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 $\cos \theta _{f}=d^{2}/{\sqrt {d^{2}+x^{2}}}$, and $\cos ^{2}\theta _{f}=d^{4}/(d^{2}+x^{2})$ so:

${\begin{alignedat}{2}{\frac {1}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}&={\frac {1}{\sqrt {d^{2}+z^{2}\left(d^{4}/(d^{2}+x^{2})\right)}}}\\&={\frac {1}{{\sqrt {d^{2}}}{\sqrt {((d^{2}+x^{2})+z^{2}d^{2})/(d^{2}+x^{2})}}}}\\&={\frac {\sqrt {d^{2}+x^{2}}}{d{\sqrt {d^{2}+x^{2}+z^{2}d^{2}}}}}\\\end{alignedat}}$

And:

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

As a check:

Working results 2 (contains errors)

As a check of these results, consider:

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

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:

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