[NOTES/ME-02004]-The structure of rotation matrices

The set of all rotations about a fixed axis $\hat{n}$ form a one parameter group. A well known result, Stone's Theorem, implies that the matrices $R_{\hat{n}}(\theta)$ may be written as \begin{equation} R_{\hat{n}}(\theta)=\exp(-i\theta J_{\hat{n}}) \end{equation} where $J_{\hat{n}}$ is a (numerical) hermitian matrix independent of $\theta$. If we introduce $I_{\hat{n}}= -i J_{\hat{n}}$, the matrix $I_{\hat{n}}$ is anti-hermitian, and also real because $R=\exp(\theta I_{\hat{n}})$ is real. In fact $R_{\hat{n}}(\theta) = \exp(\theta I_{\hat{n}})$ and \begin{equation} I_{\hat{n}}=\lim_{\theta\longrightarrow 0}\frac{R_{\hat{n}}(\theta)-I}{\theta}= \frac{d}{d\theta}R_{\hat{n}}(\theta)\Big|_{\theta=0}. \end{equation} Since $R_{\hat{n}}$ are orthogonal, \begin{equation} R_{\hat{n}}^{T}(\theta)=R_{\hat{n}}^{-1}(\theta)=R_{\hat{n}}(-\theta) \Longrightarrow \exp( I_{\hat{n}}^{T}\theta)=\exp(- I_{\hat{n}} \theta) \end{equation} Thus $I_{\hat{n}}^{T}=-I_{\hat{n}}$ and $I_{\hat{n}}$ is a real antisymmetric matrix. For a given $\hat{n}$, therefore, $I_{\hat{n}}$ has the form \begin{equation} I_{\hat{n}}=\left[\begin{array}{clc}0 &a &b\\ -a &0 &c\\\label{EQ22} -b &-c &0 \end{array}\right]. \end{equation} and \begin{equation} R_{\hat{n}}(\theta)=\exp(\theta I_{\hat{n}}). \end{equation} Here $a,b,c$ are real numbers depending on $\hat{n}=(n_1,n_2,n_3)$ and are to be fixed.

Determination of $a,b,c$:
$R_{\hat{n}}(\theta)$ is a rotation about axis along unit vector $\hat{n}$, and therefore the components of every vector parallel to $\hat{n}$ do not change under rotations. Hence % \begin{eqnarray} {R_{\hat{n}}}(\theta)\hat{n}&=&\hat{n},\\ \text{or} \quad \exp(\theta I_{\hat{n}})\hat{n}&=&\hat{n},\\ \text{or}\quad \Big[1+\theta I_{\hat{n}}+\frac{\theta^2}{2}I_{\hat{n}}^2+\dots\Big]\hat{n}=\hat{n}\label{ EQ40 } \end{eqnarray} should hold for all $\theta$. If we compare powers of $\theta$ on both sides of Eq.(\eqRef{EQ40}), we see that a necessary and sufficient condition for Eq.(\eqRef{EQ40}) to hold is \begin{equation} I_{\hat{n}} \hat{n}=0 . \end{equation} Hence we get \begin{equation} \left[\begin{array}{ccc} 0 &a &b\\ -a &0 &c\\ -b &-c &0 \end{array}\right]\left[\begin{array}{c} n_1\\n_2\\n_3 \end{array}\right]=0 . \end{equation} This equation implies \begin{eqnarray} a.n_2+b.n_3&=&0\Longrightarrow \frac{a}{n_3}=-\frac{b}{n_2},\\ -a.n_1+c.n_3&=&0\Longrightarrow \frac{a}{n_3}=\frac{c}{n_1},\\ -b.n_1-c.n_2&=&0\Longrightarrow \frac{-b}{n_2}=\frac{c}{n_1}, \end{eqnarray} or, \begin{equation} -\frac{a}{n_3}=\frac{b}{n_2}=-\frac{c}{n_1}=\mu.\label{EQ48} \end{equation} Using \EqRef{EQ48}, we get \begin{equation} a=\mu n_3,~~ b=-\mu n_2,~~ c=\mu n_1. \end{equation} Therefore, we obtain the form of matrix $I_n$: \begin{equation} I_{\hat{n}}=\mu \left[\begin{array}{ccc} 0 &-n_3 &n_2\\ n_3 &0 &-n_1\\ -n_2 &n_1 &0 \end{array}\right], \end{equation} and \begin{equation} R_{\hat{n}}(\theta)=\exp(\mu \theta I_n). \end{equation} % Writing \begin{equation} I_n=n_1 I_1+n_2 I_2+n_3 I_3, \end{equation} we get $R_{\hat{n}}(\theta)=\exp(\mu \theta \hat{n}\cdot \vec{I})$, where \begin{equation} I_1=\left[\begin{array}{rrr} 0 &0 &0\\ 0 &0 &-1\\ 0 &1 &0 \end{array}\right],I_2=\left[\begin{array}{rrr} 0 &0 &1\\ 0 &0 &0\\ -1 &0 &0 \end{array}\right],I_3=\left[\begin{array}{rrr} 0 &-1 &0\\ 1 &0 &-1\\ 0 &1 &0 \end{array}\right]. \end{equation} It is straight forward to check that \begin{equation} (\hat{n}\cdot\vec{I})^{3}=-(\hat{n}\cdot I) \end{equation} holds if $\hat{n}$ is a unit vector. Hence \begin{eqnarray} R_{n}(\theta) &=&\exp(\mu \theta \hat{n}\cdot \hat{I})\\ &=& 1+\mu \theta \hat{n}\cdot \vec{I}+\frac{(\mu \theta)^2}{2!} (\hat{n}\cdot \vec{I})^{3}+\frac{(\mu \theta)^3}{3!} (n\cdot I)^{3}+\frac{(\mu \theta)^4}{4!}(\hat{n}\cdot\vec{I})^4+\ldots,\\ &=& I+(\mu \theta -\frac{(\mu \theta)^3}{3!}+\frac{(\mu\theta)^5}{5!}+\dots) (\hat{n}\cdot I)+\nonumber \\ &&\quad (\frac{(\mu \theta)^2}{2!}-\frac{(\mu\theta)^4}{4!}+\dots)(\hat{n}\cdot \vec{I})^{2}. \end{eqnarray} %Hence \begin{equation} R_{\hat{n}}(\theta)=I+\sin (\mu \theta) \hat{n}\cdot \vec{I}+\big(1-\cos (\mu \theta)\big)(\hat{n}\cdot\vec{I})^{2}.\label{EQ58} \end{equation} Remembering \begin{equation} (\hat{n}\cdot\vec{I}A)=B \Longrightarrow \vec{B}=\hat{n}\times \vec{A}, \end{equation} and using \EqRef{EQ58} in $\acute{x}=R_{n}(\theta) x$ we get \begin{equation} {\vec{x}}'=\vec{x}+\sin \mu \theta \, \hat{n}\times \vec{x}+(1-\cos\mu\theta)\,\hat{n}\times \big(\hat{n}\times \vec{x}\big)\\ \end{equation} % This equation can also be written as \begin{eqnarray} {\vec{x}}' &=&\vec{x}+\sin (\mu \theta )\hat{n}\times \vec{x}+(1-\cos\mu\theta)\big((\hat{n}\cdot\vec{x})\hat{n}-\vec{x})\big) \\ &=&(\hat{n}\cdot\vec{x})\hat{n}+\cos\mu\theta \big(\vec{x}-(\hat{n}\cdot\vec{x})\hat{n}\big)+\sin\mu \theta \big(\hat{n}\times \vec{x}\big).\label{EQ62} \end{eqnarray} The constant $\mu$ can now be fixed by taking $\hat{n}$ to have special value$(0,0,1)$ and comparing the answer with the known result on rotation about $x_3$axis. For $\hat{n}=(0,0,1)$, \EqRef{EQ62} gives \begin{eqnarray} \acute{x}_1&=&\cos\mu \theta{x_1}-\sin\mu\theta{x_2}\\ \acute{x}_2&=&\sin\mu\theta{x_1}+\cos\mu\theta{x_2}\\ \acute{x}_3&=& x_3. \end{eqnarray} Comparing this with the known result for rotations about $x_3$ axis we get $\mu=-1$. Hence \begin{equation} R_{\hat{n}}(\theta)=\exp(-\theta \hat{n}\cdot\vec{I}) \end{equation} and \begin{eqnarray} {\vec{x}}{'} &=& \vec{x}-\sin\theta (\hat{n}\times \vec{x})+(1-\cos\theta)\big((\hat{n}\cdot\vec{x})\hat{n}-\vec{x})\big)\\ &=& \vec{x}-\sin\theta (\hat{n}\times \vec{x})+(1-\cos\theta)\big(\hat{n}\times(\hat{n}\times\vec{x}) \big) \end{eqnarray} Also, the inverse relation is obtained by changing \(\theta \to -\theta\): \begin{equation} \vec{x}=\vec{x}{'}+\sin \theta \big(\hat{n}\times\vec{x}{'}\big) +(1-\cos\theta)\big(\hat{n}\times(\hat{n}\times\vec{x}{'})\big). \end{equation}