Relativistic beaming transforms the dipole pattern of Larmor radiation in the electron frame (dotted curve) to a narrow searchlight beam in the observer's frame.
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 $\phi'$ 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
$$ \cos\phi' = {v_{\rm
x}' \over c} \qquad \sin\phi' = {v_{\rm y}' \over c}$$
In the observer's frame
we have for the same photons,
$$ \cos\phi = {v_{\rm x} \over c} \qquad
\sin\phi = {v_{\rm y} \over c}$$ Using our equations for velocities, we
get
$$\cos\phi
= {v_{\rm x}' + v \over 1 + \beta v_{\rm x}' / c } = {c \cos \phi' + v
\over 1 + \beta c \cos \phi' / c } = {c \cos \phi' + v \over 1 + \beta
\cos \phi'}$$
$$\cos \phi = { \cos \phi' + \beta
\over 1 + \beta \cos \phi'}$$
$$\sin\phi = {v_{\rm y}' \over c
\gamma ( 1 + \beta v_{\rm x}'/ c ) }$$
$$\sin\phi = { \sin \phi' \over
\gamma( 1 + \beta \cos \phi')}$$
In the frame moving with the
electron, the Larmor equation implies a power pattern proportional to
$\cos^2 \phi'$ with nulls at $\phi' = \pm \pi/2$. In
the observer's frame, these nulls appear at much smaller angles
$$\sin\phi = {\sin \phi' \over \gamma (1 + \beta \cos
\phi')} \approx { 1 \over \gamma} \approx \phi$$
since $1/\gamma \ll 1$. We see the
radiation confined to a very narrow beam of width $2/\gamma$ between
nulls:
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 is nearly keeping up with the photons
that it is emitting.
$$\Delta t = t{\rm
(end~of~observed~pulse)} -
t{\rm (start~of~observed~pulse)}$$
$$\Delta t = {\Delta x \over v} + {( x -
\Delta x) \over c} - {x \over c}$$ The first term in this equation
represents the time taken by the electron to cover the
distance $\Delta x$, the second is the light travel time from the
electron position at the end
of the pulse to the observer, and the
third is the light travel
time from the electron position at the beginning of the pulse to the
observer. Note that the duration
$$ \Delta t = {\Delta x \over v} -
{\Delta x \over c} = {\Delta x \over v} \biggl(1 - {v \over c} \biggr)
\ll {\Delta x \over v}$$
of the observed pulse is much less than the time it takes the electron
to
move a distance $\Delta x$ because, in the observer's frame, the
electron nearly keeps up with its radiation. In the limit $v
\rightarrow c$,
$$\biggl( 1
- {v \over c}
\biggr) = \biggl( 1 - {v \over c} \biggr) { 1 + v/c \over 1 + v/c} = {1
- v^2/c^2
\over 1 + v/
c} \approx { \gamma^{-2} \over 2} = { 1 \over 2 \gamma^2 }$$
so
$$
\Delta t = { \Delta x \over v} { 1 \over 2 \gamma^2} ={ \Delta \phi
\over \omega_{\rm B} } { 1 \over 2 \gamma^2}~.$$
Recall that $\Delta
\phi \approx 2/\gamma$ so
$$\Delta t = { 2 \over \gamma \omega_{\rm B} 2
\gamma^2} = {1 \over \gamma^3 \omega_{\rm B} } = {1 \over \gamma^2
\omega_{\rm G}} $$ is the full observed duration of a pulse.
Example: Just how short is the observed duration of one synchrotron
pulse? In the typical interstellar
magnetic field of our Galaxy, $B \approx 5 \times 10^{-6}$ G so
$\omega_{\rm G} \approx 2 \pi \times
14$ rad s$^{-1}$. For an electron having $\gamma \approx 10^4$,
$$\Delta t \approx {1 \over \gamma^2 \omega_{\rm G}} \approx {1 \over
(10^4)^2 \times 2 \pi \times 14}
\approx 10
^{-10}{\rm ~s}$$
Allowing for the motion of the
electron parallel to the magnetic field, we replace the total magnetic
field by its perpendicular component $B \sin \alpha$, yielding
$$ \Delta t = {1 \over \gamma^2 \omega_{\rm G} \sin \alpha}~,$$
where $\alpha$ is the pitch angle of the electron.
Thus a plot of the power received as
a function of time is very spiky:
Synchrotron
radiation is a very spiky series of widely spaced narrow pulses.
The numerical values indicated on the plot of power versus time
correspond to an electron
with $\gamma \approx 10^4$ in a magnetic field $B \approx 5 \mu$G
typical of the interstellar medium in our Galaxy.
If $\gamma \approx 10^4$, the duration of each pulse is $\Delta t \sim 10^{- 10}$ s and if $B \approx 5\mu$G, the spacing between pulses is $\gamma / \nu_{\rm G} \sim 10^3$ s.
The observed synchrotron power
spectrum is the Fourier transform of this time series of pulses.
Instead of calculating the Fourier transform directly, we note first
that the pulse train is the
convolution
of the profile p(t) of an individual pulse with the Shah
function
(see Appendix B of
Rohlfs & Wilson or Bracewell's valuable reference book The Fourier Transform and Its Applications):
$$ I I I (t / \Delta t) \equiv \sum_{n = -\infty}^\infty \delta [(t /
\Delta t)] - n) $$
where each delta
function $\delta$ is an infinitesimally narrow spike at
integer $t / \Delta t=
n$ whose
integral is unity. Then we use the convolution
theorem to show that
that the Fourier
transform of the pulses is the product of the Fourier transform of one
pulse times the Fourier
transform of the shah function.
From appendix B of Rohlfs &
Wilson, the Fourier transform
of the shah function is also the shah function, so by the
similarity theorem, the Fourier
transform of
$$III \biggl( {t \nu_{\rm G} \over \gamma} \biggr)$$
is
proportional to
$$III \biggl( { \nu \gamma \over \nu_{\rm G} }
\biggr)~,$$
a nearly continuous series of spikes in the frequency domain separated
in frequency
by only
$$ \Delta\nu = { \nu_{\rm G} \over \gamma} \sim 10
^{-3}{\rm ~Hz}~.$$
Although this is not formally a continuous spectrum, the distortions
caused by even tiny fluctuations in electron energy, magnetic field
strength, or pitch angle cause frequency shifts larger than $\Delta
\nu$, so the spectrum of synchrotron radiation is effectively
continuous.
Thus the synchrotron spectrum of a
single electron is fairly flat at
low frequencies and tapers off at
$$\nu_{\rm max} \approx { 1 \over 2 \Delta t} \approx
\pi \gamma^2 \nu_{\rm G} \sin \alpha~.$$
It isn't really necessary to calculate
the Fourier transform of the pulse shape precisely because
astrophysical sources don't contain electrons with
just one energy and one pitch angle in a uniform magnetic field. The
actual energy distribution of cosmic rays in our Galaxy is a very broad
power law, and
this smears out the details of the spectrum from each energy range.
Just for the record, the synchrotron power spectrum of a single
electron is
$$\bbox[border:3px blue solid,7pt]{P(\nu) = {\sqrt{3} e^3 B \sin \alpha
\over m c^2} \biggl( { \nu \over \nu_{
\rm c}} \biggr) \int_{\nu/\nu_{\rm c}}^\infty K_{5/3} (\eta) d
\eta}\rlap{\quad \rm {(5C1)}}$$
where $K_{
5/3}$ is a modified Bessel function and $\nu_{\rm c}$ is the
critical frequency whose value is
$$\bbox[border:3px blue solid,7pt]{\nu_{\rm c} = {3 \over 2} \gamma^2
\nu_{\rm G}
\sin \alpha}\rlap{\quad \rm {(5C2)}}$$
(For the full mathematical derivation of this result, see Pacholczyk's Radio Astrophysics, a useful
reference book for those who must get into the nitty gritty of
radiation processes.)
The synchrotron spectrum of a
single electron plotted in terms of
$$F(x) \equiv x \int_x^\infty K_{5/3} (\eta) d
\eta$$
where $x \equiv \nu / \nu_{\rm c}$.
The spectrum of a single electron has
a logarithmic slope
$$d \log P(\nu) / d \log \nu \approx {1 \over 3}$$
at low frequencies, a broad peak near the critical frequency $\nu_{\rm
c}$, and falls of sharply at higher frequencies.
One way to look at $\nu_{\rm c}$ is
$$\nu_{\rm c} = \biggl( { 3 \over
2} \sin \alpha \biggr) {E^2 \over m c^2} {e B \over 2 \pi m_{\rm e} c}
\propto E^2
B_\bot~.$$
That is, the frequency at which an electron emits most strongly is
proportional
to
the square of the electron energy multiplied by the strength of the
perpendicular component of the magnetic field.