Rutherford Scattering Cross-Section

Consider a species-$s$ particle, incident with relative velocity $u_{ss'}$ onto an ensemble of species-$s'$ particles with number density $n_{s'}$. If $p_s({\mit\Omega})\,d{\mit\Omega}$ is the probability per unit time of the particle being scattered into the range of solid angle ${\mit\Omega}$ to ${\mit\Omega}+d{\mit\Omega}$ then the differential scattering cross-section, $d\sigma/d{\mit\Omega}$, is defined via (Reif 1965)

$$p_s({\mit\Omega})\,d{\mit\Omega} = n_{s'}\,u_{ss'}\,\frac{d\sigma}{d{\mit\Omega}}\,d{\mit\Omega}.$$ (3.65)

Assuming that the scattering is azimuthally symmetric (i.e., symmetric in $\phi$), we can write $d{\mit\Omega}=2\pi\,\sin\chi\,d\chi$. Now, the probability per unit time of a collision having an impact parameter in the range $b$ to $b+db$ is

$$p_s(b)\,db = n_{s'}\,u_{ss'}\,2\pi\,b\,db.$$ (3.66)

Furthermore, we can write

$$p_s({\mit\Omega})\,\left\vert\frac{d{\mit\Omega}}{db}\right\vert = p_s(b),$$ (3.67)

provided that $\chi$ and $b$ are related according to the two-particle scattering law, Equation (3.64). (The absolute value of $d{\mit\Omega}/db$ is taken because $\chi$ is a monotonically decreasing function of $b$.) It follows that

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{2\pi\,b}{\vert d{\mit\Omega}/db\vert}.$$ (3.68)

Equation (3.64) yields

$$\frac{d{\mit\Omega}}{db} = 2\pi\,\sin\chi\,\frac{d\chi}{db} = -2\... ...4\pi\,\epsilon_0\,\mu_{ss'}\,u_{ss'}^{2}}{e_s\,e_{s'}}\right)2\,\sin^2(\chi/2).$$ (3.69)

Finally, Equations (3.64), (3.68), and (3.69) can be combined to give the so-called Rutherford scattering cross-section,

$$\frac{d\sigma}{d{\mit\Omega}} = \frac{1}{4}\left(\frac{e_s\,e_{s'}}{4\pi\,\epsilon_0\,\mu_{ss'}\,u_{ss'}^{2}}\right)^2 \frac{1}{\sin^4(\chi/2)}$$ (3.70)

(Rutherford 1911). It is immediately apparent, from the previous formula, that two-particle Coulomb collisions are dominated by small-angle (i.e., small $\chi$) scattering events.