Talk:Geometry:WAXS 3D

Check of Total Magnitude #1: Doesn't work

${\begin{alignedat}{2}\left({\frac {q}{k}}\right)^{2}d^{\prime 2}&={\begin{alignedat}{2}[&\left(x\cos \phi _{g}-\sin \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\right)^{2}\\&+\left(x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})-d^{\prime }\right)^{2}\\&+\left(d\sin \theta _{g}+z\cos \theta _{g}\right)^{2}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}\cos ^{2}\phi _{g}-x\cos \phi _{g}\sin \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})+\sin ^{2}\phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})^{2}\\&+x^{2}\sin ^{2}\phi _{g}+x\sin \phi _{g}\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})-d^{\prime }x\sin \phi _{g}\\&+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})x\sin \phi _{g}+\cos ^{2}\phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})^{2}-d^{\prime }\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\\&-d^{\prime }x\sin \phi _{g}-d^{\prime }\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+2d\sin \theta _{g}z\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}\cos ^{2}\phi _{g}-x\cos \phi _{g}\sin \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})+\sin ^{2}\phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})^{2}\\&+x^{2}\sin ^{2}\phi _{g}+2x\sin \phi _{g}\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})-2d^{\prime }x\sin \phi _{g}\\&+\cos ^{2}\phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})^{2}-2d^{\prime }\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+2d\sin \theta _{g}z\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}-x\sin \phi _{g}\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})+(d\cos \theta _{g}-z\sin \theta _{g})^{2}\\&+2x\sin \phi _{g}\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})-2d^{\prime }x\sin \phi _{g}\\&-2d^{\prime }\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+2dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}+d^{2}\cos ^{2}\theta _{g}-2dz\cos \theta _{g}\sin \theta _{g}+z^{2}\sin ^{2}\theta _{g}\\&+(-x\sin \phi _{g}\cos \phi _{g}+2x\sin \phi _{g}\cos \phi _{g}-2d^{\prime }\cos \phi _{g})(d\cos \theta _{g}-z\sin \theta _{g})\\&-2d^{\prime }x\sin \phi _{g}\\&+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+2dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&={\begin{alignedat}{2}[&d^{\prime 2}+x^{2}+d^{2}+z^{2}-2dz\cos \theta _{g}\sin \theta _{g}\\&+(x\sin \phi _{g}\cos \phi _{g}-2d^{\prime }\cos \phi _{g})(d\cos \theta _{g}-z\sin \theta _{g})\\&+2dz\sin \theta _{g}\cos \theta _{g}-2d^{\prime }x\sin \phi _{g}]\end{alignedat}}\\&=2d^{\prime 2}-2d^{\prime }x\sin \phi _{g}+(x\sin \phi _{g}\cos \phi _{g}-2d^{\prime }\cos \phi _{g})(d\cos \theta _{g}-z\sin \theta _{g})\\&=2d^{\prime 2}-2d^{\prime }x\sin \phi _{g}+(x\sin \phi _{g}-2d^{\prime })\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\\&=?\\&=?\\&=2d^{\prime 2}-2d^{\prime }x\sin \phi _{g}+2d^{\prime }\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\\&=2d^{\prime }\left(d^{\prime }-x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\right)\\\left({\frac {q}{k}}\right)^{2}&=2\left(1-{\frac {x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})}{d^{\prime }}}\right)\end{alignedat}}$

Check of Total Magnitude #2: Doesn't work

We define:

${\begin{alignedat}{2}d^{\prime }&={\sqrt {x^{2}+d^{2}+z^{2}}}=\|\mathbf {v} _{1}\|\\(v_{2y})&=(d\cos \theta _{g}-z\sin \theta _{g})\\(v_{2y})^{2}&=(d\cos \theta _{g}-z\sin \theta _{g})^{2}\\&=d^{2}\cos ^{2}\theta _{g}-dz\cos \theta _{g}\sin \theta _{g}+z^{2}\sin ^{2}\theta _{g}\end{alignedat}}$

And calculate:

${\begin{alignedat}{2}q^{2}&=[(q_{x})^{2}+(q_{y})^{2}+(q_{z})^{2}]\\\left({\frac {q}{k}}\right)^{2}d^{\prime 2}&={\begin{alignedat}{2}[&\left(x\cos \phi _{g}-\sin \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\right)^{2}\\&+\left(x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})-d^{\prime }\right)^{2}\\&+\left(d\sin \theta _{g}+z\cos \theta _{g}\right)^{2}]\end{alignedat}}\\&={\begin{alignedat}{2}[&\left(x\cos \phi _{g}-\sin \phi _{g}(v_{2y})\right)^{2}\\&+\left(x\sin \phi _{g}+\cos \phi _{g}(v_{2y})-d^{\prime }\right)^{2}\\&+\left(d\sin \theta _{g}+z\cos \theta _{g}\right)^{2}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}\cos ^{2}\phi _{g}-2x\cos \phi _{g}\sin \phi _{g}(v_{2y})+\sin ^{2}\phi _{g}(v_{2y})^{2}\\&+x^{2}\sin ^{2}\phi _{g}+x\sin \phi _{g}\cos \phi _{g}(v_{2y})-d^{\prime }x\sin \phi _{g}\\&+x\sin \phi _{g}\cos \phi _{g}(v_{2y})+\cos ^{2}\phi _{g}(v_{2y})^{2}-d^{\prime }\cos \phi _{g}(v_{2y})\\&-d^{\prime }x\sin \phi _{g}-d^{\prime }\cos \phi _{g}(v_{2y})+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+2dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\\end{alignedat}}$

Grouping and rearranging:

${\begin{alignedat}{2}\left({\frac {q}{k}}\right)^{2}d^{\prime 2}&={\begin{alignedat}{2}[&x^{2}+(v_{2y})^{2}\\&-2d^{\prime }x\sin \phi _{g}\\&+x\sin \phi _{g}\cos \phi _{g}(v_{2y})-d^{\prime }\cos \phi _{g}(v_{2y})\\&-d^{\prime }\cos \phi _{g}(v_{2y})+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}+(d^{2}\cos ^{2}\theta _{g}-dz\cos \theta _{g}\sin \theta _{g}+z^{2}\sin ^{2}\theta _{g})\\&-2d^{\prime }x\sin \phi _{g}\\&+(x\sin \phi _{g}-d^{\prime }-d^{\prime })\cos \phi _{g}(v_{2y})\\&+d^{\prime 2}+d^{2}\sin ^{2}\theta _{g}+dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&={\begin{alignedat}{2}[&d^{\prime 2}+x^{2}+d^{2}+z^{2}\\&-2d^{\prime }x\sin \phi _{g}\\&+(x\sin \phi _{g}-2d^{\prime })\cos \phi _{g}(v_{2y})]\end{alignedat}}\\&=2d^{\prime 2}-2d^{\prime }x\sin \phi _{g}+(x\sin \phi _{g}-2d^{\prime })\cos \phi _{g}(v_{2y})\\&=?\\&=?\\&=?\\&=2d^{\prime 2}-2d^{\prime }x\sin \phi _{g}+2d^{\prime }\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\\&=2d^{\prime }\left(d^{\prime }-x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\right)\\\left({\frac {q}{k}}\right)^{2}&=2\left(1-{\frac {x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})}{d^{\prime }}}\right)\end{alignedat}}$