🗒️ Note: we work with $c=1$ such that $E_\gamma = p_\gamma$
Before
$$ \begin{aligned} \tilde p_\gamma &= (p_\gamma, p_\gamma,0,0) \\ \tilde p _e &= (m,0,0,0) \end{aligned} $$
Because the electron is initially at rest and the collision is along the $x^1$ axis
🗒️ Note: $\gamma$ means photon here
After
$$ \begin{aligned} \tilde p'\gamma & = (p'\gamma, \vec p'_\gamma )\\ \tilde p_e'&= (E'_e, \vec p'_e)
\end{aligned} $$
⚽ Goal: work out $p_\gamma '$ as a function of $\theta$
$$ \begin{align*} \tilde{p}\gamma + \tilde{p}e &= \tilde p\gamma' + \tilde p_e' \\[10pt] (\tilde{p}e)^2 &= \left( \tilde{p}\gamma + p_e - p\gamma' \right)^2 =m^2\\[5pt] &= \tilde{p}\gamma^2 + \tilde p_e^2 + \tilde p\gamma'^2 + 2 \left( \tilde{p}\gamma \cdot \tilde p_e - \tilde{p}\gamma \cdot p_\gamma' - \tilde p_e \cdot \tilde p_\gamma' \right) \\[10pt] m^2 &= 0 + m^2 + 0 + 2 \left( p_\gamma m - \left[ p_\gamma p_\gamma' - \vec p_\gamma \cdot \vec p_\gamma' \right] - m p_\gamma' \right) \\[10pt] p_\gamma^m &= p_\gamma' \left[ p_\gamma (1 - \cos \theta) + m \right] \end{align*} $$
We get the Compton scattering formula
$$ \boxed{\frac{p_\gamma'}{p_\gamma}=\frac{m}{p_\gamma [1-\cos+m]}} $$
💼 Case:
the classical ie non-quantum mechanical and non-relativistic limit of Compton scattering
This condition is valid when $\hbar \omega \ll mc^2$ where $m$ is the rest mass of the electron
Incoming linearly polarized e.m. wave scattering elastically off of a free electron eg
$$ \vec E = \hat y E_0 \cos(\omega t-kx) $$
🧠 Remember: in radiation fields we found $|B|=|E|/c$ therefore
$$ \vec F=q(\vec E + \underbrace{\vec v \times \vec B}_{\vec B \propto \frac 1c \;\& \; v\ll c} )\sim q \vec E $$
Acceleration of electron
$$ \dot \beta c= \frac{eE}{m} $$
We are more interested in the time averages which give
$$ \braket {\dot \beta ^2} = \frac{e^2}{m^2 c^2 } \braket{\vec E ^2} = \frac{e^2}{m^2 c^2} \frac{E_0^2}{2} $$
Now we find the Larmor radiation
$$ \begin{aligned} \left <\frac{\text d P}{\text d \Omega} \right > & = \frac{\mu_0 c e^2}{16 \pi^2} \sin ^2 \theta \braket{\dot \beta ^2} \\ &= \frac{\mu_0}{c} \frac{e^4 E_0 ^2}{32\pi^2 m^2} \sin ^2 \theta
\end{aligned} $$
🗒️ Note: the expression is independent of $\omega$ of incoming radiation
Total scatter power
$$ \braket{P}=\frac{\mu_0}{c} \frac{e^4 E^2_0}{12 \pi m^2} $$