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| − | ===[[TSAXS]] 3D===
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| − | The ''q''-vector in fact has three components:
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| − | :<math>
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| − | \mathbf{q} = \begin{bmatrix} q_x & q_y & q_z \end{bmatrix}
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| − | </math>
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| − | Consider that the [[x-ray]] beam points along +''y'', so that on the [[detector]], the horizontal is ''x'', and the vertical is ''z''. We assume that the x-ray beam hits the flat 2D area detector at 90° at detector (pixel) position <math>\scriptstyle (x,z) </math>. The scattering angles are then:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | \theta_f & = \arctan\left[ \frac{x}{d} \right] \\
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| − | \alpha_f ^\prime & = \arctan\left[ \frac{z}{d} \right] \\
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| − | \alpha_f & = \arctan \left[ \frac{z }{d / \cos \theta_f} \right]
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| − | \end{alignat}
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| − | </math>
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| − | where <math>\scriptstyle d</math> is the sample-detector distance, <math>\scriptstyle \alpha_f ^{\prime} </math> is the out-of-plane component (angle w.r.t. to ''y''-axis, rotation about x-axis), and <math>\scriptstyle \theta_f </math> is the in-plane component (rotation about ''z''-axis). The alternate angle, <math>\scriptstyle \alpha_f </math>, is the elevation angle in the plane defined by <math>\scriptstyle \theta_f </math>. Also note that the full scattering angle is:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | 2 \theta_s = \Theta & = \arctan\left[ \frac{ \sqrt{x^2 + z^2}}{d} \right] \\
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| − | & = \arctan\left[ \frac{ \sqrt{(d \tan \theta_f)^2 + (d \tan \alpha_f^\prime )^2}}{d} \right] \\
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| − | & = \arctan\left[ \sqrt{\tan^2 \theta_f + \tan^2 \alpha_f^\prime } \right] \\
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| − | & = \arctan\left[ \sqrt{\tan^2 \theta_f + \frac{ \tan^2 \alpha_f }{ \cos^2 \theta_f } } \right] \\
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| − | \end{alignat}
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| − | </math>
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| − |
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| − | The [[momentum transfer]] components are:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | q_x & = \frac{2 \pi}{\lambda} \sin \theta_f \cos \alpha_f \\
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| − | q_y & = \frac{2 \pi}{\lambda} \left ( \cos \theta_f \cos \alpha_f - 1 \right ) \\
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| − | q_z & = \frac{2 \pi}{\lambda} \sin \alpha_f
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| − | \end{alignat}
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| − | </math>
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| − |
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| − | ====Check====
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| − | As a check of these results, consider:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | q & = \sqrt{ q_x^2 + q_y^2 + q_z^2 } \\
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| − | & = \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 } \\
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| − | \frac{q}{k}
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| − | & = \sqrt{ (\sin \theta_f)^2 (\cos \alpha_f)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + (\sin \alpha_f)^2 } \\
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| − | \frac{q^2}{k^2}
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| − | & = \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 \\
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| − | & = \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 \\
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| − | & = \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 } \\
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| − | & = \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 }
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| − | \end{alignat}
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| − | </math>
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| − | Where we used:
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| − | ::<math>
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| − | \begin{alignat}{2}
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| − | \sin( \arctan[u]) & = \frac{u}{\sqrt{1+u^2}} \\
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| − | \sin \theta_f & = \sin( \arctan [x/d] ) \\
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| − | & = \frac{x/d}{\sqrt{1 + (x/d)^2}} \\
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| − | & = \frac{x}{\sqrt{d^2+x^2}}
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| − | \end{alignat}
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| − | </math>
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| − |
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| − | And, we further note that:
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| − | ::<math>
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| − | \begin{alignat}{2}
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| − | \cos( \arctan[u]) & = \frac{1}{\sqrt{1+u^2}} \\
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| − | \cos \theta_f & = \frac{1}{\sqrt{1 + (x/d)^2}} \\
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| − | & = \frac{d^2}{\sqrt{d^2+x^2}}
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| − | \end{alignat}
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| − | </math>
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| − | ====cont====
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| − | Continuing:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | \frac{q^2}{k^2}
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| − | & = \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} \\
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| − | & = 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 \\
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| − | & = \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 \\
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| − | & = \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 ) \\
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| − | & = \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 \\
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| − | & = \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} \\
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| − | & = \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} \\
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| − | & = \frac{ (x^2 + z^2) } {(d^2 + x^2 + z^2)} \frac{(d^2 + x^2 + z^2)}{ (x^2 + z^2) } \frac{d^6 + d^2(1+x^2+2z^2) + x^2 -2d^3\sqrt{d^2(1+z^2)+x^2}}{d^2(1+z^2)+x^2} \\
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| − | & = ? \\
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| − | & = \frac{ x^2 + z^2 } {d^2 + x^2 + z^2} \\
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| − | \frac{q}{k} & = \sqrt{ \frac{ x^2 + z^2 } {d^2 + x^2 + z^2} } \\
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| − | & = \frac{ \sqrt{x^2 + z^2} } {\sqrt{d^2 + x^2 + z^2 }} \\
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| − | & = \frac{ \left[ \sqrt{x^2 + z^2}/d \right ] } {\sqrt{1 + \left[ \sqrt{x^2 + z^2}/d \right ]^2 }} \\
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| − | & = \sin \left( \arctan\left [ \frac{\sqrt{x^2 + z^2}}{d} \right ] \right) \\
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| − | q & = \frac{2 \pi}{\lambda} \sin \left( 2 \theta_s \right)
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| − | \end{alignat}
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| − | </math>
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