= Compton recoil kinematics
{c}
{title2=$c\mathbf p_e=E\hat{\mathbf x}-E'\hat{\mathbf n}$}
Choose the incident <photon> direction as $+x$ and the plane of the <collision> as the $xy$ plane, with the outgoing <photon> above the $x$ axis. The <Electron> initially has zero <momentum>. <Conservation of energy> and <momentum conservation>, using the relativistic <energy–momentum relation>, give
$$
E+m_ec^2=E'+\sqrt{m_e^2c^4+c^2|\mathbf p_e|^2},
\qquad
c\mathbf p_e=(E-E'\cos\theta)\hat{\mathbf x}-E'\sin\theta\hat{\mathbf y}.
$$
Squaring the <Electron> <energy> and the <momentum> magnitude in these two equations yields
$$
(m_ec^2+E-E')^2
=m_e^2c^4+E^2+E'^2-2EE'\cos\theta.
$$
Cancel common terms to obtain $m_ec^2(E-E')=EE'(1-\cos\theta)$. Division by $EE'>0$ gives $1/E'-1/E=(1-\cos\theta)/(m_ec^2)$. Substitution of $\lambda=hc/E$ proves both <Compton scattering> formulas. Using the nonrelativistic approximation $T_e=|\mathbf p_e|^2/(2m_e)$ here would lose the exact relativistic result.
Let $\phi$ be the nonnegative <Electron> recoil angle below the incident direction. For $0<\theta<\pi$, the <vector components> above imply
$$
\tan\phi=\frac{E'\sin\theta}{E-E'\cos\theta}
=\frac{\cot(\theta/2)}{1+E/(m_ec^2)}.
$$
This assumes exactly the same initially stationary free <Electron> and single outgoing <photon>. At $\theta=\pi$, the <Electron> moves forward with $\phi=0$; at $\theta=0$ it stays at rest, so its recoil direction is undefined even though the limiting angle as $\theta\downarrow0$ is $\pi/2$. At $\theta=90^\circ$ in the chosen example, $c\mathbf p_e=(70,-61.566)\,\mathrm{keV}$ and $\phi\simeq41.33^\circ$.
\Image[/compton-photon-and-electron-momentum-vectors.png]
{title=<Photon> deflection and <Electron> recoil in <Compton scattering>}
{description=Original $90^\circ$ example. Left: directions in the plane of the <collision>, with a labeled incident <photon> source and an initially stationary <Electron>; ray lengths are schematic. Right: <momentum> vectors multiplied by $c$, in $\mathrm{keV}$, with equal horizontal and vertical scales. The translated outgoing <photon> <momentum> closes the triangle: $\mathbf p=\mathbf p_e+\mathbf p'$. No experimental trace is shown.}
Back to article page
