# Momentum-transfer cross section

In physics, and especially scattering theory, the momentum-transfer cross section (sometimes known as the momentum-transport cross section) is an effective scattering cross section useful for describing the average momentum transferred from a particle when it collides with a target. Essentially, it contains all the information about a scattering process necessary for calculating average momentum transfers but ignores other details about the scattering angle.
The momentum-transfer cross section $\sigma _{\mathrm {tr} }$ is defined in terms of an (azimuthally symmetric and momentum independent) differential cross section ${\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )$ by ${\begin{aligned}\sigma _{\mathrm {tr} }&=\int (1-\cos \theta ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\,\mathrm {d} \Omega \\&=\iint (1-\cos \theta ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \phi .\end{aligned}}$
The momentum-transfer cross section can be written in terms of the phase shifts from a partial wave analysis as $\sigma _{\mathrm {tr} }={\frac {4\pi }{k^{2}}}\sum _{l=0}^{\infty }(l+1)\sin ^{2}[\delta _{l+1}(k)-\delta _{l}(k)].$

## Explanation

The factor of $1-\cos \theta$ arises as follows. Let the incoming particle be traveling along the $z$-axis with vector momentum ${\vec {p}}_{\mathrm {in} }=q{\hat {z}}.$
Suppose the particle scatters off the target with polar angle $\theta$ and azimuthal angle $\phi$ plane. Its new momentum is ${\vec {p}}_{\mathrm {out} }=q'\cos \theta {\hat {z}}+q'\sin \theta \cos \phi {\hat {x}}+q'\sin \theta \sin \phi {\hat {y}}.$
For collision to much heavier target than striking particle (ex: electron incident on the atom or ion), $q'\backsimeq q$ so ${\vec {p}}_{\mathrm {out} }\simeq q\cos \theta {\hat {z}}+q\sin \theta \cos \phi {\hat {x}}+q\sin \theta \sin \phi {\hat {y}}$
By conservation of momentum, the target has acquired momentum $\Delta {\vec {p}}={\vec {p}}_{\mathrm {in} }-{\vec {p}}_{\mathrm {out} }=q(1-\cos \theta ){\hat {z}}-q\sin \theta \cos \phi {\hat {x}}-q\sin \theta \sin \phi {\hat {y}}.$
Now, if many particles scatter off the target, and the target is assumed to have azimuthal symmetry, then the radial ($x$ and $y$) components of the transferred momentum will average to zero. The average momentum transfer will be just $q(1-\cos \theta ){\hat {z}}$. If we do the full averaging over all possible scattering events, we get ${\begin{aligned}\Delta {\vec {p}}_{\mathrm {avg} }&=\langle \Delta {\vec {p}}\rangle _{\Omega }\\&=\sigma _{\mathrm {tot} }^{-1}\int \Delta {\vec {p}}(\theta ,\phi ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\,\mathrm {d} \Omega \\&=\sigma _{\mathrm {tot} }^{-1}\int \left[q(1-\cos \theta ){\hat {z}}-q\sin \theta \cos \phi {\hat {x}}-q\sin \theta \sin \phi {\hat {y}}\right]{\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\,\mathrm {d} \Omega \\&=q{\hat {z}}\sigma _{\mathrm {tot} }^{-1}\int (1-\cos \theta ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\,\mathrm {d} \Omega \\[1ex]&=q{\hat {z}}\sigma _{\mathrm {tr} }/\sigma _{\mathrm {tot} }\end{aligned}}$ where the total cross section is $\sigma _{\mathrm {tot} }=\int {\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\mathrm {d} \Omega .$
Here, the averaging is done by using expected value calculation (see ${\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )/\sigma _{\mathrm {tot} }$ 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 $\sigma _{\mathrm {tr} }$.

## Application

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