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

$\sigma _{\mathrm {tr} }=\int (1-\cos \theta ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\mathrm {d} \Omega$

$=\int \int (1-\cos \theta ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\sin \theta \mathrm {d} \theta \mathrm {d} \phi$.

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

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

By conservation of momentum, the target has acquired momentum

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

Now, if many particles scatter off the target, and the target is assumed 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

$\Delta {\vec {p}}_{\mathrm {avg} }=\langle \Delta {\vec {p}}\rangle _{\Omega }$.

$=\sigma _{\mathrm {tot} }^{-1}\int \Delta {\vec {p}}{\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\mathrm {d} \Omega$.

$=\sigma _{\mathrm {tot} }^{-1}\int \left[q(1-\cos \theta ){\hat {z}}-q\sin \theta \cos \phi {\hat {x}}-q\sin \theta \cos \phi {\hat {y}}\right]{\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\mathrm {d} \Omega$

$=q{\hat {z}}\sigma _{\mathrm {tot} }^{-1}\int (1-\cos \theta ){\frac {\mathrm {d} \sigma }{\mathrm {d} \Omega }}(\theta )\mathrm {d} \Omega$

$=q{\hat {z}}\sigma _{\mathrm {tr} }/\sigma _{\mathrm {tot} }$

where the total cross section is

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

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 $\sigma _{\mathrm {tr} }$.