Mechanics of Continua and Structures
I am going to replace the \(\u{e}\) vector with the \(\u{a}\) vector. Since we also use the \(\u{e}\) vector as the Cartesian basis vectors.
\[\begin{align} \u{u} &\sim u_i \\ \u{a} &\sim a_i \\ (\u{u}\cdot\u{a})\u{a} &\sim (u_{k}a_{k})a_{i} \\ \u{u} - (\u{u}\cdot\u{a})\u{a} &\sim u_{i}-(u_{k}a_{k})a_{i} \\ \u{a} \times \u{u} &\sim \epsilon_{ijk}a_{j}u_{k} \end{align}\]Using the above substitutions, we get
\[\begin{align} \u{R}\u{u} \sim (u_{k}a_{k})a_{i} - (u_{i}-(u_{k}a_{k})a_{i})\cos\theta + \epsilon_{ijk}a_{j}u_{k}\sin\theta. \end{align}\]Let us simplify the above expression on the right hand side:
\[\begin{align} (\u{R}\u{u})_{i} &= (a_{i}a_{k})u_{k} - (\delta_{ik}u_{k}-a_{k}a_{i}u_{k})\cos\theta + \epsilon_{ijk}a_{j}\sin\theta u_{k} \\ &=\underbrace{\left[a_{i}a_{k}-(\delta_{ij}-a_{i}a_{k})\cos\theta + \epsilon_{ijk}a_{j}\sin\theta\right]}_{R_{ij}}u_{k} \\ (\u{R}\u{u})_{i} &= R_{ij}u_{k} \\ R_{ij} &= \left[a_{i}a_{j}-(\delta_{ij}-a_{i}a_{j})\cos\theta + \epsilon_{ijk}a_{j}\sin\theta\right] \end{align}\]Here, * is the Hodge Star operator, which to any vector \(\u{a}\) associates the second order tensor that represents the linear mapping defined by the cross-product with \(\u{a}\). In indicial notation, it is \((*a)_{ij}=\epsilon_{ikj}a_{k}\).
In summary, for any vector \(\u{u}\), we can obtain its rotated version \(\u{v}\) under a rotation \(\u{R}\) as \(\u{v} = \u{R}\u{u}\), where we have obtained the explicit representation of \(\u{R}\) in both indicial notation and the invariant notation.
Example: We would like to construct a rotation tensor that corresponds to a \(\frac{\pi}{3}\) rotation around the axis \(\u{a} = (\u{e}_1+\u{e}_2+\u{e}_3)\frac{1}{\sqrt{3}}\). We will do this in component form.
\[\begin{align} \left[\u{a}\right] =\begin{bmatrix}\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\end{bmatrix}, \quad \cos(\frac{\pi}{3})=\frac{1}{2}, \quad \sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2} \end{align}\] \[\begin{align} (*a)_{ij} = \epsilon_{ikj}a_{k} = (\epsilon_{i1j} + \epsilon_{i2j} + \epsilon_{i3j})\frac{1}{\sqrt{3}}; \\ (*a)_{11} = (*a)_{22} = (*a)_{33} = 0; \\ (*a)_{12} = -\frac{1}{\sqrt{3}}; (*a)_{23} = -\frac{1}{\sqrt{3}}; \\ (*a)_{21}= +\frac{1}{\sqrt{3}}; (*a)_{32} = +\frac{1}{\sqrt{3}}; \\ (*a)_{13}= + \frac{1}{\sqrt{3}}; (*a)_{31}= -\frac{1}{\sqrt{3}} \end{align}\]The values can then be written as
\[\begin{align} \left[*\u{a}\right]\sin\theta= \left[*\u{a}\right]\sin(\frac{\pi}{3}) = \begin{bmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\\frac{1}{2} & 0 & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & 0\end{bmatrix} \end{align}\] \[\begin{align} \left[\u{e}\otimes\u{e}\right] = \begin{bmatrix}\frac{1}{\sqrt{3}}\times\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\times\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\times\frac{1}{\sqrt{3}} \\ '' & '' & '' \\ '' & '' & ''\end{bmatrix} = \begin{bmatrix}\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{bmatrix} \end{align}\]Final answer:
\[\begin{align} \left[\u{R}\right] = \begin{bmatrix}\frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\end{bmatrix} \end{align}\]