Calculation of the twin element

Koch, E.

doi:10.1107/97809553602060000574

International
Tables for
Crystallography
Volume C
Mathematical, physical and chemical tables
Edited by E. Prince

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International Tables for Crystallography (2006). Vol. C. ch. 1.3, p. 14

Section 1.3.5. Calculation of the twin element

E. Koch^a

^a Institut für Mineralogie, Petrologie und Kristallographie, Universität Marburg, Hans-Meerwein-Strasse, D-35032 Marburg, Germany

1.3.5. Calculation of the twin element

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If the twin element cannot be recognized by direct macroscopic or microscopic inspection, it may be calculated as described below. Given are two analogous bases a, b, c and a′, b′, c′ referring to the two twin components. If possible, both basis systems should be chosen with the same handedness. If no such bases exist, the twin is a reflection twin and one of the bases has to be replaced by its centrosymmetrical one, e.g. a′, b′, c′ by −a′, −b′, −c′. The relation between the two bases is described by $[\eqalign{ {\bf a}' &=e_{11}{\bf a}+e_{12}{\bf b}+e_{13}{\bf c}, \cr {\bf b}'& =e_{21}{\bf a}+e_{22}{\bf b}+e_{23}{\bf c}, \cr {\bf c}' &=e_{31}{\bf a}+e_{32}{\bf b}+e_{33}{\bf c}.}]$ The coefficients e_ij have to be obtained by measurement.

Basis a, b, c may be mapped onto a′, b′, c′ by a pure rotation that brings a to a′, b to b′, and c to c′. To derive the direction of the rotation axis, calculate the three vectors $[{\bf a}_1={\bf a}+{\bf a}', \quad {\bf b}_1={\bf b}+{\bf b}', \quad {\bf c}_1={\bf c}+{\bf c}'.]$ a₁, b₁, c₁ bisect the angles $[\sigma_a={\bf a}\wedge{\bf a}']$ , $[\sigma_b={\bf b}\wedge{\bf b}']$ , and $[\sigma_c={\bf c}\wedge{\bf c}']$ , respectively. Calculate three further vectors of arbitrary length a₂, b₂, c₂ which are perpendicular to the planes defined by a and a′, b and b′, and c and c′, respectively, from the scalar products $[\eqalign{ {\bf a}_2\cdot{\bf a}&={\bf a}_2\cdot{\bf a}'=0, \cr {\bf b}_2\cdot{\bf b}&={\bf b}_2\cdot{\bf b}'=0, \cr {\bf c}_2\cdot{\bf c}&={\bf c}_2\cdot\,{\bf c}'=0.}]$ The plane defined by a₁ and a₂ is perpendicular to the plane defined by a and a′ and bisects the angle $[{\bf a}\wedge{\bf a}']$ . Analogous planes refer to b₁ and b₂, and c₁ and c₂. Vectors $[{\bf r}_a]$ , $[{\bf r}_b]$ , and $[{\bf r}_c]$ lying within one of these planes may be described as linear combinations of a₁ and a₂, b₁ and b₂, or c₁ and c₂, respectively: $[\eqalign{ {\bf r}_a&=\lambda_a{\bf a}_1+\mu_a{\bf a}_2, \cr {\bf r}_b &=\lambda_b{\bf b}_1+\mu_b{\bf b}_2, \cr {\bf r}_c &=\lambda_c{\bf c}_1\,+\mu_c{\bf c}_2.}]$ The common intersection line of these three planes is parallel to the twin axis. It may be calculated by solving any of the three equations $[{\bf r}_a={\bf r}_b, \quad {\bf r}_a={\bf r}_c, \quad \hbox{or} \quad {\bf r}_b={\bf r}_c.]$ $[{\bf r}_a={\bf r}_b]$ : choose $[\lambda_a]$ arbitrarily equal to 1. $[{\bf a}_1+\mu_a{\bf a}_2=\lambda_b{\bf b}_1+\mu_b{\bf b}_2.]$ Solve the inhomogeneous system of three equations that corresponds to this vector equation for the three variables $[\mu_a]$ , $[\lambda_b]$ , and $[\mu_b]$ . Calculate the vector $[{\bf r}={\bf a}_1+\mu_a{\bf a}_2]$ . Its components with respect to a, b, c describe the direction of the twin axis.

The angle τ of the twin rotation may then be calculated by $[ \sin\textstyle{1\over2}\tau = \displaystyle{\sin{1\over2}\sigma_a \over\sin\delta_a} =\displaystyle{\sin{1\over2}\sigma_b\over\sin\delta_b} = \displaystyle{\sin{1\over2}\sigma_c\over\sin\delta_c}]$ with $[\delta_a={\bf r}\wedge{\bf a}, \delta_b={\bf r}\wedge{\bf b}, \delta_c={\bf r}\wedge{\bf c}]$ .

If the basis a, b, c is orthogonal, τ may be obtained from $[\cos\tau= \textstyle{1\over2}(\cos\sigma_a+\cos\sigma_b+\cos\sigma_c-1).]$ If the coefficients of r are rational and τ equals 180°, then r describes the direction either of the twofold twin axis or of the normal of the twin plane. If r is rational and τ equals 60, 90 or 120°, r is parallel to the twin axis. If r is irrational, but τ equals 180° and there exists, in addition, a net plane perpendicular to r, this net plane describes the twin plane.

If none of these conditions is fulfilled, one has to repeat the calculations with a differently chosen basis system for one of the twin components. The number of possibilities for this choice depends on the lattice symmetry. The following list gives all equivalent basis systems for all descriptions of Bravais lattices used in IT A (2005):

aP : a, b, c;
mP, mS (unique axis b): a, b, c; −a, b, −c;
mP, mS (unique axis c): a, b, c; −a, −b, c;
oP, oS, oI, oF : a, b, c; −a, −b, c; −a, b, −c; a, −b, −c;
tP , tI: a, b, c; −a, −b, c; −a, b, −c; a, −b, −c; b, −a, c; −b, a, c; b, a, −c; −b, −a, −c;
hP : a, b, c; b, −a − b, c; −a − b, a, c; b, a, −c; −a − b, b, −c; a, −a − b, −c; −a, −b, c; −b, a + b, c; a + b, −a, c; −b, −a, −c; a + b, −b, −c; −a, a + b, −c;
hR (hexagonal description): a, b, c; b, −a − b, c; −a − b, a, c; b, a, −c; −a − b, b, −c; a, −a − b, −c;
hR (rhombohedral description): a, b, c; b, c, a; c, a, b; −b, −a, −c; −a, −c, −b; −c, −b, −a;
cP , cI, cF: a, b, c; b, c, a; c, a, b; −a, −b, c; −b, c, −a; c, −a, −b; −a, b, −c; b, −c, −a; −c, −a, b; a, −b, −c; −b, −c, a; −c, a, −b; −b, −a, −c; −a, −c, −b; −c, −b, −a; b, a, −c; a, −c, b; −c, b, a; b, −a, c; −a, c, b; c, b, −a; −b, a, c; a, c, −b; c, −b, a.

References

International Tables for Crystallography (2005). Vol. A, Space-group symmetry, edited by Th. Hahn. Heidelberg: Springer.Google Scholar

International Tables for Crystallography (2006). Vol. C. ch. 1.3, p. 14