Talk:Geometry:TSAXS 3D

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Compute q_y


\begin{alignat}{2}
\mathbf{q} & = \begin{bmatrix} q_x \\ q_y \\ q_z \end{bmatrix} \\
    & = k \begin{bmatrix} \sin \theta_f \cos \alpha_f  \\ \cos \theta_f \cos \alpha_f - 1 \\ \sin \alpha_f \end{bmatrix}
\end{alignat}

So:


\begin{alignat}{2}
\alpha_f & = \sin^{-1} \left[ \frac{q_z}{k} \right] \\
\frac{q_x}{k} & = \sin \theta_f \cos \alpha_f \\
\theta_f & = \sin^{-1} \left[ \frac{q_x}{k} \frac{1}{\cos \alpha_f} \right] \\
\frac{q_y}{k} & = \cos \theta_f \cos \alpha_f - 1 \\
q_y & = k \left ( \cos \left( \sin^{-1} \left[ \frac{q_x}{k} \frac{1}{\cos \alpha_f} \right] \right ) \cos \left ( \sin^{-1} \left[ \frac{q_z}{k} \right] \right ) - 1 \right )\\
 & = k \left ( \sqrt{ 1 - \left[ \frac{q_x}{k} \frac{1}{\cos \alpha_f} \right]^2 } \sqrt{ 1 - \left[ \frac{q_z}{k} \right]^2 } - 1 \right )
\end{alignat}

Or equivalently:


\begin{alignat}{2}
q_y & = k \left ( \sqrt{ 1 - \left[ \frac{q_x}{k} \frac{1}{\sqrt{1-[q_z/k]^2}} \right]^2 } \sqrt{ 1 - \left[ \frac{q_z}{k} \right]^2 } - 1 \right ) \\
    & = k \sqrt{ 1 - \frac{q_x^2}{k^2 (1-q_z^2/k^2) } } \sqrt{ 1 - \frac{q_z^2}{k^2} } - k
\end{alignat}

Scratch/working (contains errors)

As a check of these results, consider:


\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{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:


\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{alignat}{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^4x^2+d^4z^2}{d^4+d^2x^2+d^4z^2}  + \left ( \frac{d^4}{\sqrt{d^4+d^2x^2+d^4z^2}} - 1 \right )^2 \\
    & = \frac{d^2x^2+d^2z^2}{d^2+x^2+d^2z^2}  + \left ( \frac{d^8}{d^4+d^2x^2+d^4z^2} -2 \frac{d^4}{\sqrt{d^4+d^2x^2+d^4z^2}} + 1 \right ) \\
    & = \frac{d^2x^2+d^2z^2}{d^2+x^2+d^2z^2}  + \frac{d^6}{d^2+x^2+d^2z^2} -2 \frac{d^3}{\sqrt{d^2+x^2+d^2z^2}} + 1 \\
    & = \frac{d^2x^2+d^2z^2 + d^6 -2d^3\sqrt{d^2+x^2+d^2z^2} + d^2+x^2+d^2z^2}{d^2+x^2+d^2z^2}  \\
    & = \frac{d^6 + d^2 + d^2x^2 + x^2 + 2d^2z^2 -2d^3\sqrt{d^2+x^2+d^2z^2}}{d^2+x^2+d^2z^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{alignat}