Geometry:TSAXS 3D

In transmission-SAXS (TSAXS), the x-ray beam hits the sample at normal incidence, and passes directly through without refraction. TSAXS is normally considered in terms of the one-dimensional momentum transfer (q); however the full 3D form of the q-vector is necessary when considering scattering from anisotropic materials. The q-vector in fact has three components:

$\mathbf {q} ={\begin{bmatrix}q_{x}\\q_{y}\\q_{z}\end{bmatrix}}$

This vector is always on the surface of the Ewald sphere. 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 $\scriptstyle (x,z)$. The scattering angles are then:

${\begin{alignedat}{2}\theta _{f}&=\arctan \left[{\frac {x}{d}}\right]\\\alpha _{f}^{\prime }&=\arctan \left[{\frac {z}{d}}\right]\\\alpha _{f}&=\arctan \left[{\frac {z}{d/\cos \theta _{f}}}\right]\end{alignedat}}$

where $\scriptstyle d$ is the sample-detector distance, $\scriptstyle \alpha _{f}^{\prime }$ is the out-of-plane component (angle w.r.t. to y-axis, rotation about x-axis), and $\scriptstyle \theta _{f}$ is the in-plane component (rotation about z-axis). The alternate angle, $\scriptstyle \alpha _{f}$, is the elevation angle in the plane defined by $\scriptstyle \theta _{f}$.

Total scattering

The full scattering angle is defined by a right-triangle with base d and height $\scriptstyle {\sqrt {x^{2}+d^{2}}}$:

${\begin{alignedat}{2}2\theta _{s}=\Theta &=\arctan \left[{\frac {\sqrt {x^{2}+z^{2}}}{d}}\right]\\&=\arctan \left[{\frac {\sqrt {(d\tan \theta _{f})^{2}+(d\tan \alpha _{f}^{\prime })^{2}}}{d}}\right]\\&=\arctan \left[{\sqrt {\tan ^{2}\theta _{f}+\tan ^{2}\alpha _{f}^{\prime }}}\right]\\&=\arctan \left[{\sqrt {\tan ^{2}\theta _{f}+{\frac {\tan ^{2}\alpha _{f}}{\cos ^{2}\theta _{f}}}}}\right]\\\end{alignedat}}$

The total momentum transfer is:

${\begin{alignedat}{2}q&={\frac {4\pi }{\lambda }}\sin \left(\theta _{s}\right)\\&={\frac {4\pi }{\lambda }}\sin \left({\frac {1}{2}}\arctan \left[{\frac {\sqrt {x^{2}+z^{2}}}{d}}\right]\right)\end{alignedat}}$

Given that:

${\begin{alignedat}{2}\cos(\arctan[u])&={\frac {1}{\sqrt {1+u^{2}}}}\\\cos(2\theta _{s})&={\frac {1}{\sqrt {1+({\sqrt {x^{2}+z^{2}}}/d)^{2}}}}\\&={\frac {d}{\sqrt {d^{2}+x^{2}+z^{2}}}}\end{alignedat}}$

We can also write:

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

Where we take for granted that q must be positive.

In-plane only

If $\scriptstyle \alpha _{f}=0$ (and $\scriptstyle \alpha _{f}^{\prime }=0$), then $\scriptstyle q_{z}=0$, $\scriptstyle 2\theta _{s}=\theta _{f}$, and:

${\begin{alignedat}{2}q&=2k\sin \left(\theta _{f}/2\right)\\&=2k{\sqrt {\frac {1-\cos(\theta _{f})}{2}}}\\&=2k{\sqrt {{\frac {1}{2}}\left(1-{\frac {1}{\sqrt {1+(x/d)^{2}}}}\right)}}\\&=2k{\sqrt {{\frac {1}{2}}\left(1-{\frac {d}{\sqrt {d^{2}+x^{2}}}}\right)}}\end{alignedat}}$

The other component can be thought of in terms of the sides of a right-triangle with angle $\scriptstyle \theta _{f}=0$:

${\begin{alignedat}{2}q_{x}&=q\cos(\theta _{f}/2)\\&=2k\sin(\theta _{f}/2)\cos(\theta _{f}/2)\\&=k\sin(\theta _{f})\\q_{y}&=-q\sin(\theta _{f}/2)\\&=-2k\sin(\theta _{f}/2)\sin(\theta _{f}/2)\\&=-k\left(1-\cos \theta _{f}\right)\\&=k\left(\cos \theta _{f}-1\right)\\\end{alignedat}}$

Summarizing:

$\mathbf {q} ={\frac {2\pi }{\lambda }}{\begin{bmatrix}\sin \theta _{f}\\\cos \theta _{f}-1\\0\end{bmatrix}}$

Out-of-plane only

If $\scriptstyle \theta _{f}=0$, then $\scriptstyle q_{x}=0$, $\scriptstyle \alpha _{f}^{\prime }=\alpha _{f}=2\theta _{s}$, and:

${\begin{alignedat}{2}q&=2k\sin \left(\alpha _{f}/2\right)\\&=2k{\sqrt {\frac {1-\cos(\alpha _{f})}{2}}}\\&=2k{\sqrt {{\frac {1}{2}}\left(1-{\frac {d}{\sqrt {d^{2}+z^{2}}}}\right)}}\end{alignedat}}$

The components are:

${\begin{alignedat}{2}q_{z}&=q\cos(\alpha _{f}/2)\\&=2k\sin(\theta _{f}/2)\cos(\theta _{f}/2)\\&=k\sin(\alpha _{f})\\q_{y}&=-q\sin(\alpha _{f}/2)\\&=k\left(\cos \alpha _{f}-1\right)\\\end{alignedat}}$

Summarizing:

$\mathbf {q} ={\frac {2\pi }{\lambda }}{\begin{bmatrix}0\\\cos \alpha _{f}-1\\\sin \alpha _{f}\end{bmatrix}}$

Components

The momentum transfer components are:

${\begin{alignedat}{2}q_{x}&={\frac {2\pi }{\lambda }}\sin \theta _{f}\cos \alpha _{f}\\q_{y}&={\frac {2\pi }{\lambda }}\left(\cos \theta _{f}\cos \alpha _{f}-1\right)\\q_{z}&={\frac {2\pi }{\lambda }}\sin \alpha _{f}\end{alignedat}}$

In vector form:

$\mathbf {q} ={\frac {2\pi }{\lambda }}{\begin{bmatrix}\sin \theta _{f}\cos \alpha _{f}\\\cos \theta _{f}\cos \alpha _{f}-1\\\sin \alpha _{f}\end{bmatrix}}$

Total magnitude

Note that this provides a simple expression for q total:

${\begin{alignedat}{2}q&={\sqrt {q_{x}^{2}+q_{y}^{2}+q_{z}^{2}}}\\&={\frac {2\pi }{\lambda }}{\sqrt {\sin ^{2}\theta _{f}\cos ^{2}\alpha _{f}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\sin ^{2}\alpha _{f}}}\\\left({\frac {q}{k}}\right)^{2}&=\sin ^{2}\theta _{f}\cos ^{2}\alpha _{f}+\cos ^{2}\theta _{f}\cos ^{2}\alpha _{f}-2\cos \theta _{f}\cos \alpha _{f}+1+\sin ^{2}\alpha _{f}\\&=\cos ^{2}\alpha _{f}(\sin ^{2}\theta _{f}+\cos ^{2}\theta _{f})+\sin ^{2}\alpha _{f}-2\cos \theta _{f}\cos \alpha _{f}+1\\&=\cos ^{2}\alpha _{f}(1)+\sin ^{2}\alpha _{f}-2\cos \theta _{f}\cos \alpha _{f}+1\\&=2-2\cos \theta _{f}\cos \alpha _{f}\\q&={\sqrt {2}}k{\sqrt {1-\cos \theta _{f}\cos \alpha _{f}}}\\\end{alignedat}}$

Check

As a check of these results, consider:

${\begin{alignedat}{2}q&={\frac {4\pi }{\lambda }}\sin \left(\theta _{s}\right)\\&={\frac {4\pi }{\lambda }}{\sqrt {\frac {1-\cos 2\theta _{s}}{2}}}\\\left({\frac {q}{k}}\right)^{2}&={\frac {4}{2}}\left(1-\cos 2\theta _{s}\right)\\&=2\left(1-{\frac {1}{\sqrt {1+\left({\sqrt {\tan ^{2}\theta _{f}+{\frac {\tan ^{2}\alpha _{f}}{\cos ^{2}\theta _{f}}}}}\right)^{2}}}}\right)\\&=2\left(1-{\frac {1}{\sqrt {1+\tan ^{2}\theta _{f}+{\frac {\tan ^{2}\alpha _{f}}{\cos ^{2}\theta _{f}}}}}}\right)\\&=2-{\frac {2}{\sqrt {1+{\frac {\sin ^{2}\theta _{f}}{\cos ^{2}\theta _{f}}}+{\frac {\sin ^{2}\alpha _{f}}{\cos ^{2}\alpha _{f}\cos ^{2}\theta _{f}}}}}}\\\end{alignedat}}$

And: