Talk:Scattering

TSAXS 3d

The q-vector in fact has three components:

$\mathbf {q} ={\begin{bmatrix}q_{x}&q_{y}&q_{z}\end{bmatrix}}$

Consider that the x-ray beam points along +y, so that on the detector, the horizontal is x, and the vertical is z. We assume that the x-ray beam hits the flat 2D area detector at 90° at detector (pixel) position $\scriptstyle (x,z)$. The scattering angles are then:

${\begin{alignedat}{2}\theta _{f}&=\arctan \left({\frac {x}{d}}\right)\\\alpha _{f}^{\prime }&=\arctan \left({\frac {z}{d}}\right)\\\alpha _{f}&=\arctan \left({\frac {z}{d/\cos \theta _{f}}}\right)\end{alignedat}}$

where $\scriptstyle d$ is the sample-detector distance, $\scriptstyle \alpha _{f}^{\prime }$ is the out-of-plane component (angle w.r.t. to y-axis, rotation about x-axis), and $\scriptstyle \theta _{f}$ is the in-plane component (rotation about z-axis). The alternate angle, $\scriptstyle \alpha _{f}$, is the elevation angle in the plane defined by $\scriptstyle \theta _{f}$.

The momentum transfer components are:

${\begin{alignedat}{2}q_{x}&={\frac {2\pi }{\lambda }}\sin \theta _{f}\cos \alpha _{f}\\q_{y}&={\frac {2\pi }{\lambda }}\left(\cos \theta _{f}\cos \alpha _{f}-1\right)\\q_{z}&={\frac {2\pi }{\lambda }}\sin \alpha _{f}\end{alignedat}}$

Or:

${\begin{alignedat}{2}q_{x}&={\frac {2\pi }{\lambda }}\sin \theta _{f}\cos \alpha _{f}^{\prime }\\q_{y}&={\frac {2\pi }{\lambda }}\left(\cos \theta _{f}\cos \alpha _{f}^{\prime }-1\right)\\q_{z}&={\frac {2\pi }{\lambda }}\sin \alpha _{f}^{\prime }\cos \theta _{f}\end{alignedat}}$

And, of course:

${\begin{alignedat}{2}q&={\sqrt {q_{x}^{2}+q_{y}^{2}+q_{z}^{2}}}\end{alignedat}}$