[NOTES/CM-08006] Axis Angle Parametrization of Rotation Matrix

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}\label{EQ01}R_{\hat{n}},(\theta)=\exp(\theta I_{\hat{n}})\end{equation}where $I_{\hat{n}}$ is a real numerical matrix independent of $\theta$. We now prove that \(I_n\) must be antisymmetric. Since $R_{\hat{n}}$ are orthogonal \begin{equation} R_{\hat{n}}^{T}(\theta)=R_{\hat{n}}^{-1}(\theta)=R_{\hat{n}}(-\theta)\Longrightarrow \exp(\theta I_{\hat{n}}^{T})=\exp(-\theta I_n).\label{EQ02}\end{equation}Thus $I_{\hat{n}}^{T}=-I_{\hat{n}}$ and the matrix $I_{\hat{n}}$ is anti symmetric. Hence, for a given $\hat{n}$, $I_n$ can be written in the form\begin{equation} I=\left[\begin{array}{clc} 0 &a &b\\ -a &0 &c\\ -b &-c &0 \end{array}\right]\label{EQ03} \end{equation}where a,b,c are real numbers depending on $\hat{n}=(n_1,n_2,n_3)$ and are to befixed.Determination of a,b,c$R_{\hat{n}}(\theta)$ is a rotation about axis along unit vector $\hat{n}$, the components of every vector parallel to $\hat{n}$ do not change under rotations. Hence\begin{equation}{R_{\hat{n}}}(\theta)\hat{n}=\hat{n}\end{equation}Thus we get\begin{equation}\exp(\theta I_{\hat{n}})\hat{n}=\hat{n}\end{equation}or\begin{equation}\left[1+\theta I_{n}+\frac{\theta^2}{2}I_{n}^2+\dots\right]\hat{n}=\hat{n}\label{EQ06}.\end{equation}This equation should hold for all $\theta$. Comparing powers of $\theta$ on both sides of \eqref{EQ06}, wesee that it is necessary and sufficient that we must have\begin{equation}I_{\hat{n}} \hat{n}=0 .\end{equation}The above equation written in a full form gives\begin{equation}\left[\begin{array}{clc} 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 gives us the following relations\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}Hence\begin{equation}-\frac{a}{n_3}=\frac{b}{n_2}=-\frac{c}{n_1}=\mu.\end{equation}Therefore$$a=-\mu n_3,\quad b=\mu n_2, \quad c=-\mu n_1$$where \(\mu\) is a constant. Thus we can write the matrix \(I_n\) in the form\begin{equation} I_{\hat{n}}=\mu \left[\begin{array}{clc} 0 &-n_3 &n_2\\ n_3 &0 &-n_1\\ -n_2 &n_1 &0 \end{array}\right].\label{EQ12} \end{equation}We introduce three matrices \((I_1, I_2, I_3)(\equiv \vec I)\) defined by

\begin{equation} I_1=\left[\begin{array}{clc} 0 &0 &0\\ 0 &0 &-1\\ 0 &1 &0 \end{array}\right],I_2=\left[\begin{array}{clc} 0 &0 &1\\ 0 &0 &0\\ -1 &0 &0 \end{array}\right],I_3=\left[\begin{array}{clc} 0 &-1 &0\\ 1 &0 &-1\\ 0 &1 &0 \end{array}\right],\label{EQ13} \end{equation}and write \begin{equation}I_{\hat{n}}=n_1 I_1+n_2 I_2+n_3 I_3 \label{EQ14}\end{equation}

We have $R_{\hat{n}}(\theta)=\exp(\mu \theta \hat{n}\cdot \vec{I})$, with \(\vec I_n =(I_1,I_2,I_3)\).

Closed form for $R_{\hat n}(\theta)$It is straight forward to check that\begin{equation}(\hat{n}\cdot\vec{I})^{3}=-(\hat{n}\cdot\vec I)\label{EQ15}\end{equation}holds if $\hat{n}$ is a unit vector. We now expand \(R_n(\theta)\), and use the above relation, \eqref{EQ15}, to sum the series expansion. This gives the rotation matrix \(R_{\hat n}(\theta)\) in a closed form, \GBX{i}{cm-lec-08008}{(verify this)},

\begin{equation}R_{\hat{n}}(\theta)=\widehat {Id}+\sin (\mu \theta) (\hat{n}\cdot \vec{I})+(1-\cos \mu \theta)(\hat{n}\cdot\vec{I})^{2}\label{EQ16}\end{equation}were \(\widehat {Id}\) is the \(3\times3\) identity matrix.

Transformations of components of position vector

For a vector \(\vec V= (V_1, V_2, V_3)\) we use the notation \(\U{V}\) to denote column vector have components of \(\vec V\). It is easy to verify that \begin{equation} (\hat{n}\cdot\vec{I})\U{A}=\U{B} \Longrightarrow \vec{B}=\hat{n}\times \vec{A}. \end{equation} Using \eqref{EQ16} in $\U{x}\Prime=R_{n}(\theta) \U{x}$, we get \begin{equation}\label{EQ18} \vec{x}\Prime=\vec{x}+\sin (\mu \theta) \hat{n}\times \vec{x}+\big(1-\cos(\mu\theta)\big)\big(\hat{n}\times (\hat{n}\times \vec{x}\big)\big). \end{equation}

Fixing the constant $\mu$The constant $\mu$ can now be fixed by taking $\hat{n}=(0,0,1)$ and comparing the answer with the known result on rotation about $Z$-axis. For $\hat{n}=(0,0,1)$, \eqref{EQ18} gives\begin{eqnarray} x_1\Prime&=&\cos(\mu \theta){x_1}-\sin(\mu\theta){x_2}\\ x_2\Prime&=&\sin(\mu\theta){x_1}+\cos(\mu\theta){x_2}\\ x_3\Prime&=&x_3 \end{eqnarray}

Comparing the above equations with the known result for rotations about $Z$ axis we get \(\mu=-1\).Hence

\begin{equation}R_{\hat{n}}(\theta)=\exp(-\theta \hat{n}\cdot\vec{I})\end{equation}The transformation equations for the components of the position vector can be written as

\begin{equation}{\vec{x}}\Prime=\vec{x}+(1-\cos\theta)\big(\hat n\times(\hat n \times\vec{x})\big))-\sin\theta (\hat{n}\times \vec{x})\end{equation}`