International
Tables for
Crystallography
Volume D
Physical properties of crystals
Edited by A. Authier

© International Union of Crystallography 2006
International Tables for Crystallography (2006). Vol. D. ch. 1.3, pp. 72-73

Section 1.3.1.2.3. Cubic dilatation

A. Authiera* and A. Zarembowitchb

a Institut de Minéralogie et de la Physique des Milieux Condensés, Bâtiment 7, 140 rue de Lourmel, 75015 Paris, France, and bLaboratoire de Physique des Milieux Condensés, Université P. et M. Curie, 75252 Paris CEDEX 05, France
Correspondence e-mail:  aauthier@wanadoo.fr

1.3.1.2.3. Cubic dilatation

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Let [{\bf e}_{i}] be the basis vectors before deformation. On account of the deformation, they are transformed into the three vectors [{\bf e}'_{i} = B_{ij}{\bf e}_{j}.]The parallelepiped formed by these three vectors has a volume V′ given by [V' = ({\bf e}'_{1},{\bf e}'_{2},{\bf e}'_{3}) = \Delta (B) ({\bf e}_{1},{\bf e}_{2},{\bf e}_{3}) = \Delta (B)V, ]where [\Delta (B)] is the determinant associated with matrix B, V is the volume before deformation and [({\bf e}_{1},{\bf e}_{2},{\bf e}_{3}) = ({\bf e}_{1} \wedge {\bf e}_{2})\cdot {\bf e}_{3} ]represents a triple scalar product.

The relative variation of the volume is [{V' - V \over V} = \Delta (B) - 1. \eqno(1.3.1.5)]It is what one calls the cubic dilatation. [\Delta (B)] gives directly the volume of the parallelepiped that is formed from the three vectors obtained in the deformation when starting from vectors forming an orthonormal base.








































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