Difference between revisions of "Geometry:WAXS 3D"
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The final scattering vector depends on: | The final scattering vector depends on: | ||
− | * <math> | + | * <math> x </math>: Pixel position on detector (horizontal). |
− | * <math> | + | * <math> z </math>: Pixel position on detector (vertical). |
− | * <math> | + | * <math> d </math>: Sample-detector distance. |
− | * <math> | + | * <math> \theta_g </math>: Elevation angle of detector. |
− | * <math> | + | * <math> \phi_g </math>: In-plane angle of detector. |
Note that <math>\scriptstyle x </math> and <math>\scriptstyle z </math> are defined relative to the direct-beam. That is, for <math>\scriptstyle \theta_g = 0 </math> and <math>\scriptstyle \phi_g =0 </math>, the direct beam is at position <math>\scriptstyle (x,z)=(0,0) </math> on the area detector. | Note that <math>\scriptstyle x </math> and <math>\scriptstyle z </math> are defined relative to the direct-beam. That is, for <math>\scriptstyle \theta_g = 0 </math> and <math>\scriptstyle \phi_g =0 </math>, the direct beam is at position <math>\scriptstyle (x,z)=(0,0) </math> on the area detector. | ||
Line 21: | Line 21: | ||
:<math> | :<math> | ||
\begin{alignat}{2} | \begin{alignat}{2} | ||
− | \mathbf{v} | + | \mathbf{v}_2 & = R_x(\theta_g) \mathbf{v}_i \\ |
& = \begin{bmatrix} | & = \begin{bmatrix} | ||
1 & 0 & 0 \\ | 1 & 0 & 0 \\ | ||
Line 30: | Line 30: | ||
\end{alignat} | \end{alignat} | ||
</math> | </math> | ||
+ | And then rotated about the <math>\scriptstyle z</math>-axis by <math>\scriptstyle \phi_g</math>: | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | \mathbf{v}_f & = R_z(\phi_g) \mathbf{v}_2 \\ | ||
+ | & = \begin{bmatrix} | ||
+ | \cos \phi_g & -\sin \phi_g & 0 \\ | ||
+ | \sin \phi_g & \cos \phi_g & 0\\ | ||
+ | 0 & 0 & 1\\ | ||
+ | \end{bmatrix} \begin{bmatrix} 0 \\ d \cos \theta_g \\ d \sin \theta_g \end{bmatrix} \\ | ||
+ | & = d \begin{bmatrix} -\sin \phi_g \cos \theta_g \\ \cos \phi_g \cos \theta_g \\ \sin \theta_g \end{bmatrix} | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | ===Total scattering=== | ||
+ | The point <math>\scriptstyle (x,z)=(0,0) </math> on the detector probes the total scattering angle <math>\scriptstyle \Theta = 2 \theta_s</math>, which is simply the angle between <math>\scriptstyle \mathbf{v}_i</math> and <math>\scriptstyle \mathbf{v}_f</math>: | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | \cos \Theta & = \frac{ \mathbf{v}_i \cdot \mathbf{v}_f }{ \left\| \mathbf{v}_i \right\| \left\| \mathbf{v}_f \right\|} \\ | ||
+ | & = \cos \phi_g \cos \theta_g \\ | ||
+ | 2 \theta_s & = \arccos \left[ \cos \phi_g \cos \theta_g \right] | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | Thus: | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | q & = \frac{4 \pi}{\lambda} \sin \left( \theta_s \right) \\ | ||
+ | & = \pm \frac{4 \pi}{\lambda} \sqrt{ \frac{1-\cos 2\theta_s }{2} } \\ | ||
+ | & = \frac{4 \pi}{\lambda} \sqrt{ \frac{1}{2}\left(1 - \cos \phi_g \cos \theta_g \right) } \\ | ||
+ | & = \sqrt{2} k \sqrt{ 1 - \cos \phi_g \cos \theta_g } | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | |||
+ | ===Components=== | ||
+ | The [[momentum transfer]] vector is (for elastic [[scattering]]): | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | \mathbf{q} & = \mathbf{k}_f - \mathbf{k}_i \\ | ||
+ | & = \frac{2 \pi}{\lambda} \begin{bmatrix} -\sin \phi_g \cos \theta_g \\ \cos \phi_g \cos \theta_g \\ \sin \theta_g\end{bmatrix} - \frac{2 \pi}{\lambda} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \\ | ||
+ | & = \frac{2 \pi}{\lambda} \begin{bmatrix} -\sin \phi_g \cos \theta_g \\ \cos \phi_g \cos \theta_g - 1 \\ \sin \theta_g\end{bmatrix} | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | This vector is of course the surface of the [[Ewald sphere]]. | ||
+ | |||
+ | ==Arbitrary Point== | ||
+ | For other points on the detector face, we can combine the above result with the known results for the [[Geometry:TSAXS 3D|Geometry of TSAXS]]. The incident beam is: | ||
+ | :<math> | ||
+ | \mathbf{v}_i = \begin{bmatrix} 0 \\ d \\ 0 \end{bmatrix} | ||
+ | </math> | ||
+ | :<math> | ||
+ | \mathbf{k}_i = \frac{2 \pi}{\lambda} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} | ||
+ | </math> | ||
+ | :<math> | ||
+ | \mathbf{q}_i = \frac{2 \pi}{\lambda} \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} | ||
+ | </math> | ||
+ | |||
+ | For <math>\scriptstyle \phi_g = 0</math> and <math>\scriptstyle \theta_g = 0</math>, we can compute the vector onto the detector face: | ||
+ | :<math> | ||
+ | \mathbf{v}_1 = \begin{bmatrix} x \\ d \\ z \end{bmatrix} | ||
+ | </math> | ||
+ | |||
+ | :<math> | ||
+ | \mathbf{k}_1 | ||
+ | = \frac{2 \pi}{\lambda} \begin{bmatrix} | ||
+ | \frac{x}{ \sqrt{x^2 + d^2 + z^2 }} \\ | ||
+ | \frac{d }{\sqrt{x^2 + d^2 + z^2 }} \\ | ||
+ | \frac{z }{\sqrt{x^2 + d^2 + z^2 }} \end{bmatrix} | ||
+ | </math> | ||
+ | |||
+ | :<math> | ||
+ | \mathbf{q}_1 | ||
+ | = \frac{2 \pi}{\lambda} \begin{bmatrix} | ||
+ | \frac{x}{ \sqrt{x^2 + d^2 + z^2 }} \\ | ||
+ | \frac{d }{\sqrt{x^2 + d^2 + z^2 }} - 1 \\ | ||
+ | \frac{z }{\sqrt{x^2 + d^2 + z^2 }} \end{bmatrix} | ||
+ | </math> | ||
+ | This vector is then rotated about the <math>\scriptstyle x</math>-axis by <math>\scriptstyle \theta_g</math>: | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | \mathbf{v}_2 & = R_x(\theta_g) \mathbf{v}_1 \\ | ||
+ | & = \begin{bmatrix} | ||
+ | 1 & 0 & 0 \\ | ||
+ | 0 & \cos \theta_g & -\sin \theta_g \\ | ||
+ | 0 & \sin \theta_g & \cos \theta_g \\ | ||
+ | \end{bmatrix} \begin{bmatrix} x \\ d \\ z \end{bmatrix} \\ | ||
+ | & = \begin{bmatrix} x \\ d \cos \theta_g - z \sin \theta_g \\ d \sin \theta_g + z \cos \theta_g \end{bmatrix} | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | |||
+ | :<math> | ||
+ | \mathbf{k}_2 | ||
+ | = \frac{2 \pi}{\lambda} \frac{1}{\sqrt{x^2 + d^2 + z^2} } \begin{bmatrix} x \\ d \cos \theta_g - z \sin \theta_g \\ d \sin \theta_g + z \cos \theta_g \end{bmatrix} | ||
+ | </math> | ||
+ | |||
+ | :<math> | ||
+ | \mathbf{q}_2 | ||
+ | = \frac{2 \pi}{\lambda} \begin{bmatrix} \frac{x}{\sqrt{x^2 + d^2 + z^2}} \\ \frac{ d \cos \theta_g - z \sin \theta_g}{\sqrt{x^2 + d^2 + z^2}} - 1 \\ \frac{d \sin \theta_g + z \cos \theta_g}{\sqrt{x^2 + d^2 + z^2}} \end{bmatrix} | ||
+ | </math> | ||
+ | |||
+ | The vector is then rotated about the <math>\scriptstyle z</math>-axis by <math>\scriptstyle \phi_g</math>: | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | \mathbf{v}_f & = R_z(\phi_g) \mathbf{v}_2 \\ | ||
+ | & = \begin{bmatrix} | ||
+ | \cos \phi_g & -\sin \phi_g & 0 \\ | ||
+ | \sin \phi_g & \cos \phi_g & 0\\ | ||
+ | 0 & 0 & 1\\ | ||
+ | \end{bmatrix} \begin{bmatrix} x \\ d \cos \theta_g - z \sin \theta_g \\ d \sin \theta_g + z \cos \theta_g \end{bmatrix} \\ | ||
+ | & = \begin{bmatrix} | ||
+ | x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ | ||
+ | x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ | ||
+ | d \sin \theta_g + z \cos \theta_g \end{bmatrix} | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | |||
+ | :<math> | ||
+ | \mathbf{k}_f | ||
+ | = \frac{2 \pi}{\lambda} \frac{1}{\sqrt{x^2 + d^2 + z^2} } \begin{bmatrix} | ||
+ | x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ | ||
+ | x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ | ||
+ | d \sin \theta_g + z \cos \theta_g \end{bmatrix} | ||
+ | </math> | ||
+ | |||
+ | :<math> | ||
+ | \mathbf{q} | ||
+ | = \frac{2 \pi}{\lambda} \begin{bmatrix} \frac{x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g )}{\sqrt{x^2 + d^2 + z^2}} \\ \frac{ x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )}{\sqrt{x^2 + d^2 + z^2}} - 1 \\ \frac{d \sin \theta_g + z \cos \theta_g}{\sqrt{x^2 + d^2 + z^2}} \end{bmatrix} | ||
+ | </math> | ||
+ | ===Components=== | ||
+ | :<math> | ||
+ | \mathbf{q} | ||
+ | = \frac{2 \pi}{\lambda} \frac{1}{d^{\prime}} \begin{bmatrix} x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) - d^{\prime} \\ d \sin \theta_g + z \cos \theta_g \end{bmatrix} | ||
+ | </math> | ||
+ | Where: | ||
+ | ::<math> | ||
+ | d^{\prime} = \sqrt{x^2 + d^2 + z^2} | ||
+ | </math> | ||
+ | ===Total magnitude=== | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | \cos \Theta & = \frac{ \mathbf{k}_i \cdot \mathbf{k}_f }{ \left\| \mathbf{k}_i \right\| \left\| \mathbf{k}_f \right\|} \\ | ||
+ | & = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \cdot \frac{1}{\sqrt{x^2 + d^2 + z^2} } \begin{bmatrix} | ||
+ | x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ | ||
+ | x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ | ||
+ | d \sin \theta_g + z \cos \theta_g \end{bmatrix} \\ | ||
+ | & = \frac{x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )}{\sqrt{x^2 + d^2 + z^2} } \\ | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | Thus: | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | q | ||
+ | & = \sqrt{2} k \sqrt{ 1 - \frac{x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )}{\sqrt{x^2 + d^2 + z^2} } } | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | |||
+ | ====Check==== | ||
+ | We define: | ||
+ | ::<math> | ||
+ | \begin{alignat}{2} | ||
+ | d^{\prime} & = \sqrt{x^2 + d^2 + z^2} = \| \mathbf{v}_1 \| \\ | ||
+ | ( v_{2y} ) & = ( d \cos \theta_g - z \sin \theta_g ) \\ | ||
+ | ( v_{2y} )^2 & = ( d \cos \theta_g - z \sin \theta_g )^2 \\ | ||
+ | & = d^2 \cos^2 \theta_g - 2dz \cos \theta_g \sin\theta_g + z^2 \sin^2 \theta_g | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | |||
+ | And calculate: | ||
+ | :<math> | ||
+ | \begin{alignat}{2} | ||
+ | q^2 & = [ (q_x)^2 + (q_y)^2 + (q_z)^2 ] \\ | ||
+ | \left ( \frac{q}{k} \right )^2 d^{\prime 2} | ||
+ | & = \begin{alignat}{2} [ & \left( x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \right)^2 \\ & + \left( x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) - d^{\prime} \right)^2 \\ & + \left( d \sin \theta_g + z \cos \theta_g \right)^2 ] \end{alignat} \\ | ||
+ | & = \begin{alignat}{2} [ | ||
+ | & \left( x \cos \phi_g -\sin \phi_g ( v_{2y} ) \right)^2 \\ | ||
+ | & + \left( x \sin \phi_g + \cos \phi_g ( v_{2y} ) - d^{\prime} \right)^2 \\ | ||
+ | & + \left( d \sin \theta_g + z \cos \theta_g \right)^2 ] \end{alignat} \\ | ||
− | + | & = \begin{alignat}{2} [ | |
+ | & x^2 \cos^2 \phi_g - 2 x \cos \phi_g \sin \phi_g ( v_{2y} ) + \sin^2 \phi_g ( v_{2y} )^2 \\ | ||
+ | & + x^2 \sin^2 \phi_g + x \sin \phi_g \cos \phi_g ( v_{2y} ) - d^{\prime} x \sin \phi_g \\ | ||
+ | & + x \sin \phi_g \cos \phi_g ( v_{2y} ) + \cos^2 \phi_g ( v_{2y} )^2 - d^{\prime} \cos \phi_g ( v_{2y} ) \\ | ||
+ | & - d^{\prime} x \sin \phi_g - d^{\prime} \cos \phi_g ( v_{2y} ) + d^{\prime 2} \\ | ||
+ | & + d^2 \sin^2 \theta_g + 2 d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g ] \end{alignat} \\ | ||
+ | \end{alignat} | ||
+ | </math> | ||
+ | Grouping and rearranging: | ||
:<math> | :<math> | ||
\begin{alignat}{2} | \begin{alignat}{2} | ||
− | 2 \ | + | \left ( \frac{q}{k} \right )^2 d^{\prime 2} |
− | & = 1 | + | & = \begin{alignat}{2} [ |
+ | & x^2 + ( v_{2y} )^2 \\ | ||
+ | & - 2 d^{\prime} x \sin \phi_g \\ | ||
+ | & - 2 d^{\prime} \cos \phi_g ( v_{2y} ) \\ | ||
+ | & + d^{\prime 2} \\ | ||
+ | & + d^2 \sin^2 \theta_g + 2 d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g ] \end{alignat} \\ | ||
+ | |||
+ | & = \begin{alignat}{2} [ | ||
+ | & d^{\prime 2} + x^2 + ( d^2 \cos^2 \theta_g - 2dz \cos \theta_g \sin\theta_g + z^2 \sin^2 \theta_g ) \\ | ||
+ | & - 2 d^{\prime} x \sin \phi_g - 2 d^{\prime} \cos \phi_g ( v_{2y} ) \\ | ||
+ | & + d^2 \sin^2 \theta_g + 2 d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g ] \end{alignat} \\ | ||
+ | |||
+ | & = \begin{alignat}{2} [ | ||
+ | & d^{\prime 2} + x^2 + d^2 + z^2 \\ | ||
+ | & - 2 d^{\prime} x \sin \phi_g - 2 d^{\prime} \cos \phi_g ( v_{2y} ) ] \end{alignat} \\ | ||
+ | |||
+ | & = 2 d^{\prime 2} - 2 d^{\prime} x \sin \phi_g - 2 d^{\prime} \cos \phi_g ( v_{2y} )\\ | ||
+ | |||
+ | & = 2 d^{\prime 2} - 2 d^{\prime} x \sin \phi_g - 2 d^{\prime} \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ | ||
+ | & = 2 d^{\prime} \left( d^{\prime} - x \sin \phi_g - \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \right) \\ | ||
+ | \left( \frac{q}{k} \right)^2 | ||
+ | & = 2 \left( 1 - \frac{x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )}{d^{\prime} } \right) | ||
\end{alignat} | \end{alignat} | ||
</math> | </math> | ||
− | ==See Also | + | =Area Detector on Goniometer Arm, with offsets= |
+ | In the most general case, the sample may not sit at the exact center of the goniometer rotation. In such a case, corrections must be applied. | ||
+ | |||
+ | TBD | ||
+ | |||
+ | =See Also= | ||
* [[Geometry:TSAXS 3D]] | * [[Geometry:TSAXS 3D]] | ||
− |
Latest revision as of 08:06, 10 August 2017
In wide-angle scattering (WAXS), one cannot simply assume that the detector plane is orthogonal to the incident x-ray beam. Converting from detector pixel coordinates to 3D q-vector is not always trivial, and depends on the experimental geometry.
Contents
Area Detector on Goniometer Arm
Consider a 2D (area) detector connected to a goniometer arm. The goniometer has a center of rotation at the center of the sample (i.e. the incident beam passes through this center, and scattered rays originate from this point also). Let be the in-plane angle of the goniometer arm (rotation about -axis), and be the elevation angle (rotation away from plane and towards axis).
The final scattering vector depends on:
- : Pixel position on detector (horizontal).
- : Pixel position on detector (vertical).
- : Sample-detector distance.
- : Elevation angle of detector.
- : In-plane angle of detector.
Note that and are defined relative to the direct-beam. That is, for and , the direct beam is at position on the area detector.
Central Point
The point can be thought of in terms of a vector that points from the source-of-scattering (center of goniometer rotation) to the detector:
This vector is then rotated about the -axis by :
And then rotated about the -axis by :
Total scattering
The point on the detector probes the total scattering angle , which is simply the angle between and :
Thus:
Components
The momentum transfer vector is (for elastic scattering):
This vector is of course the surface of the Ewald sphere.
Arbitrary Point
For other points on the detector face, we can combine the above result with the known results for the Geometry of TSAXS. The incident beam is:
For and , we can compute the vector onto the detector face:
This vector is then rotated about the -axis by :
The vector is then rotated about the -axis by :
Components
Where:
Total magnitude
Thus:
Check
We define:
And calculate:
Grouping and rearranging:
Area Detector on Goniometer Arm, with offsets
In the most general case, the sample may not sit at the exact center of the goniometer rotation. In such a case, corrections must be applied.
TBD