To derive the equations describing
inverse-Compton scattering, we
begin by considering non-relativistic Thomson scattering in the rest
frame of an electron. If the Poynting flux (power per unit area) of a
plane wave incident on the electron is

$$\vec{S} = {c \over 4 \pi}
\vec{E} \times \vec{H} = {c \over 4 \pi} \vert \vec{E} \vert^2~,$$ the
electric field of the incident radiation will accelerate the electron,
and the accelerated electron will in turn emit
radiation according to Larmor's equation. The net result is simply to
scatter a portion of the
incoming radiation with no net transfer of energy between the radiation
and the electron. The scattered radiation
has power

$$P = \vert \vec{S} \vert \sigma_{\rm T}~,$$ where

$$\sigma_{\rm T} \equiv {8 \pi \over 3} \biggl( { e^2 \over m_{\rm e}
c^2}
\biggr)^2 \approx 6.65 \times 10^{-25} {\rm ~cm}^2$$ is called the Thomson
cross section of an electron. In other words, the electron will
extract
from the incident radiation the amount of power flowing through the
area
$\sigma_{\rm
T}$ and reradiate that power over the doughnut-shaped pattern given by
Larmor's equation. The scattered power can be rewritten as

$$\bbox[border:3px blue solid,7pt]{P =
\sigma_{\rm T} c U_{\rm rad}}\rlap{\quad \rm {(5E1)}}$$ where $U_{\rm rad} = \vert \vec{S} \vert
/ c$ is the energy density of the incident radiation.

Next consider radiation scattering by
an ultrarelativistic electron.
The Thomson scattering formula above is valid only in the primed frame
instantaneously moving with the electron:

$$ P' = \sigma_{\rm T} c U'_{\rm rad}~.$$ We want to transform this
nonrelativistic result to the unprimed rest frame of an
observer. We already showed that $P = P'$ so

$$P = \sigma_{\rm T} c U'_{\rm
rad}~.$$ We only need to transform $U'_{\rm rad}$ into $U_{\rm rad}$.
Suppose that the electron is moving with speed $v = v_{\rm x}$ in the
rest frame of the observer and it is hit successively by two low-energy
photons approaching from an angle $\theta$ in the observer's frame
($\theta'$ in the
electron frame) from the $x$-axis as shown below.

If the coordinates corresponding to
the arrival of the first and
second photons at the electron (which is always located at $x' = y' =
z' = 0$)
are

$$(x_1, 0, 0, t_1) {\rm ~and~} (x_2, 0, 0, t_2) $$ in the observer's
frame, then the Lorentz
transform gives their coordinates as $$(\gamma v t'_1, 0, 0, \gamma
t'_1) {\rm ~and~} (\gamma v t'_2, 0, 0, \gamma t'_2)~,$$ also in the
observer's frame, as shown below.