Talk:Geometry:TSAXS 3D

Working results 1

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

Note that $\cos \theta _{f}=d/{\sqrt {d^{2}+x^{2}}}$, and $\cos ^{2}\theta _{f}=d^{2}/(d^{2}+x^{2})$ so:

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

And:

${\begin{alignedat}{2}\mathbf {q} &={\frac {2\pi }{\lambda }}{\begin{bmatrix}{\frac {xd}{\sqrt {d^{2}+x^{2}}}}{\frac {\sqrt {d^{2}+x^{2}}}{d{\sqrt {d^{2}+x^{2}+z^{2}}}}}\\{\frac {d}{\sqrt {d^{2}+x^{2}}}}{\frac {d{\sqrt {d^{2}+x^{2}}}}{d{\sqrt {d^{2}+x^{2}+z^{2}}}}}-1\\{\frac {z\left(d/{\sqrt {d^{2}+x^{2}}}\right){\sqrt {d^{2}+x^{2}}}}{d{\sqrt {d^{2}+x^{2}+z^{2}}}}}\end{bmatrix}}\\&={\frac {2\pi }{\lambda }}{\begin{bmatrix}{\frac {x}{\sqrt {x^{2}+d^{2}+z^{2}}}}\\{\frac {d}{\sqrt {x^{2}+d^{2}+z^{2}}}}-1\\{\frac {z}{\sqrt {x^{2}+d^{2}+z^{2}}}}\end{bmatrix}}\\\end{alignedat}}$

As a check:

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} \left( \frac{q}{k} \right)^2 & = \left( \frac{x}{ \sqrt{x^2 + d^2 + z^2 d^2 }} \right)^2 + \left( \frac{d - \sqrt{x^2 + d^2 + z^2 d^2 } }{\sqrt{x^2 + d^2 + z^2 d^2 }} \right)^2 + \left( \frac{z d }{\sqrt{x^2 + d^2 + z^2 d^2 }} \right)^2 \\ & = \frac{x^2 + \left( d - \sqrt{x^2 + d^2 + z^2 d^2 }\right)^2 + z^2d^2 }{x^2 + d^2 + z^2d^2} \\ & = \frac{x^2 + \left( d^2 - 2d \sqrt{x^2 + d^2 + z^2 d^2} + x^2 + d^2 + z^2 d^2 \right) + z^2d^2 }{x^2 + d^2 + z^2d^2} \\ & = \frac{2 x^2 + 2 d^2 + 2 z^2d^2 - 2d \sqrt{x^2 + d^2 + z^2 d^2} }{x^2 + d^2 + z^2d^2} \\ & = 2 \frac{( x^2 + d^2 + z^2d^2 ) - d \sqrt{x^2 + d^2 + z^2 d^2} }{x^2 + d^2 + z^2d^2} \\ & = 2 \left( 1 - \frac{d}{\sqrt{x^2 + d^2 + z^2d^2}} \right) \end{alignat} }

Working results 2 (contains errors)

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:

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} \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{alignat} }

And, we further note that:

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} \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{alignat} }

Continuing:

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