Momentum-transfer cross section

The factor of ${\displaystyle 1-\cos \theta }$ arises as follows. Let the incoming particle be traveling along the ${\displaystyle z}$-axis with vector momentum

$${\displaystyle {\vec {p}}_{\mathrm {in} }=q{\hat {z}}.}$$

Suppose the particle scatters off the target with polar angle ${\displaystyle \theta }$ and azimuthal angle ${\displaystyle \phi }$ plane. Its new momentum is

$${\displaystyle {\vec {p}}_{\mathrm {out} }=q'\cos \theta {\hat {z}}+q'\sin \theta \cos \phi {\hat {x}}+q'\sin \theta \sin \phi {\hat {y}}.}$$

For collision to much heavier target than striking particle (ex: electron incident on the atom or ion), ${\displaystyle q'\backsimeq q}$ so

$${\displaystyle {\vec {p}}_{\mathrm {out} }\simeq q\cos \theta {\hat {z}}+q\sin \theta \cos \phi {\hat {x}}+q\sin \theta \sin \phi {\hat {y}}}$$

By conservation of momentum, the target has acquired momentum

$${\displaystyle \Delta {\vec {p}}={\vec {p}}_{\mathrm {in} }-{\vec {p}}_{\mathrm {out} }=q(1-\cos \theta ){\hat {z}}-q\sin \theta \cos \phi {\hat {x}}-q\sin \theta \sin \phi {\hat {y}}.}$$

Now, if many particles scatter off the target, and the target is assumed to have azimuthal symmetry, then the radial (${\displaystyle x}$ and ${\displaystyle y}$) components of the transferred momentum will average to zero. The average momentum transfer will be just ${\displaystyle q(1-\cos \theta ){\hat {z}}}$. If we do the full averaging over all possible scattering events, we get

$${\displaystyle {\begin{aligned}\Delta {\vec {p}}_{\mathrm {avg} }&=\langle \Delta {\vec {p}}\rangle _{\Omega }\\&=\sigma _{\mathrm {tot} }^{-1}\int \Delta {\vec {p}}(\theta ,\phi ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\,\mathrm {d} \Omega \\&=\sigma _{\mathrm {tot} }^{-1}\int \left[q(1-\cos \theta ){\hat {z}}-q\sin \theta \cos \phi {\hat {x}}-q\sin \theta \sin \phi {\hat {y}}\right]{\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\,\mathrm {d} \Omega \\&=q{\hat {z}}\sigma _{\mathrm {tot} }^{-1}\int (1-\cos \theta ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\,\mathrm {d} \Omega \\[1ex]&=q{\hat {z}}\sigma _{\mathrm {tr} }/\sigma _{\mathrm {tot} }\end{aligned}}}$$

where the total cross section is

$${\displaystyle \sigma _{\mathrm {tot} }=\int {\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\mathrm {d} \Omega .}$$

Here, the averaging is done by using expected value calculation (see ${\displaystyle {\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )/\sigma _{\mathrm {tot} }}$ as a probability density function). Therefore, for a given total cross section, one does not need to compute new integrals for every possible momentum in order to determine the average momentum transferred to a target. One just needs to compute ${\displaystyle \sigma _{\mathrm {tr} }}$.

This concept is used in calculating charge radius of nuclei such as proton and deuteron by electron scattering experiments.

To this purpose a useful quantity called the scattering vector q having the dimension of inverse length is defined as a function of energy E and scattering angle θ:

$${\displaystyle q={\frac {{\frac {2E}{\hbar c}}\sin(\theta /2)}{\left[1+{\frac {2E}{Mc^{2}}}\sin ^{2}(\theta /2)\right]^{1/2}}}}$$