Thomson scattering

Remarks

Let $r_q = \dfrac{q^2}{4\pi\varepsilon_0 m c^2}$, the classical radius of the particle. For a linearly polarized incident wave, the differential cross-section is $\dfrac{d\sigma}{d\Omega} = r_q^2 \sin^2 \chi$, where $\chi$ is the angle between the incident electric field and the direction of observation. For unpolarized light, it is $\dfrac{d\sigma}{d\Omega} = r_q^2 \dfrac{1 + \cos^2\theta}{2}$, where $\theta$ is the scattering angle. The total cross-section is the Thomson cross-section $\sigma_T = \dfrac{8\pi}{3} r_q^2$.

Example: for an electron, $r_e \approx 2.82 \times 10^{-15}$ m and $\sigma_T \approx 6.65 \times 10^{-29}$ m$^2$. Check: $\frac{8\pi}{3} (2.82 \times 10^{-15})^2 \approx 8.38 \times 2.82^2 \times 10^{-30} \approx 6.66 \times 10^{-29}$.

Non-example: X-rays or gamma rays with $\hbar \omega$ comparable to $m_e c^2$ on electrons. Clause 3 fails: the photon loses energy to the recoiling electron, and the process is Compton scattering, with the Klein–Nishina cross-section in place of $\sigma_T$.