Case Study: Choice of Basis and the UB Matrix#

This page works through the two questions posed at the end of the case study diffractometer — the six-circle equipment described there. The objective is to demonstrate that the choice of basis is arbitrary: different basis assignments produce numerically different axis vectors and U/UB matrices that are related by a fixed rotation, but the underlying physics — the conversion of motor angles to (h, k, l) and back — is invariant. The package’s flexibility around BASIS_YOU, BASIS_BL, and any other right-handed basis exists because of this invariance, not in spite of it.

For background on the B, U, UB matrices themselves and the You (1999) full diffraction equation cited below, see Concepts.

References:

W.R. Busing & H.A. Levy, Acta Cryst. 22, 457–464 (1967). DOI: 10.1107/S0365110X67000970
H. You, J. Appl. Cryst. 32, 614–623 (1999). DOI: 10.1107/S0021889899001223

Question 1 — assignment using the You (1999) basis#

The You (1999) convention assigns the three observable physical directions to Cartesian unit vectors as:

xHat: vertical (along the mu and nu rotation axes)
yHat: longitudinal (along the incoming beam direction)
zHat: transverse

This is a valid right-handed system: xHat × yHat = zHat. From You (1999) §2: “the x axis is defined along the vertical mu and nu axes and the y axis is defined along the incoming beam direction.”

The You (1999) paper describes the same 4S+2D six-circle geometry as the case-study equipment: four sample-orienting stages (mu, eta, chi, phi) and two detector stages (nu, delta). Mapping the case-study stages onto the You convention:

You angle	You matrix	Case-study stage	Physical axis	Axis vector
mu	U	sample stage 1	vertical	+xHat
eta	X	sample stage 2	transverse	−zHat
chi	H	sample stage 3	longitudinal	+yHat
phi	M	sample stage 4	transverse	−zHat
nu	P	detector stage 1	vertical	+xHat
delta	D	detector stage 2	transverse	−zHat

Notes on this assignment:

mu and nu share a colinear vertical axis (+xHat) with the same right-handed sense of rotation; the stages are mechanically independent.
eta, phi, and delta share the transverse axis (−zHat) — that is, left-handed rotation about +zHat. The signed axis vector encodes the handedness; see Concepts for the sign-convention identity.
chi is the only stage with a longitudinal axis (+yHat), right-handed.

Computing U in the You basis#

Let \(\mathsf{U}_\text{You}\) denote the orthogonal matrix that relates the crystal Cartesian frame to the phi-axis frame in this basis (the “U matrix” of Busing & Levy 1967, eq. 4). The procedure to determine \(\mathsf{U}_\text{You}\) from the geometry above is:

Construct the B matrix from the sample lattice parameters \((a, b, c, \alpha, \beta, \gamma)\) as described in Direct Lattice in Crystallography and Busing & Levy (1967, eq. 3). B encodes the reciprocal-lattice metric only and is independent of the basis assignment for the diffractometer axes.
Measure two non-collinear reflections at known motor angles — each provides a Q-vector in the phi-axis frame computed from the product of the four sample-stage rotation matrices in the You basis (mu about +xHat, eta about −zHat, chi about +yHat, phi about −zHat) acting on the lab-frame scattering vector.
Solve for the orthogonal \(\mathsf{U}_\text{You}\) that aligns the crystal Cartesian frame onto the phi-axis frame using Busing & Levy (1967, eqs. 27–28), or equivalently use the package helper ub_from_two_reflections().

The resulting \(\mathsf{UB}_\text{You} = \mathsf{U}_\text{You}\, \mathsf{B}\) maps Miller indices directly to the phi-axis frame in the You basis.

Question 2 — assignment using a different basis#

Any right-handed orthogonal mapping of the three physical directions to Cartesian axes is also valid. The Busing & Levy (1967) convention is the natural alternative:

xHat: transverse
yHat: longitudinal
zHat: vertical

This is again a valid right-handed system: xHat × yHat = zHat. This is the convention used in BASIS_BL and in the SPEC control system.

The physical equipment is unchanged — the same six rotation stages, the same axes of rotation, the same handedness. Only the labels change. The same case-study stages now have these axis vectors:

Stage	Physical axis	You-basis vector	BL-basis vector
sample stage 1	vertical	+xHat (You)	+zHat (BL)
sample stage 2	transverse	−zHat (You)	−xHat (BL)
sample stage 3	longitudinal	+yHat (You)	+yHat (BL)
sample stage 4	transverse	−zHat (You)	−xHat (BL)
detector stage 1	vertical	+xHat (You)	+zHat (BL)
detector stage 2	transverse	−zHat (You)	−xHat (BL)

The longitudinal direction is +yHat in both bases (so chi’s axis vector reads identically as +yHat); only the assignments of the vertical and transverse directions to xHat and zHat are swapped.

Computing U in the BL basis#

The procedure is structurally identical to Question 1 — same B matrix, same two reflections, same package helper — but every sample- stage rotation is now constructed about the BL-basis axis vector listed in the table above. Call the resulting orthogonal matrix \(\mathsf{U}_\text{BL}\) and the product \(\mathsf{UB}_\text{BL} = \mathsf{U}_\text{BL}\,\mathsf{B}\).

Numerically \(\mathsf{U}_\text{BL} \neq \mathsf{U}_\text{You}\), and \(\mathsf{UB}_\text{BL} \neq \mathsf{UB}_\text{You}\). The two matrices differ by exactly the rotation that carries one basis into the other — see the next section.

The invariance result#

Define \(\mathsf{R}_\text{You}^\text{BL}\) as the orthogonal matrix that re-expresses any vector from the You Cartesian frame to the BL Cartesian frame. Because both bases are right-handed and assign the same three physical directions to some permutation of \(\pm\hat{x}\), \(\pm\hat{y}\), \(\pm\hat{z}\), \(\mathsf{R}_\text{You}^\text{BL}\) is a fixed signed-permutation matrix that depends only on the two basis dictionaries — not on the sample, the lattice, the wavelength, or any motor angle.

The U and UB matrices in the two bases are related by a single similarity-style transformation:

\[ \mathsf{U}_\text{BL} \;=\; \mathsf{R}_\text{You}^\text{BL}\, \mathsf{U}_\text{You}\, \bigl(\mathsf{R}_\text{You}^\text{BL}\bigr)^{\mathsf{T}}. \]

The B matrix is unchanged because it depends only on the reciprocal lattice, not on the diffractometer-axis basis, so

\[ \mathsf{UB}_\text{BL} \;=\; \mathsf{R}_\text{You}^\text{BL}\, \mathsf{UB}_\text{You}\, \bigl(\mathsf{R}_\text{You}^\text{BL}\bigr)^{\mathsf{T}}. \]

The physical Bragg condition — the equation \(\mathsf{h}^M = M\,H\,X\,U_\mu \cdot \mathsf{UB} \cdot \mathsf{h}\) of You (1999, eqs. 10–11) — is therefore invariant under the basis change: every motor rotation matrix on the left-hand side picks up the same conjugation by \(\mathsf{R}_\text{You}^\text{BL}\), and the conjugations cancel. In words:

The motor angles produced by forward(h, k, l) and the (h, k, l) recovered by inverse(angles) are the same regardless of which right-handed basis was used to construct the geometry.

The choice of basis is therefore a representational convenience. It fixes which Cartesian symbols appear in the printed axis vectors and in the printed U/UB matrices, but it does not affect the diffraction calculation. This is why AdHocDiffractometer exposes a basis constructor argument and ships several pre-made choices (BASIS_YOU, BASIS_BL): users can pick the convention they are most comfortable comparing against published literature, without changing what the package computes.