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%TCIDATA{Created=Thu Dec 05 11:21:07 1996}
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%TCIDATA{Language=American English}

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\markright{{\sf Synchrotron Radiation   }}
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\begin{document}


\section{14.6\quad Synchrotron Radiation.}

\subsection{14.6.1}

\quad \thinspace \textbf{a. }As examples, of great practical importance, of
the application of the radiation formulae, we work out in detail the
radiation from a linear harmonic oscillator and from a charge in circular
motion (synchrotron radiation).

\smallskip\ 

\textbf{b}. We must recognize first of all that for a periodic motion the $%
\mathbf{E}$ field and the radiated power spectrum will contain only
multiples of the fundamental frequency $\omega _0.$ Hence we write, instead
of (\ref{1}) and (\ref{3}) of 5.1

\[
\mathbf{E(x,\,}t)=\sum_{m}e^{-im\omega _{0}t}\mathbf{E}_{m}\mathbf{(x}%
)\,,\quad \quad \quad \mathbf{E}_{m}\mathbf{(x})=\frac{\omega _{0}}{2\pi }%
\int_{0}^{2\pi /\omega _{0}}\mathbf{E(x,\,}t)e^{im\omega _{0}t}dt\,. 
\]

\begin{equation}
\frac{dW}{d\Omega }=\sum_m\frac{dP_m}{d\Omega }\,,\quad \quad \quad \frac{%
dP_m}{d\Omega }=2\frac c{4\pi }R^2\left| \mathbf{E}_m\mathbf{(x)}\right|
^2\,.  \label{1}
\end{equation}

\smallskip\ 

\textbf{c. }Carrying through the corresponding changes we arrive at the
analogue of (\ref{6}) 
\begin{equation}
\frac{dP_{m}}{d\Omega }=\frac{e^{2}m^{2}\omega _{0}^{4}}{\left( 2\pi
c\right) ^{3}}\left| \int_{0}^{2\pi /\omega _{0}}dte^{im\omega _{0}\left[ t-%
\mathbf{n\cdot \QTR{mathbf}{x}\left( t\right) /}c\right] }\,\mathbf{n\times }%
\left[ \mathbf{n}\times \mathbf{v}\left( t\right) \right] \right| ^{2}
\label{2}
\end{equation}

\smallskip\ 

\textbf{d. }The vector $\mathbf{n}=\mathbf{R/}R$ is conveniently reckoned
from the center of the orbit and can be regarded as constant. Then in (\ref
{2}) $\mathbf{n\times }\left( \mathbf{n}\times \mathbf{v}\right) $ can be
replaced by $\mathbf{n}\times \mathbf{v}$ (see problem 14.6 of Jackson). The
polarization, however, is given by $\mathbf{v}_{\perp }=\mathbf{v}-\mathbf{n}%
\left( \mathbf{n}\cdot \mathbf{v}\right) =-\,\mathbf{n\times }\left( \mathbf{%
n}\times \mathbf{v}\right) $ and the power emitted with polarization $%
\mathbf{\epsilon }$ is obtained by using $\mathbf{\epsilon }^{*}\cdot 
\mathbf{v}$ instead of $\mathbf{v}_{\perp }$ in the integrand of (\ref{2}).
Recall that $\mathbf{\epsilon }$ is a unit vector perpendicular to $\mathbf{n%
}.$

\newpage

\subsection{14.6.2}

\quad \thinspace \textbf{a. }For a \textsl{linear harmonic oscillator}, we
take $\mathbf{x}(t)=(0,\,0,\,a\cos \omega _{0}t),$ where $a$ is the
amplitude, and we find $\mathbf{v}_{\perp }=-\omega _{0}a\sin \omega _{0}t$%
sin$\vartheta \,\mathbf{\epsilon },$where $\mathbf{\epsilon }=(-\cos \theta
\cos \phi ,\,-\cos \theta \sin \phi ,\,\sin \theta )$ is the polarization
vector. Since $\mathbf{\epsilon }$ does not vary in time, all we need to
compute is 
\[
I_{m}=\int_{0}^{2\pi /\omega _{0}}\exp \left( im\omega _{0}t-iz\cos \omega
_{0}t\right) \left( -\omega _{0}a\sin \omega _{0}t\sin \vartheta \right) dt
\]
with $z=m(\omega _{0}a/c)\cos \vartheta .$ Using the integral representation
of the Bessel function 
\[
J_{m}\left( z\right) =\frac{1}{2\pi i^{m}}\int_{0}^{2\pi }e^{iz\cos \alpha
\,-im\alpha }d\alpha 
\]
we obtain 
\[
I_{m}=-\frac{a\sin \vartheta }{2i}\,\left[ 2\pi i^{m-1}J_{m-1}\left(
z\right) -2\pi i^{m+1}J_{m+1}\left( z\right) \right] 
\]
This can be simplified by using the recursion relation (see Jackson, (3.87))
\[
J_{m-1}\left( z\right) +J_{m+1}\left( z\right) =\frac{2m}{z}J_{m}\left(
z\right) 
\]
and eq. (\ref{2}) gives the result of Jackson's Problem 14.7:
\begin{equation}
\frac{dP_{m}}{d\Omega }=\frac{e^{2}m^{2}\omega _{0}^{4}}{\left( 2\pi
c\right) ^{3}}\,\left| I_{m}\right| ^{2}=\frac{e^{2}m^{2}\omega _{0}^{2}}{%
2\pi c}\tan ^{2}\vartheta \,\,J_{m}^{2}\left( m\frac{\omega _{0}a}{c}\cos
\vartheta \right) \,.  \label{3}
\end{equation}

\smallskip\ 

\textbf{b}. For \textsl{small} $\omega _0a/c$ this gives the dipole
approximation result 
\begin{equation}
\frac{dP_1}{d\Omega }=\frac{e^2\omega _0^4a^2}{2\pi c^3}\sin ^2\vartheta
\label{4}
\end{equation}
in agreement with (1) of 14.2. We note that the multipolar expansion in
powers of $ka=\omega _0a/c$ is equivalent to an expansion in powers of $v/c;$
thus the classical result corresponds to the lowest multipole.

\smallskip\ 

\textbf{c. }For \textsl{large }$\omega _{0}a$ (extreme case $\omega _{0}a=c)$
many values of $m$ will contribute and most of the radiated power is at
large $m$ values. In fact, $J_{m}\left( m\dfrac{\omega _{0}a}{c}\cos
\vartheta \right) $ peaks for $m\dfrac{\omega _{0}a}{c}\cos \vartheta
\approx m.$ 

\newpage

\subsection{14.6.3}

\quad \thinspace \textbf{a. }For a charge in \textsl{uniform circular motion}
in the $xy$ plane, it is sufficient to compute the power radiated in the $xz$
plane $\left( \varphi =0\right) ,$ due to cylindrical symmetry. To effect
the calculation, transport $\mathbf{v}=\left( -\omega _{0}a\sin \omega
_{0}t,\,\omega _{0}a\cos \omega _{0}t,\,0\right) $ to the point of
observation and take spherical polar components (see sketch). Then $%
v_{\vartheta }=v_{x}\cos \vartheta $ and $v_{\varphi }=v_{y}$ give the
components of $\mathbf{v}_{\perp }$ . Using also $\mathbf{n\cdot x}=\omega
_{0}a\cos \omega _{0}t\sin \vartheta ,$ one sees that $v_{\vartheta }$ gives
the same expression as $\mathbf{v}_{\perp }$ for the harmonic oscillator,
but with $\cos \vartheta \rightarrow \sin \vartheta .$ Similarly, the
integral of $v_{\varphi }$ leads to the combination of Bessel functions 
\[
J_{m-1}\left( z\right) -J_{m+1}\left( z\right) =2\frac{dJ_{m}\left( z\right) 
}{dz},
\]
$\quad \quad $with $z=m\dfrac{\omega _{0}a}{c}\sin \vartheta \,.$ The
complete result is then (compare Jackson, Problem 14.8) 
\begin{equation}
\frac{dP_{m}}{d\Omega }=\frac{e^{2}\omega _{0}^{2}}{2\pi c}m^{2}\left[
\left( \dfrac{\omega _{0}a}{c}\frac{dJ_{m}\left( z\right) }{dz}\right)
^{2}+\cot ^{2}\vartheta \,\left( J_{m}\left( z\right) \right) ^{2}\right] \,.
\label{5}
\end{equation}

\smallskip\ 

\textbf{b. }For small $\dfrac{\omega _0a}c$ this gives the dipole
approximation result 
\begin{equation}
\frac{dP_1}{d\Omega }=\frac{e^2\omega _0^4a^2}{8\pi c^3}\left( 1+\cos
^2\vartheta \right) \,.  \label{6}
\end{equation}

\smallskip\ 

\textbf{c. }For $\omega _0 a$ approaching $c$, we can use asymptotic formulas
for $J_m\left( z\right) $ and we can also regard the index $m$ as a
continuous variable, or introduce the frequency variable $\omega =m\omega
_0. $ Physically, the radiation comes in bursts of duration $2\pi /\omega
_c\ll 2\pi /\omega _0,$ where $\omega _c$ is a ``cutoff frequency'' (This
means that the power spectrum of each burst involves frequencies up to $%
\omega _c). $

\newpage

\subsection{14.6.4}

\quad \thinspace \textbf{a. }There is an asymptotic formula for $J_{m}\left(
z\right) $ which is valid for $m\rightarrow \infty ,$ $z\rightarrow \infty ,$
with $z/m=\left( \omega _{0}a/c\right) \sin \vartheta $, a constant. It
involves Airy functions, or modified Bessel functions of order $1/3:$%
\[
\pi Ai\left( x\right) =\sqrt{\frac{x}{3}}K_{1/3}\left( \frac{2}{3}%
x^{3/2}\right) 
\]
\[
\pi Ai\left( x\right) =\frac{1}{2}\int_{-\infty }^{\infty }e^{i\left( \frac{%
\tau ^{3}}{3}+x\tau \right) }d\tau ,\quad \text{or\quad \quad }\frac{1}{%
\sqrt{3}}K_{1/3}\left( \xi \right) =\frac{1}{2}\int_{-\infty }^{\infty
}e^{i\left[ \frac{3}{2}\xi \left( t+\frac{1}{3}t^{3}\right) \right] }dt
\]
(the equivalence of these two standard definitions can be shown by the
variable change $\tau ^{3}/3=\xi t^{3}/2$ and a little algebra). The
asymptotic formula we need is 
\begin{equation}
J_{m}\left( z\right) \sim \frac{1}{\pi }\left( 1-\left( \frac{z}{m}\right)
^{2}\right) ^{-1/4}\xi ^{1/2}K_{1/3}\left( m\xi ^{\prime }\right)   \label{7}
\end{equation}
with 
\[
\xi ^{\prime }=\cosh ^{-1}\left( \frac{m}{z}\right) -\sqrt{1-\left( \frac{m}{%
z}\right) ^{2}}\,.
\]
It is obtained by applying the method of steepest descent (or stationary
phase) to the integral representation of $J_{m}\left( z\right) $ given in
14.6.2a: $i\xi $ is the point of stationary phase (note that $z/m\le 1).$

\smallskip\ 

\textbf{b. }Formula (\ref{7}) is the so called uniform asymptotic
approximation: it is valid for all values of $z/m=\left( \omega
_{0}a/c\right) \sin \vartheta $, as $m\rightarrow \infty .$ In the
relativistic limit $\omega _{0}a\rightarrow c,$ we know already that
radiation is emitted mostly in the plane of the orbit, $\vartheta =\pi /2.$
Putting $\left( z/m\right) ^{2}=1-\varepsilon $, with $\varepsilon \ll 1,$
we find $\xi ^{\prime }\approx \varepsilon ^{3/2}/3.$ It is then possible
for $m\xi $ to be small, even though $m$ is large enough that (\ref{7})
holds to a good approximation.

\smallskip\ 

\textbf{c. }With $\varepsilon =1-\left( z/m\right) ^{2}=1-\left( \omega
_{0}a/c\right) ^{2}\sin ^{2}\vartheta =1-(1-\gamma ^{-2})\left( 1-\sin
^{2}\Theta \right) \approx \gamma ^{-2}+\Theta ^{2},$ we have 
\[
m\xi ^{\prime }=\frac{m\omega _{0}}{\omega _{c}}\left( 1+\gamma ^{2}\Theta
^{2}\right) ^{-3/2}
\]
$\quad \quad $with $\omega _{c}=3\omega _{0}\gamma ^{3}.$Since $K_{1/3}$
decays exponentially, $\omega _{c}$ is the cutoff-frequency of the spectrum.
Note that $\gamma $ is the Lorentz factor pertaining to velocity $\omega
_{0}a/c$ and $\Theta $ the angle from the plane of the orbit.

\end{document}
