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

Transpose, Symmetric, and Skew Symmetric parts of Rotation Tensor \(\u{R}\)

Transpose

\[\begin{align} \u{R} &= \u{a}\otimes\u{a} + (\u{I}-\u{a}\otimes\u{a})\cos\theta + (*\u{a})\sin\theta \\ (\u{R})^{\t{T}} &= (\u{a}\otimes\u{a})^{\t{T}} + (\u{I}-\u{a}\otimes\u{a})^{\t{T}}\cos\theta + (*\u{a})^{\t{T}}\sin\theta \end{align}\]

The dyadic product is a symmetric tensor:

\[\begin{align} (\u{a}\otimes\u{a})_{,ij} &= a_{i}a_{j} \\ (\u{a}\otimes\u{a})^{\t{T}}_{,ij} &= a_{j}a_{i} = a_{i}a_{j} = (\u{a}\otimes\u{a})_{,ij} \\ (\u{a}\otimes\u{a})^{\t{T}} &= \u{a}\otimes\u{a} \end{align}\]

While on the other hand, the Hodge Star operating on \(\u{a}\) produces a skew symmetric tensor.

\[\begin{align} (*\u{a})_{,ij} &= \epsilon_{ikj}a_{k} \\ (*\u{a})_{,ij}^{\t{T}} &= \epsilon_{jki}a_{k} = -\epsilon_{ikj}a_{k} = -(*\u{a})_{,ij} \\ (*\u{a})^{T} &= - (*\u{a}) \end{align}\]

So what we find is that

\[\begin{align} \Rightarrow\u{R}^{\t{T}} = \u{a}\otimes\u{a} + (\u{I}-\u{a}\otimes\u{a})\cos\theta - (*\u{a})\sin\theta \end{align}\]

The Symmetric and Skew Symmetric parts of \(\u{R}\) follow from their definitions as,

\[\begin{align} \t{sym}(\u{R}) &= \u{a}\otimes\u{a} + (\u{I}-\u{a}\otimes\u{a})\cos\theta \\ \t{skew}(\u{R}) &= (*\u{a})\sin\theta \end{align}\]

In Indicial Notation

1) We proved earlier that that \(\u{R}^{-1} = \u{R}^{\t{T}}\).

\(\u{R}^{-1}\) is a rotation about the same axis and the rotation with the opposite angle.

\[\begin{align} R_{ij} &= \hat{e}_{i}\hat{e}_{j} + (\delta_{ij} - \hat{e}_{i}\hat{e}_{j})\cos(\theta) - \epsilon_{ijk}\hat{e}_{k}\sin(\theta) \\ R_{ij}^{-1} &= \hat{e}_{i}\hat{e}_{j} + (\delta_{ij} - \hat{e}_{i}\hat{e}_{j})\cos(-\theta) - \epsilon_{ijk}\hat{e}_{k}\sin(-\theta) \\ R_{ij}^{-1} &= \hat{e}_{i}\hat{e}_{j} + (\delta_{ij} - \hat{e}_{i}\hat{e}_{j})\cos(\theta) + \epsilon_{ijk}\hat{e}_{k}\sin(\theta) \\ R_{ij}^{\t{T}} &= \hat{e}_{i}\hat{e}_{j} + (\delta_{ij} - \hat{e}_{i}\hat{e}_{j})\cos(\theta) - \epsilon_{ijk}\hat{e}_{k}\sin(\theta) \end{align}\]

2) Using the definition of symmetric tensor,

\[\begin{align} \t{sym}(\u{R}) &= \frac{1}{2}(\u{R}^{\t{T}}+\u{R}) \\ \t{sym}(\u{R})_{ij} &= \frac{1}{2}((\u{R}^{\t{T}})_{ij}+(\u{R})_{ij}) \\ \t{sym}(\u{R})_{ij} &= \frac{1}{2}((\u{R})_{ji}+(\u{R})_{ij}) \\ \t{sym}(\u{R})_{ij} &= \frac{1}{2}(R_{ji}+R_{ij}) \end{align}\]

where

\[\begin{align} R_{ji} &= \hat{e}_{j}\hat{e}_{i} + (\delta_{ji} - \hat{e}_{j}\hat{e}_{i})\cos(\theta) - \epsilon_{jik}\hat{e}_{k}\sin(\theta) \\ R_{ij} + R_{ji} &= \hat{e}_{i}\hat{e}_{j} + (\delta_{ij} - \hat{e}_{i}\hat{e_{j}})\cos(\theta) - \epsilon_{ijk}\hat{e}_{k}\sin(\theta) + \hat{e}_{j}\hat{e_{i}} + (\delta_{ji} - \hat{e}_{j}\hat{e}_{i})\cos(\theta) - \epsilon_{jik}\hat{e}_{k}\sin(\theta) \\ &= 2\hat{e}_{i}\hat{e}_{j} + 2(\delta_{ij} - \hat{e}_{i}\hat{e}_{j})\cos(\theta) \end{align}\]

and therefore

\[\begin{align} \t{sym}(\u{R})_{ij} &= \hat{e}_{i}\hat{e}_{j} + (\delta_{ij} - \hat{e}_{i}\hat{e}_{j})\cos(\theta) \\ \t{sym}(\u{R}) &= \hat{e}\cdot\hat{e} + (\u{I}-\hat{e}\cdot\hat{e})\cos(\theta) \end{align}\]

3) Using the definition of skew-symmetric tensor,

\[\begin{align} \t{skew}(\u{R}) &= \frac{1}{2}(\u{R}-\u{R}^{\t{T}}) \\ \t{skew}(\u{R})_{ij} &= \frac{1}{2}((\u{R})_{ij}-(\u{R}^{\t{T}})_{ij}) \\ \t{skew}(\u{R})_{ij} &= \frac{1}{2}((\u{R})_{ij}-(\u{R})_{ji}) \\ \t{skew}(\u{R})_{ij} &= \frac{1}{2}(R_{ij}-R_{ji}) \end{align}\]

expanding will result in

\[\begin{align} &\t{skew}(\u{R})_{ij} = \frac{1}{2}(\hat{e}_{i}\hat{e}_{j} + (\delta_{ij} - \hat{e}_{i}\hat{e}_{j})\cos(\theta) - \epsilon_{ijk}\hat{e}_{k}\sin(\theta) - \hat{e}_{j}\hat{e_{i}} - (\delta_{ji} - \hat{e}_{j}\hat{e}_{i})\cos(\theta) + \epsilon_{jik}\hat{e}_{k}\sin(\theta)) \\ &\t{skew}(\u{R})_{ij} = -\epsilon_{ijk}\hat{e}_{k}\sin(\theta) \\ &\t{skew}(\u{R}) = (*\hat{\u{e}})\sin(\theta) \end{align}\]

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