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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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| − | ====now====
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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{x^2 + z^2}}{d} \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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| − | ====later====
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| − | As a check:
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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} & = \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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| − | & = \sqrt{ \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(\sin \alpha_f \right)^2 } \\
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| − | & = ? \\
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| − | & = ? \\
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| − | & = ? \\
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| − | & = ? \\
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| − | & = ? \\
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| − | & = ? \\
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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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| − | & = \sin \left( 2 \theta_s \right)
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| − | \end{alignat}
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| − | </math>
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