Talk:Scattering

TSAXS 3D

The q-vector in fact has three components:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

${\begin{alignedat}{2}2\theta _{s}=\Theta &=\arctan \left[{\frac {\sqrt {x^{2}+z^{2}}}{d}}\right]\\&=\arctan \left[{\frac {\sqrt {(d\tan \theta _{f})^{2}+(d\tan \alpha _{f}^{\prime })^{2}}}{d}}\right]\\&=\arctan \left[{\sqrt {\tan ^{2}\theta _{f}+\tan ^{2}\alpha _{f}^{\prime }}}\right]\\&=\arctan \left[{\sqrt {\tan ^{2}\theta _{f}+{\frac {\tan ^{2}\alpha _{f}}{\cos ^{2}\theta _{f}}}}}\right]\\\end{alignedat}}$

The momentum transfer components are:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{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{alignat} }

Check

As a check of these results, consider:

${\begin{alignedat}{2}q&={\sqrt {q_{x}^{2}+q_{y}^{2}+q_{z}^{2}}}\\&={\frac {2\pi }{\lambda }}{\sqrt {\sin ^{2}\theta _{f}\cos ^{2}\alpha _{f}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\sin ^{2}\alpha _{f}}}\\{\frac {q}{k}}&={\sqrt {(\sin \theta _{f})^{2}(\cos \alpha _{f})^{2}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+(\sin \alpha _{f})^{2}}}\\{\frac {q^{2}}{k^{2}}}&=\left({\frac {x/d}{\sqrt {1+(x/d)^{2}}}}\right)^{2}\left(\cos \alpha _{f}\right)^{2}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\left({\frac {z\cos \theta _{f}/d}{\sqrt {1+(z\cos \theta _{f}/d)^{2}}}}\right)^{2}\\&=\left({\frac {x}{\sqrt {d^{2}+x^{2}}}}\right)^{2}\left(\cos \alpha _{f}\right)^{2}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\left({\frac {z\cos \theta _{f}}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}\right)^{2}\\&={\frac {x^{2}}{d^{2}+x^{2}}}\left(\cos \alpha _{f}\right)^{2}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+{\frac {z^{2}\cos ^{2}\theta _{f}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}\\&={\frac {x^{2}}{d^{2}+x^{2}}}{\frac {d^{4}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}+\left(\cos \theta _{f}{\frac {d^{2}}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}-1\right)^{2}+{\frac {z^{2}\cos ^{2}\theta _{f}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}\end{alignedat}}$

We note that:

${\begin{alignedat}{2}\cos \theta _{f}&={\frac {1}{\sqrt {1+(x/d)^{2}}}}\\&={\frac {d^{2}}{\sqrt {d^{2}+x^{2}}}}\end{alignedat}}$

Continuing:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \frac{q^2}{k^2} & = ? \\ & = ? \\ & = ? \\ & = ? \\ & = ? \\ & = \frac{ x^2 + z^2 } {d^2 + x^2 + z^2} \\ \frac{q}{k} & = \sqrt{ \frac{ x^2 + z^2 } {d^2 + x^2 + z^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) \\ q & = \frac{2 \pi}{\lambda} \sin \left( 2 \theta_s \right) \end{alignat} }