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 ${\sigma }_{\mathrm{t}\mathrm{r}}$ is defined in terms of an (azimuthally symmetric and momentum independent) differential cross section $\frac{\mathrm{d}\sigma }{\mathrm{d}\mathrm{\Omega }}(\theta )$ by $$\begin{array}{rl}{\sigma }_{\mathrm{t}\mathrm{r}}& =\int (1-\cos \theta )\frac{\mathrm{d}\sigma }{\mathrm{d}\mathrm{\Omega }}(\theta )\mathrm{d}\mathrm{\Omega }\\ & =\iint (1-\cos \theta )\frac{\mathrm{d}\sigma }{\mathrm{d}\mathrm{\Omega }}(\theta )\sin \theta \mathrm{d}\theta \mathrm{d}\varphi .\end{array}$$ The momentum-transfer cross section can be written in terms of the phase shifts from a partial wave analysis as $${\sigma }_{\mathrm{t}\mathrm{r}}=\frac{4\pi }{k^2}\sum _{l=0}^{\mathrm{\infty }}(l+1){\sin }^2[{\delta }_{l+1}(k)-{\delta }_l(k)].$$

Explanation

The factor of $1-\cos \theta$ arises as follows. Let the incoming particle be traveling along the $z$-axis with vector momentum $${\overrightarrow{p}}_{\mathrm{i}\mathrm{n}}=q\hat{z}.$$ Suppose the particle scatters off the target with polar angle $\theta$ and azimuthal angle $\varphi$ plane. Its new momentum is $${\overrightarrow{p}}_{\mathrm{o}\mathrm{u}\mathrm{t}}=q^{\prime }\cos \theta \hat{z}+q^{\prime }\sin \theta \cos \varphi \hat{x}+q^{\prime }\sin \theta \sin \varphi \hat{y}.$$ For collision to much heavier target than striking particle (ex: electron incident on the atom or ion), $q^{\prime }\backsimeq q$ so $${\overrightarrow{p}}_{\mathrm{o}\mathrm{u}\mathrm{t}}\simeq q\cos \theta \hat{z}+q\sin \theta \cos \varphi \hat{x}+q\sin \theta \sin \varphi \hat{y}$$ By conservation of momentum, the target has acquired momentum $$\mathrm{\Delta }\overrightarrow{p}={\overrightarrow{p}}_{\mathrm{i}\mathrm{n}}-{\overrightarrow{p}}_{\mathrm{o}\mathrm{u}\mathrm{t}}=q(1-\cos \theta )\hat{z}-q\sin \theta \cos \varphi \hat{x}-q\sin \theta \sin \varphi \hat{y}.$$ 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 $$\begin{array}{rl}\mathrm{\Delta }{\overrightarrow{p}}_{\mathrm{a}\mathrm{v}\mathrm{g}}& =⟨\mathrm{\Delta }\overrightarrow{p}{⟩}_{\mathrm{\Omega }}\\ & ={\sigma }_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}\int \mathrm{\Delta }\overrightarrow{p}(\theta ,\varphi )\frac{\mathrm{d}\sigma }{\mathrm{d}\mathrm{\Omega }}(\theta )\mathrm{d}\mathrm{\Omega }\\ & ={\sigma }_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}\int [q(1-\cos \theta )\hat{z}-q\sin \theta \cos \varphi \hat{x}-q\sin \theta \sin \varphi \hat{y}]\frac{\mathrm{d}\sigma }{\mathrm{d}\mathrm{\Omega }}(\theta )\mathrm{d}\mathrm{\Omega }\\ & =q\hat{z}{\sigma }_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}\int (1-\cos \theta )\frac{\mathrm{d}\sigma }{\mathrm{d}\mathrm{\Omega }}(\theta )\mathrm{d}\mathrm{\Omega }\\ & =q\hat{z}{\sigma }_{\mathrm{t}\mathrm{r}}/{\sigma }_{\mathrm{t}\mathrm{o}\mathrm{t}}\end{array}$$ where the total cross section is $${\sigma }_{\mathrm{t}\mathrm{o}\mathrm{t}}=\int \frac{\mathrm{d}\sigma }{\mathrm{d}\mathrm{\Omega }}(\theta )\mathrm{d}\mathrm{\Omega }.$$ Here, the averaging is done by using expected value calculation (see $\frac{\mathrm{d}\sigma }{\mathrm{d}\mathrm{\Omega }}(\theta )/{\sigma }_{\mathrm{t}\mathrm{o}\mathrm{t}}$ 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 ${\sigma }_{\mathrm{t}\mathrm{r}}$.

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: $$q=\frac{\frac{2E}{\hslash c}\sin (\theta /2)}{{[1+\frac{2E}{M{c}^2}{\sin }^2(\theta /2)]}^{1/2}}$$