# Geometry:TSAXS 3D

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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:

{\displaystyle {\begin{alignedat}{2}\mathbf {q} &={\begin{bmatrix}q_{x}\\q_{y}\\q_{z}\end{bmatrix}}\\&={\frac {2\pi }{\lambda }}{\begin{bmatrix}\sin \theta _{f}\cos \alpha _{f}\\\cos \theta _{f}\cos \alpha _{f}-1\\\sin \alpha _{f}\end{bmatrix}}\end{alignedat}}}

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 ${\displaystyle \scriptstyle (x,z)}$. The scattering angles are then:

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

## Total scattering

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

{\displaystyle {\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:

{\displaystyle {\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:

{\displaystyle {\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:

{\displaystyle {\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)}}\\&={\sqrt {2}}{\frac {2\pi }{\lambda }}{\sqrt {1-{\frac {d}{\sqrt {x^{2}+d^{2}+z^{2}}}}}}\end{alignedat}}}

Where we take for granted that q must be positive.

## In-plane only

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

{\displaystyle {\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 ${\displaystyle \scriptstyle \theta _{f}=0}$:

{\displaystyle {\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:

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

## Out-of-plane only

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

{\displaystyle {\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:

{\displaystyle {\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:

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

## Components (angular)

For arbitrary 3D scattering vectors, the momentum transfer components are:

{\displaystyle {\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:

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

### Total magnitude

Note that this provides a simple expression for q total:

{\displaystyle {\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:

{\displaystyle {\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:

{\displaystyle {\begin{alignedat}{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{alignedat}}}

## Components (distances)

{\displaystyle {\begin{alignedat}{2}\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}}\\&={\frac {2\pi }{\lambda }}{\begin{bmatrix}\sin \left(\arctan \left[{\frac {x}{d}}\right]\right)\cos \left(\arctan \left[{\frac {z}{d/\cos \theta _{f}}}\right]\right)\\\cos \left(\arctan \left[{\frac {x}{d}}\right]\right)\cos \left(\arctan \left[{\frac {z}{d/\cos \theta _{f}}}\right]\right)-1\\\sin \left(\arctan \left[{\frac {z}{d/\cos \theta _{f}}}\right]\right)\end{bmatrix}}\\&={\frac {2\pi }{\lambda }}{\begin{bmatrix}{\frac {x/d}{\sqrt {1+\left(x/d\right)^{2}}}}{\frac {d}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}\\{\frac {1}{\sqrt {1+\left(x/d\right)^{2}}}}{\frac {d}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}-1\\{\frac {z\cos \theta _{f}}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}\end{bmatrix}}\\&={\frac {2\pi }{\lambda }}{\begin{bmatrix}{\frac {xd}{\sqrt {d^{2}+x^{2}}}}{\frac {1}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}\\{\frac {d}{\sqrt {d^{2}+x^{2}}}}{\frac {d}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}-1\\{\frac {z\cos \theta _{f}}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}\end{bmatrix}}\\\end{alignedat}}}

Note that ${\displaystyle \cos \theta _{f}=d/{\sqrt {d^{2}+x^{2}}}}$, and ${\displaystyle \cos ^{2}\theta _{f}=d^{2}/(d^{2}+x^{2})}$ so:

{\displaystyle {\begin{alignedat}{2}{\frac {1}{\sqrt {d^{2}+z^{2}\cos ^{2}\theta _{f}}}}&={\frac {1}{\sqrt {d^{2}+z^{2}\left(d^{2}/(d^{2}+x^{2})\right)}}}\\&={\frac {1}{{\sqrt {d^{2}}}{\sqrt {((d^{2}+x^{2})+z^{2})/(d^{2}+x^{2})}}}}\\&={\frac {\sqrt {d^{2}+x^{2}}}{d{\sqrt {d^{2}+x^{2}+z^{2}}}}}\\\end{alignedat}}}

And:

{\displaystyle {\begin{alignedat}{2}\mathbf {q} &={\frac {2\pi }{\lambda }}{\begin{bmatrix}{\frac {xd}{\sqrt {d^{2}+x^{2}}}}{\frac {\sqrt {d^{2}+x^{2}}}{d{\sqrt {d^{2}+x^{2}+z^{2}}}}}\\{\frac {d}{\sqrt {d^{2}+x^{2}}}}{\frac {d{\sqrt {d^{2}+x^{2}}}}{d{\sqrt {d^{2}+x^{2}+z^{2}}}}}-1\\{\frac {z\left(d/{\sqrt {d^{2}+x^{2}}}\right){\sqrt {d^{2}+x^{2}}}}{d{\sqrt {d^{2}+x^{2}+z^{2}}}}}\end{bmatrix}}\\&={\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}}\\\end{alignedat}}}

### Total magnitude

{\displaystyle {\begin{alignedat}{2}\left({\frac {q}{k}}\right)^{2}&=\left({\frac {x}{\sqrt {x^{2}+d^{2}+z^{2}}}}\right)^{2}+\left({\frac {d-{\sqrt {x^{2}+d^{2}+z^{2}}}}{\sqrt {x^{2}+d^{2}+z^{2}}}}\right)^{2}+\left({\frac {z}{\sqrt {x^{2}+d^{2}+z^{2}}}}\right)^{2}\\&={\frac {x^{2}+\left(d-{\sqrt {x^{2}+d^{2}+z^{2}}}\right)^{2}+z^{2}}{x^{2}+d^{2}+z^{2}}}\\&={\frac {x^{2}+\left(d^{2}-2d{\sqrt {x^{2}+d^{2}+z^{2}}}+x^{2}+d^{2}+z^{2}\right)+z^{2}}{x^{2}+d^{2}+z^{2}}}\\&={\frac {2x^{2}+2d^{2}+2z^{2}-2d{\sqrt {x^{2}+d^{2}+z^{2}}}}{x^{2}+d^{2}+z^{2}}}\\&=2{\frac {(x^{2}+d^{2}+z^{2})-d{\sqrt {x^{2}+d^{2}+z^{2}}}}{x^{2}+d^{2}+z^{2}}}\\&=2\left(1-{\frac {d}{\sqrt {x^{2}+d^{2}+z^{2}}}}\right)\\q&={\sqrt {2}}k{\sqrt {1-{\frac {d}{\sqrt {x^{2}+d^{2}+z^{2}}}}}}\end{alignedat}}}