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 $\scriptstyle \theta _{f}$. Also note that the full scattering angle is:

${\begin{alignedat}{2}2\theta _{s}=\Theta &=\arctan \left[{\frac {\sqrt {x^{2}+z^{2}}}{d}}\right]\end{alignedat}}$

The momentum transfer components are:

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

And, of course:

${\begin{alignedat}{2}q&={\sqrt {q_{x}^{2}+q_{y}^{2}+q_{z}^{2}}}\\&={\frac {2\pi }{\lambda }}{\sqrt {\sin ^{2}\theta _{f}\cos ^{2}\alpha _{f}\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\sin ^{2}\alpha _{f}}}\\{\frac {q}{k}}&={\sqrt {(\sin \theta _{f})^{2}(\cos \alpha _{f})^{2}\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+(\sin \alpha _{f})^{2}}}\\&={\sqrt {\left({\frac {x/d}{\sqrt {1+(x/d)^{2}}}}\right)^{2}\left(\cos \alpha _{f}\right)^{2}\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\left(\sin \alpha _{f}\right)^{2}}}\\&=?\\&=?\\&=?\\&=?\\&=?\\&=?\\&={\frac {\sqrt {x^{2}+z^{2}}}{\sqrt {d^{2}+x^{2}+z^{2}}}}\\&={\frac {\left[{\sqrt {x^{2}+z^{2}}}/d\right]}{\sqrt {1+\left[{\sqrt {x^{2}+z^{2}}}/d\right]^{2}}}}\\&=\sin \left(\arctan \left[{\frac {\sqrt {x^{2}+z^{2}}}{d}}\right]\right)\\&=\sin \left(2\theta _{s}\right)\end{alignedat}}$