Geometry:WAXS 3D

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.

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 $\scriptstyle \phi _{g}$ be the in-plane angle of the goniometer arm (rotation about $\scriptstyle z$-axis), and $\scriptstyle \theta _{g}$ be the elevation angle (rotation away from $\scriptstyle xy$ plane and towards $\scriptstyle z$ axis).

The final scattering vector depends on:

• $\scriptstyle x$: Pixel position on detector (horizontal).
• $\scriptstyle z$: Pixel position on detector (vertical).
• $\scriptstyle d$: Sample-detector distance.
• $\scriptstyle \theta _{g}$: Elevation angle of detector.
• $\scriptstyle \phi _{g}$: In-plane angle of detector.

Note that $\scriptstyle x$ and $\scriptstyle z$ are defined relative to the direct-beam. That is, for $\scriptstyle \theta _{g}=0$ and $\scriptstyle \phi _{g}=0$, the direct beam is at position $\scriptstyle (x,z)=(0,0)$ on the area detector.

Central Point

The point $\scriptstyle (x,z)=(0,0)$ can be thought of in terms of a vector that points from the source-of-scattering (center of goniometer rotation) to the detector:

$\mathbf {v} _{i}={\begin{bmatrix}0\\d\\0\end{bmatrix}}$

This vector is then rotated about the $\scriptstyle x$-axis by $\scriptstyle \theta _{g}$:

${\begin{alignedat}{2}\mathbf {v} _{1}&=R_{x}(\theta _{g})\mathbf {v} _{i}\\&={\begin{bmatrix}1&0&0\\0&\cos \theta _{g}&-\sin \theta _{g}\\0&\sin \theta _{g}&\cos \theta _{g}\\\end{bmatrix}}{\begin{bmatrix}0\\d\\0\end{bmatrix}}\\&={\begin{bmatrix}0\\d\cos \theta _{g}\\d\sin \theta _{g}\end{bmatrix}}\end{alignedat}}$

And then rotated about the $\scriptstyle z$-axis by $\scriptstyle \phi _{g}$:

${\begin{alignedat}{2}\mathbf {v} _{f}&=R_{z}(\phi _{g})\mathbf {v} _{1}\\&={\begin{bmatrix}\cos \phi _{g}&-\sin \phi _{g}&0\\\sin \phi _{g}&\cos \phi _{g}&0\\0&0&1\\\end{bmatrix}}{\begin{bmatrix}0\\d\cos \theta _{g}\\d\sin \theta _{g}\end{bmatrix}}\\&=d{\begin{bmatrix}-\sin \phi _{g}\cos \theta _{g}\\\cos \phi _{g}\cos \theta _{g}\\\sin \theta _{g}\end{bmatrix}}\end{alignedat}}$

The point $\scriptstyle (x,z)=(0,0)$ on the detector probes the total scattering angle $\scriptstyle \Theta =2\theta _{s}$, which is simply the angle between $\scriptstyle \mathbf {v} _{i}$ and $\scriptstyle \mathbf {v} _{f}$:

${\begin{alignedat}{2}\cos \Theta &={\frac {\mathbf {v} _{i}\cdot \mathbf {v} _{f}}{\left\|\mathbf {v} _{i}\right\|\left\|\mathbf {v} _{f}\right\|}}\\&=\cos \phi _{g}\cos \theta _{g}\\2\theta _{s}&=\arccos \left[\cos \phi _{g}\cos \theta _{g}\right]\end{alignedat}}$

Thus:

${\begin{alignedat}{2}q&={\frac {4\pi }{\lambda }}\sin \left(\theta _{s}\right)\\&=\pm {\frac {4\pi }{\lambda }}{\sqrt {\frac {1-\cos 2\theta _{s}}{2}}}\\&={\frac {4\pi }{\lambda }}{\sqrt {{\frac {1}{2}}\left(1-\cos \phi _{g}\cos \theta _{g}\right)}}\end{alignedat}}$

Arbitrary Point

TBD

See Also

• Geometry:TSAXS 3D