Difference between revisions of "Talk:Geometry:TSAXS 3D"

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}