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\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 
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\title{Chapter Fourteen\\Radiation by Moving Charges}
\author{Sir Joseph Larmor\\(1857 - 1942)}
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We have already calculated the radiation produced by some known charge and
current distribution. Now we are going to do it again. This time, however,
we shall consider that the source is a single charge moving in some fairly
arbitrary, possibly relativistic, fashion. Here the
methods of chapter 9, e.g., multipole expansions, are impractical and there
are better ways to approach the problem.
\section{Li\'enard-Wiechert potentials}
The current and charge densities produced by a charge $e$ in motion are
\beqa
\rhxt&=&e\de(\x-\x(t))\nonumber\\
\Jxt&=&e\v(t)\de(\x-\x(t))
\eeqa
if $\x(t)$ is the position of the particle at time $t$ and $\v(t)\equiv d\x
(t)/dt\equiv\xd(t)$ is its velocity. In four-vector notation,
\beq
J^\mu(\x,t)=ec\be^\mu\de(\x-\x(t))
\eeq
where $\be^\mu\equiv(1,\vbe$) is {\bf not} a four-vector; $\vbe=\v/c$.

From $\J$ and $\rh$, one finds $\A$ and $\Ph$. We can do this in an
infinite space by making use of the retarded Green's function $\gxt=
\de(t-t'-\xxpa/c)/\xxpa$. From here, we can evaluate the electromagnetic field
as appropriate derivatives of the potentials. All of these manipulations are
straightforward. Furthermore, the integrations are relatively easy because
there are many delta functions. The problem becomes interesting and unfamiliar,
however, for highly relativistic particles which produce large retardation
effects.

Let us start from the integral expression for the potentials:
\beqa
A^\mu(\x,t)&=&\int d^3x'dt'\,\gxt J^\mu(\xp,t')\nonumber\\
&=&\frac1c\int d^3x'dt'\frac{\de(t-t'-\xxpa/c)J^\mu(\xp,t')}{\xxpa}\nonumber\\
&=&e\int d^3x'dt'\,\be^\mu(t')\frac{\de(\xp-\x(t'))\de(t-t'-\xxpa/c)}\xxpa
\nonumber\\
&=&e\int dt'\,\be^\mu(t')\frac{\de(t-t'-|\x-\x(t')|/c)}{|\x-\x(t')|}
\eeqa
This form reflects the retarded nature of the problem.

\centerline{\psfig{figure=fig1.ps,height=3.5in,width=8.5in}}

\noindent The evaluation of this integral is not totally simple because 
the argument
of the $\de$-function is not a simple function of the time $t'$. In general
when one faces an integral of this form, one invokes the rule
\beq
\int dt'\,f(t')\de[g(t')]=f(t_0)\les\lel\der g{t'}\ril_{t_0}\right.
\eeq
where $t_0$ is the zero (there may be more than one) of $g$\footnote{
More generally $\de[g(x)]=\sum_j \de(x-x_j)/|g'(x_j)|$ where $x_j$ are
the simple zeroes of $g(x)$.  If $g'(x_j)=0$ (a complex zero), then
$\de[g(x)]$ makes no sense.}, i.e., $g(t_0)=0$. Applying this to the 
present case, we have
\beq
g(t')=t'+|\x-\x(t')|/c-t=t'+[(\x-\x(t'))\cdot(\x-\x(t'))]^{1/2}/c-t,
\eeq
so
\beq
\der g{t'}=1+\frac1{2c}\lep-2\der{\x(t')}{t'}\rip\cdot\frac{(\x-\x(t'))}{|
\x-\x(t')|}=1-\frac{\v(t')\cdot(\x-\x(t'))}{c|\x-\x(t')|}\equiv1-\vbe(t')
\cdot\nn(t').
\eeq
The unit vector $\nn$ points from the point $\x(t')$ on the particle's path
toward the field point $\x$; it, and $\vbe$, must be evaluated at a time
$t'$ which is earlier than $t$, the time at which the field is evaluated,
by some amount which is determined by solving the equation
\beq
t'+|\x-\x(t')|/c=t.
\eeq
Combining \eqss{3}{4}{6} we find that the potentials are given simply by
\beq
A^\mu(\x,t)=e\leb\frac{\be^\mu}{|\x-\x(t')|(1-\vbe\cdot\nn)}\rib_{ret}
=\leb\frac{e\be^\mu}{R\ka}\rib_{ret}
\eeq
where $R\equiv|\x-\x(t')|$, $\ka=1-\vbe\cdot\nn$, and the subscript $ret$ means
that the quantity in brackets [...] must be evaluated at the retarded time
$t'$ determined from \eq{7}.

Our potentials, \eq{8}, are known as the {\em Li\'enard-Wieckert
potentials}. Probably their most significant feature is the fact that they vary
inversely as $1-\vbe\cdot\nn$ or $\ka$; this factor can be very close to zero
for $\nn\parallel\vbe$ if $\be$ is close to one, i.e., for highly relativistic
particles, meaning that there is a strong maximum in the potentials
produced by a relativistic particle in the direction of the particle's
velocity (at some retarded time).

We wish next to find the electromagnetic field. One may do this in a variety
of ways. One is simply {\bf carefully} to take derivatives of the
Li\'enard-Wieckert potentials, a procedure followed in, e.g., Landau and
Lifshitz' book, {\em The Classical Theory of Fields}. Another, much more
elegant, is to find the fields in the instantaneous rest frame of the
particle and to Lorentz-transform them to the frame interest. Another, not
very elegant at all, and about to be employed here, is to go back to the
integrals, \eq{3}, for the potentials and take derivatives of these
expressions. Consider just the vector potential,
\beq
\Axt=e\int dt'\,\frac{\vbe(t')\de(t'+R(t')/c-t)}{R(t')}.
\eeq
If we wish to find $\Bxt$, we have to take derivatives of $R$ with respect
to various components of $\x$. Consider, for example,
\beq
\grad f(R)=\der fR\grad R=\der fR \frac\R R=\nn\der fR.
\eeq
Application of this simple rule gives (since $\curl (\psi\a)=\grad\psi\times\a
+\psi\curl\a$, and $\curl\vbe(t')=0$ due to the lack of an $\x$ dependence.)
\beqa
\Bxt=\curl\Axt=e\int dt'\,\grad\lep\frac{\de(t'+R/c-t)}R\rip\times\vbe(t')
\nonumber\\=e\int dt'\,(\nn\times\vbe)\leb-\frac1{R^2}\de(t'+R/c-t)+
\frac1{cR}\de'(t'+R/c-t)\rib
\eeqa
where the prime on the delta function denotes differentiation with respect
to the argument. Hence
\beq
\Bxt=e\lec\leb\frac{\vbe\times\nn}{\ka R^2}\rib_{ret}+\int dt'\,\frac\ka{cR}
\lep\frac{\de'(t'+R/c-t)}\ka\rip(\nn\times\vbe)\ric
\eeq
Now, from \eqs{5}{6}
\beq
\der{(t'+R/c-t)}{t'}=\ka\orh \ka dt'=d(t'+R/c-t)
\eeq
and so
\beqa
\Bxt&=&e\lec\leb\frac{\vbe\times\nn}{\ka R^2}\rib_{ret}+\int d(t'+R/c-t)\lep
\frac{\nn\times\vbe}{c\ka R}\rip\de'(t'+R/c-t)\ric\nonumber\\
&=&e\lec\leb\frac{\vbe\times\nn}{\ka R^2}\rib_{ret}-\int d(t'+R/c-t)\pde{[
(\nn\times\vbe)/c\ka R]}{(t'+R/c-t)}\de(t'+R/c-t)\ric\nonumber\\
&=&e\lec\leb\frac{\vbe\times\nn}{\ka R^2}\rib_{ret}-\leb\frac1\ka\pde{}{t'}
\lep\frac{\nn\times\vbe}{cR\ka}\rip\rib_{ret}\ric\nonumber\\
&=&e\lec\leb\frac{\vbe\times\nn}{\ka R^2}\rib_{ret}+\frac1c\leb\frac1\ka
\pde{}{t'}\lep\frac{\vbe\times\nn}{\ka R}\rip\rib_{ret}\ric.
\eeqa

The electric field can be found by similar manipulations:
\beqa
\Ext&=&-\grad\Phxt-\frac1c\pde\Axt t\nonumber\\
&=&-e\int dt'\,\nn\der{}R\lep\frac{\de(
t'+R/c-t)}R\rip+\frac ec\int dt'\,\frac{\vbe\de'(t'+R/c-t)}R\nonumber\\
&=&e\int dt'\lec\nn\frac{\de(t'+R/c-t)}{R^2}+\frac1{cR}(\vbe-\nn)\de'(t'+R/
c-t)\ric\nonumber\\
&=&e\leb\frac\nn{\ka R^2}\rib_{ret}-\frac ec\leb\frac1\ka\pde{}{t'}\lep
\frac{\vbe-\nn}{\ka R}\rip\rib_{ret}.
\eeqa

We now have expressions for $\E$ and $\B$, but they involve time
derivatives of retarded quantities. We can work out each of these
derivatives. First, we consider just the derivative of $\nn=\R/R$. One has
\beq
\pde\R{t'}=-\xd(t')=-\vbe c
\eeq
and
\beq
\pde R{t'}=\pde{[(\x-\x(t'))\cdot(\x-\x(t'))]^{1/2}}{t'}=
\frac1{2R}[-2(\x-\x(t'))\cdot\xd(t')]=-\nn\cdot\xd(t')=-\nn\cdot\vbe
c\eeq
Hence,
\beq
\frac1c\der\nn{t'}=\frac1{cR}\pde\R{t'}-\frac1{cR^2}\R\pde R{t'}=
-\frac1R\leb\vbe-(\nn\cdot\vbe)\nn\rib.
\eeq
The quantity in brackets is can be written more concisely:
\beq
(\nn\cdot\vbe)\nn-\vbe=(\nn\cdot\vbe)\nn-(\nn\cdot\nn)\vbe=\nn\times(\nn
\times\vbe);
\eeq
hence we find
\beq
\frac1c\der\nn{t'}=\frac1R\nn\times(\nn\times\vbe).
\eeq
Employing this useful result we have, for the electric field,
\beqa
\Ext&=&e\leb\frac\nn{\ka R^2}+\frac1{\ka^2R^2}\nn\times(\nn\times\vbe)+\frac{
\nn}{c\ka}\pde{}{t'}\lep\frac1{\ka R}\rip-\frac1{c\ka}\pde{}{t'}\lep
\frac\vbe{\ka R}\rip\rib_{ret}\nonumber\\
&=&e\leb\frac{\nn(1-\nn\cdot\vbe)}{\ka^2R^2}+\frac1{\ka^2R^2}[(\nn\cdot\vbe)
\nn-\vbe]\rib_{ret}\nonumber\\
&&+\leb\frac\nn{c\ka}\pde{}{t'}\lep\frac1{\ka R}\rip-\frac1{c\ka}\pde{}{
t'}\lep\frac\vbe{\ka R}\rip\rib_{ret}
\eeqa
or
\beq
\Ext=e\leb\frac{\nn-\vbe}{\ka^2R^2}+\frac\nn{c\ka}\pde{}{t'}\lep\frac1{\ka
R}\rip-\frac1{c\ka}\pde{}{t'}\lep\frac\vbe{\ka R}\rip\rib_{ret}.
\eeq
This expression may be related to that for the magnetic induction,
\beqa
\Bxt&=&e\leb\frac{\vbe\times\nn}{\ka R^2}\rib_{ret}+\frac ec\leb\frac1\ka
\pde{}{t'}\lep\frac{\vbe\times\nn}{\ka R}\rip\rib_{ret}\nonumber\\
&=&e\leb\frac{\vbe\times\nn}{\ka R^2}\rib_{ret}+\frac ec\leb\frac1\ka\pde{}{
t'}\lep\frac\vbe{\ka R}\rip\times\nn\rib_{ret}\nonumber\\
&&+e\leb\frac\vbe{\ka^2R}
\times\lep\frac1R\nn\times(\nn\times\vbe)\rip\rib_{ret}\nonumber\\
&=&e\leb\frac{\vbe\times\nn}{\ka R^2}\rib_{ret}+\frac ec\leb\frac1\ka\pde{}{
t'}\lep\frac\vbe{\ka R}\rip\times\nn\rib_{ret}\nonumber\\
&&+e\leb\frac1{\ka^2R^2}[\nn\lep(\nn\times\vbe)\cdot\vbe\rip
-(\nn\times\vbe)(\nn\cdot\vbe)]\rib_{ret}
\eeqa
Next, put all terms proportional to $\nn\times\vbe$ over the same
denominator, which is $\ka^2R^2$, and also make use of the fact that
$(\nn\times\vbe)\cdot\vbe=0$, to find
\beq
\Bxt=e\leb\frac{\vbe\times\nn}{\ka^2R^2}\rib_{ret}+\frac ec\leb\frac1\ka\pde{}{
t'}\lep\frac\vbe{\ka R}\rip\times\nn\rib_{ret}
\eeq
This is our final expression. Comparing it with \eq{22} for $\Ext$, we can
see that
\beq
\Bxt=[\nn]_{ret}\times\Ext.
\eeq
Thus we can find $\Bxt$ quite easily provided we can find $\Ext$.
Henceforth, we won't spend more time on $\B$ but will examine only $\E$
from which $\B$ then follows trivially.

To have an explicit expression for $\E$ with no time derivatives, we need
to evaluate
\beqa
\pde{}{t'}\lep\frac1{\ka R}\rip&=&-\frac1{\ka^2R^2}\leb\ka(-\nn\cdot\vbe c)+R
\lep-\nn\cdot\pde\vbe{t'}\rip-R\vbe\cdot[(\nn\cdot\vbe)\nn-\vbe]\frac cR
\rib\nonumber\\
&=&-\frac c{\ka^2R^2}\leb(1-\nn\cdot\vbe)(-\nn\cdot\vbe)-\frac Rc(\nn\cdot
\vbed)+\vbe\cdot[\vbe-(\nn\cdot\be)\nn]\rib\nonumber\\
&=&-\frac c{\ka^2R^2}\leb\be^2-\nn\cdot\vbe-\frac Rc(\nn\cdot\vbed)\rib
\eeqa
and so
\beqa
\Ext&=&e\leb\frac{\nn-\vbe}{\ka^2R^2}+\frac\nn{c\ka}\lep\frac{-c}{\ka^2R^2}
\rip\lep\be^2-\nn\cdot\vbe-\frac Rc\nn\cdot\vbed\rip-\frac\vbed{c\ka^2R}
\rib_{ret}\nonumber\\&&
-e\leb\frac\vbe{c\ka}\lep\frac{-c}{\ka^2R^2}\rip\lep\be^2-\nn\cdot\vbe-\frac Rc
\nn\cdot\vbed\rip\rib_{ret}\nonumber\\
&=&e\leb\frac1{\ka^3R^2}\nn\lep1-\be^2+\frac Rc\nn\cdot\vbed\rip+\frac\vbe{
\ka^3R^2}\lep-1+\be^2-\frac Rc\nn\cdot\vbed\rip-\frac1{c\ka^2R}\vbed
\rib_{ret}\nonumber\\
&=&e\leb\frac{(\nn-\vbe)(1-\be^2)}{\ka^3R^2}\rib_{ret}+e\leb\frac1{c\ka^3R}
\lep(\nn-\vbe)(\nn\cdot\vbed)-(1-\nn\cdot\vbe)\vbed\rip\rib_{ret}.
\eeqa
The second bracket in the final expression contains the quantity
\beq
(\nn-\vbe)(\nn\cdot\vbed)-\nn\cdot(\nn-\vbe)\vbed=\nn\times[(\nn-\vbe)
\times\vbed],
\eeq
so
\beq
\Ext=e\leb\frac{(\nn-\vbe)(1-\be^2)}{\ka^3R^2}\rib_{ret}+
\frac ec\leb\frac{\nn\times[(\nn-\vbe)\times\vbed]}{\ka^3R}\rib_{ret}.
\eeq

If there is no acceleration, then $\vbed=0$ and only the first term in
$\E$ and the corresponding term in $\B$ are finite.  These terms fall off
with distance as $1/R^2$, and hence cannot give rise to a net flux
of radiation to infinity.  If there is an acceleration, then the
second term of $\E$, and the corresponding term in $\B$ are finite.
These fall off as $1/R$, and hence will give rise to radiation,
meaning that the
charged particle will emit radiation only if it is accelerated.
Hence the two terms are interpreted as the non-radiation and radiation
parts respectively.

	From these results when $\vbed=0$, we should be able to recover
the results for the fields of a uniformly moving charge which we derived 
in chapter 11 (Jackson Eq.~11.152).  With the particle moving along the
x-direction with a constant velocity $v$, the fields felt by an observer
a distance $b$ away are
\beq\barr{c}
E_{\parallel}=-\ga qvt/[b^2+(\ga vt)^2]^{3/2}\\
E_{\perp}=\ga qb/[b^2+(\ga vt)^2]^{3/2}.\earr
\eeq
where the observer and particle are closest at time $t=0$.  It is
a reasonably simple task to show that this is the same result as \eq{29} 
above when $\vbed=0$.
Consider the diagram below in which the path of a charged particle
is along the abscissa.  At time $t$, the retarded and real location
of the particle are P' and P respectively, while O is the observation
point.  The time required for light to travel from P' to the observer
O is $t=R/c$, in which time the particle travels $\beta R=$PP'.  Thus
the distance P'Q is $\beta R\cth=\vbe\cdot\nn R$.  From this it follows that 
the distance OQ is $R-\vbe\cdot\nn R$.  Then $\leb(1-\vbe\cdot\nn)R\rib^2
=r^2-(PQ)^2=r^2-\beta^2R^2\sin^2(\th)$.  Thus, as $R\sth=b$
\beq
\leb(1-\vbe\cdot\nn)R\rib^2=b^2+v^2t^2-\beta^2b^2=
\frac{1}{\gamma^2}\lep b^2 + \gamma^2v^2t^2\rip\,,
\eeq
and transverse component of \eq{29} when $\vbed=0$ is
\beq
e\leb\frac{(R\sth)}{\ga^2\lep 1-\vbe\cdot\nn\rip^3 R^3}\rib_{ret}
= \frac{eb\ga}{[b^2+(\ga vt)^2]^{3/2}}
\eeq

\centerline{\psfig{figure=fig3.ps,height=1.75in,width=4.25in}}

\section{Radiation from an Accelerated Charge; the Larmor Formula}

Even given \eqs{25}{29}, the problem of calculating the fields
emitted from a charge moving along an arbitrary trajectory is
non-trivial.  This is largely due to the effect of retardation.
The problem is greatly simplified if the particle is not moving
too fast.

	In order to consider this limit, let's consider the trajectory
of a particle shown below.

\centerline{\psfig{figure=fig4.ps,height=2.0in,width=8.5in}}

\noindent We will assume that the origin is located in the center of the
region of interest of the particles trajectory, and that this
region of interest is of linear dimension $a$ (an example
would be an electron bound to a classical atom, where
$a$ would be the Bohr radius).

There are two ways in which the problem simplifies in the nonrelativistic 
limit, $\be<<1$.  First, we may approximate 
\beq
\ka\approx 1\;\;\;\;\; \nn-\vbe\approx \nn \;\;\;\;\; 1-\beta^2\approx 1\,.
\eeq
Second, and much more importantly, we can approximate functions of the
retarded time
\beq
f(t-|\x-\x(t')|/c)
\eeq
by making a Taylor series expansion around the origin
\beq
f(t-|\x-\x(t')|/c)\approx f(t-r/c)+\x(t')\cdot\grad f(t-r/c) + \cdots
\eeq
where $r=|\x|$.  The ratio of the second term to the first is roughly
$f' x(t')/fc$, where the prime means differentiation of $f$ with respect
to $t-r/c$.  If this is small, as it must be for $\beta\ll 1$, then
the second term in the series may be neglected, then
\beq
f(t-|\x-\x(t')|/c)\approx f(t-r/c)\,.
\eeq
This approximation is sometimes called the dipole approximation (we
will see why directly).  Here the effects of retardation are simple
but not insignificant.  They are simply due to the separation of
the charge from the observer, not due to the motion of the charge.
Thus, the only way they are significant is if the acceleration of
the charge changes abruptly and it takes a while for the emitted
fields to reach the observer.  \footnote{For example, suppose a particle
starts to oscillate back and forth at time $t=0$, there will be no signal
felt at the observer's location $\x$ for what may be a very long time
$t=r/c$.}

With the approximations described above, the electric field becomes
\beq
\E_{nr}(\x,t)=e\leb\frac\nn{R^2}\rib_{ret}+\frac ec\leb\frac{\nn\times(
\nn\times\vbed)}R\rib_{ret}.
\eeq
The Poynting vector is given by
\beqa
\S&=&\frac c\fpi(\E\times\B)=\frac c\fpi\E\times([\nn]_{ret}\times\E)\nonumber\\
&=&\frac c\fpi\lec[\nn]_{ret}\E\cdot\E-\E([\nn]_{ret}\cdot\E)\ric.
\eeqa
If we take the limit of large $R$ and keep just the radiation terms, then
we find that
\beq
\E(\x,t)=\frac{e}{c} \leb
\frac{\nn(\nn\cdot\vbed) -\vbed}{R}
\rib_{ret}
\;;\;
\B(\x,t)=[\nn]_{ret}\times\E=\leb -\frac{e}{cR} \nn\times\vbed\rib_{ret}
\eeq
Note that $\nn\cdot\E=\nn\cdot\B=0$ for these radiation fields, thus
(as usual) the radiation is transverse.  The Poynting vector due
to the radiation is then given by
\beq
\S=\frac {ce^2}{\fpi R^2c^2}\leb\nn\lel\nn\times(\nn\times\vbed)\ril^2
\rib_{ret}
\eeq
The angular distribution of radiated power is
\beq
\dpdo= R^2(\S\cdot\nn)
\eeq
which is easily evaluated by expanding the cross products. The result is
\beq
\dpdo=\frac{e^2}{\fpi c}(\bed^2\sin^2\th)=\frac{e^2}{\fpi c^3}(\vd^2
\sin^2\th)
\eeq
where $\th$ is the angle between $\nn$ and $\vbed$ (at the retarded time).

\centerline{\psfig{figure=fig2.ps,height=2.5in,width=8.5in}}

\noindent In these last expressions we have dropped the reminder that 
$\nn$ and $R$ must be evaluated at the retarded time. 

The radiation pattern is characteristic of dipole radiation with the reference
$z$ axis parallel to the direction of the particle's acceleration. The shape
and magnitude of the distribution are independent of the particle's velocity
and proportional to the square of the acceleration. The total power radiated is
the integral over all directions of the distribution,
\beq
{\cal P}=\int d\Om\dpdo=\frac23\frac{e^2\vd^2}{c^3}
\eeq
which is known as the {\em Larmor formula} for the power radiated by an
accelerated particle.

\subsection{Relativistic Larmor Formula}
We turn now to the relativistic generalization of the Larmor formula. One
can determine this generalization by consideration of how power transforms
under Lorentz transformations. There are actually two ways (that I know of)
to do this, of which Jackson does one.
One can also find it from direct computation
and that is what we shall do. The first step is to consider just how we
shall define the power radiated by a particle. The point is that the rate
at which energy crosses a closed surface surrounding the particle depends
on that surface because of retardation. We are going to calculate the
power as a function not of the time $t$ at which the fields are measured on the
surface but rather as a function of the retarded time $t'$. Consider a
surface $S$ which encloses the particle at all times during which it is
radiating. The power crossing unit area at $\x$ on $S$ at time $t$ is
$\S(\x,t)\cdot\nn$, where $\nn$ is a unit outward normal, and so the total
energy crossing this unit area is
\beq
W=\intmp dt\,\S(\x,t)\cdot\nn.
\eeq
Now let us transform to the retarded time $t'$:
\beq
W=\intmp dt'\,\der t{t'}\S(\x,t(t'))\cdot\nn=\intmp dt'\,\ka[\S(\x,t(t'))
\cdot\nn].
\eeq
The integrand of this expression we identify as $dW/dt'$, the rate at which
the particle radiates what eventually passes through the unit area on $S$
at $\x$. This is the {\em instantaneous} radiated power, and if we
multiply it by $R^2$ we get $d{\cal P}(t')/d\Om$:
\beq
\der{{\cal P}(t')}\Om=\ka R^2\S\cdot\nn.
\eeq
If we suppose that $R$ is large enough that only the radiation fields need
be retained, then we find, from our results for the fields and the
definition of the Poynting vector, that
\beq
\der{{\cal P}(t')}\Om=\left.\frac{e^2}{\fpi c}\frac{[\nn\times((\nn-\vbe)\times
\vbed)]^2}{(1-\nn\cdot\vbe)^5}\ril_{t'}
\eeq

Of course, if one wants to know the radiated power
per unit area at the time that it actually gets there, then he will have
to address the calculation of the retardation.  In other words, the
radiated intensity at time $t$ depends upon the behavior of the particle
at $t'$, and the differential time elements are different as well:
$dt=dt'(1-\nn\cdot\vbe)_{ret}$.  For example, if a particle of
velocity $\vbe$ is impulsively accelerated for a time $\tau$, and
then brought to rest, a pulse of radiation will appear at the observer
at time $t=r/c$, of duration $\tau(1-\nn\cdot\vbe)_{ret}$.  The total
energy lost of the particle, of course, must eventually equal the
energy radiated, but the energy lost by the charge per unit time 
will differ from the energy radiated per unit time by the factor
$(1-\nn\cdot\vbe)_{ret}$

{\em{However, by calculating the instantaneous power radiated, we have 
avoided having to calculate the retardation}}.  Thus we can proceed with 
a calculation of the
energy lost.  Our result, \eq{47}, depends in a complicated way on both 
$\vbe$ and
$\vbed$ making it possible for rather remarkable angular distributions to
occur. The total power radiated can also be computed by integrating the
distribution over directions:
\beqa
{\cal P}(t')&=&\frac{e^2}{\fpi c}\int \frac{d\Om}{\ka^5}[\nn\times((\nn-
\vbe)\times\vbed)]^2\nonumber\\
&=&\frac{e^2}{\fpi c}\int\frac{d\Om}{\ka^5}\lec\leb(\nn-\vbe)(\nn\cdot\vbed
)-\vbed(1-\vbe\cdot\nn)\rib^2\ric\nonumber\\
&=&\frac{e^2}{\fpi c}\int\frac{d\Om}{\ka^5}\lec(1-2\nn\cdot\vbe+\be^2)(\nn
\cdot\vbed)^2\right.\nonumber\\
&-&\left.2(\nn\cdot\vbed-\vbe\cdot\vbed)(\vbed\cdot\nn)(1-\nn\cdot\vbe
)+\bed^2(1-2\vbe\cdot\nn+(\vbe\cdot\nn)^2)\ric
\eeqa
Let the angle between $\vbe$ and $\nn$ be $\th$; further, let $\cos\th=u$;
also, let the angle between $\vbe$ and $\vbed$ be $\th_0$. Then
\beqa
{\cal P}(t')&=&\frac{e^2\bed^2}{2c}\int_{-1}^1\frac{du}{\ka^5}\lec(\be^2-1)
\leb u^2\cos^2\th_0+\frac12(1-u^2)\sin^2\th_0\rib\right.\nonumber\\
&&\left.+2\cos^2\th_0\,u\be(1-\be u)+(1-2\be u+\be^2u^2)\ric\nonumber\\
&=&\frac{\bed^2e^2}{2c}\int_{-1}^1\frac{du}{(1-\be u)^5}\lec-u^2(1-\be^2)+
2u\be(1-\be u)+1-2\be u+\be^2u^2\right.\nonumber\\
&&\left.+\sin^2\th_0\leb(1-\be^2)u^2-(1-u^2)(1-\be^2)/2-2\be u+2
\be^2u^2\rib\ric\\
&=&\frac{\bed^2e^2}{2c}\int_{-1}^1\frac{du}{(1-\be u)^5}\lec(1-u^2)+\sin^2
\th_0\leb\lep\frac32+\frac12\be^2\rip u^2-2\be u-\frac12(1-\be^2)\rib\ric.
\nonumber
\eeqa
Now introduce $x\equiv1-\be u$; then
\beqa
{\cal P}(t')&=&\frac{\bed^2e^2}{2c\be^3}\int_{1-\be}^{1+\be}\frac{dx}{x^5}
\lec(\be^2-1+2x-x^2)\right.\nonumber\\
&&\left.+\sin^2
\th_0\leb\lep3+\be^2\rip\lep1-2x+x^2\rip/2-2\be^2\lep1-x\rip-
\be^2\lep1-\be^2\rip/2\rib\ric\nonumber\\
&=&\frac{\bed^2e^2}{2c\be^3}\int_{1-\be}^{1+\be}\frac{dx}{x^5}\lec\lep\be^2-1+
2x-x^2\rip\right.\nonumber\\
&&+\left.\sin^2\th_0\leb\lep3-4\be^2+\be^4\rip/2+(-2+\be^2)x+\lep3+
\be^2\rip x^2/2\rib\ric\nonumber\\
&=&\frac{\bed^2e^2}{2c\be^3}\lec\lep\frac{\be^2-1}{4x^4}+\frac2{3x^3}-
\frac1{2x^2}\rip\right.\nonumber\\
&&\left.+\sin^2\th_0\lep\frac{(3/2-\be^2/2)(1-\be^2)}{4x^4}+\frac{
\be^2-3}{3x^3}+\frac{3+\be^2}{4x^2}\rip\ric\nonumber\\
&=&\frac{\bed^2e^2\ga^6}{c\be^3}\lec\lep-(\be+\be^3)+2(3\be+\be^3)/3-
\be(1-\be^2)\rip\right.\nonumber\\
&&+\left.\sin^2\th_0\leb(3-\be^2)(\be+3\be^3)/2+(-3+\be^2)(\be+\be^3/3)+
(3+\be^2)(\be-\be^3)/2\rib\ric\nonumber\\
&=&\frac{\bed^2e^2\ga^6}{c\be^3}\lec\frac23\be^3-\frac23\be^5\sin^2\th_0
\ric=\frac{2\bed^2e^2\ga^6}{3c}(1-\be^2\sin^2\th_0).
\eeqa
Putting the result back in terms of vectors and their products, we have
\beq
{\cal P}=\frac{2e^2\ga^6}{3c}[\bed^2-(\vbe\times\vbed)^2].
\eeq
This is the relativistic generalization of the Larmor formula.

It is instructive to look at some simple examples.

\subsubsection{ Example: Synchrotron}
An electron moves in a circle of radius $R$ at constant speed. 

\centerline{\psfig{figure=fig5.ps,height=1.5in,width=8.5in}}

\noindent The acceleration is then entirely centripetal and has magnitude
$\bed=c\be^2/R$. Then
\beq
\bed^2-(\vbe\times\vbed)^2=\frac{c^2\be^4}{R^2}(1-\be^2)=\frac{c^2
\be^4}{R^2\ga^2},
\eeq
and so
\beq
{\cal P}=\frac{2e^2c\ga^4\be^4}{3R^2}
\eeq
which can be rewritten in terms of the particle's kinetic energy as
\beq
{\cal P}=\frac{2e^2c}{3R^2}\lep\frac E{mc^2}\rip^4\be^4.
\eeq
The energy emitted per cycle of the motion is $\De E={\cal P}\ta$ where
$\ta$ is the period of the motion, $\ta=2\pi R/\be c$. Hence,
\beq
\De E=\frac{4\pi e^2\be^3}{3R}\lep\frac E{mc^2}\rip^4.
\eeq
An electron of energy $E=500\,Mev$ in a synchrotron of radius $R=10^2\,cm$
will radiate in each cycle and energy $\De E\approx10^{-8}\,erg\,\sim10^4\,
ev$ which is a non-trivial amount if one wants to increase the electron's
energy or even to maintain it. Basically, one must apply an accelerating
voltage of at least ten thousand volts during each cycle to break even, i.e.,
to maintain the electron's energy. This radiation is the reason why
circular very-high-energy electron accelerators don't exist; however,
the synchrotron is a great Xray source.

But if one wants to produce high-energy protons in a circular accelerator,
that is a lot easier, especially if one is willing to make $R$ rather large.
For example, a $10\,Tev$ proton in a machine of radius $R=3\times10^6\,cm$,
which is about nineteen miles, radiates away considerably less than $10^4\,ev$
in one cycle.

\subsubsection{Example: Linear Acceleration}
An electron is accelerated in the direction of its velocity,
$\vbe\parallel\vbed$. 

\centerline{\psfig{figure=fig6.ps,height=1.5in,width=8.5in}}

\noindent In this instance one trivially finds from \eq{51}
that
\beq
{\cal P}=\frac{2e^2\ga^6}{3c}\bed^2.
\eeq
This doesn't look very encouraging (for an accelerator design) because of
the factor of $\ga^6$, but against this one has the fact that the only
acceleration the particle feels is the one produced by the fields acting to
increase the particle's kinetic energy; in the case of the round accelerator, 
there
is a large acceleration even for a particle of constant energy. Thus, in
the case of a linear accelerator, there is much less acceleration to
produce radiation. To make this point more clearly, let's write the
acceleration in terms of the time rate of change of the particle's momentum,
\beq
\vd=\der\v t=\frac1{m\ga^3}\der\p t
\eeq
and so
\beq
{\cal P}=\frac{2e^2}{3m^2c^3}\lep\der\p t\rip^2.
\eeq
Now relate the rate of change of momentum to the rate of change of energy
of the particle,
\beq
\der pt=F=\der Ex,
\eeq
and so
\beq
{\cal P}=\frac{2e^2}{3m^2c^3}\lep\der Ex\rip^2
\eeq
or
\beq
\frac{{\cal P}}{dE/dt}=\frac{2e^2}{3m^2c^3}\frac{dE/dx}{dx/dt}=
\frac2{3\be}\frac{(e^2/mc^2)}{mc^2}\der Ex,
\eeq
which says that the power radiated away is quite negligible in
comparison with the rate at which energy is being pumped into the particle
unless one pumps the energy in at a rate (in space) $dE/dx$
comparable to the rest energy of the particle, $mc^2$, in a distance
$r_c=e^2/mc^2$. This distance is rather small, being about $3\times10^{-13}
\,cm$ for an electron, and it is difficult to put so much energy into the
particle in such a small distance ($500,000\,volts$ in a distance of $10^{-
13}\,cm$!). Hence particles in linear accelerators lose an insignificant amount
of energy to radiation. The difficulty with such accelerators is that, like
all things excepting round ones, they must sooner or later end.

\section{Angular distribution of radiation}
In this section we consider the angular distribution of the radiation emitted
by an accelerated charge. On the basis of what we said about the potentials,
we expect to find that it is strongly focussed in the forward direction,
or parallel to the velocity, in a frame where the particle has a speed close to
$c$. As at the end of the preceding section, we shall consider some particular
examples. The basic equation is
\beq
\dpdo=\frac{e^2}{\fpi c}\frac{\lec\nn\times\leb(\nn-\vbe)\times\vbed\rib
\ric^2}{(1-\nn\cdot\vbe)^5};
\eeq
the right-hand side of this expression is to be evaluated at the retarded
time $t'$.

\subsection{Example: Parallel acceleration and velocity}
Let a particle have parallel velocity and acceleration. 
If these define the $z$ direction, and $\nn$ points at an angle 
$\th$ to the $z$ axis,

\centerline{\psfig{figure=fig7.ps,height=1.5in,width=8.5in}}

\noindent then one finds
\beq
\dpdo=\frac{e^2\bed^2}{\fpi c}\frac{\sin^2\th}{(1-\be\cos\th)^5}=\frac{e^2
\vd^2}{\fpi c^3}\frac{\sin^2\th}{(1-\be\cos\th)^5},
\eeq
\noindent and the total power is 
\[
{\cal P}=\frac{2e^2\vd^2}{3c^3}\gamma^6
\]
\centerline{\psfig{figure=fig8.ps1,height=3.0in,width=6.0in}}

\noindent For $\be$ close to one, this expression provides 
dramatic confirmation of
the inadequacy of the multipole expansion for describing radiation from
accelerated charges. This radiation pattern is sharply peaked close to the
$z$ axis and is actually zero precisely on the axis. One would have to keep
many multipole terms to produce such a distribution. 

	We can find various
basic properties which characterize the distribution. It has a maximum at a
value of $\th$ or of $u\equiv\cos\th$ which we can find by differentiating:
\beq
0=\der{}u\lep\frac{1-u^2}{(1-\be u)^5}\rip=-\frac{2u}{(1-\be u)^5}+\frac{5\be
(1-u^2)}{(1-\be u)^6},
\eeq
or
\beq
-2u(1-\be u)+5\be(1-u^2)=0.
\eeq
This quadratic equation has the solution
\beq
u=\frac1{3\be}[\sqrt{1+15\be^2}-1]\orh\th_m=\arccos\leb\frac1{3\be}(\sqrt{1+
15\be^2}-1)\rib.
\eeq
Further, $\be^2=1-1/\ga^2$, so for $\be$ close to one, $\be\approx1-1/2\ga^2$
and
\beqa
\th_m&\approx&\arccos\lec\frac13\lep1+\frac1{2\ga^2}\rip\leb4\lep1-\frac{15}{32
\ga^2}\rip-1\rib\ric\nonumber\\
&\approx&\arccos\lep1-\frac1{8\ga^2}\rip\approx\frac1{2\ga}.
\eeqa
Also, the power distribution becomes, in the relativistic limit,

\centerline{\psfig{figure=power.ps,height=2.5in,width=4.5in}}

\noindent \beqa
\dpdo&=&\frac{e^2\dot v^2}{\fpi c^3}\frac{\th^2}{(1-\be+\be\th^2/2)^5}=
\frac{e^2\dot v^2}{\fpi c^3}\frac{32\th^2}{[2(1-\be)+\be\th^2]^5}\nonumber\\
&=&\frac{e^2\dot v^2}{\fpi c^3}\frac{32\ga^5\th^2}{(1+\ga^2\th^2)^5}=
\frac{8e^2\dot v^2}{\pi c^3}\ga^3\frac{\ga^2\th^2}{(1+\ga^2\th^2)^5}
\eeqa
This may be integrated to find the rms angle
\beq
<\theta^2>=\frac{1}{\ga}=\frac{mc^2}{E}
\eeq
%\pagebreak
\subsection{Example: Acceleration Perpendicular to Velocity}

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

A particle in instantaneously circular motion has its
acceleration perpendicular to its velocity. Let $\vbe=\be\ept$ and $\vbed=
\bed\epu$. Then, letting $\th$ and $\ph$ specify the direction from the
particle (at time $t'$) to the observation point $\x$, 
we have
\beq
\ept=\cos\th\,\nn-\sin\th\,\thu\andh\epu=\sin\th\,\cos\ph\,\nn+\cos\th\,\cos
\ph\,\thu-\sin\ph\,\phu.
\eeq
Using these conventions we can work out the relevant vector products:
\beq
(\nn-\vbe)\times\vbed=\bed[\cos\th\,\cos\ph\,\phu+\sin\ph\,\thu
-\vbe(\sin\th\sin\ph\nn+\cos\th\sin\ph\thu+\cos\ph\phu)]
\eeq
and so
\beqa
[\nn\times((\nn-\vbe)\times\vbed)]^2&=&\bed^2[\cos^2\ph(\be-\cos\th)^2+\sin^2
\ph(1-\be\cos\th)^2]\hsph\nonumber\\
&=&\bed^2[\be^2\cos^2\ph-2\be\cos\th\cos^2\ph+\cos^2\th\cos^2\ph+\sin^2\ph
\nonumber\\
&&-2\be\cos\th\sin^2\ph+\be^2\cos^2\th\sin^2\ph]\nonumber\\
&=&\bed^2[\cos^2\ph(\be^2+\cos^2\th-1-\be^2\cos^2\th)+(1-\be\cos\th)^2]\hsph.
\eeqa
Hence
\beqa
\dpdo=\frac{e^2\dot v^2}{\fpi c^3}\lec1-\frac{\cos^2\ph(1-\be^2)\sin^2\th}{(1-
\be\cos\th)^2}\ric\frac1{(1-\be\cos\th)^3}\nonumber\\
=\frac{e^2\dot v^2}{\fpi c^3(1-\be\cos\th)^3}\lec1-\frac{\cos^2\ph\sin^2\th}{
\ga^2(1-\be\cos\th)^2}\ric.
\eeqa
In this case we find that there is radiation in the forward direction, or
$\th=0$. In fact the main peak in the radiation is in this direction. If
one makes a small angle expansion of the power distribution in the case of
highly relativistic particles, he will find the result
\beq
\dpdo=\frac{2e^2\dot v^2}{\pi c^3}\frac{\ga^6}{(1+\ga^2\th^2)^3}\lec1-
\frac{4\ga^2\th^2\cos^2\ph}{(1+\ga^2\th^2)^2}\ric.
\eeq

\centerline{\psfig{figure=fig9.ps,height=4.0in,width=8.5in}}

\noindent Further, the total radiated power is
\beq
{\cal P}=\frac{2e^2\dot v^2}{\fpi c^3}\ga^4.
\eeq

\subsection{Comparison of Examples}
From the expressions for the total power produced in each of our two
examples, we see that for a given magnitude of acceleration, there is a
factor of $\ga^2$ more radiation produced when the acceleration is parallel
to the velocity than when it is perpendicular. This is a misleading
statement in some sense because what is actually applied to a particle is
not an acceleration but a force, and a given force will produce quite
different accelerations when applied perpendicular and parallel to the
velocity. For a force applied parallel to the velocity,
\beq
F=\der pt=m\ga^3\der vt
\eeq
and for a force applied perpendicular to $\v$,
\beq
F=\der pt=m\ga\der vt.
\eeq
Hence the powers produced in the two cases, expressed in terms of the
forces or $dp/dt$, are
\beq
{\cal P}\per=\frac{2e^2}{3m^2c^3}\ga^2\lep\der pt\rip^2\andh{\cal P}\pll
=\frac{2e^2}{3m^2c^3}\lep\der pt\rip^2.
\eeq
{\em Thus, for a given applied force, $\ga^2$ more radiation is produced if it
is perpendicular to $\v$ than if it is parallel}. Of course, if the force
comes from a magnetic field, then it has to be perpendicular to the
velocity.

\subsection{Radiation of an Ultrarelativistic Charged Particle}

	The radiation emitted at any instant from an accelerating charged particle may be decomposed into components coming from the parallel and 
perpendicular accelerations of the particle.  From the discussion above,
it is clear that the radiation of an ultrarelativistic $\ga\gg 1$  particle is
dominated by the perpendicular acceleration component.  Thus the
radiation is approximately the same as that emitted by a particle moving
instantaneously in a circle of radius
\beq
\rho=\frac{v^2}{{\dot{v}_{\perp}}}
\eeq
Since the angular width of the pulse is $\sim 1/\ga$, the particle will
travel a distance
\beq
d\sim \rho\Delta\th\sim\frac{\rho}{\ga}
\eeq
while illuminating the observer for a time 
\beq
\Delta t\sim \frac{d}{v}\sim \frac{\rho}{\ga v}.
\eeq
If we assume that pattern of radiation
is roughly that of a coiled beam of rectangular cross-section
\vspace{3.0in}
%insert Jackson 14.7 magnification (1:1) here
then the front edge of the beam will travel
\beq
D=c\Delta t\sim\frac{\rho c}{\ga v}=\frac{\rho}{\ga\beta}
\eeq
in time $\Delta t$; whereas, the trailing edge will be a distance
\beq
L=D-d\approx \lep\frac{\beta-\beta^2}{\beta^2}\rip\frac{\rho}{\ga}
\approx \lep 1-\frac{1}{2\ga^2}-1+\frac{1}{\ga^2}\rip \frac{\rho}{\ga}
=\frac{\rho}{2\ga^3}
\eeq
behind the front edge
since the charge moves the distance $d$ in the same time interval.
Thus the pulse width is roughly $L$ in space or $L/c$ in time.

	This derivation (straight out of Jackson) raises about
as many questions as it answers.  Perhaps the most fundamental is this: why is
$\Delta t$ different than the observed width of the pulse $L/c$.
The difference is that $\Delta t$ is really a difference
of retarded times; whereas, $L/c$ is in the time frame of the observer.
Lets repeat this calculation more carefully, distinguishing between
the observer's time and the particles (retarded) time.

	The pulse width for a distant observer would be 
\beq
\frac{L}{c}=t_2-t_1=\lep t_2'+ R_2/c\rip - \lep t_1'+ R_1/c\rip =
\Delta t- \lep R_1-R_2\rip/c\,.
\eeq

\centerline{\psfig{figure=fig10.ps,height=2.0in,width=8.5in}}

\noindent Since $R_1-R_2\approx v'\Delta t$, we have
\beq
\frac{L}{c}=t_2-t_1= \Delta t(1-v/c)\approx \frac{\Delta t}{2\ga^2}=
\frac{\rho}{2\ga^3},.
\eeq
where the right hand side is to be evaluated in the retarded time.

	An observer would receive periodic pulses of width $L/c$ in time.

\centerline{\psfig{figure=pulses.ps,height=2.2in,width=8.5in}}

\noindent From the principles of Fourier decomposition, we can also
estimate the type of radiation the observer would receive.  From
the uncertainty principle $\Delta t\Delta \om \sim 1$, the pulse
would contain components up to a cutoff\footnote{Higher frequency
components would yield a narrower pulse} of roughly
\beq
\om_c\sim\frac{c}{L}\sim \frac{c}{\rho}\ga^3
\eeq
However, the frequency $\om_0$ of the circular motion is 
$v/\rho\approx c/\rho$.
Thus, a broad spectrum of radiation is emitted up to $\ga^3$ times
the fundamental frequency of the rotation.  This kind of radiation, 
called {\em synchrotron radiation} provides a good, more or less
continuous, source of radiation in the range from visible and ultraviolet 
to soft X-rays.

	Thus in a 200 MeV electron synchrotron where $\gamma\approx 400$ and
the fundamental frequency is $\om_0\approx 3\times 10^{8}$, 
the frequency of the emitted radiation extends to $2\times 10^{16}$.
In a 10 GeV synchrotron, x-rays can be produced.  Thus synchrotrons
can be used as high intensity radiation sources. Synchrotrons are now
even being used in industry as radiation sources for x-ray 
lithography\footnote{{\bf{Physics Today}}, October 1991.}. 

\section{Frequency Distribution of the Radiated Energy}

The radiation produced by rotating 
charges is predominantly at the fundamental frequency
$\om_0$ of the motion with much smaller amounts at integral multiples, or
harmonics, of this frequency. The expansion parameter in the problem is
$k_0a=\om_0a/c\sim v/c=\be$, and the energy emitted at higher frequencies
than the fundamental is proportional to some power of this parameter. For $
\be<<1$, this energy will be relatively small. But if $\be\sim1$, there will be
a significant fraction of the total radiated energy appearing at higher
frequencies.

There is a simple and instructive way to see roughly how may
harmonics will contribute to the radiation. For a relativistic particle, $
d{\cal P}/d\Om$ is a peaked distribution which has a width in angle
$\de\th\sim1/\ga$. If the particle is, e.g., travelling in a circle with a
frequency of
motion $\om_0$, then the particle sweeps through an angle $\de\th$ in a
time $\de t'\sim\de\th/\om_0=1/\ga\om_0$. This is of the order of the
duration of the pulse observed at some fixed point in space, but measured
in units of the time at the source. The time which passes at the location
of the observer is
\beq
\de t=\de t'\der t{t'}=\ka\de t'.
\eeq
Further, during the pulse, $\nn$ is parallel to $\vbe$, $\nn\cdot\vbe=\be$
with corrections of order $1/\ga^2$ and so $\ka\sim1-\be[1+{\cal O}(1/\ga^2)]
\sim1/\ga^2$. Hence $\de t\sim1/\ga^3\om_0$. Now, a pulse which lasts a
time $\de t$ at a point must contain in it frequencies of order\footnote{In
quantum theory this statement would be called the uncertainty principle.}
 $1/\de t$. Thus the pulse which our observer sees must have in particular
frequencies of order $\om\sim\ga^3\om_0$. If the particle is highly
relativistic, $\ga>>1$, then the typical frequencies in the pulse will be much
larger than the frequency of the particle's motion, or $\om_0$; they will be
$\ga^3$ times as large, meaning that many harmonics must contribute to the
pulse.

\subsection{Continuous Frequency Distribution}
To make our analysis of the radiation more quantitative we are going to
consider the Fourier transforms in time of the fields at an observation
point $\x$. Start from the expression for the angular distribution of
radiated power:
\beq
\dpdo=\frac c\fpi\lep\Ext\times\Bxt\rip\cdot[\nn R^2]_{ret}.
\eeq
Using just the radiation fields, noting that they are transverse to
$\nn$ far away from the source, and making use the fact that
$\Bxt=[\nn]_{ret}\times\Ext$, we can write
\beq
\dpdo=\frac c\fpi\lec[R]_{ret}\Ext\ric^2.
\eeq
For simplicity of notation, define
\beq
\at\equiv\sqrt{\frac c\fpi}[R]_{ret}\Ext.
\eeq
Then, in terms of $\at$ the angular distribution of radiated power is
\beq
\dpdo=[\at]^2.
\eeq
Further, from \eq{29},
\beq
\at=\sqrt{\frac{e^2}{\fpi c}}\leb\frac{\nn\times[(\nn-\vbe)\times\vbed]}{\ka^3}
\rib_{ret}.
\eeq

Introduce also the Fourier transform\footnote{The field $\at$ depends on
$\x$ as well as on $t$; we suppress the former dependence as it is not of
interest at present.} $\ao$ of $\at$:
\beq
\at\equiv\frac1\stp\int d\om\emot\ao;\hsph\ao=\frac1\stp\int dt\eot\at.
\eeq
Using the Fourier integral (and its complex conjugate) for $\at$ in \eq{91}, we
can write the power distribution as
\beq
\der{{\cal P}(t)}\Om=\frac1\tpi\int d\om d\om'\,\ao\emot\cdot\a^*(\om')e^{i
\om't}.
\eeq
Integrating over all $t$, and thereby generating an expression for the
angular distribution of the total radiated energy, we find
\beqa
\der W\Om&\equiv&\intmp dt\,\der{{\cal P}(t)}\Om=\int d\om d\om'\,\de(\om-\om')
\ao\cdot\a^*(\om')\nonumber\\
&=&\intmp d\om\,|\ao|^2=\intzp d\om\,\lec|\ao|^2+|\a(-\om)
|^2\ric.
\eeqa
The reality of $\at$ demands that $\a^*(-\om)=\ao$, and so the two terms in
the integrand are identical:
\beq
\der W\Om=2\intzp d\om\,|\ao|^2\equiv\intzp d\om\,\der{I(\om)}\Om.
\eeq
The integrand, $dI/d\Om$, is interpreted as the total radiation received
per unit frequency per unit solid angle during the entire pulse of radiation.
It is simply the square of $\ao$; further $\ao$ is
\beqa
\ao&=&\sqrt{\frac{e^2}{\fpi c}}\frac1\stp\int dt\,\eot\leb\frac{\nn\times[(
\nn-\vbe)\times\vbed]}{\ka^3}\rib_{ret}\nonumber\\
&=&\sqrt{\frac{e^2}{8\pi^2c}}\int dt'\,\leb\frac{\nn\times[(\nn-\vbe)\times
\vbed]}{\ka^2}\rib e^{i\om(t'+R(t')/c)}
\eeqa
It's {\em d\'eja vu} all over again. Recalling the typical equations
generated in Chapter~9, we can see that this one probably can be obtained
without great difficulty from what we did there. Proceeding in familiar
fashion, then, let us look at this integral in the far zone which means
approximate $R$ in the exponent by
\beq
R=|\x-\x(t')|\approx r-\nn\cdot\x(t')
\eeq
and so
\beq
\ao=\sqrt{\frac{e^2}{8\pi^2c}}\int dt'\,e^{i\om t'}e^{i\om r/c}
e^{-i\om\nn\cdot\x(t')/c}\leb\frac{\nn\times[(\nn-\vbe)\times\vbed]}{\ka^2}
\rib
\eeq
or
\beq
\ao=e^{i\om r/c}\sqrt{\frac{e^2}{8\pi^2c}}\int dt'\,e^{i\om(t'-\nn\cdot\x(t')
/c)}\leb\frac{\nn\times[(\nn-\vbe)\times\vbed]}{\ka^2}\rib.
\eeq

\centerline{\psfig{figure=fig11.ps,height=2.0in,width=8.5in}}

If we confine the integration to times when the particle is accelerating,
i.e. confine it to the shaded region above, then we can
put this integral into a form such that the integrand involves
$\vbe$ but not $\vbed$; that can be done by, in essence, a parts integration.
First, consider
\beq
\der{}{t'}\lep\frac{\nn\times(\nn\times\vbe)}\ka\rip=-\frac{\nn\times(\nn
\times\vbe)}{\ka^2}(-\nn\cdot\vbed)+\frac{\nn\times(\nn\times\vbed)}{\ka^2}
(1-\nn\cdot\vbe)
\eeq
where we have not kept derivatives of $\nn$ because they give corrections
of relative order $|\x(t')/R|$.
Now group terms as follows:
\beqa
\der{}{t'}\lep\frac{\nn\times(\nn\times\vbe)}\ka\rip&=&\frac{\nn\times(\nn
\times\vbed)}{\ka^2}+\frac{\nn\times\lec\nn\times[(\nn\cdot\vbed)\vbe-(\nn\cdot
\vbe)\vbed]\ric}{\ka^2}\nonumber\\
&=&\frac{\nn\times(\nn\times\vbed)}{\ka^2}+\frac{\nn\times\lec\nn\times[\nn
\times(\vbe\times\vbed)]\ric}{\ka^2}\nonumber\\
&=&\frac{\nn\times[(\nn-\vbe)\times\vbed]}{\ka^2}
\eeqa
where we have used the fact that
\beq
\nn\times\lec\nn\times\leb\nn\times(\vbe\times\vbed)\rib\ric=-\nn\times(
\vbe\times\vbed).
\eeq
Using this identity to do a parts integration of the expression for
$\ao$ we find (again, drop terms proportional to powers of $1/R$)
\beqa
\ao&=&-e^{i\om r/c}\sqrt{\frac{e^2}{8\pi^2c}}\int dt'\,i\om(1-\nn\cdot\vbe)
\lep\frac{\nn\times(\nn\times\vbe)}\ka\rip e^{i\om(t'-\nn\cdot\x(t')/c)}
\nonumber\\
&=&-\sqrt{\frac{e^2}{8\pi^2c}}i\om e^{i\om r/c}\int dt'[\nn\times(\nn
\times\vbe)]e^{i\om(t'-\nn\cdot\x(t')/c)}.
\eeqa
Combining this result and \eq{96}, we find for the radiated energy per unit
frequency per unit solid angle
\beq
\der{I(\om)}\Om=\frac{e^2\om^2}{4\pi^2c}\lel\int dt'\,[\nn\times(\nn\times
\vbe)]e^{i\om(t'-\nn\cdot\x(t')/c)}\ril^2.
\eeq

In this derivation we were not careful when doing the parts integration,
meaning that we did not worry about whether the terms involving the
integrand evaluated at the endpoints of the interval of integration
contribute to the result. Consequently, in any application of \eq{105}, one
should check to see whether this assumption is justified; there are occasions
when it is not and then, naturally, the contributions from the endpoint(s) must
be included.
Our formulation of $dI/d\Om$ is most appropriate for a source
with an open orbit in which case the natural limits on the integration are
$\pm\infty$, and there is usually no difficulty in ignoring the
contributions from the endpoints where the particle is far away and
unaccelerated; however, that is not enough to guarantee that the endpoints
contribute nothing.

\subsection{Discrete Frequency Distribution}
For truly cyclic motion, it is more convenient and perhaps more sensible
from a physical point of view to set up a Fourier series to express the
frequency distribution of the radiation. 

\centerline{\psfig{figure=fig13.ps,height=2.0in,width=8.5in}}

\noindent The point is that for non-cyclic
motion, the distribution of radiation in frequency will be continuous and
the integral formula we derived above is appropriate for expressing this
distribution. But for cyclic motion, the radiation will be distributed in
frequency space only at harmonics or multiples of the fundamental frequency
of the motion and so a sum or series expansion of $dI(\om)/d\Om$ is
more appropriate for expressing the distribution. We shall now set up this sum.
To get started, suppose that the period of the motion is $\ta'=2\pi/\om_0$.
Then, letting the period measured by an observer at a point $\x$ be $\ta$,
one can show that $\ta=\ta'$. That is, a time $t$ for the observer and the
corresponding retarded time $t'$ are related by
\beq
t=t'+R(t')/c.
\eeq
One period later in the life of the source, its time has increased to $t'+
\ta'$ and the signal emitted by the source at this time will reach the
observer at a time $t+\ta$ which is
\beq
t+\ta=t'+\ta'+R(t'+\ta')/c.
\eeq
But $R(t'+\ta')=R(t')$ for the cyclic motion so
\beq
t+\ta=t'+\ta'+R(t')/c.
\eeq
Comparing \eqs{106}{108}, we see that $\ta=\ta'$. 

The radiation fields produced by this cyclic motion of a charged particle
will also be periodic.  Hence where we formerly had
a Fourier integral for $\at$, we now have a Fourier series, a sum over
frequencies which are integral multiples of the sources's frequency,
\beq
\at=\sqrt{\frac c\fpi}[R\E(t)]\equiv\sum_{n=-\infty}^\infty\a_n e^{-in\om_0t}
\eeq
with
\beq
\a_n=\frac{\om_0}\tpi\int_0^{\tpi/\om_0}dt\,\at e^{in\om_0 t}.
\eeq

The energy received during one cycle, per unit solid angle, at some point
$\x$ is the integral of $d{\cal P}/d\Om$ over one period. We shall write it
as $dW/d\Om$,
\beqa
\der W\Om\equiv\int_0^{\tpi/\om_0}dt\der{{\cal P}(t)}\Om&=&\int_0^{\tpi/\om_0}
dt\,|\at|^2\nonumber\\
&=&\sum_{n,m}\int_0^{\tpi/\om_0}dt\,(\a_n\cdot\a_m^*)e^{-i(n-m)\om_0t}
=\frac\tpi{\om_0}\sum_{n=-\infty}^\infty|\a_n|^2\nonumber\\
&=&\frac\fpi{\om_0}\sum_{n=1}^\infty|\a_n|^2.
\eeqa
Notice that the $n=0$ term has been discarded in the final expression; that
is okay because there is no radiation at zero frequency. The radiated energy at
the frequency $n\om_0$ is determined by $\a_n$, which is
\beqa
\a_n&=&\frac{\om_0}\tpi\int_0^{\tpi/\om_0}dt\leb\frac{\nn\times[(\nn-\vbe)
\times\vbed]}{\ka^3}\rib_{ret}\sqrt{\frac{e^2}{\fpi c}}e^{in\om_0 t}\nonumber\\
&=&\frac{\om_0}\tpi\sqrt{\frac{e^2}{\fpi c}}\int_0^{\tpi/\om_0}dt'\,
\frac{\nn\times[(\nn-\vbe)\times\vbed]}{\ka^2}e^{in\om_0(t'+R(t')/c)}.
\eeqa
We can now do the same integration by parts that led to \eq{104} and find
\beq
\a_n=-\frac{\om_0}\tpi\sqrt{\frac{e^2}{\fpi c}}in\om_0\int_0^{\tpi/\om_0}
dt'\,[\nn\times(\nn\times\vbe)]e^{in\om_0(t'+R(t')/c)},
\eeq
with no contribution ever from the end points of the interval as they
correspond to the same point on the periodic orbit of the particle.
From \eq{113}, $|\a_n|^2$ follows easily and hence $dW/d\Om$ from \eq{111}.
Notice also that the time-averaged power distribution is just
\beq
\lep\dpdo\rip_{ave}=\der W\Om\les\lep\frac\tpi{\om_0}\rip\right.
\eeq

\subsection{Examples}

\subsubsection{A Particle in Instantaneous Circular Motion}

Suppose we have a particle in instantaneous circular motion,
meaning that its acceleration is, at least temporarily, perpendicular to
its velocity. Any particle subjected to a magnetic field but no electric
field will satisfy this condition. We might do the calculation by supposing
that the motion is truly periodic and circular and using the relatively
easily applied Fourier series approach just developed. It is more difficult
and therefore more challenging to use the Fourier integral approach. Let's
try the latter.

First, we want to characterize the orbit of the particle. A circular orbit
at constant speed can be described as
\beq
\vbe=\be[(\cos\om_0t')\epu+(\sin\om_0t')\epd]
\eeq
and, for sufficiently small times $t'$, meaning $\om_0t'<<1$, we have
\beq
\vbe\approx\be(\epu+\om_0t'\epd)
\eeq
with corrections of order $(\om_0 t')^2$. Let the observer be located in
the $x$-$z$ plane; then $\nn=\cos\th\,\epu+\sin\th\,\ept$ where $\th<<1$ if
the observer is to experience the strong pulse of radiation that the
particle emits in the forward direction. 
We know that the times of
importance at the source for this pulse at the position of the observer are
of order $t'\sim1/\om_0\ga$. Consequently, in our approximation for $\vbe$, we
lose corrections of relative order $(\om_0t')^2\sim1/\ga^2$ in each of the
components.

Let's work out $\nn\times(\nn\times\vbe)$. Define a unit vector $\epv\per
\equiv\nn\times\epd=\cos\th\,\ept-\sin\th\,\epu$; it will prove to be
useful.

\centerline{\psfig{figure=fig13a.ps,height=1.75in,width=6.0in}}

\noindent
\beq
\nn\times\vbe=\be[(\nn\times\epu)+\om_0t'(\nn\times\epd)]=\be(\sin\th\,\epd
+\om_0t'\epv\per);
\eeq
and
\beq
\nn\times(\nn\times\vbe)=\be(\sin\th\,\epv\per-\om_0t'\epd).
\eeq
Further, the interesting range of $\th$ is of order $1/\ga<<1$, so let us
replace $\sin\th$ by $\th$. Further, let $\epd$ be designated $\epv\pll$.
Then, using this equation in \eq{105} and setting $n\om_0=\om$, we find
\beq
\der{I(\om)}\Om=\frac{e^2\om^2}{4\pi^2c}\lel\int dt'\,(\be\th\epv\per
-\om_0t'\epv\pll)e^{i\om(t'-\nn\cdot\x(t')/c)}\ril^2.
\eeq
We need to evaluate the exponent in order to complete the integral.
Consider
\beqa
\nn\cdot\x(t')&=&\cos\th\,(\epu\cdot\x(t'))=\frac{c\be}{\om_0}\cos\th\,
\sin\om_0t'\nonumber\\
&\approx&\frac{c\be}{\om_0}\lep1-\frac{\th^2}2\rip\lep\om_0t'-
\frac16(\om_0t')^3\rip;
\eeqa
Notice that we have kept the leading term and corrections to it of order $1
/\ga^2$. Basically, we have kept all phases of order unity when $\om_0t'\sim1/
\ga$, $\th\sim1/\ga$, and $\om\sim\om_0\ga^3$, these being what we believe to
be the important ranges of $t'$, $\om$, and $\th$. Hence the total phase is,
to the order indicated,
\beqa
\om[t'-\nn\cdot\x(t')/c]&=&\om t'-\frac{\om\be}{\om_0}\om_0t'\lep1-\frac{\th^2}
2\rip+\frac{\om\be}{\om_0}\frac16(\om_0t')^3\lep1-\frac{\th^2}2\rip\nonumber\\
&=&\om t'\lep1-\be+\frac{\be\th^2}2\rip+\frac{\om\be}{6\om_0}(\om_0t')^3\lep1-
\frac{\th^2}2\rip\nonumber\\
&\approx&\om t'\lep\frac1{2\ga^2}+\frac{\th^2}2\rip+
\frac\om{6\om_0}(\om_0t')^3\nonumber\\
&=&\lep\frac\om{2\om_0}\rip\leb\om_0t'\lep\frac1{\ga^2}+\th^2\rip+\frac13(
\om_0t')^3\rib.
\eeqa

With the foregoing expression for the phase, we are able to write the
frequency and angle distribution of the intensity as
\beq
\der{I(\om)}\Om=\frac{e^2\om^2}{4\pi^2c}\lel\intmp dt'\,(\th\epv\per-\om_0t'
\epv\pll)e^{i\frac\om{2\om_0}\leb\om_0t'\lep\frac1{\ga^2}+\th^2\rip+\frac13
(\om_0t')^3\rib}\ril^2.
\eeq
The integration has been extended to $\pm\infty$ even though
the integrand is accurate only for $|\om_0t'|\sim1/\ga$ (or less).
The extended range of integration is sensible only if there is no important
contribution coming from other regimes of $t'$. That is the case because
the term in the phase proportional to $t'^3$ produces rapid oscillations of
the integrand at larger $|t'|$ which yield a very small net contribution to
the integral.

Introduce $x$ such that $\om_0t'\equiv x(1/\ga^2+\th^2)^{1/2}$. The important
range of $t'$ corresponds to $|x|\sim 1$ and the frequency distribution of the
intensity is given by
\beq
\der{I(\om)}\Om=\frac{e^2\om^2}{4\pi^2c}\lel\frac1{\om_0}\sqrt{\frac1{\ga^2}+
\th^2}\intmp dx\,\leb\th\epv\per-\sqrt{\frac1{\ga^2}+\th^2}\,x\epv\pll\rib
e^{i\frac\om{2\om_0}\lep\frac1{\ga^2}+\th^2\rip^{3/2}\lep x+\frac13x^3\rip}\ril
^2.
\eeq
Let
\beq
\et=\frac\om{3\om_0}\lep\frac1{\ga^2}+\th^2\rip^{3/2}.
\eeq
Then
\beqa
\der{I(\om)}\Om&=&\frac{e^2\om^2}{4\pi^2c\,\om_0^2}\lel\intmp dx\,\leb\sqrt{
\frac1{\ga^2}+\th^2}\,\th\epv\per+\lep\frac1{\ga^2}+\th^2\rip x\epv
\pll\rib e^{i\frac32\et(x+x^3/3)}\ril^2\nonumber\\
&\equiv&\frac{e^2\om^2}{4\pi^2c\,\om_0^2}\lel\sqrt{\frac1{\ga^2}+\th^2}\,
\th I\per\epv\per+\lep\frac1{\ga^2}+\th^2\rip I\pll\epv\pll\ril^2.
\eeqa
where
\beq
I\per=\intmp dt\,e^{i(3\et t+\et t^3)/2}
\eeq
and
\beq
I\pll=\intmp dt\,te^{i(3\et t+\et t^3)/2}.
\eeq
These integrals may be expressed in terms of {\em Airy functions} which are
modified Bessel functions of order 1/3 and 2/3. An integral
representation\footnote{Abramowitz and Stegun, 10.4.32.} of the function
$Ai$ is
\beq
\frac\pi{(3a)^{1/3}}Ai[x/(3a)^{1/3}]=\intzp dt\,\cos(xt+at^3)=
\frac12\intmp dt\,e^{i(xt+at^3)}.
\eeq
From this representation one can see that
\beq
I\per=\frac\tpi{(3\et/2)^{1/3}}Ai[(3\et/2)^{2/3}].
\eeq
As for $I\pll$, it is
\beqa
I\pll&=&\left.\intmp dt\,te^{i(3\et t+\et t^3)/2}=\intmp dt\,te^{i(xt+\et
t^3/2)}\ril_{x=3\et/2}\nonumber\\
&=&\frac1i\der{}x\left.\lep\intmp dt e^{i(xt+\et t^3/2)}\rip\ril_{x=3\et/2}
=\frac1i\der{}x\left.\lep\frac\tpi{(3\et/2)^{1/3}}Ai[x/(3\et/2)^{1/3}]\rip\ril_{
x=3\et/2}\nonumber\\
&=&\frac\tpi{i(3\et/2)^{2/3}}Ai'[(3\et/2)^{2/3}].
\eeqa
The prime on the Airy function denotes differentiation with respect to the
argument.

The connection between Airy functions and modified Bessel functions
is\footnote{Abramowitz and Stegun, 10.4.14.}
\beq
Ai[(3\et/2)^{2/3}]=\frac1\pi\leb\frac{(3\et/2)^{2/3}}3
\rib^{1/2}K_{1/3}(\et).
\eeq
Also\footnote{Abramowitz and Stegun, 10.4.16.},
\beq
-Ai'[(3\et/2)^{2/3}]=\frac1\pi\frac{(3\et/2)^{2/3}}{\sqrt3}
K_{2/3}(\et).
\eeq
Thus we may express the result for the frequency distribution of intensity
in terms of modified Bessel functions as
\beqa
\der{I(\om)}\Om&=&\frac{e^2}{4\pi^2c}\lep\frac\om{\om_0}\rip^2\lel\sqrt{\frac1{
\ga^2}+\th^2}\th\frac2{\sqrt3}K_{1/3}(\et)\epv\per+\lep\frac1{\ga^2}
+\th^2\rip\frac2{i\sqrt3}K_{2/3}(\et)\epv\pll\ril^2\nonumber\\
&=&\frac{e^2}{3\pi^2c}\lep\frac\om{\om_0}\rip^2\lep\frac1{\ga^2}+\th^2
\rip^2\leb K_{2/3}^2(\et)+\frac{\th^2}{1/\ga^2+\th^2}K_{1/3}^2(\et)\rib
\nonumber\\
&=&\frac{3e^2\ga^2}{\pi^2c}\lep\frac\om{\om_c}\rip^2(1+\ga^2\th^2)^2
\leb K_{2/3}^2(\et)+\frac{\ga^2\th^2}{1+\ga^2\th^2}K_{1/3}^2(\et)\rib
\eeqa
with
\beq
\om_c=3\ga^3\om_0\;\;\;{\rm{and}}\;\;\;
\et=\frac\om{3\om_0}\lep\frac1{\ga^2}+\th^2\rip^{3/2}=
\frac\om{\om_c}(1+\th^2\ga^2)^{3/2}.
\eeq

This is a fairly transparent, if not simple, result. The variable $\et$ is
proportional to $\om$ and is scaled by $3\ga^3\om_0\equiv\om_c$ which we
believe, on the basis of arguments given earlier, to be the appropriate
or characteristic scale of frequency in this radiating system. For $\et<<1$,
\beq
K_\nu(\et)\sim\frac{\Ga(\nu)}2\lep\frac2\et\rip^\nu
\eeq
and for $\et>>1$,
\beq
K_\nu(\et)\sim\sqrt{\frac\pi{2\et}}e^{-\et}.
\eeq
Thus, for $\et<<1$ and at $\th=0$,
\beq
\der{I(\om)}\Om=\frac{3e^2}{\pi^2c}\ga^2\lep\frac\om{\om_c}\rip^2\lep
\frac{\Ga(2/3)}2\rip^2\lep\frac2{\om/\om_c}\rip^{4/3}=\frac{e^2}c\lep\frac{\Ga(2
/3)}\pi\rip^2\lep\frac34\rip^{1/3}\lep\frac\om{\om_0}\rip^{2/3}.
\eeq
If $\th\ne0$, we pick up a contribution proportional to $K_{1/3}^2$,
leading to an additional term which is proportional to $\om^{4/3}$. Note also
that the term we do have is produced by waves with the electric field polarized
in the plane of the particle's orbit.

In the large frequency regime, $\om>>\om_c$ and $\th=0$, we find
\beq
\der{I(\om)}\Om=\frac{e^2}{\pi^2c}3\ga^2\lep\frac\om{\om_c}\rip^2\frac\pi{2
\om/\om_c}e^{-2\om/\om_c}=\frac{e^2}{2\pi c\ga}\frac\om{\om_c}e^{-2\om/3
\ga^3\om_0}.
\eeq
The accompanying figure shows the angular distribution of radiation at
several values of $\om/\om_c$. 
%\vspace{3.0in}
\centerline{\psfig{figure=fig13b.ps,height=2.5in,width=3.5in}}
This figure also very clearly makes the point
that at a given frequency, except for the very low ones, the intensity is
greatest at the smallest angles.

\subsubsection{A Particle in Circular Motion}

Let's do the same sort of calculation for truly
circular motion. Then, from \eqs{111}{113}, we find that the energy emitted per
cycle per unit solid angle is
\beq
\der W\Om=\sum_{n=0}^\infty\der{W_n}\Om=\frac\fpi{\om_0}\sum_{n=1}^\infty
|\a_n|^2
\eeq
with
\beq
|\a_n|=\sqrt{\frac{e^2}{16\pi^3c}}n\om_0^2\lel\int_0^{2\pi/\om_0}dt'\,
(\nn\times\vbe)e^{in\om_0(t'-\nn\cdot\x(t')/c)}\ril.
\eeq
An individual term in the sum, $dW_n/d\Om$, is the energy per cycle per unit
solid angle radiated at frequency $\om_n=n\om_0$. It can be written as
\beq
\der{W_n}\Om=\frac\fpi{\om_0}\frac{e^2n^2\om_0^4}{16\pi^3c}\lel\int_0^{2\pi
/\om_0}dt'\,(\nn\times\vbe)e^{in\om_0(t'-\nn\cdot\x(t')/c)}\ril^2.
\eeq
If we divide by the period, $2\pi/\om_0$, we find the time-averaged power
at frequency $n\om_0$ per unit solid angle,
\beq
\der{{\cal P}_n}\Om=\frac{e^2n^2\om_0^4}{8\pi^3c}\lel\int_0^{2\pi/\om_0}dt'
\,(\nn\times\vbe)e^{in\om_0(t'-\nn\cdot\x(t')/c)}\ril^2.
\eeq
Let the motion be in the $x$-$y$ plane as before,
\beq
\x(t')=a[\epu\cos(\om_0t')+\epd\sin(\om_0t')]
\eeq

\centerline{\psfig{figure=fig14.ps,height=2.5in,width=8.5in}}

\noindent so that
\beq
\vbe(t')=\frac{\om_0a}c[-\epu\sin(\om_0t')+\epd\cos(\om_0t')].
\eeq
Also, write $\nn$ in terms of the usual\footnote{But be aware that the
angle $\th$ in this example is the polar angle and not the latitude as in
the previous example; the angle $\al$ introduced below is the latitude, i.e.,
the same as the $\th$ of the previous example.} spherical coordinates,
\beq
\nn=\ept\cos\th+\epu\sin\th\,\cos\ph+\epd\sin\th\,\sin\ph.
\eeq
Then
\beq
\nn\cdot\x(t')=a[\sin\th\,\cos\ph\,\cos(\om_0t')+\sin\th\,\sin\ph\,\sin(\om_0t'
)]=a\sin\th\,\cos(\om_0t'-\ph),
\eeq
and
\beqa
\nn\times\vbe(t')&=&\frac{a\om_0}c[-\epd\cos\th\sin(\om_0t')-\epu\cos\th\cos(\om_0
t')\nonumber\\
&&+\ept\sin\th(\cos\ph\cos\om_0t'+\sin\ph\sin\om_0t')]\nonumber\\
&=&\frac{a\om_0}c[-\epd\cos\th\sin(\om_0t')-\epu\cos\th\cos(\om_0t')+\ept\sin
\th\cos(\om_0t'-\ph)].
\eeqa
We must do the integral
\beq
{\bf I}=\int_0^{2\pi/\om_0}dt'(\nn\times\vbe)e^{in\om_0(t'-\nn\cdot\x(t')/c)}
\equiv\frac\be{\om_0}[-\epu K\cos\th-\epd J\cos\th+\ept L'\sin\th]
\eeq
where $\om_0 t'=y$, and
\beqa
K&=&\int_0^\tpi dy\,\cos(y+\ph)e^{i(ny-n\be\sin\th\cos y)}e^{in\ph}\nonumber\\
J&=&\int_0^\tpi dy\,\sin(y+\ph)e^{i(ny-n\be\sin\th\cos y)}e^{in\ph}\nonumber\\
L'&=&\int_0^\tpi dy\,\cos ye^{i(ny-n\be\sin\th\cos y)}e^{in\ph}
\eeqa
with $\be=a\om_0/c$.
Thus
\beqa
\lec\barr{l}J\\K\earr\ric&=&\int_0^\tpi dy\lec\barr{c}\sin(y+\ph)\\ \cos(y+
\ph)\earr\ric e^{in\ph}e^{iny}e^{-in\be\sin\th\cos y}\nonumber\\
&=&e^{in\ph}\int_0^\tpi dy\,\leb\sin\ph\lec\barr{c}\cos y\\-\sin y\earr\ric
+\cos\ph\lec\barr{c}\sin y\\\cos y\earr\ric\rib e^{i(ny-n\be\sin\th\cos y)}
\nonumber\\
&\equiv&e^{in\ph}\leb\sin\ph\lec\barr{c}L\\M\earr\ric+\cos\ph\lec\barr{c}M\\L
\earr\ric\rib
\eeqa
where
\beqa
M&=&\int_0^\tpi dy\,\sin y\,e^{iny-in\be\sin\th\cos y}\nonumber\\
&=&\frac1{in\be\sin\th}\int_0^\tpi dy\,e^{iny}\der{}y\lep e^{-in\be\sin\th
\cos y}\rip\nonumber\\
&=&-\frac1{\be\sin\th}\int_0^\tpi dy\,e^{in y}e^{-in\be\sin\th\cos y}=-
\frac\tpi{\be\sin\th}\frac{J_n(n\be\sin\th)}{i^n},
\eeqa
and
\beqa
L&=&\int_0^\tpi dy\,\cos y\,e^{in y}e^{-in\be\sth\cos y}\nonumber\\
&=&\frac1{-i}\der{}{(n\be\sth)}\int_0^\tpi dy\,e^{i(ny-n\be\sth\cos y)}
\nonumber\\
&=&-\frac\tpi i\frac1{i^n}\der{J_n(n\be\sth)}{(n\be\sth)}
\eeqa
Hence
\beqa
|{\bf I}|^2&=&\frac{\be^2}{\om_0^2}\leb\cos^2\th\,(|K|^2+|J|^2)+
\sin^2\th\,|L|^2\rib\nonumber\\
&=&\frac{4\pi^2\be^2}{\om_0^2}\leb\cos^2\th\lep\frac{J_n(n\be\sth)}{\be
\sth}\rip^2+(\cos^2\th+\sin^2\th)\lep\der{J_n(n\be\sth)}{(n\be\sth)}\rip^2
\rib.
\eeqa
Also,
\beq
\der{{\cal P}_n}\Om=\frac{e^2n^2\be^2\om_0^2}{\tpi c}\leb\lep\der{J_n(n\be
\sth)}{(n\be\sth)}\rip^2+\frac{\cot^2\th}{\be^2}\lep J_n(n\be\sth)\rip^2\rib.
\eeq

For better comparison of this result with things we already know, let us
look at some limiting cases. First, in the nonrelativistic limit, $\be<<1$,
we expect that only the $n=1$ term will contribute appreciably and, 
in addition, we can
use the small argument approximation to the Bessel function,
\beq
J_1(x)\approx x/2\andh J_1'(x)\approx1/2
\eeq  so that
\beq
\der{{\cal P}_1}\Om=\frac{e^2\be^2\om_0^2}{\tpi c}\lep\frac14+\frac{\cot^2
\th\sin^2\th}4\rip=\frac{e^2\om_0^4a^2}{8\pi c^3}(1+\cos^2\th)
\eeq
which may be compared with the result of a completely nonrelativistic
calculation as in, e.g., Jackson, Problem 14.2(b). In the highly
relativistic limit, on the other hand, we have $\be\approx1$ and we also
know that most of the radiation is close to the equatorial plane or $\th
\approx\pi/2$. Let us introduce $\al\equiv\pi/2-\th<<1$. Then
\beq
\der{{\cal P}_n}\Om=\frac{e^2n^2\be^2\om_0^2}{\tpi c}\left.\leb\lep\der{J_n(x)}x
\rip^2+\frac{\tan^2\al}{\be^2}\lep J_n(x)\rip^2\rib\ril_{x=n\be\cos\al}.
\eeq
The argument of the Bessel function $J_n$ is comparable to $n$ but is
always less than $n$. For this particular range of argument, it is the 
case\footnote{See, e.g., Watson, {\em Bessel
Functions}, p.~249.} that
\beq
J_n(x)=\frac1\pi\sqrt{\frac{2(n-x)}{3x}}K_{1/3}\lep\frac{2\sqrt2(n-x)^{3
/2}}{3\sqrt x}\rip,
\eeq
or
\beq
J_n(x)=\frac1\pi\frac{y^{1/3}}{3^{1/6}x^{1/3}}K_{1/3}(y)\wheh y=\frac{2
\sqrt2(n-x)^{3/2}}{3\sqrt x}.
\eeq
Also,
\beq
\der{J_n(x)}x=\frac1{\pi3^{1/6}}\der{}x\lep\frac{y^{1/3}K_{1/3}(y)}{x^{1
/3}}\rip.
\eeq
As shown in the figure below, for the interesting values of 
$x$, $K_{1/3}(y)$ varies much more than $x$.

\centerline{\psfig{figure=fig15.ps,height=3.0in,width=4.5in}}

\noindent Thus, the preceding derivative may be approximated by
\beqa
\der{J_n(x)}x&\approx&\frac1{\pi3^{1/6}x^{1/3}}\der{[y^{1/3}K_{1/3}(y)]}y\der
yx\nonumber\\
&=&-\frac{y^{1/3}K_{2/3}(y)}{\pi3^{1/6}x^{1/3}}\der yx=\frac{y^{1/3}K_{2/3}(
y)}{3^{1/6}x^{1/3}}\frac{\sqrt2(n-x)^{1/2}}{3x^{3/2}}(2x+n).
\eeqa
Hence
\beq
\der{{\cal P}_n}\Om=\frac{e^2\be^2n^2\om_0^2y^{2/3}}{\tpi c\pi^23^{1/3}x^{2
/3}}\leb\frac{n^2\sin^2\al}{x^2}K_{1/3}^2(y)+\frac{2(n-x)(2x+n)
^2}{9x^3}K_{2/3}^2(y)\rib.
\eeq
Now let's see what can be said about $y$ and $x$. Expanding in powers of $
\al$ and $1/\ga$, we have
\beq
y\approx\frac n3\lep\frac1{\ga^2}+\al^2\rip^{3/2}\andh x\approx n\lep1-
\frac1{2\ga^2}-\frac{\al^2}2\rip,
\eeq
 so
\beq
\der{{\cal P}_n}\Om=\frac{e^2\om_0^2n^2}{6\pi^3c}\lep\frac1{\ga^2}+\al^2
\rip^2\leb K_{2/3}^2(y)+\frac{\al^2\ga^2}{1+\al^2\ga^2}K_{1/3}^2(y)\rib.
\eeq
This term describes radiation at the particular frequency $\om=n\om_0$, so
we can also write it as
\beq
\der{{\cal P}_n}\Om=\frac{\om_0^2}\tpi\frac{e^2(\om^2/\om_0^2)}{3\pi^2c
\ga^6}(1+\al^2\ga^2)^2\ga^2\leb K_{2/3}^2(y)+\frac{\al^2\ga^2}{1+\al^2\ga^2}
K_{1/3}^2(y)\rib
\eeq
with
\beq
y=\frac\om{3\om_0\ga^3}(1+\al^2\ga^2)^{3/2}\equiv\frac\om{\om_c}(1+\al^2
\ga^2)^{3/2}.
\eeq
By grouping terms appropriately in \eq{165} we can write
\beq
\der{{\cal P}_n}\Om=\frac{\om_0^2}\tpi\frac{3e^2}{\pi^2c}\lep\frac\om{
\om_c}\rip^2(1+\al^2\ga^2)^2\ga^2\leb K_{2/3}^2(y)+\frac{\al^2\ga^2}{1+\al^2
\ga^2}K_{1/3}^2(y)\rib.
\eeq
This is the time-averaged power received at latitude $\al$, at frequency
$\om=n\om_0$, per unit solid angle. The energy/cycle received at this
frequency is obtained if we multiply by $2\pi/\om_0$, and, finally, the
energy per unit frequency is found it we multiply by a factor of the
inverse spacing of the harmonics, or $1/\om_0$. Hence we conclude that for
a single pulse, meaning one cycle of the particle,
\beq
\der{I(\om)}\Om=\frac\tpi{\om_0^2}\der{{\cal P}_n}\Om=\frac{3e^2}{\pi^2c}
\lep\frac\om{\om_0}\rip^2(1+\al^2\ga^2)^2\ga^2\leb K_{2/3}^2(y)+\frac{\al^2
\ga^2}{1+\al^2\ga^2}K_{1/3}^2(y)\rib
\eeq
which is precisely what we found when we calculated the radiated intensity
from a particle in instantaneous circular motion. 

\section{Thomson Scattering; Blue Sky}
It's time for a change of pace to something with less nineteenth-century
analysis. {\em Thomson scattering} provides just such a respite. It is the
scattering of radiation by a free charge. The mechanism involved,
classically, is the coupling of the charge to the incident electric field
$\E_0$; this produces acceleration of the charge and consequent radiation by
the accelerated charge which becomes the ``scattered'' radiation. We saw
in Chapter~9 how one can describe scattering of electromagnetic radiation
in this way. We didn't actually do Thomson scattering at that time, so
we'll do it now.

Thomson scattering is to be distinguished from {\em Compton scattering} in
which the same phenomenon is treated using quantum theory. One must use 
quantum theory, when $\la$, the wavelength of the radiation, is
comparable to the {\em Compton wavelength} of the scatterer, $\la=2\pi c/\om
\sim\hbar/mc\equiv\la_c$; $\la_c$ is the Compton wavelength. This condition
may also be written as $\hbar\om\sim2\pi mc^2$ which says that if the
photon energy $\hbar\om$ ($\sim$ eV) is comparable to the particle's 
rest energy $mc^2$ ($\sim$ MeV),
then the calculation must be done using quantum theory. For, e.g., visible
light the photon energy is much smaller than an electron's rest energy so
the Thomson scattering, or classical, calculation is quite adequate.

Assuming a nonrelativistic particle, we can make use of the Larmor formula
for the scattered radiation,
\beq
\dpdo=\frac{e^2}{4\pi c^3}a^2\sin^2\Th\wheh\Th=\angle(\nn,\a)
\eeq
for the instantaneous radiated power. To find the acceleration $\a$ of the
particle, we must solve the equation of motion of the scatterer. The force
on it is provided by the incident electric field,
\beq
\E_0=E_0\epv\ekxot
\eeq
where for convenience we shall let $\k=k\ept$ and assume linear
polarization of the incident plane wave $\epv=\cos\ps\,\epu+\sin\ps\,\epd$.

\centerline{\psfig{figure=fig16.ps,height=2.0in,width=3.0in}}

Similarly, we shall suppose for simplicity that the charged particle has no
velocity component along the direction of $\k$ and shall ignore the
magnetic force which is reasonable so long as the particle is nonrelativistic.
Letting the particle be located in the $z=0$ plane, we find that it experiences
an incident field
\beq
\E_0=\epv E_0\emot;
\eeq
given that it has a mass $m$ and charge $e$, the force acting on it is $\F
=e\E_0=m\a$, so
\beq
\a=\epv\frac{eE_0}m\emot,
\eeq
and
\beq
\a\cdot\nn=\frac{eE_0}m\sin\th(\cos\ps\cos\ph+\sin\ps\sin\ph)\emot.
\eeq
where $\th$ and $\ph$ specify, in spherical coordinates, the direction to the
observation point.

The time average of the power emitted in some particular direction will be
proportional to
\beq
<a^2\sin^2\Th>=<a^2(1-\cos^2\Th)>=<a^2-(\a\cdot\nn)^2>=
\frac{e^2E_0^2}{2m^2}[1-\sin^2\th\cos^2(\ps-\ph)]
\eeq
where the brackets $<...>$ denote a time average. If the incident radiation
is not polarized, then we must average over $\ps$ with the result that
\beq
<a^2\sin^2\Th>=\frac{e^2E_0^2}{2m^2}\lep1-\frac12\sin^2\th\rip=
\frac{e^2E_0^2}{4m^2}(1+\cos^2\th)
\eeq
and
\beq
\dpdo=\frac{e^4}{16\pi c^3m^2}|E_0|^2(1+\cos^2\th)
\eeq
is the time-averaged power distribution when the incident wave is
unpolarized.

We define the scattering cross-section in the usual way, i.e., the
time-averaged power per unit solid angle divided by the time-averaged incident
power per unit area,
\beq
\der\si\Om=\frac{d{\cal P}/d\Om}{(c/8\pi)|E_0|^2}=\frac12\lep
\frac{e^2}{mc^2}\rip^2(1+\cos^2\th).
\eeq
This is J.~J.~Thomson's formula for the scattering of light by a charged
particle. The total cross-section is
\beq
\si=\int d\Om\,\lep\der\si\Om\rip=\frac12\lep\frac{e^2}{mc^2}\rip^24\pi(1+1
/3)=\frac{8\pi}3\lep\frac{e^2}{mc^2}\rip^2\equiv\frac{8\pi}3r_c^2
\eeq
where $r_c$ is the classical radius of the particle. For an
electron it \mbox{is $\sim3\times10^{-13}\,cm$} and $\si$ is about
$0.7\times10^{-24}\,cm^2$.

We may also calculate the scattering of radiation by a bound charge using
the model of a damped harmonic oscillator. Let a charge $e$ with mass $m$ be
bound at the origin of coordinates with a natural frequency of oscillation
$\om_0$ and damping constant $\Ga$. Then, given an applied electric field
\beq
\E_0(\x,t)=\epv E_0\ekxot
\eeq
with $\k=k\ept$, we know from earlier calculations that the charge will
respond with a displacement
\beq
\xt=\x_0\emot\wheh\x_0=\frac{eE_0/m}{\om_0^2-\om^2-i\om\Ga}\epv
\eeq
if we approximate $\E_0(\x,t)$ by $\E_0(0,t)$ as is reasonable when the
particle's displacement from the origin is small compared to the wavelength
of the incident radiation. That is certainly true for visible light and an
atomic electron.

From $\x(t)$ it is a simple matter to compute the acceleration,
\beq
\a=-\epv\frac{eE_0\om^2/m}{\om_0^2-\om^2-i\om\Ga}\emot,
\eeq
and then to find the radiated, or scattered, time- and
incident-polarization-averaged power per unit solid angle,
\beq
\dpdo=\frac{e^2}{16\pi c^3}\lep\frac{eE_0}m\rip^2\lep\frac{1+\cos^2\th}{(1
-\om^2_0/\om^2)^2+\Ga^2/\om^2}\rip;
\eeq
the scattering cross-section follows:
\beq
\der\si\Om=\frac12r_c^2\lep\frac{1+\cos^2\th}{(1-\om_0^2/\om^2)^2+\Ga^2/
\om^2}\rip.
\eeq
For sufficiently large $\om$, $\om>>\om_0,\Ga$, the cross-section reduces to
the Thomson result as it should since for very large $\om$ the particle
will respond to the field as though it were a free particle. Also, for
$\om<<\om_0$ and $\om_0^2/\Ga$, we find the Rayleigh scattering result,
\beq
\der\si\Om=\frac12r_c^2\lep\frac\om{\om_0}\rip^4(1+\cos^2\th)
\eeq
with the characteristic $\om^4$ behavior indicating dipole scattering.

	Why is the sky blue?  Why is the light polarized when we look
at the sky perpendicular to the line of sight from us to the sun?

\centerline{\psfig{figure=fig17.ps,height=2.5in,width=8.5in}}


\section{Cherenkov Radiation Revisited}
While studying the energy loss of a charged particle traversing a material,
we derived an expression for the rate of energy loss through Cherenkov
radiation. Specifically, for a charge $q$ moving at speed $\be$ through a
medium with a real dielectric function $\epo$, we found that the energy
loss per unit path length is
\beq
\der Ex=\frac{q^2}{c^2}\int_{\ep\be^2>1}d\om\,\om(1-\frac1{\epo\be^2}).
\eeq
The form of this expression suggests that the radiation per unit length of
path per unit frequency is given by the integrand,
\beq
\der{I(\om)}x=\om\frac{q^2}{c^2}\lep1-\frac1{\epo\be^2}\rip.
\eeq
What we did not do is look at the form of the potentials and fields in real
space and time. That is interesting and revealing, so we are going to do it
now using retarded potentials. 

Before starting, lets be sure that we
understand the physical mechanism giving rise to the radiation. It is not the
incident particle, which may be reasonably described as having constant
velocity, that is doing the radiating. Rather, the incident particle produces
fields which act on the particles in the medium, causing them to be accelerated
in various ways. They then produce radiation fields which, when the incident
particle moves more rapidly than the speed of light in the medium, but not when
it moves more slowly, add in a coherent fashion to give Cherenkov radiation.

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

\noindent It is not easy to see why the fields must cancel when the
particle is moving slower than the speed of light, but it {\em is}
easy to see why they must not when it is moving faster than light in the
medium.  Consider the right hand side of the figure above.  Before
the wake of the radiation hits a particle in the medium, it does not
feel the incident particle.  Once the wake hits a particular particle
in the medium, not only it, but all of its neighbors accelerate in a 
direction perpendicular to the wake (depending upon the relative charge).
Clearly these particles will radiate coherently.

If we are going to produce a calculation of Cherenkov radiation, then, we
have to find the total field produced by all of the particles and not just
the incident particle. We in fact did that in Chapter 13 in the space of
$\k$ and $\om$. Here we want to determine appropriate Li\'enard-Wieckert
potentials for this field so as to find it in real space and time. That turns
out not to be very hard if we ignore the frequency-dependence of the dielectric
function.

Consider the Fourier-transformed potentials $\Ako$ and $\Phko$ produced by
the incident charge in the medium where it is taken to have constant velocity.
As we have seen, these obey the equations
\beqa
\lep k^2-\frac{\om^2}{c^2}\ep\rip\Phko=\fpi\frac{\rhko}\ep\nonumber\\
\lep k^2-\frac{\om^2}{c^2}\ep\rip\Ako=\frac\fpi c\Jko
\eeqa
where $\rh$ and $\J$ are the macroscopic sources, i.e., the charge and
current density of the incident particle. Given that $\ep$ is independent
of $\om$, then these are the same as the equations obeyed by the potentials
of a fictitious system consisting of a point particle with charge $q/\sqrt\ep$
moving at constant velocity $\v$ in a `vacuum' where the speed of light is
$c'=c/\sqrt\ep$. Let's rewrite them in such a way as to see this
correspondence more clearly:
\beqa
\lep k^2-\frac{\om^2}{c'^2}\rip[\sqrt\ep\Phko]=\fpi\frac\rhko{\sqrt\ep}
\nonumber\\
\lep k^2-\frac{\om^2}{c'^2}\rip\Ako=\frac\fpi{c'}\frac\Jko{\sqrt
\ep}.
\eeqa

Now let's think about these potentials in real space and time. Because they
are the same as a particle with a renormalized charge $q/\sqrt\ep$ moving
in a vacuum with a renormalized speed of light is $c'$, we can write them as
Li\'enard-Wieckert potentials using the renormalized charge and velocity
of signal propagation,
\beq
\sqrt\ep\Phxt=\frac q{\sqrt\ep}\leb\frac1{\ka R}\rib_{ret}
\eeq
and similarly for $\Axt$. In this case the retardation means that $\ka R$
should be evaluated at the time $t'=t-R(t')/c'$. Further, if $\x(t')=\v t'$,
$\v=v\ept$, and $\x=z\ept+\rh\epv\per$,
then
\beq
R(t')=(z-vt')\ept+\rh\epv\per.
\eeq
Also, for this system, $\ka=1-\nn\cdot\v/c'$. Notice that $\ka$ can be
negative; the absolute value of $\ka$ should be employed in evaluating the
potential because the potential really involves $|\ka R|$ as one may see by
going back to its derivation, especially \eq{4}.

\centerline{\psfig{figure=fig18.ps,height=2.5in,width=8.5in}}

It is useful to introduce a vector $\X\equiv\x-\v t$ which is the relative
displacement of the observation point and the particle at time $t$. Then
\beq
\R=\x-\v t'=\x-\v t+\v(t-t')=\X+\v(t-t'),
\eeq
and
\beq
t-t'=R/c'=|\X+\v(t-t')|/c'.
\eeq
Square this relation to find
\beq
(t-t')^2=\frac{X^2}{c'^2}+\frac{v^2}{c'^2}(t-t')^2+\lep\frac{2\X\cdot
\v}{c'^2}\rip(t-t').
\eeq
This is a quadratic equation that can be solved for $t-t'$; there are two
solutions which are
\beq
t-t'=\frac{-\X\cdot\v\pm\sqrt{(\X\cdot\v)^2-(v^2-c'^2)X^2}}{v^2-c'^2}.
\eeq
Acceptable solutions must be real and positive. Given that $v^2>c'^2$,
which we know to be the regime where there is Cherenkov radiation, we find
that there are either no such solutions or there are two of them. The
conditions under which there are two are
\beq
\X\cdot\v<0\andh(\X\cdot\v)^2>(v^2-c'^2)X^2.
\eeq
Let the angle between $\X$ and $\v$ be $\al$. Then we require, first, that
$\al$ be larger than $\pi/2$ and, second, that $X^2v^2\cos^2\al>(v^2-c'^2)
X^2$ or $\cos^2\al>1-c'^2/v^2$. Hence there is a cutoff angle $\al_0$ given
by
\beq
\al_0=\arccos(-\sqrt{1-c'^2/v^2})
\eeq
such that for $\al<\al_0$ there is no potential. There can thus be
potentials and fields at time $t$ only within a cone whose apex is the
current position of the particle and which has an apex angle of $\pi-\al_0$.
Within this cone the potential is the sum of two terms, $\Ph=\Ph_1+\Ph_2$,
corresponding to the two allowed values of $t-t'$. Making use of \eq{191},
and the fact that $\R\parallel\nn$, we see that we can write, for 
either case,
\beq
[\ka R]_{ret}=|(1-\nn\cdot\v/c)\cdot\R|= R -\v\cdot\R/c=|\X+\v(t-t')|-\X\cdot\v/c'-v^2(t-t')/c'.
\eeq
Using \eqs{189}{197} with \eq{194} for $t-t'$, one finds that
$\Ph_1\equiv\Ph_2$ and that
\beq
\Phxt=\lep\frac{2q}\ep\rip\frac1{X\sqrt{1-(v^2/c'^2)\sin^2\al}}.
\eeq

A similar expression can be found for $\Axt$ and with a bit more work one
can compute the radiated power, recovering the same equations as found in
chapter 13 for the particular case that $\epo$ is a constant. Such a
dielectric function never exists, of course, and so our conclusions are
flawed in some respects. One of them has to do with the form of $\Ph$ close
to $\al=\al_0$; \eq{198} indicates that it in fact diverges here. That would
indeed happen if signals with all frequencies traveled at speed $c'$ so
that such a singular non-dispersing wave front could be built by superposing
waves with many different wavelengths including ones approaching zero. In
reality, there is no such singularity although the amplitude of the wave does
have a strong maximum at the leading edge.

\section{Cherenkov Radiation; Transition Radiation}

This time we will do a calculation using perturbation theory much the way
we did scattering {\em via} perturbation theory. We will learn a little bit
more about the character of Cherenkov radiation and will also derive a new
(to us) phenomenon. The basic requirement for validity of the calculation
is to have
$|\epo-1|<<1$ so that we can get away with calculating the true macroscopic
fields as a correction to the fields produced by the incident particle in
vacuum. We have shown that, when expressed as a function of $\x$ and $\om$,
the electric field of a particle with charge $q$ moving at constant velocity
$\v=v\ept$ on a trajectory $\x(t)=\v t$ is, in vacuum,
\beq
\E_i(\x',\om)=\sqrt{\frac2\pi}\frac{q\om}{\ga v^2}e^{i\om z'/v}[K_1(\om\rh'
/\ga v)\epv_{\rh'}-(i/\ga)K_0(\om\rh'/\ga v)\ept].
\eeq

\centerline{\psfig{figure=fig19.ps,height=2.5in,width=8.5in}}

\noindent This field produces a polarization in the medium which is
\beq
\P(\x'\om)=\frac{\epo-1}\fpi\E_i(\xp,\om)
\eeq
and a dipole moment $\P(\xp,\om)d^3x'$ in a volume element $d^3x'$. From
chapter 9 (\eq{43}), we know that such a
harmonic dipole moment gives rise to a radiation field
\beq
d\E_{rad}(\x,\om)=\frac\ekR Rk^2[\nn\times\{\P(\xp,\om)d^3x'\}]\times\nn.
\eeq
In the radiation zone we can expand $R=\xxpa$ as $R=r-\nn\cdot\xp$, leading
to
\beq
\E_{rad}(\x,\om)=\frac\ekr r\lep\frac{\epo-1}\fpi\rip k^2\invp e^{-ik
\nn\cdot\xp}\{[\nn\times\E_i(\xp,\om)]\times\nn\}\nonumber
\eeq
\beq
=\frac\ekr r\lep\frac{\epo-1}\fpi\rip k^2\int d^2x'\per\{[\nn\times\E_i(
\x'\per,0,\om)]\times\nn\}e^{-ikx'\sin\th}\int dz'e^{i(\om z'/v-kz'\cos\th)}
\eeq
where we have specified $\k=k(\ept\cos\th+\epu\sin\th)$ without loss of
generality since the fields are invariant under rotation around the
direction of $\v$. Also, $k=\om\sqrt{\epo}/c=\om/v_p$ where $v_p$ is the
phase velocity of a wave with frequency $\om$. If the medium is infinite in
the $z$ direction, we can complete the integration over $z'$ with ease and find
\beq
\E_{rad}(\x,\om)=\frac\ekr r\lep\frac{\epo-1}\fpi\rip k^2\de(\om/v-k\cos\th
)\int d^2x'\per\{[\nn\times\E_i(\x'\per,\om)]\times\nn\} e^{ikx'\sin\th}.
\eeq
Notice that $k$ and $\om$ are related by $k=\om\sqrt{\epo}/c=\om/v_p$ so that
we find no radiation field unless $\cos\th=v_p/v$ which can only happen if
$v>v_p$. We seem to be on the right track. The implication is that there is
radiation coming out of this system in a direction $\nn$ which makes an
angle $\al_0$ with the $z$ axis where $\al_0=\arccos(v_p/v)$. What we have
shown in chapter 13 is that this is the direction of the 
outcoming radiation, which is Cherenkov
radiation, is perpendicular to the ``bow wave'' shown in the figure earlier.

We can get something new out of this calculation also. Suppose that $v<v_p$
and that the medium is finite in extent in the $z$ direction. 
If it begins
at $z=0$ and ends at some arbitrary, large, value of $z$, then the integral
over $z'$ is not from $-\infty$ to $+\infty$ but rather starts from zero
and will be non-zero even though $v<v_p$. 

\centerline{\psfig{figure=fig20.ps,height=2.5in,width=8.5in}}

\noindent The corresponding radiation,
known as {\em transition radiation}, comes about because of the presence
of the boundary, in this case between vacuum and dielectric.
The integral we must do can be evaluated with the use of a convergence
factor (which emulates the damping present in most materials):
\beqa
\intzp dz'\,e^{iz'(\om/v-k\cos\th)}=\lim_{\et\rta0}\intzp dz'\, e^{[-\et+i(
\om/v-k\cos\th)]z'}\nonumber\\
=\lim_{\et\rta0}\lep\frac{-1}{-\et+i(\om/v-k\cos\th)}\rip=\frac i{\om/v-k
\cos\th},
\eeqa
and so
\beq
\E_{rad}(\x,\om)=\frac\ekr r\frac{\epo-1}\fpi k^2\lep\frac i{\om/v-k\cos
\th}\rip\int d^2x'\per\,\{[\nn\times\E_i(\x'\per,\om)]\times\nn\}
e^{-ikx'\sin\th}.
\eeq
More generally, a slab of material of finite thickness $d$ produces an integral
over $z'$ which is
\beq
\int_0^d dz'\,e^{i(\om/v-k\cos\th)z'}=\frac{i\lep1-e^{i(\om/v-k\cos\th)d}
\rip}{\om/v-k\cos\th}.
\eeq

Whether we treat a finite slab or no, we have to do the integral over the
other two coordinates. The integrand involves $(\nn\times\E_i)\times\nn$.
For a given $\nn=\k/k$, $\E_i$, and $\xp$, and noting that
$\epv_{\rh'}=\cos\ph'\,\epu+\sin\ph'\,\epd$, we have
\beq
(\nn\times\E_i)\times\nn=(E_\rh\cos\ph'\cos\th-E_z\sin\th)(\epd\times\nn)+
E_\rh\sin\ph'\epd.
\eeq
The second term is an odd function of $\ph'$ and will give zero when
integrated over $\x'\per$. Thus we must do the integral
\beq
\I=\int d^2x'\per(\nn\times\E_i)\times\nn e^{-ikx'\sin\th}\hsph\hsph\hsph
\nonumber
\eeq
\beq
=\sqrt{\frac2\pi}\frac{q\om}{\ga v^2}(\epd\times\nn)\int dx'dy'\,[K_1(
\om\rh'/\ga v)\cos\th\cos\ph'+(i/\ga)K_0(\om\rh'/\ga v)\sin\th]e^{-ikx'\sin
\th}
\eeq
Now, $\cos\ph'=x'/\rh'$ and
\beq
\cos\ph'\,K_1(\om\rh'/\ga v)=\frac{x'}{\rh'}K_1(\om\rh'/\ga v)=-\frac{\ga v}\om
\pde{K_0(\om\rh'/\ga v)}{x'}.
\eeq
Using this in the expression for $\I$, we can do an integration by parts
and find
\beqa
\I&=&\sqrt{\frac2\pi}\frac{q\om}{\ga v^2}(\epd\times\nn)\int dx'dy'\,\leb
-\frac{ik\ga v\sin\th}\om K_0(\om\rh'/\ga v)\cos\th+\frac i\ga K_0(\om\rh'
/\ga v)\sin\th\rib\nonumber\\
&=&\sqrt{\frac2\pi}\frac qv(-i\sin\th)\lep k\cos\th-\frac\om{\ga^2v}\rip(
\epd\times\nn)\int dx'dy'\,K_0(\om\rh'/\ga v)e^{-ikx'\sin\th}.
\eeqa
The integral over $x'$ can be done using the identity
\beq
\intzp dz\,K_0(\be\sqrt{z^2+t^2})\cos(\al z)=\frac\pi{2\sqrt{\al^2+
\be^2}}e^{-|t|\sqrt{\al^2+\be^2}}.
\eeq
Thus,
\beqa
\I&=&\sqrt{\frac2\pi}\frac qv\epd\times\nn(-i\sin\th)\lep k\cos\th-\frac\om{
\ga^2v}\rip\intmp\frac{\pi dy'\exp(-|y'|\sqrt{k^2\sin^2\th+
\om^2/\ga^2v^2})}{\sqrt{k^2\sin^2\th+\om^2/\ga^2v^2}}
\nonumber\\
&=&\sqrt{\frac2\pi}\frac qv\epd\times\nn(-i\sin\th)\lep k\cos\th-\frac\om{
\ga^2v}\rip\frac\tpi{k^2\sin^2\th+\om^2/\ga^2v^2}.
\eeqa
Hence
\beq
\E_{rad}=\ekr\frac{\epo-1}{\fpi r}\,\frac{k^2\sin\th}{\om/v-k\cth}\,\frac{2
\sqrt\tpi(q/v)\epd\times\nn}{k^2\sin^2\th+\om^2/\ga^2v^2}\lep k\cth-\frac\om{
\ga^2v}\rip.
\eeq

\subsection{Cherenkov Radiation in a Dilute Collisionless Plasma}

This is a general result. Let us look at some specific example in a
simple limit in order to extract the salient qualitative features of
transition radiation. First, we shall choose a dielectric function. A very
simple one, which has the added virtue that for some frequencies $\ep<1$ so
that it is impossible to have Cherenkov radiation, is that for a dilute
collisionless plasma, $\epo=1-\om_p^2/\om^2$. 

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

\noindent For $\om$ large enough, this
function is less than, but close to, unity so that it satisfies the
criteria for validity of the perturbation theory.
Then
\beq
(\epo-1)k^2/\fpi=-\om_p^2k^2/\fpi\om^2=-\om_p^2/\fpi c^2
\eeq
where we approximate $\om$ by $ck$ wherever it is not important.\footnote{This
can be done in some places because $\epo\approx1$; in other places the
difference $\epo-1$ is needed and here we cannot set $\epo$ equal to 1.}
Suppose also that $\ga>>1$, i.e., that the particle is highly relativistic.
These conditions allow us to approximate as follows:
\beq
\frac\om v-k\cos\th=\frac\om v-\frac \om{v_p}\cos\th=\frac\om{\be c}-\frac
\om c\sqrt\ep\cos\th\approx\frac\om{c(1-1/2\ga^2)}-\frac\om c\lep1-\frac{
\om_p^2}{2\om^2}\rip\lep1-\frac{\th^2}2\rip\nonumber
\eeq
\beq
\approx\frac\om c\lep\frac1{2\ga^2}+\frac{\om_p^2}{2\om^2}+\frac{
\th^2}2\rip=\frac\om{2\ga^2c}\lep1+\frac{\ga^2\om_p^2}{\om^2}+\ga^2\th^2\rip
\equiv\frac{\om}{2\ga^2c}\lep1+\frac1{\nu^2}+\et\rip
\eeq
where
\beq
\nu\equiv\om/\ga\om_p \;\;\; \et\equiv\ga^2\th^2 \;\;\;
\beta=\sqrt{1-\gamma^{-2}}\approx 1-1/2\gamma^2 \andh
\sqrt{\ep}=\sqrt{1-\frac{\om^2_p}{\om^2}}\approx 1-\frac{\om^2_p}{2\om^2}.
\eeq
Also,
\beq
\frac{\om^2}{\ga^2v^2}+k^2\sin^2\th=\frac{\om^2}{\ga^2\be^2c^2}+\frac{\om^2
\ep\sin^2\th}{c^2}=\frac{\om^2}{\ga^2\be^2c^2}[1+\ga^2\be^2\ep\sin^2\th]
\approx\frac{\om^2}{\ga^2c^2}(1+\et)
\eeq
while
\beq
k\cos\th-\frac\om{v\ga^2}=\frac\om c\sqrt{\epo}\cth-\frac\om{\be c\ga^2}=
\frac\om c\lep\sqrt{\epo}\cos\th-\frac1{\be\ga^2}\rip\approx\frac\om c.
\eeq
Hence,
\beqa
\E_{rad}&=&\frac\ekr r\lep\frac{-\om_p^2}{\fpi c^2}\rip\frac{\th(2\sqrt\tpi)(q
/c)(\om/c)\epd\times\nn}{(\om/2\ga^2c)(1+\et+1/\nu^2)(\om^2/\ga^2 c^2)(1+\et
)}\nonumber\\
&=&-\frac\ekr r\,\frac{\om_p^2}\fpi\,\frac{2\ga^4\th2\sqrt{\tpi} q\epd
\times\nn\om}{c\om(1+\et+1/\nu^2)\om^2(1+\et)}\nonumber\\
&=&-\frac\ekr r\sqrt{\frac2\pi}\,\frac{\ga\sqrt\et}{\nu^2}\frac{q\epd\times
\nn}{c(1+\et+1/\nu^2)(1+\et)}.
\eeqa
The radiated energy per unit frequency per unit solid angle
is\footnote{Previously, we just called this $dI/d\Om$.}
\beq
\csde{I(\om)}\Om\om=2\frac c\fpi|r\E_{rad}(\x,\om)|^2=\frac{q^2\ga^2}{
\pi^2c}\,\frac\et{\nu^4(1+\et+1/\nu^2)^2(1+\et)^2}.
\eeq
We can write $d\Om=\sth d\th d\ph\approx\th d\th d\ph=d\et d\ph/2\ga^2$ and
integrate over $\ph$ to find the distribution per unit $\et$. Also, let's
write the frequency in terms of $\nu$, $d\nu=d\om/\ga\om_p$, to find
\beq
\csde I\et\nu=\int d\ph\csde I\Om\om\der\om\nu\der\Om\et=\csde I\Om\om\frac{
\ga\om_p\pi}{\ga^2}=\frac{q^2\ga\om_p}{\pi c}\,\frac\et{\nu^4(1+\et+1/\nu^2)^2
(1+\et)^2}.
\eeq
This expression fails at small $\nu$ ($\om\sim\om_p$). It falls off as $
\et^{-3}$ at large $\et$. It peaks as a function of $\et$ around $\et=1$ for
small $\nu$ and at $\et=1/3$ for large $\nu$. If one integrates over $\et$
the result is
\beq \der I\nu=\frac{q^2\ga\om_p}{\pi c}[(1+2\nu^2)\ln(1+1/\nu^2)-2].
\eeq
Further, the total energy radiated is
\beq
I=\intzp d\nu \der I\nu=\frac{q^2\ga\om_p}{3c}=\frac{(q/e)^2\ga\hbar
\om_p}3\frac{e^2}{\hbar c}=\frac{(q/e)^2}{3(137)}(\ga\hbar\om_p).
\eeq
A typical photon energy is $\ga\hbar\om_p/3$ ($\nu=1/3$), so the number of
photons emitted on average is $(q/e)^2/137$ which is quite a bit smaller
than one. However, a sizable amount of transition radiation can be obtained
by employing a stack of thin slabs of material with adjacent slabs having
significantly different dielectric constants. Then there is some radiation
produced at each interface between different materials.

\section{Example Problems}
\subsection{A Relativistic Particle in a Capacitor}

A particle of charge $e$ and mass $m$ initially
at rest is accelerated across a parallel plate capacitor held at
(stat) voltage $V$; the distance between the plates is $d$.  Assuming
nonrelativistic motion, find the total energy radiated by the particle
during this process.  Then, {\em{without calculation, answer or
estimate the following:}}
\begin{enumerate}
\item The angular distribution of the radiated energy of the particle,
\item the order of magnitude of typical frequencies emitted from the
charge, and
\item the angular distribution of radiated energy from the charge
if it were highly relativistic.
\end{enumerate}

{\bf{Solution.}} The field in a parallel plate capacitor is constant,
thus so is the acceleration of the charged particle and its radiation.
\[
P=\frac23 \frac{e^2}{c^3} a^2\;\;\;\; W=\int Pdt =\frac23 \frac{e^2}{c^3} t
\]
where t is the duration of the pulse.
\[
a=F/m=eE/m=eV/md\;\;\;{\rm{and}}\;\;\;t=\sqrt{2d/a}
\]
Hence, solving for $W$
\[
W=\frac{2\sqrt2 e^{7/2} V^{3/2}}{3c^3 m^{3/2}d}
\]

In the non relativistic limit, the angular distribution of the radiation
is given by $\sin^2(\theta)$ where the angle is measured relative
to the velocity vector.  The typical frequencies can be found using
the Fourier uncertainty principle.  Since the retarded duration of the
pulse is the same as the observer's duration for a nonrelativistic
particle, we have $\omega\sim 1/t=\sqrt{eV/md^2}$.  In the relativistic
limit the angular distribution of the radiation will be strongly pitched
in the direction of the velocity/acceleration, but will be zero
along the axis.  The maximum of the pulse of radiation will be at
an angle of $\theta\sim 1/\gamma$ away from the axis defined
by the velocity vector.

\subsection{Relativistic Electrons at SLAC}

At the Stanford linear accelerator, devices have been added 
at the end of the accelerator to guide electrons and positrons around 
roughly semicircular paths until they collide head-on as shown in the
sketch below.  If each particle has a total energy of 50GeV and
rest energy of 0.5 MeV, while the circular paths have radii of 
about 1km, roughly what is the fraction of the particles energy
lost to radiation before the collision takes place?

% source /home/wanderer/jarrell/text/courses/jackson/exams/q3/
\centerline{\psfig{figure=slac.ps,height=3.0in,width=8.5in}}

{\bf{Solution.}}  To determine the power radiated, we must use the relativistic
Larmor formula.  For $\vbed\perp\vbe$, $|\vbe\times\vbed|=|\be\bed|$,
so the power radiated is
\[
P=\frac{2e^2}{3c} \gamma^6 \leb \vbed^2-\lep\vbe\times\vbed\rip^2\rib
=\frac{2e^2}{3c} \gamma^6 \bed^2\lep 1- \be^2\rip=
\frac{2e^2}{3c} \gamma^4 \bed^2
\]
The acceleration is centripetal, so $\bed=v^2/cr\approx c/r$, so
\[
P=\frac{2e^2}{3c} \gamma^4 \frac{c^2}{r^2}
\]
Counting the initial deflection of the particles into the circular
region, both the electron and the positron travel about 3/4 of a circle.
The time it takes to do this is roughly $\tau=3\pi r/2c$ (assuming
that the particles travel roughly at velocity $c$), so the energy
radiated is
\[
\Delta E=\frac{\pi e^2}{r} \gamma^4
\]
Since $\gamma=E/mc^2$, the relative energy loss is
\[
\frac{\Delta E}{E} =\frac{\pi e^2/r}{mc^2} \gamma^3
\]
For $E=50$ GeV and $mc^2= 0.511$ MeV, $\gamma \approx 10^5$,
and the relative energy loss is
\[
\frac{\Delta E}{E} =\frac{\pi 23 \;(10^{-20})\; (10^{15})}
{0.911 \;(10^{-27}) \;(9) \;(10^{20})\;
(10^{5})}\approx 0.9\times 10^{-2} 
\]
or roughly only one percent of the energy is lost.



\edo
