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 $$\begin{align} \sigma_{\mathrm{tr}} &= \int (1 - \cos \theta) \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} (\theta) \, \mathrm{d} \Omega \\ &= \iint  (1 - \cos \theta) \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} (\theta) \sin \theta \, \mathrm{d} \theta \, \mathrm{d} \phi. \end{align}$$

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 $$\begin{align} \Delta \vec{p}_\mathrm{avg} &= \langle \Delta \vec{p} \rangle_\Omega \\ &= \sigma_\mathrm{tot}^{-1} \int \Delta \vec{p}(\theta,\phi) \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} (\theta) \, \mathrm{d} \Omega \\ &= \sigma_\mathrm{tot}^{-1} \int \left[ q (1 - \cos \theta) \hat{z} - q \sin \theta \cos \phi\hat{x} - q \sin \theta \sin \phi\hat{y} \right ] \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} (\theta) \, \mathrm{d} \Omega \\ &= q \hat{z} \sigma_\mathrm{tot}^{-1} \int  (1 -  \cos \theta)  \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} (\theta) \, \mathrm{d} \Omega \\[1ex] &= q \hat{z} \sigma_\mathrm{tr} / \sigma_\mathrm{tot} \end{align}$$ 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 $θ$: $$q = \frac{\frac{2E}{\hbar c} \sin (\theta/2)}{\left[1+ \frac{2E}{Mc^2} \sin^2 (\theta/2)\right]^{1/2}}$$