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

International Tables for Crystallography (2013). Vol. D, ch. 1.1, p. 13

Section Symmetric tensors

A. Authiera*

aInstitut de Minéralogie et de Physique des Milieux Condensés, 4 Place Jussieu, 75005 Paris, France
Correspondence e-mail: Symmetric tensors

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A bilinear form is symmetric if [T({\bf x},{\bf y}) = T({\bf y},{\bf x}).]Its components satisfy the relations [t_{ij}= t_{ji}.]

The associated matrix, T, is therefore equal to its transpose [T^{T}]: [T = \pmatrix { t_{11} & t_{12} & t_{13}\cr t_{21} & t_{22} & t_{23}\cr t_{31} & t_{32} & t_{33}\cr } = T^T = \pmatrix { t_{11} & t_{21} & t_{31}\cr t_{12} & t_{22} & t_{32}\cr t_{13} & t_{23} & t_{33}\cr}.]In a space with n dimensions, the number of independent components is equal to [(n^{2}- n)/2 + n = (n^{2} + n)/2.]


  • (1) The metric tensor (Section[link]) is symmetric because the scalar product is commutative.

  • (2) The tensors representing one of the physical properties associated with the leading diagonal of the matrix of physical properties (Section[link]), such as the dielectric constant. Let us take up again the demonstration of this case and consider a capacitor being charged. The variation of the stored energy per unit volume for a variation dD of the displacement is [\hbox{d}W = {\bf E}\cdot\hbox{d}{\bf D},]where [equation ([link]] [D^{i}= \varepsilon ^i_{j}E\hskip1pt^{j}.]

    Since both [D^{i}] and [E\hskip1pt^{j}] are expressed through contravariant components, the expression for the energy should be written [\hbox{d}W = g_{ij}E\hskip1pt^{j}\,\,\hbox{d}D^{i}.]If we replace [D^{i}] by its expression, we obtain [\hbox{d}W = g_{ij}\varepsilon ^i_{k}E\hskip1pt^{j} \,\,\hbox{d}E^{k} = \varepsilon _{jk}E\hskip1pt^{j}\,\,\hbox{d}E^{k},]where we have introduced the doubly covariant form of the dieletric constant tensor, [\varepsilon _{jk}]. Differentiating twice gives [{\partial ^{2}W\over \partial E^{k}\partial E\hskip1pt^{j}} = \varepsilon _{jk}.]

    If one can assume, as one usually does in physics, that the energy is a `good' function and that the order of the derivatives is of little importance, then one can write [{\partial ^{2}W\over\partial E^{k}\partial E\hskip1pt^{j}} = {\partial ^{2}W\over \partial E\hskip1pt^{j}\partial E^{k}}.]As one can exchange the role of the dummy indices, one has [\partial ^{2}W/(\partial E\hskip1pt^{j}\partial E^{k}) = \varepsilon _{kj}.]Hence one deduces that [\varepsilon _{jk}= \varepsilon _{kj}.]

    The dielectric constant tensor is therefore symmetric. One notes that the symmetry is conveyed on two indices of the same variance. One could show in a similar way that the tensor representing magnetic susceptibility is symmetric.

  • (3) There are other possible causes for the symmetry of a tensor of rank 2. The strain tensor (Section 1.3.1[link] ), which is a field tensor, is symmetric because one does not take into account the rotative part of the deformation; the stress tensor, also a field tensor (Section 1.3.1[link] ), is symmetric because one neglects body torques (couples per unit volume); the thermal conductivity tensor is symmetric because circulating flows do not produce any detectable effects etc. Tensors of higher rank

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A tensor of rank higher than 2 may be symmetric with respect to the indices of one or more couples of indices. For instance, by its very nature, the demonstration given in Section[link] shows that the tensors representing principal physical properties are of even rank. If n is the rank of the associated square matrix, the number of independent components is equal to [(n^{2} + n)/2]. In the case of a tensor of rank 4, such as the tensor of elastic constants relating the strain and stress tensors (Section[link] ), the number of components of the tensor is [3^{4} = 81]. The associated matrix is a [ 9 \times 9] one, and the number of independent components is equal to 45.

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