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%TCIDATA{Created=Fri Dec 06 09:07:07 1996}
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\markright{{\sf Frequency Distribution of Emitted Radiation   }}
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\section{14.5\quad Frequency Distribution of Emitted Radiation.}

\subsection{14.5.1}

\quad \thinspace \textbf{a. }In this section the distribution of emitted
radiation as a function of angles and frequency is obtained, first by a
method that shows directly the connection with the treatment of previous
sections, then by a more direct way that establishes the equivalence to the
treatment of Ch. 9.

\smallskip\ 

\textbf{b}. It must be clearly understood that the frequency distribution $%
d^{2}I/d\Omega d\omega $ and the time distribution $dP\left( t\right)
/d\Omega $ are \textsl{not} the fourier transforms of each other. It is the
fields (i.e. the amplitudes) that are Fourier conjugates, not the
intensities. Explicitly, 
\begin{equation}
\mathbf{E(x,\,}t)=\int \frac{d\omega }{2\pi }e^{-i\omega t}\mathbf{E(x,\,}%
\omega )\,.  \label{1}
\end{equation}
\begin{equation}
\frac{dP\left( t\right) }{d\Omega }=\frac{c}{4\pi }\,R^{2}\mathbf{E}^{2}%
\mathbf{(x,\,}t)\quad \quad \quad \quad \frac{d^{2}I}{d\Omega d\omega }=2\,%
\frac{c}{4\pi }\,R^{2}\left| \mathbf{E(x,\,}\omega )\right| ^{2}\,.
\label{2}
\end{equation}

\smallskip\ 

\textbf{c. }The factor of $2$ in the definition of $d^2I/d\Omega d\omega $
is introduced so that only positive frequencies need to be considered. (For
formal manipulations it is better to define the intensity without this
factor and consider all frequencies). The energy radiated per unit solid
angle is the same in the time or in the frequency descriptions: 
\begin{equation}
\frac{dW}{d\Omega }=\int_{-\infty }^{+\infty }dt\frac{dP\left( t\right) }{%
d\Omega }=\int_{-\infty }^{+\infty }\frac{d\omega }{2\pi }\frac c{4\pi
}R^2\left| \mathbf{E(x,\,}t)\right| ^2=\int_0^{+\infty }\frac{d\omega }{2\pi 
}\frac{d^2I}{d\Omega d\omega }  \label{3}
\end{equation}
where in the last equality we use the fact that $\mathbf{E(x,\,-}\omega )=%
\mathbf{E}^{*}\mathbf{(x,\,}\omega ).$

\smallskip\ 

\textbf{d. }The definition given here for $d^2I/d\Omega d\omega $ still
differs from Jackson's by a factor $2\pi $ because we \textsl{always} use $%
\int d\omega /2\pi $ in fourier integrals.

\newpage

\subsection{14.5.2}

\quad \thinspace \textbf{a. }The formal parallelism between the radiation
formulae and those of linear response theory is not accidental. The Green
function $D\left( k\right) =D\left( \omega ,\,\mathbf{k}\right) $ can be
properly regarded as describing the linear response of the vacuum to the
driving term $\mathsf{J}\left( \mathsf{k}\right) $: the dissipated power is
given by $\mathbf{J\cdot E}$, where $\mathbf{E}$ is the driving term and $%
\mathbf{J}$ the response or vice versa. The new ingredient here is that the
relation between $\mathbf{E}$ and $\mathbf{J}$ is strongly non-local and
that, through the conservation theorems, the power can be expressed solely
in terms of the field $\mathbf{E}$ in the asymptotic region.

\smallskip\ 

\textbf{b}. To proceed, we insert in (\ref{2}) the inverse of (\ref{1}) 
\[
\mathbf{E(x,\,}\omega )=\int dt\,e^{i\omega t}\mathbf{E(x,\,}t)\, 
\]
with 
\[
\mathbf{E(x,\,}t)=\frac{\mathbf{n\times }\left[ (\mathbf{n-\beta )}\times 
\mathbf{\dot{\beta}}\right] }{\left( 1-\mathbf{\beta \cdot n}\right) ^{3}}%
\frac{e}{cR}\,, 
\]
according to 3b.

\smallskip\ 

\textbf{c. }We change variable to $t^{\prime },$ with $t=t^{\prime }+R\left(
t^{\prime }\right) /c,$ $dt/dt^{\prime }=\left( 1-\mathbf{\beta \cdot n}%
\right) .$ Further, in the asymptotic region, we can put 
\[
R\left( t^{\prime }\right) =\left| \mathbf{x-x}^{\prime }\left( t^{\prime
}\right) \right| \simeq \left| \mathbf{x}\right| -\mathbf{n\cdot 
\QTR{mathbf}{x}^{\prime }\left( t^{\prime }\right) } 
\]
so that 
\[
\mathbf{E(x,\,}\omega )=\frac{e}{c}\,\frac{e^{i\omega \left| \mathbf{x}%
\right| /c}}{\left| \mathbf{x}\right| }\int_{-\infty }^{+\infty }dt^{\prime
}e^{i\omega t^{\prime }-\mathbf{n\cdot \QTR{mathbf}{x}^{\prime }\left(
t^{\prime }\right) }/c)}\,\frac{\mathbf{n\times }\left[ \left( \mathbf{%
n-\beta }\right) \times \mathbf{\dot{\beta}}\right] }{\left( 1-\mathbf{\beta
\cdot n}\right) ^{2}}\,. 
\]
Note that this has the form of an amplitude times an outgoing spherical
wave, as could have been anticipated.

\smallskip\ 

\textbf{d}. Substituting in the definition (\ref{2}) we have finally 
\begin{equation}
\frac{d^{2}I}{d\Omega d\omega }=\frac{e^{2}}{2\pi c}\left| \int_{-\infty
}^{+\infty }dt^{\prime }e^{i\omega t^{\prime }-\mathbf{n\cdot \QTR{mathbf}{x}%
^{\prime }\left( t^{\prime }\right) }/c)}\,\frac{\mathbf{n\times }\left[
\left( \mathbf{n-\beta }\right) \times \mathbf{\dot{\beta}}\right] }{\left(
1-\mathbf{\beta \cdot n}\right) ^{2}}\right| ^{2}.  \label{4}
\end{equation}

\newpage

\subsection{14.5.3}

\quad \thinspace \textbf{a. }We now show that (\ref{4}) is equivalent to the
formula obtained from 
\[
\mathbf{A(x,\,}\omega )=\frac 1c\int d^3x^{\prime }\mathbf{J(x}^{\prime }%
\mathbf{,\,}\omega )\frac{e^{ikR}}R\,,\quad \quad k=\omega /c 
\]
by using $\mathbf{B}=\mathbf{\nabla \times A,}$ $\mathbf{E}=\left( i/\hbar
\right) \mathbf{\nabla \times B.}$ Since we are in the asymptotic region, we
can replace once again $R$ by $\left| \mathbf{x}\right| -\mathbf{n\cdot 
\QTR{mathbf}{x}^{\prime }}$ and $\mathbf{\nabla \times }e^{ikR}\simeq
ike^{ikR}\mathbf{n\times }$ (with $\mathbf{n}=\mathbf{\nabla }R=\mathbf{R/}%
R).$ We find 
\begin{equation}
\frac{d^2I}{d\Omega d\omega }=\frac{\omega ^2}{2\pi c^3}\left| \int \mathbf{%
n\times }\left[ \mathbf{n\times J(x}^{\prime }\mathbf{,\,}\omega )\right]
e^{-i\omega \mathbf{n\cdot \QTR{mathbf}{x}^{\prime }/}c}d^3x^{\prime
}\right| ^2.  \label{5}
\end{equation}

\smallskip\ 

\textbf{b. }Even after we insert 
\[
\mathbf{J(x}^{\prime }\mathbf{,\,}\omega )=e\int dt^{\prime }e^{i\omega
t^{\prime }}\mathbf{v}\left( t^{\prime }\right) \delta \left( \mathbf{x}%
^{\prime }\mathbf{-x}^{\prime }\left( t^{\prime }\right) \right) , 
\]
as appropriate for a single particle, it is not obvious that the resulting
expression 
\begin{equation}
\frac{d^2I}{d\Omega d\omega }=\frac{e^2\omega ^2}{2\pi c}\left|
\int_{-\infty }^{+\infty }dt^{\prime }e^{i\omega \left[ t^{\prime }-\mathbf{%
n\cdot \QTR{mathbf}{x}^{\prime }\left( t^{\prime }\right) /}c\right] }%
\mathbf{n\times }\left( \mathbf{n}\times \mathbf{\beta }\right) \right| ^2
\label{6}
\end{equation}
is equivalent to (\ref{4}). One must carry out an integration by parts,
using 
\begin{equation}
\frac{\mathbf{n\times }\left[ \left( \mathbf{n-\beta }\right) \times \mathbf{%
\dot{\beta}}\right] }{\left( 1-\mathbf{\beta \cdot n}\right) ^2}=\frac
d{dt^{\prime }}\left[ \frac{\mathbf{n\times }\left( \mathbf{n}\times \mathbf{%
\beta }\right) }{1-\mathbf{\beta \cdot n}}\right] \,.  \label{7}
\end{equation}
This last equation is easily proven if one notes that $d\mathbf{n/}%
dt^{\prime }$ can be neglected, being of order $1/R.$

\smallskip\ 

\textbf{c. }The above integration by parts offers no problem if the path $%
\mathbf{\QTR{mathbf}{x}^{\prime }\left( t^{\prime }\right) }$ is confined to
a finite region. Otherwise the part at the boundaries must be treated
carefully by ``switching off'' the charge of the particle in the distant
past and future. Thus in bremsstrahlung problems, for instance,
eq.\thinspace (\ref{4}), often with (\ref{7}) inserted, is more convenient,
since $\mathbf{\dot{\beta}}$ is usually non-vanishing over a finite time
interval only.

\end{document}
