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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{q} = \frac{2 \pi}{\lambda} \begin{bmatrix} 0 \\ \cos \alpha_f - 1 \\ \sin \alpha_f \end{bmatrix} }

Components

The momentum transfer components are:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{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{alignat} }

In vector form:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{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{alignat} }

Check

As a check of these results, consider:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{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{alignat} }

And:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} & \left( 1+\frac{\sin^2 \theta_f}{\cos^2 \theta_f} + \frac{ \sin^2 \alpha_f }{ \cos^2 \alpha_f \cos^2 \theta_f } \right) ^{-1/2} \\ = & \left( \frac{\cos^2 \alpha_f \cos^2 \theta_f}{\cos^2 \alpha_f \cos^2 \theta_f}+\frac{\cos^2 \alpha_f \sin^2 \theta_f}{\cos^2 \alpha_f \cos^2 \theta_f} + \frac{ \sin^2 \alpha_f }{ \cos^2 \alpha_f \cos^2 \theta_f } \right) ^{-1/2} \\ = & \left( \frac{\cos^2 \alpha_f \cos^2 \theta_f + \cos^2 \alpha_f \sin^2 \theta_f + \sin^2 \alpha_f}{\cos^2 \alpha_f \cos^2 \theta_f} \right) ^{-1/2} \\ = & \left( \frac{\cos^2 \theta_f \cos^2 \alpha_f }{\cos^2 \alpha_f \cos^2 \theta_f + \cos^2 \alpha_f \sin^2 \theta_f + \sin^2 \alpha_f} \right) ^{+1/2} \\ = & \frac{\cos \theta_f \cos \alpha_f }{ \sqrt{ \cos^2 \alpha_f (\cos^2 \theta_f + \sin^2 \theta_f) + \sin^2 \alpha_f }} \\ = & \cos \theta_f \cos \alpha_f \end{alignat} }