CalculateUMatrix v1¶

CalculateUMatrix dialog.

Table of Contents

Summary
Properties
Description
Source

Summary ¶

Calculate the U matrix from a peaks workspace, given lattice parameters.

Properties ¶

Name	Direction	Type	Default	Description
PeaksWorkspace	InOut	PeaksWorkspace	Mandatory	An input workspace.
a	Input	number	Mandatory	Lattice parameter a
b	Input	number	Mandatory	Lattice parameter b
c	Input	number	Mandatory	Lattice parameter c
alpha	Input	number	Mandatory	Lattice parameter alpha
beta	Input	number	Mandatory	Lattice parameter beta
gamma	Input	number	Mandatory	Lattice parameter gamma

Given a set of peaks (Q in the goniometer frame, HKL values), and given lattice parameters $(a,b,c,\alpha,\beta,\gamma)$ , it will try to find the U matrix, using least squares approach and quaternions 1. Units of length are in in $\rm \AA$ , angles are in degrees.

The algorithm calculates first the B matrix according to Busing and Levi.

Given a set of peaks in the reference frame of the inner axis of the goniometer, $\rm Q_{gon}$ , indexed by $(h_i, k_i, l_i)$ , we want to find the U matrix that maps peaks in the reciprocal space of the sample to the peaks in the goniometer frame

(1) $\rm U \rm B \left( \begin{array}{c} h_i \\ k_i \\ l_i \\ \end{array} \right) = \rm Q_{gon,i}$

For simplicity, we define

(2) $\rm Q_{hkl,i} = \rm B \left( \begin{array}{c} h_i \\ k_i \\ l_i \\ \end{array} \right)$

In the real world, such a matrix is not always possible to find. Therefore we just try minimize the difference between the two sets of p

(3) $\sum_i |\rm U \rm Q_{hkl,i} - \rm Q_{gon,i}|^2 = \sum_i \left(|\rm U \rm Q_{hkl,i}|^2 + |\rm Q_{gon,i}|^2 -2 \rm U \rm Q_{hkl,i} \cdot \rm Q_{gon,i}\right)$

In equation (3), $\left|\rm U \rm Q_{hkl,i}\right|^2 = |\rm Q_{hkl,i}|^2$ , so the first two terms on the left side are U independent. Therefore we want to maximize

(4) $\sum_i \left(\rm U \rm Q_{hkl,i} \cdot \rm Q_{gon,i}\right)$

We are going to write the scalar product of the vectors in terms of quaternions 2. We define $q_{hkl,i} = \left(0, Q_{hkl,i}\right)$ , $q_{gon,i} = \left(0, Q_{gon,i}\right)$ and the rotation U is described by quaternion $u = \left(w,x,y,z\right)$

Then equation (4) will be written as

(5) $\sum_i \left(\rm U \rm Q_{hkl,i} \cdot \rm Q_{gon,i}\right) = 0.5 \cdot \left(u q_{hkl,i} u^*\right) q_{gon,i}\ + 0.5 \cdot q_{gon,i} \left(u q_{hkl,i} u^*\right)$

We define matrices

(6) $H_i= \left(\begin{array}{cccc} 0 & -q_{hkl,i,x} & -q_{hkl,i,y} & -q_{hkl,i,z} \\ q_{hkl,i,x} & 0 & q_{hkl,i,z} & -q_{hkl,i,y} \\ q_{hkl,i,y} & -q_{hkl,i,z} & 0 & q_{hkl,i,x} \\ q_{hkl,i,z} & q_{hkl,i,y} & -q_{hkl,i,x} & 0 \end{array} \right)$

and

(7) $S_i= \left(\begin{array}{cccc} 0 & -q_{gonl,i,x} & -q_{gon,i,y} & -q_{gon,i,z} \\ q_{gon,i,x} & 0 & -q_{gon,i,z} & q_{gon,i,y} \\ q_{gon,i,y} & q_{gon,i,z} & 0 & -q_{gon,i,x} \\ q_{gon,i,z} & -q_{gon,i,y} & q_{gon,i,x} & 0 \end{array} \right)$

Then, we can rewrite equation (5) using matrices 3, 4:

(8) $\sum_i \left(\rm U \rm Q_{hkl,i} \cdot \rm Q_{gon,i}\right) = \left(\begin{array}{cccc} w & x & y & z\end{array} \right) \sum_i H_i S_i \left( \begin{array}{c} w \\ x \\ y \\ z \end{array} \right)$

The problem of finding $\left(w,x,y,z\right)$ that maximizes the sum can now be rewritten in terms of eigenvectors of $HS= \sum_i \left(H_i S_i\right)$ . Let $\epsilon_j$ and $\nu_j$ be the eigenvalues and corresponding eigenvectors of $HS$ , with $\epsilon_0 > \epsilon_1 > \epsilon_2 > \epsilon_3$ . We can write any vector $(w,x,y,z)$ as a linear combination of the eigenvectors of $HS$ :

(9) $\left(w,x,y,z\right) = \delta_0 \nu_0 +\delta_1 \nu_1 +\delta_2 \nu_2 +\delta_3 \nu_3$

(10) $\left(\begin{array}{cccc} w & x & y & z\end{array} \right) HS \left( \begin{array}{c} w \\ x \\ y \\ z \end{array} \right) = \delta_0^2 \nu_0 HS \nu_0 + \delta_1^2 \nu_1 HS \nu_1 +\delta_2^2 \nu_2 HS \nu_2 +\delta_3 \nu_3 HS \nu_3$