Talk:Extra:Intersecting planes

Rotate $\mathbf {v} _{2b}$ 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 $\scriptstyle \chi _{0}$ (about the $\scriptstyle y$-axis), before the 'powder' rotation about the $\scriptstyle x$-axis (where $\scriptstyle \alpha$ goes from $\scriptstyle -\pi$ 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 $\scriptstyle y$-axis by $\scriptstyle \chi _{0}$) involves:

${\begin{alignedat}{2}\mathbf {v} _{d1}&=R_{y}(\chi _{0})\mathbf {v} _{d0}\\&={\begin{bmatrix}\cos \chi _{0}&0&\sin \chi _{0}\\0&1&0\\-\sin \chi _{0}&0&\cos \chi _{0}\\\end{bmatrix}}{\begin{bmatrix}q\sin \phi \\q\cos \phi \\0\\\end{bmatrix}}\\&={\begin{bmatrix}q\sin \phi \cos \chi _{0}\\q\cos \phi \\-q\sin \phi \sin \chi _{0}\\\end{bmatrix}}\end{alignedat}}$

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.):

${\begin{alignedat}{2}\mathbf {n} _{d1}&=R_{y}(\chi _{0})\mathbf {n} _{d0}\\&={\begin{bmatrix}\cos \chi _{0}&0&\sin \chi _{0}\\0&1&0\\-\sin \chi _{0}&0&\cos \chi _{0}\\\end{bmatrix}}{\begin{bmatrix}0\\0\\1\\\end{bmatrix}}\\&={\begin{bmatrix}\sin \chi _{0}\\0\\\cos \chi _{0}\\\end{bmatrix}}\end{alignedat}}$

So the second rotation yields:

${\begin{alignedat}{2}\mathbf {v} _{d2}&={\begin{bmatrix}u(ux+wz)(1-\cos \alpha )+x\cos \alpha +-wy\sin \alpha \\y\cos \alpha +(wx-uz)\sin \alpha \\w(ux+wz)(1-\cos \alpha )+z\cos \alpha +uy\sin \alpha \end{bmatrix}}\\&={\begin{bmatrix}\sin \chi _{0}(\sin \chi _{0}q\sin \phi \cos \chi _{0}-\cos \chi _{0}q\sin \phi \sin \chi _{0})(1-\cos \alpha )+q\sin \phi \cos \chi _{0}\cos \alpha +-\cos \chi _{0}q\cos \phi \sin \alpha \\q\cos \phi \cos \alpha +(\cos \chi _{0}q\sin \phi \cos \chi _{0}+\sin \chi _{0}q\sin \phi \sin \chi _{0})\sin \alpha \\\cos \chi _{0}(\sin \chi _{0}q\sin \phi \cos \chi _{0}-\cos \chi _{0}q\sin \phi \sin \chi _{0})(1-\cos \alpha )+-q\sin \phi \sin \chi _{0}\cos \alpha +\sin \chi _{0}q\cos \phi \sin \alpha \end{bmatrix}}\\&=q{\begin{bmatrix}\sin ^{2}\chi _{0}\sin \phi (\cos \chi _{0}-\cos \chi _{0})(1-\cos \alpha )+\sin \phi \cos \chi _{0}\cos \alpha -\cos \chi _{0}\cos \phi \sin \alpha \\\cos \phi \cos \alpha +\sin \phi (\cos ^{2}\chi _{0}+\sin ^{2}\chi _{0})\sin \alpha \\\cos \chi _{0}\sin \chi _{0}\sin \phi (\cos \chi _{0}-\cos \chi _{0})(1-\cos \alpha )-\sin \phi \sin \chi _{0}\cos \alpha +\sin \chi _{0}\cos \phi \sin \alpha \end{bmatrix}}\\&=q{\begin{bmatrix}\cos \chi _{0}(\sin \phi \cos \alpha -\cos \phi \sin \alpha )\\\cos \phi \cos \alpha +\sin \phi \sin \alpha \\-\sin \chi _{0}(\sin \phi \cos \alpha -\cos \phi \sin \alpha )\end{bmatrix}}\\\end{alignedat}}$