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 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 \scriptstyle (x,z) } . The scattering angles are then:

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} \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{alignat} }

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:

now

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} 2 \theta_s = \Theta & = \arctan\left[ \frac{ \sqrt{x^2 + z^2}}{d} \right] \\ & = \arctan\left[ \frac{ \sqrt{x^2 + z^2}}{d} \right] \\ \end{alignat} }

The momentum transfer components are:

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

later

As a check:

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