Talk:Geometry:TSAXS 3D

Alternate check

As a check of these results, consider:

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 & = \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 } \\ \left( \frac{q}{k} \right)^2 & = (\sin \theta_f)^2 (\cos \alpha_f)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + (\sin \alpha_f)^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{alignat} }

Where we used:

${\begin{alignedat}{2}\sin(\arctan[u])&={\frac {u}{\sqrt {1+u^{2}}}}\\\sin \theta _{f}&=\sin(\arctan[x/d])\\&={\frac {x/d}{\sqrt {1+(x/d)^{2}}}}\\&={\frac {x}{\sqrt {d^{2}+x^{2}}}}\end{alignedat}}$

And, we further note that:

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

Continuing:

${\begin{alignedat}{2}\left({\frac {q}{k}}\right)^{2}&={\frac {x^{2}}{d^{2}+x^{2}}}{\frac {d^{4}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}+\left({\frac {d^{2}}{\sqrt {d^{2}+x^{2}}}}{\frac {d^{2}}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}-1\right)^{2}+{\frac {z^{2}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}{\frac {d^{4}}{d^{2}+x^{2}}}\\&=d^{4}{\frac {x^{2}+z^{2}}{(d^{2}+x^{2})(d^{2}+z^{2}\cos ^{2}\theta _{f})}}+\left({\frac {d^{4}}{\sqrt {(d^{2}+x^{2})(d^{2}+z^{2}\cos ^{2}\theta _{f})}}}-1\right)^{2}\\&={\frac {d^{4}x^{2}+d^{4}z^{2}}{d^{4}+d^{2}x^{2}+d^{4}z^{2}}}+\left({\frac {d^{4}}{\sqrt {d^{4}+d^{2}x^{2}+d^{4}z^{2}}}}-1\right)^{2}\\&={\frac {d^{2}x^{2}+d^{2}z^{2}}{d^{2}+x^{2}+d^{2}z^{2}}}+\left({\frac {d^{8}}{d^{4}+d^{2}x^{2}+d^{4}z^{2}}}-2{\frac {d^{4}}{\sqrt {d^{4}+d^{2}x^{2}+d^{4}z^{2}}}}+1\right)\\&={\frac {d^{2}x^{2}+d^{2}z^{2}}{d^{2}+x^{2}+d^{2}z^{2}}}+{\frac {d^{6}}{d^{2}+x^{2}+d^{2}z^{2}}}-2{\frac {d^{3}}{\sqrt {d^{2}+x^{2}+d^{2}z^{2}}}}+1\\&={\frac {d^{2}x^{2}+d^{2}z^{2}+d^{6}-2d^{3}{\sqrt {d^{2}+x^{2}+d^{2}z^{2}}}+d^{2}+x^{2}+d^{2}z^{2}}{d^{2}+x^{2}+d^{2}z^{2}}}\\&={\frac {d^{6}+d^{2}+d^{2}x^{2}+x^{2}+2d^{2}z^{2}-2d^{3}{\sqrt {d^{2}+x^{2}+d^{2}z^{2}}}}{d^{2}+x^{2}+d^{2}z^{2}}}\\&=?\\&=?\\&={\frac {{\sqrt {d^{2}+x^{2}+z^{2}}}-d}{2{\sqrt {d^{2}+x^{2}+z^{2}}}}}\\&={\frac {1}{2}}\left(1-{\frac {d}{\sqrt {d^{2}+x^{2}+z^{2}}}}\right)\\q&={\frac {4\pi }{\lambda }}\sin \left(\theta _{s}\right)\end{alignedat}}$