Tensor and tensor transformation

A tensor is an array of numbers or functions that can be used to describe properties of a body, such as the scalar $x$ for temperature, the vector $\begin{pmatrix} a_1\\a_2 \\a_3 \end{pmatrix}$ for velocity, and the matrix $\begin{pmatrix} \sigma_{11} &\sigma_{12}&\sigma_{13}\\ \sigma_{21} &\sigma_{22}&\sigma_{23}\\ \sigma_{31} &\sigma_{32}& \sigma_{33}\end{pmatrix}$ for stress.

Tensors are categorised by their ranks (also known as order), where

1. A zeroth-order tensor is an array representing a quantity that only has magnitude and no direction, i.e. a scalar.
2. A first-order tensor is an array representing a quantity that has magnitude and a direction, i.e. a vector.
3. A second-order tensor is an array representing a quantity that has magnitude and two directions.

The above implies that the number of elements of a tensor is 3ⁿ, where n is the order of the tensor.

Next, we shall investigate how the elements of a tensor, particularly a 2^nd-order tensor, transform between two sets of coordinates. From an earlier article, u_j, which is a set of three quantities, is known as a first order tensor (i.e. a vector). Consider the product of two first order tensors u_jand v_l, whose elements with respect to another set of axes are given by:

$u'_iv'_k=\left ( a_{ij}u_j \right )\left ( a_{kl}v_l \right )\; \; \; \; \; \; \; \; 12a$

Since each term on the RHS of eq12a is a scalar, we can rewrite eq12a as

$u'_iv'_k=a_{ij}a_{kl}u_jv_l=\sum_{j}\sum_{i}a_{ij}a_{kl}u_jv_l\; \; \; \; \; \; \; \; 12b$

There are 9 possible products of $u'_iv'_k$ because i = 1,2,3 and k = 1,2,3. In other words, $u'_iv'_k$ is a set of 9 elements, each being a sum of nine quantities , e.g.

$u'_1v'_1=a_{11}a_{11}u_1v_1+a_{11}a_{12}u_1v_2+a_{11}a_{13}u_1v_3$

$+a_{12}a_{11}u_2v_1+a_{12}a_{12}u_2v_2+a_{12}a_{13}u_2v_3$

$+a_{13}a_{11}u_3v_1+a_{13}a_{12}u_3v_2+a_{13}a_{13}u_3v_3$

We can therefore represent the elements of $u'_iv'_k$ by a 3×3 matrix $\begin{pmatrix} u'_1v'_1 &u'_1v'_2&u'_1v'_3\\ u'_2v'_1&u'_2v'_2&u'_2v'_3\\ u'_3v'_1&u'_3v'_2&u'_3v'_3 \end{pmatrix}$ . Similarly, $u_jv_l$ in eq12b is a multiplication of two first order tensors and can be represented by a 3×3 matrix $\begin{pmatrix} u_1v_1 &u_1v_2&u_1v_3\\ u_2v_1&u_2v_2&u_2v_3\\ u_3v_1&u_3v_2&u_3v_3 \end{pmatrix}$ . This results in:

$T'_{ik}=a_{ij}a_{kl}T_{jl}\; \; \; \; \; \; \; (13)$

Eq13 indicates that each element of $u'_1v'_1$ in one reference frame is a result of the transformation of $u_jv_l$ in a different reference frame by $a_{ij}a_{kl}$ . In general, a set of nine quantities $T_{jl}$ (as is the case of a second-order tensor) with reference to an orthogonal set of axes is transformed to another set of nine quantities $T'_{ik}$ with reference to another orthogonal set of axes by the transformation matrix $a_{ij}a_{kl}$ .

Using the same logic, a third order tensor transforms as follows:

$u'_iv'_kw'_m=\left ( a_{ij}u_j \right )\left ( a_{kl}u_l \right )\left ( a_{mn}u_n \right )=a_{ij}a_{kl}a_{mn}u_jv_lw_n$

$T'_{ikm}=a_{ij}a_{kl}a_{mn}T_{jln}$

and a fourth order tensor:

$u'_iv'_kw'_mx'_o=\left ( a_{ij}u_j \right )\left ( a_{kl}u_l \right )\left ( a_{mn}u_n \right )\left ( a_{op}x_p \right )=a_{ij}a_{kl}a_{mn}a_{op}u_jv_lw_nx_p$