$$\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}$$ $$\renewcommand{\u}[1]{\boldsymbol{#1}} \renewcommand{\usf}[1]{\mathsf{\u{\mathsf{#1}}}}$$

Co-rotational basis

We showed in a previous section that a tensor can be written as $\begin{align} \u{T}&=T_{ij}\u{E}_i\otimes \u{E}_j, \end{align}$

where $\begin{align} T_{ij}=\u{E}_i\cdot(\u{T}\u{E}_j) \end{align}$

Let the rotation linear operator map the vectors $\u{E}_i$ to the vectors $\u{e}_i(\u{t})$, i.e.,

$$\begin{align} \u{e}_i(\u{t})&=\u{R}_{\u{t}}\u{E}_i. \end{align}$$

It can be shown that in the case of rotation tensors, the vectors $\u{e}_{i}(\u{t})$ form an orthonormal set, i.e., $\u{e}_i(\u{t}) \cdot \u{e}_j(\u{t})=\delta_{ij}$, and, in fact, that they form a basis for $\mathbb{E}$. The set $\left(\u{e}_i(\u{t})\right)_{i=1,2,3}$ is called the co-rotational basis (corresponding to $\left(\u{E}_i\right)_{i=1,2,3}$). It can be shown that,

$$\begin{align} \u{R}_{\u{t}}&=\u{e}_i(\u{t})\otimes\u{E}_i \end{align}$$

Mathematica file demonstrating co-rotational basis

[edit]