Momentum transfer

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In particle physics, wave mechanics and optics, momentum transfer is the amount of momentum that one particle gives to another particle.

In the simplest example of scattering of two colliding particles with initial momenta $\vec{p}_{i1},\vec{p}_{i2}$ , resulting in final momenta $\vec{p}_{f1},\vec{p}_{f2}$ , the momentum transfer is given by

$\vec q = \vec{p}_{i1} - \vec{p}_{f1} = \vec{p}_{f2} - \vec{p}_{i2}$

where the last identity expresses momentum conservation. Momentum transfer is an important quantity because $\Delta x = \hbar / |q|$ is a better measure for the typical distance resolution of the reaction than the momenta themselves.

Wave mechanics and optics

A wave has a momentum $p = \hbar k$ and is a vectorial quantity. The difference of the momentum of the scattered wave to the incident wave is called momentum transfer. The wave number k is the absolute of the wave vector $k = q / \hbar$ and is related to the wavelength $k = 2\pi / \lambda$ . Often, momentum transfer is given in wavenumber units in reciprocal length $Q = k_f - k_i$

Diffraction

The momentum transfer plays an important role in the evaluation of neutron, X-ray and electron diffraction for the investigation of condensed matter. Bragg diffraction occurs on the atomic crystal lattice, conserves the wave energy and thus is called elastic scattering, where the wave numbers final and incident particles, $k_f$ and $k_i$ , respectively, are equal and just the direction changes by a reciprocal lattice vector $G = Q = k_f - k_i$ with the relation to the lattice spacing $G = 2\pi / d$ . As momentum is conserved, the transfer of momentum occurs to crystal momentum.

The presentation in $Q$ -space is generic and does not depend on the type of radiation and wavelength used but only on the sample system, which allows to compare results obtained from the many different methods. Some established communities such as powder diffraction employ the diffraction angle $2\theta$ as the independent variable, which worked fine in the early years when only a few characteristic wavelengths such as Cu-K $\alpha$ were available. The relationship to $Q$ -space is

$Q = \frac {4 \pi \sin \left ( \theta \right )}{\lambda}$

and basically states that larger $2\theta$ corresponds to larger $Q$ .

Momentum transfer

Wave mechanics and optics

Diffraction

See also