Talk:Extra:Intersecting planes

Rotate 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{v}_{2b}} about $\mathbf {n} _{1}$

In general, rotation of a vector $\scriptstyle \mathbf {v} _{\mathrm {start} }={\begin{bmatrix}x&y&z\end{bmatrix}}$about an arbitrary unit-vector $\scriptstyle \mathbf {n} ={\begin{bmatrix}u&v&w\end{bmatrix}}$ gives (1, 2):

$\mathbf {v} _{\mathrm {end} }={\begin{bmatrix}u(ux+vy+wz)(1-\cos \theta )+x\cos \theta +(-wy+vz)\sin \theta \\v(ux+vy+wz)(1-\cos \theta )+y\cos \theta +(wx-uz)\sin \theta \\w(ux+vy+wz)(1-\cos \theta )+z\cos \theta +(-vx+uy)\sin \theta \end{bmatrix}}$

In this particular case, we thus expect:

${\begin{alignedat}{2}\mathbf {v} _{2}&={\begin{bmatrix}-y\sin \phi \\y\cos \phi \\z(1-\cos \phi )+z\cos \phi \end{bmatrix}}\\&={\begin{bmatrix}-q\cos \alpha \sin \phi \\q\cos \alpha \cos \phi \\q\sin \alpha (1-\cos \phi )+q\sin \alpha \cos \phi \end{bmatrix}}\\&=q{\begin{bmatrix}-\cos \alpha \sin \phi \\\cos \alpha \cos \phi \\\sin \alpha \end{bmatrix}}\end{alignedat}}$

Rotate $\mathbf {v} _{2b}$ about $\mathbf {n} _{2}$

In general, rotation of a vector $\scriptstyle \mathbf {v} _{\mathrm {start} }={\begin{bmatrix}x&y&z\end{bmatrix}}$about an arbitrary unit-vector $\scriptstyle \mathbf {n} ={\begin{bmatrix}u&v&w\end{bmatrix}}$ gives (1, 2):

$\mathbf {v} _{\mathrm {end} }={\begin{bmatrix}u(ux+vy+wz)(1-\cos \theta )+x\cos \theta +(-wy+vz)\sin \theta \\v(ux+vy+wz)(1-\cos \theta )+y\cos \theta +(wx-uz)\sin \theta \\w(ux+vy+wz)(1-\cos \theta )+z\cos \theta +(-vx+uy)\sin \theta \end{bmatrix}}$

In this particular case, we thus expect:

${\begin{alignedat}{2}\mathbf {v} _{2}&={\begin{bmatrix}(-wy+vz)\sin \phi \\v(vy+wz)(1-\cos \phi )+y\cos \phi \\w(vy+wz)(1-\cos \phi )+z\cos \phi \end{bmatrix}}\\&={\begin{bmatrix}(-\cos \alpha q\cos \alpha +-\sin \alpha q\sin \alpha )\sin \phi \\-\sin \alpha (-\sin \alpha q\cos \alpha +\cos \alpha q\sin \alpha )(1-\cos \phi )+q\cos \alpha \cos \phi \\\cos \alpha (-\sin \alpha q\cos \alpha +\cos \alpha q\sin \alpha )(1-\cos \phi )+q\sin \alpha \cos \phi \end{bmatrix}}\\&=q{\begin{bmatrix}-(\cos ^{2}\alpha +\sin ^{2}\alpha )\sin \phi \\\sin ^{2}\alpha (\cos \alpha -\cos \alpha )(1-\cos \phi )+\cos \alpha \cos \phi \\\cos ^{2}\alpha (-\sin \alpha +\sin \alpha )(1-\cos \phi )+\sin \alpha \cos \phi \end{bmatrix}}\\&=q{\begin{bmatrix}-\sin \phi \\\cos \alpha \cos \phi \\\sin \alpha \cos \phi \end{bmatrix}}\end{alignedat}}$

Generalized distance between two vectors

Warning: Errors below (this is just intermediate/working stuff)

Imagine reciprocal-space scattering that is a ring; more specifically a pseudo-toroid with Gaussian-like decay. The intensity overall is:

$I=\exp \left[-(q_{rr}-q_{0})^{2}/(2\sigma _{q}^{2})\right]\exp \left[-q_{rz}^{2}/(2\sigma _{q}^{2})\right]$

Where we use the subscript r to denote the reciprocal-space coordinate system, and $\scriptstyle q_{rr}={\sqrt {q_{rx}^{2}+q_{ry}^{2}}}$. The plane of the detector (i.e. the Ewald plane) is denoted by d:

$\mathbf {v} _{d}={\begin{bmatrix}q_{dx}&q_{dy}&0\end{bmatrix}}$

We set the symmetry axis in realspace (detector coordinate system) to be the $\scriptstyle x$-axis. The reciprocal-space is tilted 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 \chi_0} (about the 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 y} -axis), before the 'powder' rotation about the 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} -axis (where $\scriptstyle \alpha$ goes from 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 -\pi} to $\scriptstyle +\pi$). Consider an initial vector:

$\mathbf {v} _{d0}={\begin{bmatrix}q\sin \phi &q\cos \phi &0\end{bmatrix}}$

The 1st rotation (about 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 y} -axis 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 \chi_0} ) involves:

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} \mathbf{v}_{d1} & = R_y(\chi_0) \mathbf{v}_{d0} \\ & = \begin{bmatrix} \cos \chi_0 & 0 & \sin \chi_0 \\ 0 & 1 & 0 \\ -\sin \chi_0 & 0 & \cos \chi_0 \\ \end{bmatrix} \begin{bmatrix} q \sin \phi \\ q \cos \phi \\ 0 \\ \end{bmatrix} \\ & = \begin{bmatrix} q \sin \phi \cos \chi_0 \\ q \cos \phi \\ - q \sin \phi \sin \chi_0 \\ \end{bmatrix} \end{alignat} }

Consider a 2nd rotation around the vector (normal to the detector plane) (Warning: This is erroneous since the alpha rotation is just another phi rotation.):

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} \mathbf{n}_{d1} & = R_y(\chi_0) \mathbf{n}_{d0} \\ & = \begin{bmatrix} \cos \chi_0 & 0 & \sin \chi_0 \\ 0 & 1 & 0 \\ -\sin \chi_0 & 0 & \cos \chi_0 \\ \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} \\ & = \begin{bmatrix} \sin \chi_0 \\ 0 \\ \cos \chi_0 \\ \end{bmatrix} \end{alignat} }

So the second rotation yields:

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} \mathbf{v}_{d2} & = \begin{bmatrix} u(ux +wz)(1-\cos \alpha) + x \cos \alpha + -wy \sin \alpha \\ y \cos \alpha + (wx-uz) \sin \alpha\\ w(ux+wz)(1-\cos \alpha) + z \cos \alpha + uy \sin \alpha \end{bmatrix} \\ & = \begin{bmatrix} \sin \chi_0 (\sin \chi_0 q \sin \phi \cos \chi_0 - \cos \chi_0 q \sin \phi \sin \chi_0)(1-\cos \alpha) + q \sin \phi \cos \chi_0 \cos \alpha + - \cos \chi_0 q \cos \phi \sin \alpha \\ q \cos \phi \cos \alpha + (\cos \chi_0 q \sin \phi \cos \chi_0 + \sin \chi_0 q \sin \phi \sin \chi_0) \sin \alpha\\ \cos \chi_0(\sin \chi_0 q \sin \phi \cos \chi_0 - \cos \chi_0 q \sin \phi \sin \chi_0)(1-\cos \alpha) + - q \sin \phi \sin \chi_0 \cos \alpha + \sin \chi_0 q \cos \phi \sin \alpha \end{bmatrix} \\ & = q \begin{bmatrix} \sin^2 \chi_0 \sin \phi (\cos \chi_0 - \cos \chi_0 )(1-\cos \alpha) + \sin \phi \cos \chi_0 \cos \alpha - \cos \chi_0 \cos \phi \sin \alpha \\ \cos \phi \cos \alpha + \sin \phi (\cos^2 \chi_0 + \sin^2 \chi_0 ) \sin \alpha\\ \cos \chi_0 \sin \chi_0 \sin \phi (\cos \chi_0 - \cos \chi_0 )(1-\cos \alpha) - \sin \phi \sin \chi_0 \cos \alpha + \sin \chi_0 \cos \phi \sin \alpha \end{bmatrix} \\ & = q \begin{bmatrix} \cos \chi_0 ( \sin \phi \cos \alpha - \cos \phi \sin \alpha ) \\ \cos \phi \cos \alpha + \sin \phi \sin \alpha\\ -\sin \chi_0 ( \sin \phi \cos \alpha - \cos \phi \sin \alpha ) \end{bmatrix} \\ \end{alignat} }