Indifferent Tensor Field

Question

An exercise in the continuum mechanics book by Gurtin says:

A spatial tensor field $A$ is indifferent if, during any change in observer, A transforms according to $A^{*}(x^{*},t) = Q(t)A(x,t)Q(t)^{T}$.

We have that $x^{*}(p,t) = Q(t)x(p,t)$.

The book does not give a definition for $A^*.$

What does "transforms according to" mean?

Answer 1

It means the following:

Let $x^*$ be a new coordinate system related to $x$ by $x^*(p,t) = Q(t)x(p,t)$. We denote by $A^*$ the spacial vector field in the coordinate $x^*$. Then $A$ is called indifferent if for any change of coordinate system as above we have $$A^*(x^*,t) = Q(t)A(x,t)Q(t)^T.$$

Answer 2

Perhaps this will do for your wish for concreteness?

Let $\{\boldsymbol{e}_i, \boldsymbol{e}_j, \boldsymbol{e}_k\}$ be a basis for the space. Under a change of reference frame, this basis becomes $$\{\boldsymbol{e}^*_i, \boldsymbol{e}^*_j, \boldsymbol{e}^*_k\}=\{\boldsymbol{Q}\boldsymbol{e}_i, \boldsymbol{Q}\boldsymbol{e}_j, \boldsymbol{Q}\boldsymbol{e}_k\}.$$ Suppose two different observers are attempting to study the same tensor field. In the reference frame defined by the basis $\{\boldsymbol{e}_i, \boldsymbol{e}_j, \boldsymbol{e}_k\}$ call this tensor field $\boldsymbol{A}$. In the other reference frame call this tensor field, $\boldsymbol{A}^*$. Then $\boldsymbol{A}$ and $\boldsymbol{A}^*$ can be written in components. $$\boldsymbol{A} = A_{ij}\boldsymbol{e}_i\otimes\boldsymbol{e}_j,\\ \boldsymbol{A^*} = A^*_{ij}\boldsymbol{e}^*_i\otimes\boldsymbol{e}^*_j.$$ But using the relation for $\{\boldsymbol{e}^*_i, \boldsymbol{e}^*_j, \boldsymbol{e}^*_k\}$ in terms of the basis $\{\boldsymbol{e}_i, \boldsymbol{e}_j, \boldsymbol{e}_k\}$, then \begin{align} \boldsymbol{A}^* &= A^*_{ij}\boldsymbol{Q}\boldsymbol{e}_i\otimes\boldsymbol{Q}\boldsymbol{e}_j\\ &= \boldsymbol{Q}\left(A^*_{ij}\boldsymbol{e}_i\otimes\boldsymbol{e}_j\right)\boldsymbol{Q}^{T}. \end{align} Now if the tensor field under study is indifferent, then by your definiton \begin{align} \boldsymbol{A}^* &= \boldsymbol{Q}\boldsymbol{A}\boldsymbol{Q}^{T}\\ &= \boldsymbol{Q}\left(A_{ij}\boldsymbol{e}_i\otimes\boldsymbol{e}_j\right)\boldsymbol{Q}^{T}. \end{align} Comparing the two ways of writing $\boldsymbol{A}^*$, it is seen that for indifferent tensor fields $A^*_{ij}=A_{ij}$.