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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( x/d ) \\
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| − | \alpha_f & = \arctan( z/d )
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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</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 [[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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| − | And, of course:
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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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| − | \end{alignat}
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| − | </math>
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