Geometry:WAXS 3D

From GISAXS
Jump to: navigation, search

In wide-angle scattering (WAXS), one cannot simply assume that the detector plane is orthogonal to the incident x-ray beam. Converting from detector pixel coordinates to 3D q-vector is not always trivial, and depends on the experimental geometry.

Area Detector on Goniometer Arm

Consider a 2D (area) detector connected to a goniometer arm. The goniometer has a center of rotation at the center of the sample (i.e. the incident beam passes through this center, and scattered rays originate from this point also). Let \scriptstyle \phi_g be the in-plane angle of the goniometer arm (rotation about \scriptstyle z -axis), and \scriptstyle \theta_g be the elevation angle (rotation away from \scriptstyle xy plane and towards \scriptstyle z axis).

The final scattering vector depends on:

  •  x : Pixel position on detector (horizontal).
  •  z : Pixel position on detector (vertical).
  •  d : Sample-detector distance.
  •  \theta_g : Elevation angle of detector.
  •  \phi_g : In-plane angle of detector.

Note that \scriptstyle x and \scriptstyle z are defined relative to the direct-beam. That is, for \scriptstyle \theta_g = 0 and \scriptstyle \phi_g =0 , the direct beam is at position \scriptstyle (x,z)=(0,0) on the area detector.

Central Point

The point \scriptstyle (x,z)=(0,0) can be thought of in terms of a vector that points from the source-of-scattering (center of goniometer rotation) to the detector:


\mathbf{v}_i = \begin{bmatrix} 0 \\ d \\ 0 \end{bmatrix}

This vector is then rotated about the \scriptstyle x-axis by \scriptstyle \theta_g:


\begin{alignat}{2}
\mathbf{v}_2 & = R_x(\theta_g) \mathbf{v}_i \\
    & = \begin{bmatrix}
1 & 0 & 0 \\
0 & \cos \theta_g &  -\sin \theta_g \\
0 & \sin \theta_g  &  \cos \theta_g \\
\end{bmatrix} \begin{bmatrix} 0 \\ d \\ 0 \end{bmatrix} \\
    & = \begin{bmatrix} 0 \\ d \cos \theta_g \\ d \sin \theta_g \end{bmatrix}
\end{alignat}

And then rotated about the \scriptstyle z-axis by \scriptstyle \phi_g:


\begin{alignat}{2}
\mathbf{v}_f & = R_z(\phi_g) \mathbf{v}_2 \\
    & = \begin{bmatrix}
\cos \phi_g &  -\sin \phi_g & 0 \\
\sin \phi_g & \cos \phi_g & 0\\
0 & 0 & 1\\
\end{bmatrix} \begin{bmatrix} 0 \\ d \cos \theta_g \\ d \sin \theta_g \end{bmatrix} \\
    & = d \begin{bmatrix} -\sin \phi_g \cos \theta_g \\ \cos \phi_g \cos \theta_g \\ \sin \theta_g \end{bmatrix}
\end{alignat}

Total scattering

The point \scriptstyle (x,z)=(0,0) on the detector probes the total scattering angle \scriptstyle \Theta = 2 \theta_s, which is simply the angle between \scriptstyle \mathbf{v}_i and \scriptstyle \mathbf{v}_f:


\begin{alignat}{2}
\cos \Theta & = \frac{ \mathbf{v}_i \cdot \mathbf{v}_f }{ \left\| \mathbf{v}_i \right\| \left\| \mathbf{v}_f \right\|} \\
    & = \cos \phi_g \cos \theta_g \\
2 \theta_s & = \arccos \left[ \cos \phi_g \cos \theta_g \right]
\end{alignat}

Thus:


\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 - \cos \phi_g \cos \theta_g \right) } \\
    & = \sqrt{2} k \sqrt{ 1 - \cos \phi_g \cos \theta_g }
\end{alignat}

Components

The momentum transfer vector is (for elastic scattering):


\begin{alignat}{2}
\mathbf{q} & = \mathbf{k}_f - \mathbf{k}_i \\
    & = \frac{2 \pi}{\lambda} \begin{bmatrix} -\sin \phi_g \cos \theta_g \\ \cos \phi_g \cos \theta_g \\ \sin \theta_g\end{bmatrix} - \frac{2 \pi}{\lambda} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \\
    & = \frac{2 \pi}{\lambda} \begin{bmatrix} -\sin \phi_g \cos \theta_g \\ \cos \phi_g \cos \theta_g - 1 \\ \sin \theta_g\end{bmatrix}
\end{alignat}

This vector is of course the surface of the Ewald sphere.

Arbitrary Point

For other points on the detector face, we can combine the above result with the known results for the Geometry of TSAXS. The incident beam is:


\mathbf{v}_i = \begin{bmatrix} 0 \\ d \\ 0 \end{bmatrix}

\mathbf{k}_i = \frac{2 \pi}{\lambda} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}

\mathbf{q}_i = \frac{2 \pi}{\lambda} \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

For \scriptstyle \phi_g = 0 and \scriptstyle \theta_g = 0, we can compute the vector onto the detector face:


\mathbf{v}_1 = \begin{bmatrix} x \\ d \\ z \end{bmatrix}

\mathbf{k}_1 
 = \frac{2 \pi}{\lambda} \begin{bmatrix} 
\frac{x}{ \sqrt{x^2 + d^2 + z^2   }}  \\ 
\frac{d }{\sqrt{x^2 + d^2 + z^2   }}  \\ 
\frac{z }{\sqrt{x^2 + d^2 + z^2   }} \end{bmatrix}

\mathbf{q}_1 
 = \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}

This vector is then rotated about the \scriptstyle x-axis by \scriptstyle \theta_g:


\begin{alignat}{2}
\mathbf{v}_2 & = R_x(\theta_g) \mathbf{v}_1 \\
    & = \begin{bmatrix}
1 & 0 & 0 \\
0 & \cos \theta_g &  -\sin \theta_g \\
0 & \sin \theta_g  &  \cos \theta_g \\
\end{bmatrix} \begin{bmatrix} x \\ d \\ z \end{bmatrix} \\
    & = \begin{bmatrix} x \\ d \cos \theta_g - z \sin \theta_g \\ d \sin \theta_g + z \cos \theta_g \end{bmatrix}
\end{alignat}

\mathbf{k}_2 
 = \frac{2 \pi}{\lambda} \frac{1}{\sqrt{x^2 + d^2 + z^2} } \begin{bmatrix} x \\ d \cos \theta_g - z \sin \theta_g \\ d \sin \theta_g + z \cos \theta_g \end{bmatrix}

\mathbf{q}_2 
 = \frac{2 \pi}{\lambda} \begin{bmatrix} \frac{x}{\sqrt{x^2 + d^2 + z^2}} \\ \frac{ d \cos \theta_g - z \sin \theta_g}{\sqrt{x^2 + d^2 + z^2}} - 1 \\ \frac{d \sin \theta_g + z \cos \theta_g}{\sqrt{x^2 + d^2 + z^2}} \end{bmatrix}

The vector is then rotated about the \scriptstyle z-axis by \scriptstyle \phi_g:


\begin{alignat}{2}
\mathbf{v}_f & = R_z(\phi_g) \mathbf{v}_2 \\
    & = \begin{bmatrix}
\cos \phi_g &  -\sin \phi_g & 0 \\
\sin \phi_g & \cos \phi_g & 0\\
0 & 0 & 1\\
\end{bmatrix} \begin{bmatrix} x \\ d \cos \theta_g - z \sin \theta_g \\ d \sin \theta_g + z \cos \theta_g \end{bmatrix} \\
    & = \begin{bmatrix} 
      x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ 
      x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ 
      d \sin \theta_g + z \cos \theta_g \end{bmatrix}
\end{alignat}

\mathbf{k}_f
 = \frac{2 \pi}{\lambda} \frac{1}{\sqrt{x^2 + d^2 + z^2} } \begin{bmatrix} 
      x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ 
      x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ 
      d \sin \theta_g + z \cos \theta_g \end{bmatrix}

\mathbf{q}
 = \frac{2 \pi}{\lambda} \begin{bmatrix} \frac{x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g )}{\sqrt{x^2 + d^2 + z^2}} \\ \frac{ x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )}{\sqrt{x^2 + d^2 + z^2}} - 1 \\ \frac{d \sin \theta_g + z \cos \theta_g}{\sqrt{x^2 + d^2 + z^2}} \end{bmatrix}

Components


\mathbf{q}
 = \frac{2 \pi}{\lambda} \frac{1}{d^{\prime}} \begin{bmatrix} x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) - d^{\prime} \\ d \sin \theta_g + z \cos \theta_g \end{bmatrix}

Where:


d^{\prime} = \sqrt{x^2 + d^2 + z^2}

Total magnitude


\begin{alignat}{2}
\cos \Theta & = \frac{ \mathbf{k}_i \cdot \mathbf{k}_f }{ \left\| \mathbf{k}_i \right\| \left\| \mathbf{k}_f \right\|} \\
    & = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \cdot \frac{1}{\sqrt{x^2 + d^2 + z^2} } \begin{bmatrix} 
      x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ 
      x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \\ 
      d \sin \theta_g + z \cos \theta_g \end{bmatrix} \\
    & = \frac{x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )}{\sqrt{x^2 + d^2 + z^2} } \\ 
\end{alignat}

Thus:


\begin{alignat}{2}
q
    & = \sqrt{2} k \sqrt{ 1 - \frac{x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )}{\sqrt{x^2 + d^2 + z^2} } }
\end{alignat}

Check

We define:


\begin{alignat}{2}
d^{\prime} & = \sqrt{x^2 + d^2 + z^2} = \| \mathbf{v}_1 \| \\
( v_{2y} ) & = ( d \cos \theta_g - z \sin \theta_g ) \\
( v_{2y} )^2 & = ( d \cos \theta_g - z \sin \theta_g )^2 \\
    & = d^2 \cos^2 \theta_g - 2dz \cos \theta_g \sin\theta_g + z^2 \sin^2 \theta_g
\end{alignat}

And calculate:


\begin{alignat}{2}
q^2 & = [ (q_x)^2 + (q_y)^2 + (q_z)^2 ] \\
\left ( \frac{q}{k} \right )^2 d^{\prime 2}
    & = \begin{alignat}{2} [ & \left( x \cos \phi_g -\sin \phi_g ( d \cos \theta_g - z \sin \theta_g ) \right)^2 \\ & + \left( x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) - d^{\prime} \right)^2 \\ & + \left( d \sin \theta_g + z \cos \theta_g \right)^2  ] \end{alignat}  \\

    & = \begin{alignat}{2} [ 
      & \left( x \cos \phi_g -\sin \phi_g ( v_{2y} ) \right)^2 \\ 
      & + \left( x \sin \phi_g + \cos \phi_g ( v_{2y} ) - d^{\prime} \right)^2 \\ 
      & + \left( d \sin \theta_g + z \cos \theta_g \right)^2  ] \end{alignat}  \\

    & = \begin{alignat}{2} [ 
      & x^2 \cos^2 \phi_g - 2 x \cos \phi_g \sin \phi_g ( v_{2y} ) + \sin^2 \phi_g ( v_{2y} )^2  \\ 
      & + x^2 \sin^2 \phi_g + x \sin \phi_g \cos \phi_g ( v_{2y} ) - d^{\prime} x \sin \phi_g \\ 
      & + x \sin \phi_g \cos \phi_g ( v_{2y} ) + \cos^2 \phi_g ( v_{2y} )^2 - d^{\prime} \cos \phi_g ( v_{2y} ) \\ 
      & - d^{\prime} x \sin \phi_g - d^{\prime} \cos \phi_g ( v_{2y} ) + d^{\prime 2} \\ 
      & + d^2 \sin^2 \theta_g + 2 d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g  ] \end{alignat}  \\
\end{alignat}

Grouping and rearranging:


\begin{alignat}{2}
\left ( \frac{q}{k} \right )^2 d^{\prime 2}
    & = \begin{alignat}{2} [ 
      & x^2 + ( v_{2y} )^2  \\ 
      & - 2 d^{\prime} x \sin \phi_g \\ 
      & - 2 d^{\prime} \cos \phi_g ( v_{2y} ) \\ 
      & + d^{\prime 2} \\ 
      & + d^2 \sin^2 \theta_g + 2 d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g  ] \end{alignat}  \\

    & = \begin{alignat}{2} [ 
      & d^{\prime 2} + x^2 + ( d^2 \cos^2 \theta_g - 2dz \cos \theta_g \sin\theta_g + z^2 \sin^2 \theta_g )  \\ 
      & - 2 d^{\prime} x \sin \phi_g - 2 d^{\prime} \cos \phi_g ( v_{2y} ) \\ 
      & + d^2 \sin^2 \theta_g + 2 d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g  ] \end{alignat}  \\

    & = \begin{alignat}{2} [ 
      & d^{\prime 2} + x^2 + d^2 + z^2   \\ 
      & - 2 d^{\prime} x \sin \phi_g - 2 d^{\prime} \cos \phi_g ( v_{2y} ) ] \end{alignat}  \\

    & = 2 d^{\prime 2} - 2 d^{\prime} x \sin \phi_g - 2 d^{\prime} \cos \phi_g ( v_{2y} )\\

    & = 2 d^{\prime 2} - 2 d^{\prime} x \sin \phi_g - 2 d^{\prime} \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )  \\
    & = 2 d^{\prime} \left( d^{\prime} - x \sin \phi_g - \cos \phi_g ( d \cos \theta_g - z \sin \theta_g ) \right) \\
\left( \frac{q}{k} \right)^2
    & = 2 \left( 1 - \frac{x \sin \phi_g + \cos \phi_g ( d \cos \theta_g - z \sin \theta_g )}{d^{\prime} } \right)
\end{alignat}

Area Detector on Goniometer Arm, with offsets

In the most general case, the sample may not sit at the exact center of the goniometer rotation. In such a case, corrections must be applied.

TBD

See Also