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

International Tables for Crystallography (2013). Vol. D, ch. 1.1, pp. 24-25

Section 1.1.4.10.3. Reduction of the number of independent components of third-rank polar tensors due to the symmetry of the strain and stress tensors

A. Authiera*

aInstitut de Minéralogie et de Physique des Milieux Condensés, 4 Place Jussieu, 75005 Paris, France
Correspondence e-mail: aauthier@wanadoo.fr

1.1.4.10.3. Reduction of the number of independent components of third-rank polar tensors due to the symmetry of the strain and stress tensors

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Equation (1.1.4.5)[link] can be written [P_{i} = \textstyle\sum\limits_{j} d_{ijj}T_{jj} + \textstyle\sum\limits_{j\neq k} (d_{ijk}+ d_{ikj})T_{jk}.]

The sums [(d_{ijk} + d_{ikj})] for [j \neq k] have a definite physical meaning, but it is impossible to devise an experiment that permits [d_{ijk}] and [d_{ikj}] to be measured separately. It is therefore usual to set them equal: [d_{ijk} = d_{ikj}. \eqno(1.1.4.7)]

It was seen in Section 1.1.4.8.1[link] that the components of a third-rank tensor can be represented as a [9 \times 3] matrix which can be subdivided into three [3 \times 3] submatrices:[\pmatrix{&{\bf 1}&|&{\bf 2}&|&{\bf 3}&\cr}.]

Relation (1.1.4.7)[link] shows that submatrices 1 and 2 are identical.

One puts, introducing a two-index notation, [\left. \matrix{\hfill d_{ijj} = d_{i\alpha}\quad (\alpha = 1, 2, 3){\phantom.}\cr d_{ijk} + d_{ikj}\,\, (j\neq k) = d_{i\alpha}\quad (\alpha = 4, 5, 6).\cr}\right\}]Relation (1.1.4.7)[link] becomes [P_{i} = d_{i\alpha} T_{\alpha}.]

The coefficients [d_{i\alpha}] may be written as a [3 \times 6] matrix: [\left(\matrix{11 &12 &13 \cr 21 &22 &23\cr 31 &32 &33\cr}\right | \left. \matrix{14 &15 &16\cr 24 &25 &26\cr 34 &35 &36\cr}\right).]This matrix is constituted by two [3 \times 3] submatrices. The left-hand one is identical to the submatrix 1, and the right-hand one is equal to the sum of the two submatrices 2 and 3:[\pmatrix{&{\bf 1}&|&{\bf 2} + {\bf 3}&\cr}.]

The inverse piezoelectric effect expresses the strain in a crystal submitted to an applied electric field: [S_{ij} = d_{ijk}E_{k},]where the matrix associated with the coefficients [d_{ijk}] is a [9 \times 3] matrix which is the transpose of that of the coefficients used in equation (1.1.4.5)[link], as shown in Section 1.1.1.4[link].

The components of the Voigt strain matrix [S_{\alpha}] are then given by [\left. \matrix{S_{\alpha } = d_{iik}E_{k} \hfill &(\alpha = 1,2,3)\cr S_{\alpha} = S_{ij} + S_{ji} = (d_{ijk} + d_{jik})E_{k} &(\alpha = 4,5,6).\cr}\right\}]This relation can be written simply as [S_{\alpha} = d_{\alpha k}E_{k},]where the matrix of the coefficients [d_{\alpha k}] is a [6 \times 3] matrix which is the transpose of the [d_{i\alpha}] matrix.

There is another set of piezoelectric constants (see Section 1.1.5[link]) which relates the stress, [T_{ij}], and the electric field, [E_{k}], which are both intensive parameters: [T_{ij} = e_{ijk}E_{k}, \eqno(1.1.4.8)]where a new piezoelectric tensor is introduced, [e_{ijk}]. Its components can be represented as a [3 \times 9] matrix:[\pmatrix{{\bf 1}\cr-\cr{\bf 2}\cr-\cr{\bf 3}\cr}.]

Both sides of relation (1.1.4.8)[link] remain unchanged if the indices i and j are interchanged, on account of the symmetry of the stress tensor. This shows that [e_{ijk} = e_{jik}.]

Submatrices 2 and 3 are equal. One introduces here a two-index notation through the relation [e_{\alpha k} = e_{ijk}], and the [e_{\alpha k}] matrix can be written[\left({{\bf 1}\over{{\bf 2}+{\bf 3}}}\right).]

The relation between the full and the reduced matrix is therefore different for the [d_{ijk}] and the [e_{kij}] tensors. This is due to the particular property of the strain Voigt matrix (1.1.4.6)[link], and as a consequence the relations between nonzero components of the reduced matrices are different for certain point groups (3, 32, [3m], [\bar{6}], [\bar{6}2m]).








































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