The Lorentz transforms relating the event coordinates $(x,y,z,t)$ in the unprimed frame and the coordinates $(x',y',z',t')$ in the primed frame moving with velocity $v$ in the $x$ direction are: $$\bbox[border:3px blue solid,7pt]{x = \gamma (x' + vt') \qquad y = y' \qquad z = z' \qquad t = \gamma (t' + \beta x' / c)}\rlap{\quad \rm {(5B1)}}$$ $$ \bbox[border:3px blue solid,7pt]{x' = \gamma (x - vt) \qquad y' = y \qquad z' = z \qquad t' = \gamma (t - \beta x / c)}\rlap{\quad \rm {(5B2)}}$$ where $$\bbox[border:3px blue solid,7pt]{\beta \equiv v/c}\rlap{\quad \rm {(5B3)}}$$ and $$\bbox[border:3px blue solid,7pt]{\gamma \equiv (1 - \beta^2)^{-1/2}}\rlap{\quad \rm {(5B4)}}$$ is called the Lorentz factor.

If $(\Delta x, \Delta y, \Delta z,
\Delta t)$ and $(\Delta x',
\Delta y',
\Delta z', \Delta t')$ are the coordinate differences between two
events, the differential form of the (linear) Lorentz transforms is:

$$\bbox[border:3px blue
solid,7pt]{\Delta x = \gamma(\Delta x' + v
\Delta t') \qquad \Delta y =
\Delta y' \qquad \Delta z = \Delta z' \qquad \Delta t = \gamma (\Delta
t' + \beta \Delta x' / c) }\rlap{\quad \rm {(5B5)}}$$
$$\bbox[border:3px blue solid,7pt]{\Delta x' = \gamma(\Delta x - v
\Delta
t) \qquad \Delta y' = \Delta y \qquad \Delta z' = \Delta z \qquad
\Delta t' = \gamma (\Delta t - \beta \Delta x / c) }\rlap{\quad \rm
{(5B6)}}$$
Here
is a derivation of these results that you should review before
proceeding.

Using the famous equation $E = m c^2$
we can calculate the energy equivalent to the rest mass $m_{\rm e}$ of
an electron:

$$E = m_{\rm e} c^2 = 9.1 \times
10^{-28} {\rm ~g} \times (3 \times
10^{10} {\rm ~cm~s}^{-1})^2 = 8.2 \times 10^{-7} {\rm ~erg}$$ $$E =
{8.2 \times 10^{-7} {\rm ~erg} \over 1.60 \times 10^{-12} {\rm
~erg~(eV)}^{-1} } = 5.1 \times 10^5 {\rm ~eV} = 0.51 {\rm ~MeV}$$

Cosmic-ray electrons with energies in
the range $10^9$ to $10^{14}$ eV have

$$\gamma \approx {10^9 {\rm ~to~} 10^{14} \over 0.51 \times 10^6}
\approx 10^3
{\rm ~to~} 10^8 \gg 1~$$ and such cosmic-ray electrons are called**
**ultrarelativistic**.**
These electrons
still move on spiral paths along
magnetic field lines, but the angular frequencies of their orbits are
lower because the inertial masses of the electrons are higher by a
factor of $\gamma$:

$$\omega_{\rm B} = {e B
\over \gamma m_{\rm e} c} = { \omega_{\rm G} \over \gamma}$$

Example: A cosmic-ray electron with $\gamma = 10^5$ in the Galactic magnetic field $B \approx 5 \times 10^{-6}$ G will have an orbital frequency $$\nu_{\rm B} \equiv { \omega_{\rm B} \over 2 \pi} \approx 14 \times 10^{-5} {\rm ~Hz} \approx 1 {\rm ~cycle~in~two~hours.}$$

Since
$v \approx c$ whenever $\gamma \gg 1$, the orbital radius $R$ of an
ultrarelativistic electron is quite
large:

$$R \approx { c \over \omega_{\rm B} } \approx { 3 \times
10^{10} {\rm ~cm~s}^{-1} \over 2 \pi \times 14 \times 10^{-5} {\rm ~Hz}
} \approx 3.4 \times 10^{13} {\rm ~cm} \approx 2 {\rm ~AU}$$

At first glance, these results are not
very promising for the
production of radio radiation: the high relativistic masses of
cosmic-ray electrons reduce their orbital frequencies and accelerations
to extremely low values. However, the Larmor radiation formula is only
valid at low velocities; that is, in inertial frames in which the
electron is nearly at rest. In the observer's frame, two relativistic
effects account for
the strong radio radiation: (1) the total power is multiplied by
$\gamma^2$ and (2) beaming turns the slow sinusoidal radiation into a
series of sharp pulses containing power at much higher frequencies
$\sim \gamma^3 \nu_{\rm B} = \gamma^2 \nu_{\rm

G}$. We proceed to calculate these relativistic corrections.

Synchrotron Power From a Single Electron

Nonrelativistic equations such as
Larmor's equation describing the electromagnetic radiation from an
accelerated charge are correct only in inertial frames where the
electron velocity $v \ll
c$, but the results can be transformed to any other inertial frame by
the Lorentz transform. In this way, it is possible to calculate the
total power radiated by an ultrarelativistic electron in a magnetic
field parallel to the $x$-axis. We use primed coordinates to describe
an inertial frame in which the electron is (temporarily) nearly at
rest.
Then Larmor's equation correctly gives

$$ P' = {2 e^2 (a_\bot')^2 \over 3 c^3}~.$$

What is $a_\bot$, the magnetic
acceleration of the electron in the galaxy frame?

$$v
_{\rm y} \equiv {d y \over d t} = {d y \over d t'} { d t' \over d t} =
{d y' \over d t'} { d t' \over d t} = v'_{\rm y} {d t' \over d t}$$ The
differential form of the Lorentz transform yields

$$ { d t' \over d t} = {1 \over \gamma}~,$$ so

$$v_{\rm y} = {v'_{\rm y} \over \gamma}~.$$ This factor of $\gamma$ is
a consequence of relativistic time
dilation—clocks in moving
frames appear to run slow by a factor $\gamma$. Consequently,

$$a_{\rm y} \equiv {d v_{\rm y} \over d t} = {d v_{\rm y} \over d t'}
{d t' \over d t} = {1 \over
\gamma} {d v'_{
\rm y} \over d t'} {d t' \over d t} ={a'_{\rm y} \over \gamma^2}~.$$
Similarly, $a_{\rm z} = a'_{\rm z} / \gamma^2$ so

$$a_\bot = {a'_\bot \over \gamma^2}~.$$ Thus

$$P' = {2 e^2 (a'_\bot)^2 \over 3 c^3} = {2 e^2 a_\bot^2 \gamma^4 \over
3 c^3}$$

How do we transform $P'$ to $P$, the
power measured by an observer at rest in the Galaxy? The
following argument is from Rindler's * Essential Relativity*, p.
98. Imagine two identical electrons of rest mass $m_{\rm e}$, one at
rest in the
unprimed frame and the other at rest in the
primed frame. If one electron is slightly displaced from the other
along the
$y$-axis, they will interact as they pass each other and be accelerated
in the $\pm y$ direction. Observers
at rest in each frame see "their" electron move with some small $v_{\rm
y} \ll c$, but the "other" electron will appear to move in the
opposite $y$ direction by a factor $\gamma$ more slowly because of time
dilation—recall the result $v_{\rm y} = v_{\rm y'} / \gamma$
above.
Invoking momentum conservation, observers in each frame conclude that
the "other" electron
has inertial mass $m_{\rm e}
\gamma$ and hence its energy is greater by the same factor $\gamma$.
Thus

$$P \equiv { d E
\over d t} = {d E \over d t'} {d t' \over d t} = {d E \over d E'} {d E'
\over d t'} {d t' \over d t} = \gamma P' {1 \over \gamma} = P'~;$$ that
is, power is the same in all
frames. Consequently,

$$P = {2 e^2 a_\bot^2 \gamma^
4 \over 3 c^3} \qquad (a_\parallel = 0)$$

Recall that

$$\omega_{\rm B} = { e B
\over \gamma m c}$$ and, by force balance,

$$a_\bot \equiv {d v_\bot \over d t} = \omega_{\rm B} v_\bot$$ so

$$a_\bot = {e B v_\bot \over \gamma m c} = {e B v \sin\alpha \over
\gamma m c}~,$$ where the angle $\alpha$
between
$\vec{v}$ and $\vec{B}$ is called the pitch
angle. For a given pitch angle $\alpha$, the
time-averaged radiated power of a single electron is

$$P = {2 e^2 \over
3 c^3} \gamma^2 {e^2 B^2 \over m^2 c^2} v^2 \sin^2\alpha$$

The pitch angle $\alpha$ between the directions of the magnetic field $\vec B$ and the electron velocity $\vec v$.

We can express this power
in terms of the
Thomson cross
section of an electron, $\sigma_{\rm T}$. The Thomson
cross section is the classical scattering cross section for
electromagnetic radiation. If a plane wave of
electromagnetic radiation is incident on a charge at rest, the
electric field of that radiation will accelerate the charge, which
in turn will radiate power in other directions according to
Larmor's equation. This process is called scattering, not
absorption, because the total power in electromagnetic radiation
is unchanged: all of the power lost from the incident plane wave
is reradiated in other directions. In one of the problem
sets, you show that the geometric area that would intercept this
amount
of incident power is

$$\bbox[border:3px blue solid,7pt]{\sigma_{\rm T} \equiv {8 \pi \over
3} \biggl(
{ e^2 \over m_{\rm e} c^2 } \biggr)^2}\rlap{\quad \rm {(5B7)}}$$ Numerically,

$$\sigma_{\rm T} = {8 \pi
\over 3
} \biggl[ { (4.8 \times 10^{-10} {\rm ~statcoul})^2 \over 9.1 \times
10^{-28} {\rm ~g~} (3 \times 10^{10} {\rm ~cm~s}^{-1})^2 } \biggr]
\approx 6.65
\times 10^{
-25} {\rm ~cm}^2$$ The reason for using the Thomson cross section will
become clear when
we discuss inverse-Compton scattering of radiation by the same cosmic
rays that are producing synchrotron radiation.

Also, we can replace $B^2$ by the
magnetic
energy density**
**

$$\bbox[border:3px blue solid,7pt]{U_{\rm B} = {B^2 \over 8
\pi}}\rlap{\quad \rm {(5B8)}}$$ to get

$$P = \biggl[ { 8
\pi \over 3} \biggl( { e^2
\over m c^2 }\biggr)^2 \biggr] 2 \biggl( {B^2 \over 8 \pi} \biggr) c
\gamma^2 {v^2 \over c^2} \sin^2\alpha$$ $$\bbox[border:3px blue
solid,7pt]{P = 2 \sigma_{\rm T} \beta^2
\gamma^2 c\, U_{\rm B} \sin^2\alpha}\rlap{\quad \rm {(5B9)}}$$ where $\beta \equiv v / c$. The
radiated power depends only on
physical constants, the square of the electron energy
(via $\gamma^2$; $\beta^2 \approx 1$ for all $\gamma \gg 1$), the
magnetic energy density, and the pitch angle.

The relativistic electrons in radio
sources can have lifetimes of
thousands to millions
of years before losing their ultrarelativistic energies via synchrotron
radiation or other processes, so they are scattered repeatedly by
magnetic-field fluctuations and charged particles in their environment,
and the distribution of
their pitch angles $\alpha$ gradually becomes random. The average
synchrotron
power
$\langle P \rangle$ per electron in an ensemble of electrons with the
same Lorentz
factor $\gamma$ but random pitch angles is therefore

$$\langle P \rangle = 2
\sigma_{\rm T
} \beta^2 \gamma^2 c \, U_{\rm B} \langle \sin^2 \alpha \rangle~.$$
$$\langle \sin^2 \alpha \rangle \equiv \int \sin^2 \alpha d \Omega
\bigg/\int d \Omega = {1
\over 4 \pi} \int \sin^2 \alpha d \Omega$$ $$\langle \sin^2 \alpha
\rangle = {1 \over 4 \pi} \int_{\phi =
0}^{2\pi} \int_{\alpha = 0}^\pi \sin^2 \alpha \sin\alpha \, d\alpha \,
d\phi$$ $$\langle \sin^2 \alpha \rangle = {1
\over 4 \pi
} {2 \pi} {4 \over 3} = {2 \over 3}$$ $$\bbox[border:3px blue
solid,7pt]{\langle P \rangle = {4 \over 3}
\sigma_{\rm T} \beta^2 \gamma^2 c U_{\rm B}}\rlap{\quad \rm {(5B10)}}$$ This is the
average
synchrotron power emitted by a relativistic electron.
When $\gamma \gg 1$, $\beta \approx 1$ and the $\beta^2$ factor
may be ignored. Relativistic effects make the synchrotron power
a factor $\gamma^2$ larger than in the limit $v \ll c$, so for
electrons with $\gamma \sim 10^4$, the power radiated by each electron
is multiplied by $10^8$, a huge amount.