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

Total scattering

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

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} 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{alignat} }

The total momentum transfer is:

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} \sin \left( \frac{1}{2} \arctan\left [ \frac{\sqrt{x^2 + z^2}}{d} \right ] \right) \end{alignat} }

Given that:

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} \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{alignat} }

We can also write:

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) \\ & = \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{alignat} }

Where we take for granted that q must be positive.

In-plane only

If 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 \scriptstyle \alpha_f = 0 } (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 \scriptstyle \alpha_f ^{\prime} = 0} ), then 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 \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:

Alternate check

As a check of these results, consider:

${\begin{alignedat}{2}q&={\sqrt {q_{x}^{2}+q_{y}^{2}+q_{z}^{2}}}\\&={\frac {2\pi }{\lambda }}{\sqrt {\sin ^{2}\theta _{f}\cos ^{2}\alpha _{f}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\sin ^{2}\alpha _{f}}}\\\left({\frac {q}{k}}\right)^{2}&=(\sin \theta _{f})^{2}(\cos \alpha _{f})^{2}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+(\sin \alpha _{f})^{2}\\&=\left({\frac {x/d}{\sqrt {1+(x/d)^{2}}}}\right)^{2}\left(\cos \alpha _{f}\right)^{2}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\left({\frac {z\cos \theta _{f}/d}{\sqrt {1+(z\cos \theta _{f}/d)^{2}}}}\right)^{2}\\&=\left({\frac {x}{\sqrt {d^{2}+x^{2}}}}\right)^{2}\left(\cos \alpha _{f}\right)^{2}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+\left({\frac {z\cos \theta _{f}}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}\right)^{2}\\&={\frac {x^{2}}{d^{2}+x^{2}}}\left(\cos \alpha _{f}\right)^{2}+\left(\cos \theta _{f}\cos \alpha _{f}-1\right)^{2}+{\frac {z^{2}\cos ^{2}\theta _{f}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}\\&={\frac {x^{2}}{d^{2}+x^{2}}}{\frac {d^{4}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}+\left(\cos \theta _{f}{\frac {d^{2}}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}-1\right)^{2}+{\frac {z^{2}\cos ^{2}\theta _{f}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}\end{alignedat}}$

Where we used:

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

And, we further note that:

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

Continuing:

${\begin{alignedat}{2}\left({\frac {q}{k}}\right)^{2}&={\frac {x^{2}}{d^{2}+x^{2}}}{\frac {d^{4}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}+\left({\frac {d^{2}}{\sqrt {d^{2}+x^{2}}}}{\frac {d^{2}}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}-1\right)^{2}+{\frac {z^{2}}{d^{2}+z^{2}\cos ^{2}\theta _{f}}}{\frac {d^{4}}{d^{2}+x^{2}}}\\&=d^{4}{\frac {x^{2}+z^{2}}{(d^{2}+x^{2})(d^{2}+z^{2}\cos ^{2}\theta _{f})}}+\left({\frac {d^{4}}{\sqrt {(d^{2}+x^{2})(d^{2}+z^{2}\cos ^{2}\theta _{f})}}}-1\right)^{2}\\&={\frac {d^{4}x^{2}+d^{4}z^{2}}{d^{4}+d^{2}x^{2}+d^{4}z^{2}}}+\left({\frac {d^{4}}{\sqrt {d^{4}+d^{2}x^{2}+d^{4}z^{2}}}}-1\right)^{2}\\&={\frac {d^{2}x^{2}+d^{2}z^{2}}{d^{2}+x^{2}+d^{2}z^{2}}}+\left({\frac {d^{8}}{d^{4}+d^{2}x^{2}+d^{4}z^{2}}}-2{\frac {d^{4}}{\sqrt {d^{4}+d^{2}x^{2}+d^{4}z^{2}}}}+1\right)\\&={\frac {d^{2}x^{2}+d^{2}z^{2}}{d^{2}+x^{2}+d^{2}z^{2}}}+{\frac {d^{6}}{d^{2}+x^{2}+d^{2}z^{2}}}-2{\frac {d^{3}}{\sqrt {d^{2}+x^{2}+d^{2}z^{2}}}}+1\\&={\frac {d^{2}x^{2}+d^{2}z^{2}+d^{6}-2d^{3}{\sqrt {d^{2}+x^{2}+d^{2}z^{2}}}+d^{2}+x^{2}+d^{2}z^{2}}{d^{2}+x^{2}+d^{2}z^{2}}}\\&={\frac {d^{6}+d^{2}+d^{2}x^{2}+x^{2}+2d^{2}z^{2}-2d^{3}{\sqrt {d^{2}+x^{2}+d^{2}z^{2}}}}{d^{2}+x^{2}+d^{2}z^{2}}}\\&=?\\&=?\\&={\frac {{\sqrt {d^{2}+x^{2}+z^{2}}}-d}{2{\sqrt {d^{2}+x^{2}+z^{2}}}}}\\&={\frac {1}{2}}\left(1-{\frac {d}{\sqrt {d^{2}+x^{2}+z^{2}}}}\right)\\q&={\frac {4\pi }{\lambda }}\sin \left(\theta _{s}\right)\end{alignedat}}$