Operations on tensors

Authier, A.

doi:10.1107/97809553602060000628

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

pdf | chapter contents | chapter index | related articles

International Tables for Crystallography (2006). Vol. D. ch. 1.1, pp. 7-8

Section 1.1.3.3. Operations on tensors

A. Authier^a ^*

^a Institut de Minéralogie et de la Physique des Milieux Condensés, Bâtiment 7, 140 rue de Lourmel, 75015 Paris, France
Correspondence e-mail: aauthier@wanadoo.fr

1.1.3.3. Operations on tensors

| top | pdf |

1.1.3.3.1. Addition

| top | pdf |

It is necessary that the tensors are of the same nature (same rank and same variance).

1.1.3.3.2. Multiplication by a scalar

| top | pdf |

This is a particular case of the tensor product.

1.1.3.3.3. Contracted product, contraction

| top | pdf |

Here we are concerned with an operation that only exists in the case of tensors and that is very important because of its applications in physics. In practice, it is almost always the case that tensors enter into physics through the intermediary of a contracted product.

(i) Contraction. Let us consider a tensor of rank 2 that is once covariant and once contravariant . Let us write its transformation in a change of coordinate system: $[t'^{j}_{i} = A\hskip1pt_{p}^{j} B_{i}^{q}t\kern1pt_{q}^{p}.]$

Now consider the quantity $[t'^{i}_{i}]$ derived by applying the Einstein convention $[(t'^{i}_{i} = t'^{1}_{1} + t'^{2}_{2} + t'^{3}_{3})]$ . It follows that $[\eqalignno{t'^{i}_{i} &= A^{i}_{p}B\hskip1pt^{q}_{i}t\hskip1pt^{p}_{q} = \delta^{q}_{p}t\hskip1pt^{q}_{q}\cr t'^{i}_{i} &= t\hskip1pt^{p}_{p}.\cr}]$

This is an invariant quantity and so is a scalar. This operation can be carried out on any tensor of rank higher than or equal to two, provided that it is expressed in a form such that its components are (at least) once covariant and once contravariant .

The contraction consists therefore of equalizing a covariant index and a contravariant index, and then in summing over this index. Let us take, for example, the tensor $[t'^{jk}_{i}]$ . Its contracted form is $[t'^{ik}_{i}]$ , which, with a change of basis, becomes $[t'^{ik}_{i} = A^{k}_{p}t\kern1pt^{qp}_{q}.]$

The components $[t\hskip1pt^{ik}_{i}]$ are those of a vector, resulting from the contraction of the tensor $[t\hskip1pt^{jk}_{i}]$ . The rank of the tensor has changed from 3 to 1. In a general manner, the contraction reduces the rank of the tensor from n to .

Example. Let us take again the operator of symmetry O. The trace of the associated matrix is equal to $[O^{1}_{1} + O^{2}_{2} + O^{3}_{3} = O^{i}_{i}.]$ It is the resultant of the contraction of the tensor O. It is a tensor of rank 0, which is a scalar and is invariant under a change of basis.
(ii) Contracted product. Consider the product of two tensors of which one is contravariant at least once and the other covariant at least once: $[p\hskip1pt^{jk}_{i} = t\hskip1pt^{j}_{i}z^{k}.]$ If we contract the indices i and k, it follows that $[p\hskip1pt^{ji}_{i} = t\hskip1pt^{j}_{i}z^{i}.]$

The contracted product is then a tensor of rank 1 and not 3. It is an operation that is very frequent in practice.
(iii) Scalar product. Next consider the tensor product of two vectors: $[t\hskip1pt^{j}_{i} = x_{i}y\hskip1pt^{j}.]$ After contraction, we get the scalar product: $[t\hskip1pt^{i}_{i} = x_{i}y^{i}.]$