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 $\sigma_$ is defined in terms of an (azimuthally symmetric and momentum independent) differential cross section $\frac (\theta)$ by
$\begin \sigma_ &= \int (1 - \cos \theta) \frac (\theta) \, \mathrm \Omega \\ &= \iint (1 - \cos \theta) \frac (\theta) \sin \theta \, \mathrm \theta \, \mathrm \phi. \end$

The momentum-transfer cross section can be written in terms of the phase shifts from a partial wave analysis as
\sigma_ = \frac \sum_^\infty (l+1) \sin^2 delta_(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
$\vec_\mathrm = q \hat.$

Suppose the particle scatters off the target with polar angle $\theta$ and azimuthal angle $\phi$ plane. Its new momentum is
$\vec_\mathrm = q' \cos \theta \hat + q' \sin \theta \cos \phi\hat + q' \sin \theta \sin \phi\hat.$

For collision to much heavier target than striking particle (ex: electron incident on the atom or ion), $q'\backsimeq q$ so
$\vec_\mathrm \simeq q \cos \theta \hat + q \sin \theta \cos \phi\hat + q \sin \theta \sin \phi\hat$

By conservation of momentum, the target has acquired momentum
$\Delta \vec = \vec_\mathrm - \vec_\mathrm = q (1 - \cos \theta) \hat - q \sin \theta \cos \phi\hat - q \sin \theta \sin \phi\hat .$

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 $q (1 - \cos \theta) \hat$. If we do the full averaging over all possible scattering events, we get
\begin
\Delta \vec_\mathrm &= \langle \Delta \vec \rangle_\Omega \\
&= \sigma_\mathrm^ \int \Delta \vec(\theta,\phi) \frac (\theta) \, \mathrm \Omega \\
&= \sigma_\mathrm^ \int \left q (1 - \cos \theta) \hat - q \sin \theta \cos \phi\hat - q \sin \theta \sin \phi\hat \right \frac (\theta) \, \mathrm \Omega \\
&= q \hat \sigma_\mathrm^ \int (1 - \cos \theta) \frac (\theta) \, \mathrm \Omega \\ ex&= q \hat \sigma_\mathrm / \sigma_\mathrm
\end
where the total cross section is
$\sigma_\mathrm = \int \frac (\theta) \mathrm \Omega .$

Here, the averaging is done by using expected value calculation (see $\frac (\theta) / \sigma_\mathrm$ 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 $\sigma_\mathrm$.

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 having the dimension of inverse length is defined as a function of energy and scattering angle :
$q = \frac$

References

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Momentum
Scattering theory