The Synchrotron Spectrum of a Single Electron

Our next problem is to explain how the synchrotron mechanism can yield radio radiation at frequencies much higher than $\omega_{\rm B} = \omega_{\rm G} / \gamma$. To solve it, we first calculate the angular distribution of the radiation in the observer's frame.

Relativistic aberration causes
the Larmor dipole pattern in the electron frame to become beamed
sharply in the direction of motion as $v$
approaches $c$. This beaming follows directly from the relativistic
velocity equations implied by the differential form of the Lorentz
transform. For an electron moving in the $x$ direction,

$$v_{\rm x} \equiv {d x \over d t} = {d x \over d
t'} {d t' \over d t}$$ $$v_{\rm x} = \gamma \biggl({ d x' \over d t'} +
v {d t' \over
d t'} \biggr) \biggl( { d t \over d t'} \biggr)^{-1}$$ $$v_{\rm x} =
\gamma (v_{\rm x}' + v)
\biggl[\gamma \biggl( 1 + {\beta \over c} {d x' \over d t'} \biggr)
\biggr]^{-1}$$ $$v_{\rm x} = ({v_{\rm x}' + v}) \biggl( 1 + {\beta
v_{\rm x}' \over c}
\biggr)^{-1} ~.$$

In the $y$ direction perpendicular to the electron velocity, the
velocity formula gives

$$v_{\rm y} \equiv {d y \over d t} =
{d y \over d t'} {d t' \over d t} = {d y' \over d t'} \biggl( {d t
\over d t'} \biggr)^{-1}$$ $$v_{\rm y} = {
v_{\rm y}' \over \gamma} \biggl( 1 + {\beta v_{\rm x}' \over c}
\biggr)^{-1}$$

Consider the synchrotron photons
emitted with speed $c$ at an angle $\theta'$ from the
$x'$ axis. Let
$v_{\rm x}'$ and $v_{\rm y}'$ be the projections of the photon speed
onto the $x'$ and $y'$ axes. Then

$$ \cos\theta' = {v_{\rm
x}' \over c} \qquad \sin\theta' = {v_{\rm y}' \over c}$$ In the
observer's frame
we have for the same photons,

$$ \cos\theta = {v_{\rm x} \over c} \qquad
\sin\theta = {v_{\rm y} \over c}$$ Using our equations for velocities,
we
get

$$\cos\theta
= \biggl({v_{\rm x}' + v \over 1 + \beta v_{\rm x}' / c }\biggr){1
\over c} =

\biggl({c \cos
\theta' +
v
\over 1 + \beta c \cos \theta' / c }\biggr){1 \over c} $$ $$\cos \theta
= { \cos \theta' + \beta
\over 1 + \beta \cos \theta'}$$ $$\sin\theta = {v_{\rm y}' \over c
\gamma ( 1 + \beta v_{\rm x}'/ c ) }$$ $$\sin\theta = { \sin \theta'
\over
\gamma( 1 + \beta \cos \theta')}$$ In the frame moving with the
electron, the Larmor equation implies a power pattern proportional to
$\cos^2 \theta'$ with nulls at $\theta' = \pm \pi/2$. In
the observer's frame, these nulls appear at much smaller angles

$$\sin\theta = {\sin \theta' \over \gamma (1 + \beta \cos
\theta')} \approx { 1 \over \gamma} \approx \theta$$ since $1/\gamma
\ll 1$. We see the
radiation confined to a very narrow beam of width $2/\gamma$ between
nulls, as shown below.

The solid curve is the observed power pattern for $\gamma = 5$. The observed angle between the nulls of the forward beam $\approx 2 / \gamma$, and the peak power gain $\approx 2 \gamma$.

For example, a 10 Gev electron has
$\gamma \approx 2 \times 10^4$ so $2 / \gamma \approx 10^{-4} {\rm
~rad} \approx 20$ arcsec! The observer sees a
short pulse of radiation emitted during only the tiny fraction

$$ {2 \over 2 \pi \gamma} =
{1 \over \pi \gamma
}$$ of the electron orbit when the electron is moving directly toward
the
observer.

The duration $\Delta t$ of the
observed
pulse is even shorter than the time the electron needs to cover $1 /
(\pi \gamma)$
of its orbit because the electron is visible only while it is moving
directly
toward the observer with a speed approaching $c$. In the
observer's frame, the electron nearly keeps up with the radiation
that it emits.