Cross-Section

If a pointlike mass interacts with another one, it is usually scattered under a certain polar angle $\theta$. It is therefore useful to define the so-called differential cross-section $D$ as the derivative of the cross-section $\sigma$ with respect to the full solid angle $\Omega$: $$\boxed{D(\theta) = \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}}$$ Due to symmetry reasons, $D$ does only depend on the polar angle $\theta$ and not $\varphi$. In order to calculate the total cross-section from a given $D$, an integration over the full solid angle has to be performed: $$\sigma = \int_{\varphi = 0}^{2\pi} \int_{\theta = 0}^{\pi}\,D(\theta)\,\mathrm{d}\Omega$$ We can now define an impact parameter $b$ which is the distance between the trajectory of a particle and the symmetry axis as shown in the following figure. The infinitesimal small cross-section is then given as $$\mathrm{d}\sigma = b\,\mathrm{d}b\,\mathrm{d}\varphi$$ The differential $\mathrm{d}\Omega$ can be written as $$\mathrm{d}\Omega = \sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi$$ Dividing $\mathrm{d}\sigma$ by $\mathrm{d}\Omega$ leads to the following important correlation between the differential cross-section and the impact parameter: $$\boxed{\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{b}{\sin\theta}\left|\frac{\mathrm{d}b}{\mathrm{d}\theta}\right|}$$ In an experiment, only the scatter angle $\theta$ can be measured and not the impact parameter $b$. Hence, it is beneficial to make sure that the right side does only dependent on $\theta$. We want to discuss this concept deeper by calculating the total cross-section of a solid sphere. The scattering can be parameterized as illustrated in the following figure. From this figure, we can write the impact parameter as $$b = R\sin\alpha$$ where $\alpha$ is given as the difference of $\pi/2$ and $\theta/2$. This leads to $$b = R\cos\frac{\theta}{2}$$ The derivation of this equation is given by $$\frac{\mathrm{d}b}{\mathrm{d}\theta} = -\frac{R}{2}\sin\frac{\theta}{2}$$ In the next step, we have to insert these two equations into our derived formula for the differential cross section: $$\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{\frac{R^2}{2}\cos\frac{\theta}{2}\sin\frac{\theta}{2}}{\sin\theta}$$ Using now the formula $$\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$$ results in $$\frac{\mathrm{d}\sigma}{\mathrm{d}\theta} = \frac{R^2}{4}$$ It is important to mention that this formula does not contain any angular dependencies which means that the number of scattered particles per solid angle $\mathrm{d}\Omega$ is constant. This does not mean, however, that the amount of particles is the same for every direction, because the solid angle is different for every direction. We now want to calculate the total cross-section $\sigma$ by integrating over the full solid angle: $$\sigma = \int_0^{2\pi} \int_0^{\pi}\,\frac{R^2}{4}\,\mathrm{d}\Omega = \pi R^2$$ The total cross-section is therefore equal to the surface of the circle that represents the cross-section of the sphere.