_i) \\
		& = & \inv e^{i \q\cdot\x} \sum_i \de(\x-\x_i)
\end{eqnarray}
However, 
\beq
\sum_i \de(\x-\x_i)=n_{el}(\x)
\eeq
which is the electron number density at position $\x$, thus
\beq
\sum_i e^{i \q\cdot\x_i} = \inv e^{i \q\cdot\x} n_{el}(\x)=n_{el}(-\q)
\eeq
where $n_{el}(-\q)$ is the Fourier transform of the electron density.
Thus, our differential cross sections are, for polarized light:
\beq
\lep \der{\sigma}{\Om}\rip_{pol}=\leb r_0^2 \lep 1-(\nn'\cdot\epu)^2\rip\rib
\leb \left| n_{el}(-\q) \right|^2 \rib,
\eeq
and for unpolarized light
\beq
\lep \der{\sigma}{\Om}\rip_{unpol}=
\leb \frac{1}{2} r_0^2 \lep 1+(\nn'\cdot\nn)^2\rip\rib
\leb \left| n_{el}(-\q) \right|^2 \rib
\eeq
Since $\q=\k-\k'$ depends upon the scattering angle and the wave number
of the incident photon, a scan over $\th$ and/or $\k$ gives you 
information about the electron density on the target.

\subsubsection{Example: Scattering Off a Hard Sphere}

\centerline{\psfig{figure=hsphere.ps,height=3.25in,width=8.5in}}

	Consider scattering from a single molecule with a sharp 
cutoff in its electron distribution.
\beq
n_{el}(\x)= \frac{3z}{4\pi R^3} \th(R-|\x|)
\eeq
where $R$ is the radius of the sphere and $z$ is the number
of electrons, then
\beq
n_{el}(-\q)=\inv e^{i\q\cdot\x} n_{el}(\x)=
\frac{3z}{q^3 R^3} \leb \sin(qR) - qR\cos(qR) \rib\,,
\eeq
where $j_1(x)=\lep \sin x/x -\cos x \rip $ is a spherical Bessel function.
Note that this has zeroes at
\beq
qR=\tan(qR)
\eeq
so that we expect zeroes in the scattering cross section.
These are clearly related to the size of the molecule.
For forward scattering ($q=2k\sin(0)=0$), we have
\beq
n_{el}(-\q=0)=z
\eeq
This is the maximum value,
i.e. all the scattering electrons give constructive interference
for forward scattering.  

	A sketch of the unpolarized cross section 
\[
\lep \der{\sigma}{\Om}\rip_{unpol}=\frac12
\r_0^2\lep 1+\cos^2(\th)\rip
\left| n_{el}(-\q) \right|^2 
\]
looks like

\centerline{\psfig{figure=spherescat.ps,height=2.12in,width=4.0in}}

If we measured this, we could learn the spatial distribution
of electrons by inverting the transform.  It is important to
note that we could not learn the details of the electron
wavefunctions since all the phase information is lost.
I.e., the cross section just measures the modulus
squared of $n_{el}$, and so phase information is lost.

\subsubsection{Example: A Collection of Molecules}

\centerline{\psfig{figure=chsphere.ps,height=3.25in,width=8.5in}}

In a macroscopic sample, the total electron density is made up of two parts:
the electron distribution within the molecules, and the distribution of the
molecules within the sample. For simplicity, let's assume that the sample
contains just one type of molecule, each of which has an electron density
$n_0(\x)$.  The total electron number density is then
\beq
n_{el}(\x)=\sum_j n_0(\x-\R_j)
\eeq
where $\R_j$ is the position of the j'th molecule.
The Fourier transform of this is
\beq
n_{el}(-\q) = \inv e^{i\q\cdot\x} n_{el}(\x)=
\sum_j e^{i\q\cdot\R_j} n_0(-\q)
\eeq
which is a weighted sum of the Fourier transforms for each molecule.
We see that then the cross section will include the factor
\beq
\left| n_0(-\q) \right|^2
\left| \sum_j e^{i\q\cdot\R_j} \right|^2
\eeq
From above, we know that $n_0(-\q=0)=z$, the number of electrons
in a molecule, so we may write
\beq
n_0(-\q)=z F(\q)
\eeq
where $F(\q)$ is called the form factor of the molecule which is normalized
to unity for $\q=0$.  Thus the unpolarized differential cross section
becomes
\beq
\lep \der{\sigma}{\Om}\rip_{unpol}=\frac{1}{2} z^2 r_0^2(1+\cos^2(\th))
\left|F(\q)\right|^2 \left| \sum_j e^{i\q\cdot\R_j}\right|^2
\eeq
In general, we will not know where all the molecules are, nor
do we necessarily care.  Thus we will look at an average of
the term which describes the distribution of the molecules.
\beq
\left< \left| \sum_j e^{i\q\cdot\R_j}\right|^2 \right>
=\sum_{j j'} \left< e^{i\q\cdot(\R_j-\R_{j'})}\right>
\approx N\sum_j \left< e^{i\q\cdot(\R_j-\R_{0})}\right>
\equiv N S(\q)
\eeq
where $N$ is the number of molecules in the sample, $\R_0$ is the 
position of the origin, and $S(\q)$ is the structure factor of the sample.
The approximation is justified by the fact that (neglecting finite-size
effects) each link between sites occurs about $N$ times.

\centerline{\psfig{figure=links.ps,height=2.5in,width=8.5in}}


\beq
S(\q)=\sum_j \left< e^{i\q\cdot(\R_j-\R_{0})}\right>=
1+\sum_{j\neq 0} \left< e^{i\q\cdot(\R_j-\R_{0})}\right>
\eeq
Note that this goes to $1$ as $|\q|\to\infty$.
Then, using the same procedure detailed before, this becomes
\beq
N S(\q)= V\inv e^{i\q\cdot\x} \left< n(\x) n(0)\right>
\eeq
where $V$ is the sample volume, and $n(\x)$ is the number density
of molecules in the sample.

	Here, $\left< n(\x) n(0)\right>$ tell us about correlations
between molecular positions.
\beq
\left< n(\x) n(0)\right>= \lep n(0)\rip^2 g(\x)
\eeq
where $g(\x)$ is just the probability of finding a molecule at $\x$ if
there is one at the origin.  This is the normalized two-body
correlation function.  We see that if the molecular form factor
is known, then a measurement of the differential cross section
tells us about the Fourier transform of $g(\x)$.  This is the key to using
x-ray (or neutron, etc) scattering to determine the internal structure of
materials.
\[
\lep \der{\sigma}{\Om}\rip_{unpol}=\frac{1}{2} z^2 r_0^2(1+\cos^2(\th))
\left|F(\q)\right|^2 V \lep n(0)\rip^2 g(\q)
\]

\centerline{\psfig{figure=fig9.ps,height=2.5in,width=8.5in}}

\pagebreak
\section{Diffraction}

In this section we will discuss diffraction which is related
to scattering.  The prototypical setup is shown below in which
radiation from a source is diffracted through an opaque screen
with one or more holes in it.  

\centerline{\psfig{figure=fig10.ps,height=2.5in,width=8.5in}}

	Clearly, diffraction is the study of the propagation of
light or radiation, or rather the deviation of light from
rectilinear propagation.  As undergraduates we all learned that
the propagation of light was governed by {\bf{Huygen's Principle}}
that {\em{ every point on a primary wavefront serves a the source
of spherical secondary wavelets such that the primary wavefront at some 
later time is the envelope of these wavelets.  Moreover, the 
wavelets advance with a speed and frequency equal to that of the primary
wave at each point in space.}}\footnote{Hecht-Zajac,
page 60}.  This serves as the
paradigm for our study of geometric optics.  Diffraction is the first
crisis in this paradigm.

	To see this consider wave propagation in a ripple tank
through an aperture.

\centerline{\psfig{figure=diffracto.ps,height=3.0in,width=8.5in}}

\noindent In the figure, wavefronts are propagating toward the aperture
from the bottom of each box.  In case a) the wavelength $\lambda$ is
much smaller than the size of the aperture $d$.  In this case, the
diffracted waves interfere destructively except immediately
in front of the aperture.  In case b), $\lambda\gg d$, and no
such interference is is observed.

	Huygen's principle cannot explain the difference between cases a)
and b), since it is independent
of any wavelength considerations, and thus would predict the same wavefront
in each case.  The difficulty was resolved by Fresnel.  The
corresponding {\bf{Huygens-Fresnel principle}} states that {\em{every
unobstructed point of a wavefront,at a given instant of in time,
serves as a source  of spherical secondary wavelets of the same frequency as
the source.  The amplitude of the diffracted wave is the sum of the
wavelets considering their amplitudes and relative phases}}\footnote{
Hecht-Zajac, page 330}.  Applying these ideas clarifies case a).  Here what is
happening is that the wavelets from the right and left sides of the
aperture interfere constructively in front of the aperture (since
they travel the same distance and hence remain in phase), whereas these
wavelets interfere destructively to the sides of the aperture (since
they travel two paths with length differences of order $\lambda/2$).
In case b), we approach the limit of a single point source of spherical
waves.  Beside being rather hypothetical, the Huygens-Fresnel principle 
involves some approximations which we will discuss later; however, Kirchoff 
showed that the Huygen's-Fresnel principle is a direct consequence of the
wave equation.

\subsection{Scalar Diffraction Theory: Kirchoff Approximation}
This problem could be solved using the techniques we just developed
to treat scattering. I.e. by considering the dynamics of the
charged particles in the screen, and then calculating
the scattered radiation generated by these particles.
However, diffraction is conventionally treated as a boundary value
problem in which the presence of the screen is taken into
account with boundary conditions on the wave.

	Several excellent references for this problem are worth noting:
\begin{enumerate}
\item B.B. Baker and E.T. Copson, {\bf{The Mathematical Theory of
Huygens Principle}}, (Clarendon Press, Oxford, 1950).
\item Landau and Lifshitz, {\bf{The Classical Theory of Fields}}.
\item L. Eyges, {\bf{The Classical Electromagnetic Field}},
(Addison-Wesley, Reading, 1972).
\item Hecht-Zajac, {\bf{Optics}}, (Addison-Wesley, Reading, 1979).
\end{enumerate}
\noindent The Baker reference, especially, has a good discussion of
the limits of the Kirchoff approximation, of course,
Landau and Lifshitz have an excellent discussion of the physics,
but Hecht-Zajac's discussion is perhaps the most elementary, and will
often be quoted here.

The typical question we ask is, given a strictly monochromatic 
point source $S$, what resulting radiation
is observed at the point $O$ on the opposite side of the screen.
At first this approach may seem to limit us just to point sources.
The case of a real extended source which emits non-monochromatic
light does not, however, require special treatment.  This is because 
of the linearity of our equations and the complete independence 
(incoherence) of the light emitted by different points of the source.
The interference terms average to zero. Thus
the total diffraction pattern is simply the sum of the intensity 
distributions obtained from the diffraction of the independent components
of the light.

\centerline{\psfig{figure=fig10a.ps,height=2.5in,width=8.5in}}

	To treat the theory of diffraction, a number of approximations
will be necessary. 
\begin{itemize}
\item First, we assume that we can neglect the vector
nature of the electromagnetic fields, and work instead with a scalar 
complex function $\psi$ (a component of $\E$ or $\B$, or the
single polarization observed in the ripple tank discussed above).  
In principle,
$\psi$ is any of the three components of either $\E$ or $\B$.
In practice, however, the polarization of the radiation is usually
ignored and the intensity of the radiation at a point is usually taken 
as $|\psi(\x,t)|^2$.  This first
assumption limits the number of geometries we can treat.  
\item Second,
we will generally assume that $\la/d \ll 1$, where d is the linear
dimension of the aperture or obstacle.  
\item The third assumption is
that we will only look for the first correction to geometric
optics due to diffraction.  This is often called the Kirchoff
approximation, which will be discussed a bit later. (This set of
approximations are sometimes also called the Kirchoff approximation
scheme.)
\end{itemize}

We then impose the boundary condition\footnote{Note that we only impose
one boundary condition on the screen.  This is to avoid the difficulties
when both boundary conditions (Neumann and Dirichlet) are imposed, as
discussed in Jackson on page 429}
\beq
\psi(\x,t)=0\;\;\; {\rm everywhere\;on\;the\;screen}
\eeq
We will assume that $\psi$ obeys the wave equation, and the source
is harmonic so that it emits radiation of frequency $\om$, hence
\beq
\psi(\x,t)=\psi(\x)\emot\,.
\eeq
Thus the spatial wave function obeys the Helmholtz equation
\beq
\dal\psi=(\lap +k^2)\psi(\x)=0\,\;\, , {\rm with} \;\; k=\om/c
\eeq
if we assume that the waves propagate in a homogeneous medium and
restrict our observations to points away from the source $S$.  As always,
the physical quantities will be the {\em real} amplitude 
$\Re\lep\psi(\x) \emot\rip$, and the modulus squared which denotes the
time-averaged intensity.

	Let's consider just the observation points in the volume V
below which is bounded by the screen ($z=0$ plane), and a hemisphere at 
infinity.  

\centerline{\psfig{figure=fig11.ps,height=2.5in,width=8.5in}}

Then, everywhere within $V$, the Helmholtz equation $\dal\psi=0$ is obeyed
since the source lies outside of $V$.  To find $\psi$ within $V$, we
use Green's theorem
\beq
\invp\lep\psxp\lapp\phxp -\phxp\lapp\psxp\rip=
\inap \nnp\cdot\lep \psxp\grad'\phxp-\phxp\grad'\psxp\rip\, ,
\eeq
or, adding and subtracting $k^2\psxp\phxp$ from the first integrand, we
get
\beq
\invp\lep\psxp(\lapp+k^2)\phxp-\phxp(\lapp+k^2)\psxp\rip
=\inap \nnp\cdot\lep \psxp\grad'\phxp-\phxp\grad'\psxp\rip\
\eeq
This works for any two functions $\psi$ and $\phi$.  We will take 
$\psxp$ to be the wave amplitude, and take $\phxp=G(\x,\xp)$,
where the Dirichlet Green's function satisfies
\beq
(\lap + k^2)G(\x,\xp)=-4\pi\de(\x-\xp)\; {\rm in\; V}
\eeq
{\em{i.e.}}, it is the response to a unit point source so that in free
space $G(\x,\xp)=\frac{e^{ik|\x-\xp|}}{|\x-\xp|}$.  However, need to solve 
for $G$ with boundary conditions
\beq
G(\x,\xp)=0\;{\rm for} \;\xp\;{\rm on}\; S\,.
\eeq
since we will be using Dirichlet boundary conditions
on $\psi$:  We will specify the value of $\psx$ (as opposed to 
its derivative) on the boundary $S$.  Now using the facts that
\beq
(\lap + k^2)G(\x,\xp)=-4\pi\de(\x-\xp)\; {\rm in\; V}
\eeq
\beq
(\lap + k^2)\psx = 0 \; {\rm in\; V}
\eeq
and that
\beq
G(\x,\xp)=0\;{\rm for} \;\xp\;{\rm on}\; S
\eeq
this Green's theorem becomes (the Kirchoff Integral)
\beq
\psx=-\frac{1}{4\pi}\inap \nnp\cdot\lep\psxp\grad' G(\x,\xp) \rip
\eeq

To proceed further we must determine the form of $G(\x,\xp)$ and we must
now specify $\psxp$ on the surface $S$.  For an infinite planar screen
the Green's function
is given by the method of images (analogously to the electrostatic case)
\beq
G(\x,\xp)=\frac{e^{ik|\x-\xp|}}{|\x-\xp|}-\frac{e^{ik|\x-\yp|}}{|\x-\yp|}
\eeq

\centerline{\psfig{figure=green.ps,height=1.25in,width=8.5in}}

\noindent where $\xp=(x',y',z')$ is in $V$, but $\yp=(x',y',-z')$
(the image point)
is not.  This satisfies $(\lap + k^2)G(\x,\xp)=-4\pi\de(\x-\xp)$ in
$V$, and vanishes both on the plane $z'=0$ and on the hemisphere at
infinity.  Note  that it vanishes as $1/{r'}^2$ as $r'\to\infty$.
From this it is clear that the integral
\beq
\psx=-\frac{1}{4\pi}\inap \nnp\cdot\lep\psxp\grad' G(\x,\xp) \rip
\eeq
has a vanishing contribution from the hemisphere at infinity,
since $\psxp$ vanishes at least as fast as $1/r'$ as $r'\to\infty$ (since 
it is a solution of the wave equation for a finite source).
Furthermore, from the boundary condition 
\beq
\psxp=0 \;\;{\rm on\;the\;screen}
\eeq
we can see that the integral only gets a nonzero contribution from
the opening
\beq
\psx=-\frac{1}{4\pi}{\int_{opening} d^2x'\,}\nnp\cdot\lep\psxp\grad' G(\x,\xp) \rip\,.
\eeq

	So far we have made an exact evaluation of our {\em{scalar}} theory.
However, to proceed we must make an approximation.  We will assume that
the value of $\psxp$ in the opening will be the {\em same} as if
the screen was not there at all.  This means that the wavelength
we are considering must be small compared to to characteristic
size of the problem (i.e. the size of the opening).  As a result,
our formalism will yield just the the lowest-order correction
due to diffraction to the results of geometrical or ray optics.
This approximation is called the {\bf Kirchoff approximation}.

	Now to implement this approximation we need two things.
First, we need the field strength at the opening.  For a source
of unit strength at position $\x_0$, the field at  position $\xp$
in the opening is taken to be

\centerline{\psfig{figure=fig12.ps,height=2.5in,width=8.5in}}

\noindent which is just a spherical wave. Second, we need the component of 
$\grad' G(\x,\xp)$ in the direction of $\nnp$ in the opening.
Since $\nnp$ is the outward normal direction from the volume $V$,
this means we need
\beq
-\der{}{z'} G(\x,\xp)=-\der{}{z'}\lep\frac{e^{ik|\x-\xp|}}{|\x-\xp|}-
\frac{e^{ik|\x-\yp|}}{|\x-\yp|}\rip\,.
\eeq
Since we are using a short wavelength approximation, it follows
that $k|\x-\xp| \gg 1$ and $k|\x-\yp| \gg 1$.  Consequently, in taking
the derivative of $G(\x,\xp)$, the derivatives coming from the denominators
are negligible compared to those from the exponentials.  Thus
\beq
-\der{}{z'} G(\x,\xp)\approx
\frac{{ik(z-z')}}{|\x-\xp|}\frac{e^{ik|\x-\xp|}}{|\x-\xp|}
+
\frac{{ik(z+z')}}{|\x-\yp|}\frac{e^{ik|\x-\yp|}}{|\x-\yp|}.
\eeq
We only need to evaluate this in the opening ($z'=0$)
\beq
\left.-\der{}{z'} G(\x,\xp)\right|_{z'=0}
\approx
\frac{{2ikz}}{|\x-\xp|}\frac{e^{ik|\x-\xp|}}{|\x-\xp|}
\eeq
since $\xp=\yp$ here.

	Thus, our Kirchoff approximation yields the following
expression for the field observed at $\x$
\begin{eqnarray}
\psx &=& -\frac{1}{4\pi}
{\int_{opening} d^2x'\,}\nnp\cdot\lep\psxp\grad G(\x,\xp) \rip \\
   &=& -\frac{1}{4\pi}{\int_{opening} d^2x'\,}
\frac{{2ikz}}{|\x-\xp|}\frac{e^{ik|\x-\xp|}}{|\x-\xp|}
\frac{e^{ik|\x_0-\xp|}}{|\x_0-\xp|}\,.
\end{eqnarray}
We may also write
\beq
\frac{z}{|\x-\xp|}=\cos(\th')\, ,
\eeq
where $\th'$ is the angle from the normal to the opening at integration
point $\xp$ to the observation point $\x$.  

\centerline{\psfig{figure=fig13.ps,height=2.5in,width=8.5in}}

\noindent Note that $\th'$ is different for each integration point $\xp$.
Thus our expression becomes
\beq
\psx =-\frac{ik}{2\pi}{\int_{opening} d^2x'\,}
\frac{e^{ik|\x_0-\xp|}}{|\x_0-\xp|}
\frac{e^{ik|\x-\xp|}}{|\x-\xp|}
\cos(\th')
\eeq
The first term is the field from the source located at
$\x_0$ and received at point $\xp$ in the opening.  The second
term is the field from the ``source'' at $\xp$ in the opening
received at observation point $\x$.  The third term is the
inclination factor.
Thus, this integral is just an expression of the Huygens-Fresnel principle.

\subsection{Babinet's Principle}

	Let's explore the consequences of this formula.  First,
consider two different screens which are complimentary. 

\centerline{\psfig{figure=fig14.ps,height=2.0in,width=8.5in}}

\noindent One has an opening, labeled $\sigma$, while the other is just 
a disk.  For the second screen the "opening" is the entire plane $z=0$
except for the disk.  This opening is the compliment of $\sigma$,
so we will call it $\sib$.

	We notice something interesting about the amplitudes received
at the point $O$ in the two complimentary cases, we have
\begin{eqnarray}
\psx &=& -\frac{1}{4\pi}{\int_{\si} d^2x'\,}\nnp\cdot\lep\psxp\grad G(\x,\xp) \rip \\
\overline{\psx} &=& 
-\frac{1}{4\pi}{\int_{\sib} d^2x'\,}\nnp\cdot\lep\psxp\grad G(\x,\xp) \rip
\end{eqnarray}
The {\em sum} of these two amplitudes is then
\beq
\psx + \overline{\psx} =
-\frac{1}{4\pi}{\int_{\sib+\si} d^2x'\,}\nnp\cdot\lep\psxp\grad G(\x,\xp) \rip
\eeq
However, $\sib+\si$ represents the entire plane $z=0$.  Thus
$\psx + \overline{\psx}$ represents the amplitude detected at $\x$
if there was no screen at all.  In other words $\psx + \overline{\psx}$
includes no diffraction.

	To interpret this, note that $\psx$ may be written as
\beq
\psx=f \frac{e^{ik|\x-\x_0|}}{|\x-\x_0|} + \psi_{diff}(\x)
\eeq
where f is a ray optic amplitude.  If $f=0$, then there is no line-of-sight
from the observer to the source, while if $f=1$, then there is.
The second term is the amplitude due to diffraction of the wave.
Returning to our amplitudes $\psx$ and $\overline{\psx}$, we see that
one will have $f=0$ and the other $f=1$, so that the sum
will include a term ${e^{ik|\x-\x_0|}}/{|\x-\x_0|}$.  {\em{In fact,
this is all it will include, since $\psi+\overline{\psi}$ includes no
diffraction.  Thus, we see that the diffraction amplitudes for
complimentary screens cancel.}}  This is {\bf Babinet's Principle}.
It says that solving one diffraction problem is tantamount to solving
its compliment.  However, Babinet's principle does {\em not}
say that the intensities will cancel in the two cases!  We will
see some consequences of Babinet's principle in what follows

\subsection{ Fresnel and Fraunhofer Limits}
	Lets return to our expression for $\psx$ in the Kirchoff
approximation.  Since we assume that the wavelength is small,
it follows that
\beq
ka \gg 1 \;\;\; kr_0\gg1\;\;\; kr\gg1
\eeq
where $a$ is the typical aperture size, $r_0$ is the distance from the
screen to the source, and $r$ is the distance to the observer.

In general these approximations should be applied to the expression for 
$\psx$ only after the integral over the opening has been carried
out.  This can be done for some simple geometries
(as in the homework).  What we will do here is to insert the above
limits into the Kirchoff approximate expressions for 
$\psx$.  In what follows, we will restrict our attention to
apertures, not discs.  By doing this it follows that $r'$ will
have an upper limit $a$, so $kr'\gg 1$.  In practice, the restriction
to apertures is not a real limitation since Babinet's principle
allows us to calculate the the diffraction amplitude for a disk from
the diffraction amplitude for the complimentary screen.

Once again, our expression for $\psx$ is
\beq
\psx =-\frac{ik}{2\pi}{\int_{opening} d^2x'\,}
\frac{e^{ik|\x_0-\xp|}}{|\x_0-\xp|}
\frac{e^{ik|\x-\xp|}}{|\x-\xp|}
\cos(\th')\,.
\eeq
Now if $a/r_0$ and/or $a/r$ are not small, then we are limited
in what further approximations are possible since $r'$ is not
necessarily small compared to either $r$ or $r_0$.  This 
limit is called {\bf Fresnel Diffraction}, in which the source
and/or the observation points are close enough to the aperture
that we must worry about the diffraction of spherical waves.
This presents a difficult problem, which we will not treat
(but please see the homework and Landau and Lifshitz, Classical Theory 
of Fields, Sec. 60).

	In the opposite limit, which we will treat, is called
{\bf{ Fraunhofer Diffraction}}.  In this limit
\beq
ka > 1 \;\;\; kr_0\gg1\;\;\; kr\gg1 \;\;\; a\ll r_0
\;\;{\rm and}\;\; a\ll r
\eeq
Note that we specify $ka >1$ rather than $ka \gg 1$ since we want to
solve for the first finite corrections to the latter inequality,
so that
\beq
\frac{ka^2}{r_0}=ka\lep\frac{a}{r_0}\rip\ll 1 \;\;
{\rm and} \;\;
\frac{ka^2}{r'}\ll 1
\eeq
Thus we are looking at plane waves instead of spherical waves in this limit.

	It is possible to simplify our expression for $\psx$
even further in this limit.  Then
\beq
|\xp-\x_0|\approx r_0 -\frac{\x_0\cdot\xp}{r_0}=r_0+ \nn_0\cdot\xp
\eeq
where $\nn_0$ is a unit vector from the source to the origin
(taken to be the center of the aperture), so that $\x_0=-\nn_0 r_0$.
Similarly,
\beq
|\x-\xp|\approx r-\nn\cdot\xp
\eeq
where $\nn$ is a unit vector from the origin to the observer
so that $\x=\nn r$.

\centerline{\psfig{figure=fig15.ps,height=2.5in,width=8.5in}}

\noindent Thus,
\beq
\frac{e^{ik|\xp-\x_0|}}{|\xp-\x_0|}
\frac{e^{ik|\x-\xp|}}{|\x-\xp|} \approx
\frac{e^{ik(r+r_0)}}{r r_0} e^{ik(\nn_0-\nn)\cdot\xp}
\eeq
where we have dropped second-order terms coming from the denominators.
Furthermore, since $r$ is large compared to the aperture size, it is 
appropriate to take
\beq
\cos(\th')=\frac{z}{|\x-\xp|}\approx 1
\eeq
inside the integral.  Thus in the Fraunhofer limit
\beq
\psx=-\frac{ik}{2\pi}
\frac{e^{ik(r+r_0)}}{r r_0}
{\int_{opening} d^2x'\,} e^{ik(\nn_0-\nn)\cdot\xp}
\eeq
But the incident wave vector was $k\nn$, so 
\beq
k(\nn_0-\nn)=\k_{in}-\k_{diff}\equiv\q
\eeq
so we finally have
\beq
\psx=-\frac{ik}{2\pi}
\frac{e^{ik(r+r_0)}}{r r_0}
{\int_{opening} d^2x'\,} e^{i\q\cdot\xp}
\eeq

\section{Example Problems}

	Let's consider some examples.

\subsection{Example: Diffraction from a Rectangular Aperture}
	Here the opening is given by $|x'|<a$ and $|y'|<b$, so
the integral above is
\beq
{\int_{opening} d^2x'\,} e^{i\q\cdot\xp}
=\int_{-a}^{a} dx'\,e^{iq_x x'}
\int_{-b}^{b} dy'\,e^{iq_y y'}
=2\frac{\sin(q_xa)}{q_x}\,2\frac{\sin(q_yb)}{q_y}
\eeq
The amplitude at $\x$ is
\beq
\psx=-\frac{2ikab}{\pi}\frac{e^{ik(r+r_0)}}{rr_0}
\lep \frac{\sin(q_xa)}{a q_x} \rip
\lep \frac{\sin(q_yb)}{b q_y} \rip
\eeq
thus, the intensity of the diffracted wave is
\beq
\I(\x)=\left|\psx\right|^2=I_0 
\lep \frac{\sin(q_xa)}{a q_x} \rip ^2
\lep \frac{\sin(q_yb)}{b q_y} \rip ^2
\eeq
with 
\beq
I_0=\frac{4k^2a^2b^2}{\pi^2 r_0^2 r^2}
\eeq
We see that all of the angular dependence of in this is embodied in the 
dependence on $\q$, the momentum transfer.  Since we a looking at small-angle
scattering, we are taking $\q$ to lie essentially in the x-y plane.

\centerline{\psfig{figure=fig16.ps,height=1.0in,width=8.5in}}

\noindent We see that the nodes of $\I(\x)$ occur for
\beq
q_x=\frac{m\pi}{a},\;\;m\neq 0,\;\;{\rm and/or}\;\;
q_y=\frac{n\pi}{b},\;\;n\neq 0
\eeq
The global maximum ($I(\x)=I_0$) occurs for $q_x=q_y=0$, i.e. for forward
scattering.  Thus the intensity of the diffracted light
looks something like

\centerline{\psfig{figure=fig17.ps,height=3.5in,width=8.5in}}

	We can recover the formula for single-slit diffraction
by taking $b\gg a$, so that our rectangular aperture becomes a slit
of width $2a$.  Then

\centerline{\psfig{figure=fig18.ps,height=3.0in,width=8.5in}}

\beq
q_x=\lep k_x \rip_{in} - \lep k_x\rip_{diff}=
k\sin(\th_{in})-k\sin(\th_{diff})
\eeq
and the condition that there is a node is 
\beq
q_x=k\lep \sin(\th_{in})-\sin(\th_{diff})\rip=\frac{m\pi}{a}\;\;
m\neq 0\, .
\eeq
Since $k=2\pi/\la$, this may be written
\beq
2 a \lep \sin(\th_{in})-\sin(\th_{diff})\rip=m\la\;\;
m\neq 0
\eeq
which is the usual expression for a node in single-slit diffraction.

\subsection{Example: Diffraction from a Circular Aperture}
Here the opening is given by $r'<a$.  Our Fraunhofer expression for
the amplitude at $\x$ is then
\beq
\psx=-\frac{ik}{2\pi} \frac{e^{ik(r+r_0)}}{r r_0} \int_0^a dr'\, r'
\int_0^{2\pi} d\ph' e^{iqr'\cos(\ph')}
\eeq
where we have taken $\q$ to lie entirely in the plane of the aperture,
as is appropriate in small-angle diffraction.  Now
\beq
\int_0^{2\pi} d\ph' e^{iqr'\cos(\ph')}=2\pi J_0(qr')
\eeq
where $J_0$ is the zeroth-order Bessel function.  Our expression for the
diffraction amplitude is then
\beq
\psx=-\frac{ik}{2\pi} \frac{e^{ik(r+r_0)}}{r r_0} \int_0^a dr'\,{r'}\,
2\pi J_0(qr')
=-ika^2 
\frac{e^{ik(r+r_0)}}{r r_0}
\lep \frac{J_1(qa)}{qa} \rip
\eeq
where we have used $J_1(0)=0$.  The diffracted intensity is then
\beq
I(\x)=I_0
\lep \frac{J_1(qa)}{qa} \rip^2\,,\;\; I_0=\frac{k^2\,a^4}{r^2\,r_0^2}
\eeq
this reaches a maximum at $q=0$, it then has a minima of zero and
maxima which decrease as $q$ increases.  For incident light with
$\k$ normal to the opening,
\beq
q=k_{diff}\sin(\th_{diff})=k\sin(\th_{diff})
\eeq
since the magnitude of the wavevector does not change.  Thus,
\beq
I(\th_{diff})=I_0\lep \frac{J_1\lep ka\sin(\th_{diff})\rip}
{ ka\sin(\th_{diff})}\rip^2
\eeq
The first minima of $J_1(x)$ is at $x=3.83$, hence the first minima of
diffraction will occur when
\beq
ka\sin(\th_{diff})=3.83\,,\;\; {\rm or}\;\; 
sin(\th_{diff})=1.22\lep\frac{\la}{2a}\rip\,.
\eeq
Since, by assumption, $\la$ is small compared to $a$, 
$sin(\th_{diff})\approx\th_{diff}$, and so the angle of the first
node is roughly
\beq
\th_{diff} = 1.22\lep\frac{\la}{2a}\rip\,.
\eeq

\centerline{\psfig{figure=fig19.ps,height=3.0in,width=8.5in}}

	We notice from these examples of Fraunhofer diffraction
that all of the light falling on the aperture is deflected.
Granted, the intensity for forward scattering ($\q=0$) is nonzero,
but the total amount of light with $\q=0$ is
\beq
\left.|\ps(\q)|^2\,q^2\,dq\,d\Om\right|_{\q=0} \to 0
\eeq
\noindent Thus, it is legitimate to say that all of the light
is deflected.  Let us apply this observation to a disc which is
totally absorbing (i.e. a perfect black-body).  If plane waves are
incident on this disc, then all the light that falls directly upon
it is absorbed.  If there was no diffraction, then there would be a
geometrically perfect shadow cast by the object, and the total
cross section of the disk would be its area $A$.  But of course
there is diffraction, so in addition to absorbing light, some
will also be deflected by the disk.  By Babinet's
principle, as much light will be diffracted by the disc as by an aperture
of the same shape as the disc but in a screen.  But we have said that
in Fraunhofer diffraction all of the light falling on an aperture
is deflected.   Thus for our disc, the cross section  for deflection
of light will be the disc area $A$ as well.  Thus the total cross section,
including both absorption (inelastic cross section) and diffraction
(elastic cross section) is
\beq
\si_{total}=\si_{inelastic}+\si_{elastic}= A+A=2A
\eeq
This is twice the area of the disc.  Recall that in this argument we assume
that $ka \gg1$, where $a$ is the characteristic size of the disc.  
In this limit in quantum mechanics one finds
that the cross section for scattering from a sphere of radius
$a$ is $2\pi a^2$.  The factor of two has the same origins here
as it does in quantum mechanics.

\subsection{Diffraction from a Cross}
\begin{enumerate}
\item Given a normally incident wave of frequency $\om$
calculate the diffraction pattern produced by an opaque screen with an 
aperture in the shape symmetric cross of inner dimension $a$ and
outer dimension $b$
in the Fraunhofer limit of the scalar Kirchoff approximation.
Very roughly tell what the diffraction pattern will look like.
\item The correct answer to the first part of this problem also yields
the solution
to the diffraction pattern involving a quite different set of 
apertures.  What is this other set of apertures?
\end{enumerate}

%source /home/wanderer/jarrell/text/courses/jackson/exams/q3/
\centerline{\psfig{figure=cross.ps,height=2.5in,width=8.5in}}

{\bf{Solution.}}  In the Fraunhoffer limit, we can use \eq{197}
of the notes.
\[
\psx=-\frac{ik}{2\pi}
\frac{e^{ik(r+r_0)}}{r r_0}
{\int_{opening} d^2x'\,} e^{i\q\cdot\xp}
\]
where $r_0$ is the distance from the aperture to the source,
$r$ is the distance from the aperture to the observer, and $\q$
is the difference between the incident and scattered wavevector.
We will take $\q$ to lie entirely in the plane of
the aperture, as is appropriate in the small angle limit.

	For the geometry in the figure, the integrals may be performed
trivially
\[
\psx=-\frac{ik}{2\pi}
\frac{e^{ik(r+r_0)}}{r r_0}
\lep 
\int_{-a}^{a} dx' \int_{-a}^{a} dy' +
\int_{-a}^{a} dx'\lep \int_{-b}^{-a} dy' + \int_{a}^{b} dy'\rip +
\int_{-a}^{a} dy'\lep \int_{-b}^{-a} dx' + \int_{a}^{b} dx'\rip
\rip 
e^{iq_xx'} e^{iq_yy'}
\]
so that
\[
\psx=\frac{2ik}{\pi q_x q_y}
\frac{e^{ik(r+r_0)}}{r r_0}
\leb \sin(aq_x) \sin(bq_y) +\sin(aq_y) \sin(bq_x) -
\sin(aq_x) \sin(aq_y) \rib
\]
The corresponding intensity $|\psx|^2$ will have a central maximum with
$I\propto (2ab-a^2)^2$, and the whole pattern will be symmetric to 
rotations modulo $\pi/2$.  By Babinet's principle we know that the 
diffraction from a cross-shaped screen has an amplitude, that when added 
to $\psx$, is the same as if there was no diffraction at all.

\subsection{Radiation from a Reciprocating Disk}
A disc of radius $a$ lies in the $z=0$ plane and is centered at the 
origin.  It is uniformly charged with surface charge density $\sigma$, 
and it rotates around the z-axis with an angular velocity
$\Omega\cos(\omega t)$ where $\Omega$ and $\omega$ are constants.
Assuming that the motion is nonrelativistic, find the fields in the 
radiation zone, the angular distribution of radiated power, and the
total radiated power.

{\bf{Solution.}}  Here we will employ the methods developed to treat 
the radiation of harmonic current sources.  The current of
the rotation disc is (in cylindrical coordinates)
\[
\J(\x,t)=\delta(z)\sigma\Omega \cos(\omega t) \rho \phh
\]
or, in complex notation
\[
\J(\x,t)=\delta(z)\sigma\Omega e^{-i\omega t} \rho \phh = 
\J(\x)e^{-i\omega t}
\]
Then, 
\[
\f(\th,\ph)=\frac{1}{c} \invp \Jxp e^{-ik\nn\cdot\xp}=
\frac{\sigma\Omega}{c} \int_0^a \rho'd\rho'\int_0^{2\pi}d\phi' \rho'\phh'
e^{-ik\nn\cdot\xp}
\]
We can use the azimuthal symmetry of the problem, and assume that $\nn$
is in the xz plane.  Then $\nn=\cth \zh + \sth \xh $,  
$\xp=\rho'\lep \cos(\ph')\xh + \sin(\ph')\yh \rip$, and
$\phh'= \cos(\ph')\yh - \sin(\ph')\xh$.
For non-relativistic motion $\Omega a/c \ll 1$, this allows us to keep
only the first nonvanishing term in the exponential, then
\[
\f(\th,\ph)=-i\frac{\pi k a^4 \sigma \Omega}{4 c} \sth \phh
\]
Hence, in the radiation zone,
\[
\B= ik\frac{e^{ikr}}{r}\lep \nn\times\f\rip=
-\frac{\pi k^2 a^4 \sigma \Omega}{4c} \sth \frac{e^{ikr}}{r}\thh
\]
and 
\[
\E=\B\times\nn=
\frac{\pi k^2 a^4 \sigma \Omega}{4c} \sth \frac{e^{ikr}}{r}\phh
\]
The power distribution is given by
\[
\der{P}{\Omega}=\frac{ck^2}{8\pi}\leb |\f|^2-|\nn\cdot\f|^2\rib=
\frac{ck^2}{8\pi} \frac{\pi^2k^2a^8\sigma^2\Omega^2}{16c^2}\sin^2(\theta)
\]
The total power is obtained by integrating over all solid angles
\[
P=\frac{\pi^2k^4\sigma^2\Omega^2a^8}{48c}
\]



\edo

