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:


\begin{alignat}{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{alignat}

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{alignat}{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{alignat}

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{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:


\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:


\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:


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

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{alignat}{2}
q & = 2 k \sin \left( \theta_f /2 \right) \\
    & = 2 k \sqrt{ \frac{1- \cos(\theta_f)}{2} } \\
    & = 2 k \sqrt{ \frac{1}{2} \left( 1 - \frac{1}{\sqrt{1+(x/d)^2}}  \right) } \\
    & = 2 k \sqrt{ \frac{1}{2} \left( 1 - \frac{d}{\sqrt{d^2+x^2}} \right) }
\end{alignat}

The other component can be thought of in terms of the sides of a right-triangle with angle \scriptstyle \theta_f = 0 :


\begin{alignat}{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{alignat}

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{alignat}{2}
q & = 2 k \sin \left( \alpha_f /2 \right) \\
    & = 2 k \sqrt{ \frac{1- \cos(\alpha_f)}{2} } \\
    & = 2 k \sqrt{ \frac{1}{2} \left( 1 - \frac{d}{\sqrt{d^2+z^2}} \right) }
\end{alignat}

The components are:


\begin{alignat}{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{alignat}

Summarizing:


\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:


\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:


\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{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:


\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:


\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}

Components (distances)


\begin{alignat}{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{x d}{\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{alignat}

Note that \cos \theta_f = d/\sqrt{d^2+x^2}, and \cos^2 \theta_f = d^2/(d^2+x^2) so:


\begin{alignat}{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{alignat}

And:


\begin{alignat}{2}
\mathbf{q} 
& = \frac{2 \pi}{\lambda} \begin{bmatrix} 
\frac{x d}{\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{alignat}

Total magnitude


\begin{alignat}{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{2 x^2 + 2 d^2 + 2 z^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{alignat}