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

Components of the rotation tensor in terms of Euler angles

Previously, we showed that if

\[\begin{align} \bs{R}&=\uv{e}_i\otimes \uv{E}_i\\ \end{align}\]

and

\[\begin{align} \label{eq:eis} \uv{e}'_i&=\bs{R}_1 \uv{E}_i\\ \uv{e}''_i&=\bs{R}_2 \uv{e}'_i\\ \uv{e}_i&=\bs{R}_3 \uv{e}''_i\\ \end{align}\]

then \(\begin{align} \bs{R} &=\bs{R}_3\bs{R}_2 \bs{R}_1 \end{align}\)

In this section we are going to derive the componenets of $\bs{R}$ in the \(\uv{E}_i\) basis. Let \([\bs{R}_1]_{\{\uv{E}_i\}}\) be the matrix representation of $\bs{R}_1$ w.r.t the \(\uv{E}_i\) basis, \([\bs{R}_2]_{\{\uv{e}'_i\}}\) be the matrix representation of \(\bs{R}_2\) w.r.t the \(\uv{e}'_i\) basis, and \([\bs{R}_3]_{\{\uv{e}''_i\}}\) be the matrix representation of $\bs{R}_3$ w.r.t the \(\uv{e}''_i\) basis. We focus on these three matrix representations because they are easy to compute. We will demonstrate this with an example later.

Say we know \([\bs{R}_1]_{\{\uv{E}_i\}}\), \([\bs{R}_2]_{\{\uv{e}'_i\}}\), and \([\bs{R}_3]_{\{\uv{e}''_i\}}\). Can we compute \([\bs{R}]_{\{\uv{E}_i\}}\)? The answer is yes, as we show below.

From the definition of the componenets of second order tensors, we have that

\[\begin{align} \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{ji}&=(\bs{R}_1\uv{E}_i)\cdot \uv{E}_j\\ \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{ji}&=(\bs{R}_2\uv{e}'_i)\cdot \uv{e}'_j\\ \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{ji}&=(\bs{R}_3\uv{e}''_i)\cdot \uv{e}''_j\\ \end{align}\]

From $\eqref{eq:eis}$ it follows that

\[\begin{align} \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{ji}&=(\uv{e}'_i)\cdot \uv{E}_j\\ \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{ji}&=(\uv{e}''_i)\cdot \uv{e}'_j\\ \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{ji}&=(\uv{e}_i)\cdot \uv{e}''_j\\ \end{align}\]

From the above three equations we have that, \(\begin{align} \bs{e}'_i&=\left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{ji}\uv{E}_j\\ \bs{e}''_i&=\left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{ji}\uv{e}'_j\\ \bs{e}_i&=\left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{ji}\uv{e}''_j \end{align}\)

The above three equations can be combined to read \(\begin{align} \bs{e}_i&=\left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{ji}\uv{e}''_j\\ &=\left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{ji} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pj}\uv{e}'_p\\ &=\left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{ji} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pj} \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{qp}\uv{E}_q \end{align}\) which can be re-arranged to read

\[\begin{align} \bs{e}_i &= \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{qp} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pj} \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{ji} \uv{E}_q\\ \end{align}\]

Noting that $\uv{e}_i=R\uv{E}_i$, we get

\[\begin{align} \bs{R}\bs{E}_i &= \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{qp} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pj} \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{ji} \uv{E}_q\\ \end{align}\]

Replacing the dummy index $j$ with a different dummy index $m$, and then taking a dot product of the above equation w.r.t $\bs{E}_i$, we get

\[\begin{align} \left(\bs{R}\bs{E}_i\right)\cdot\uv{E}_j &= \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{qp} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pm} \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{mi} \uv{E}_q\cdot\uv{E}_j\\ \left(\bs{R}\bs{E}_i\right)\cdot\uv{E}_j &= \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{qp} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pm} \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{mi} \uv{E}_q\cdot\uv{E}_j\\ &= \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{qp} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pm} \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{mi} \delta_{qj} \\ &=\left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{jp} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pm} \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{mi} \end{align}\]

Noting that \(\left(\bs{R}\bs{E}_i\right)\cdot\uv{E}_j=\left([\bs{R}]_{\{\uv{E}_i\}}\right)_{ji}\)we get that

\[\begin{align} \left([\bs{R}]_{\{\uv{E}_i\}}\right)_{ij}&= \left([\bs{R}_1]_{\{\uv{E}_i\}}\right)_{ip} \left([\bs{R}_2]_{\{\uv{e}'_i\}}\right)_{pm} \left([\bs{R}_3]_{\{\uv{e}''_i\}}\right)_{mj} \end{align}\]

or equivalently,

\[\begin{align} [\bs{R}]_{\{\uv{E}_i\}}&= [\bs{R}_1]_{\{\uv{E}_i\}} [\bs{R}_2]_{\{\uv{e}'_i\}} [\bs{R}_3]_{\{\uv{e}''_i\}} \end{align}\]

Example : Rotation tensor components in terms of the 3-2-1 Euler angles

To elaborate further on the Euler angles, we now consider the 3–2–1 set of Euler angles. Suppose that the rotation tensor has the representation \(\begin{align} \bs{R} &= \uv{e}_i \otimes \uv{E}_i, \end{align}\) where \(\uv{E}_i\) is a fixed Cartesian basis. The first rotation is about \(\uv{E}_3\) through an angle $\psi$. This rotation transforms \(\uv{E}_i\) to \(\uv{e}'_i\). The second rotation is about the \(\uv{e}'_2\) axis through an angle $\theta$. This rotation transforms \(\uv{e}'_i\) to \(\uv{e}''_i\). The third and last rotation is through an angle $\phi$ about the axis \(\uv{e}''_1 = \uv{e}_1\). Thus, \(\begin{align} \bs{R} = \bs{R}(\phi,\uv{e}_1) \bs{R}(\theta,\uv{e}'_2) \bs{R}(\psi,\uv{E}_3), \end{align}\) Here, \(\begin{align} \uv{e}_i &= \bs{R}(\phi,\uv{e}_1) \uv{e}''_i \\ \uv{e}''_i &= \bs{R}(\theta,\uv{e}'_2) \uv{e}'_i \\ \uv{e}'_i &= \bs{R}(\psi,\uv{E}_3) \uv{E}_i. \end{align}\)

It is not difficult to express the various basis vectors as linear combinations of each other: \(\begin{align} \label{eq:E2ep} \begin{bmatrix} \uv{e}'_1 \\ \uv{e}'_2 \\ \uv{e}'_3 \end{bmatrix} = \begin{bmatrix} \cos\psi & \sin\psi & 0 \\ -\sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \uv{E}_1 \\ \uv{E}_2 \\ \uv{E}_3 \end{bmatrix}, \end{align}\)

\[\begin{align} \label{eq:ep2epp} \begin{bmatrix} \uv{e}''_1 \\ \uv{e}''_2 \\ \uv{e}''_3 \end{bmatrix} = \begin{bmatrix} \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta \end{bmatrix} \begin{bmatrix} \uv{e}'_1 \\ \uv{e}'_2 \\ \uv{e}'_3 \end{bmatrix}, \end{align}\] \[\begin{align} \label{eq:epp2e} \begin{bmatrix} \uv{e}_1 \\ \uv{e}_2 \\ \uv{e}_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\phi & \sin\phi \\ 0 & -\sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix} \uv{e}''_1 \\ \uv{e}''_2 \\ \uv{e}''_3 \end{bmatrix}, \end{align}\]

The inverses of these relationships are easy to obtain once you realize that each of the three matrices in $\eqref{eq:E2ep}$–$\eqref{eq:epp2e}$ are orthogonal. As the inverse of an orthogonal matrix is its transpose, we quickly arrive at the sought-after results: \(\begin{align} \label{eq:ep2E} \begin{bmatrix} \uv{E}_1 \\ \uv{E}_2 \\ \uv{E}_3 \end{bmatrix} = \begin{bmatrix} \cos\psi & -\sin\psi & 0 \\ \sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \uv{e}'_1 \\ \uv{e}'_2 \\ \uv{e}'_3 \end{bmatrix}, \end{align}\)

\[\begin{align} \label{eq:epp2ep} \begin{bmatrix} \uv{e}'_1 \\ \uv{e}'_2 \\ \uv{e}'_3 \end{bmatrix} = \begin{bmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{bmatrix} \begin{bmatrix} \uv{e}''_1 \\ \uv{e}''_2 \\ \uv{e}''_3 \end{bmatrix}, \end{align}\] \[\begin{align} \label{eq:e2epp} \begin{bmatrix} \uv{e}''_1 \\ \uv{e}''_2 \\ \uv{e}''_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\phi & -\sin\phi \\ 0 & \sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix} \uv{e}_1 \\ \uv{e}_2 \\ \uv{e}_3 \end{bmatrix}, \end{align}\]

By using $\eqref{eq:ep2E}$–$\eqref{eq:e2epp}$ we can find a representation for the components \(R_{ij} = (\bs{R}\uv{E}_j)\cdot\uv{E}_i\). The components are easily expressed by a matrix representation: \(\begin{align} \label{eq:Rmat} \begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix} = \begin{bmatrix} \cos\psi & -\sin\psi & 0 \\ \sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\phi & -\sin\phi \\ 0 & \sin\phi & \cos\phi \end{bmatrix}. \end{align}\)

Representation $\eqref{eq:Rmat}$ is the transpose of what one might initially expect. However, recalling that \(R_{ik} = \uv{E}_i \cdot \uv{e}_k\) will hopefully resolve this initial confusion. Indeed, it can be noted that \(\begin{align} \begin{bmatrix} \uv{E}_1 \\ \uv{E}_2 \\ \uv{E}_3 \end{bmatrix} = \begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix} \begin{bmatrix} \uv{e}_1 \\ \uv{e}_2 \\ \uv{e}_3 \end{bmatrix}, \end{align}\) and \(\begin{align} \begin{bmatrix} \uv{e}_1 \\ \uv{e}_2 \\ \uv{e}_3 \end{bmatrix} = \begin{bmatrix} R_{11} & R_{21} & R_{31} \\ R_{12} & R_{22} & R_{32} \\ R_{13} & R_{23} & R_{33} \end{bmatrix} \begin{bmatrix} \uv{E}_1 \\ \uv{E}_2 \\ \uv{E}_3 \end{bmatrix}. \end{align}\)

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