Theory

Scattering vector

The wave vector \(\vec{k}\) of a beam indicates the beam direction. Its magnitude is defined to be proportional to the spatial frequency (inverse of the wavelength \(\lambda\))

\[\|\vec{k}\|= \frac{2\pi}{\lambda}\]

The scattering vector \(\vec{q}\) is defined as the difference between the wave vector of the scattered and incident beam

\[\vec{q} = \vec{k}_s - \vec{k}_0\]

The scattering angle \(2\theta\) between incident and scattered beam is

\[\cos 2\theta = \hat{k}_s\cdot\hat{k}_0\]

For elastic scattering \(\lambda_s = \lambda_0\) the length of the scattering vector is related to the scattering angle as follows

\[\|\vec{q}\|= \frac{4\pi}{\lambda} \sin\theta\]

We define the normal vector \(\vec{n}_H\) to a family of lattice planes \(H = \{hkl\}\) as

\[\| \vec{n}_H \|= \frac{1}{d_H}\]

The scattered intensity from a crystal is maximal in the directions where the lattice planes fulfill the Laue conditions for diffraction

\[\vec{q} = 2\pi m \vec{n}_H \quad m \in \mathbb{Z}\]

When only considering the vector lengths you get Bragg’s law

\[2d\sin\theta = m\lambda = m\frac{hc}{E}\]

where \(\lambda\) and \(E\) the incident beam wavelength and energy respectively.

Strain tensor

The strain tensor in the direction of the scattering vector is

\[\epsilon^{hkl}(\phi, \psi) = \ln\left(\frac{d^{hkl}(\phi, \psi)}{d^{hkl}_0(\phi, \psi)}\right)\]

where \(\phi\) and \(\psi\) are the azimuth and polar angle of the scattering vector \(\vec{q}^{hkl}\) in the sample reference frame.

For angular dispersive diffraction (fix energy)

\[\epsilon^{hkl}(\phi, \psi) = \ln\left(\frac{\sin\theta^{hkl}_0(\phi, \psi)}{\sin\theta^{hkl}(\phi, \psi)}\right)\]

For energy dispersive diffraction (fix angle)

\[\epsilon^{hkl}(\phi, \psi) = \ln\left(\frac{E^{hkl}_0(\phi, \psi)}{E^{hkl}(\phi, \psi)}\right)\]