14.1: C.1- Infintesimal Rotations
- Page ID
- 18091
Any rotation matrix can be written as the identity plus something:
\[\underset{\sim}{C}=\underset{\sim}{\delta}+\underset{\sim}{r} \nonumber \]
The orthogonality requirement \(\underset{\sim}{C}^T \underset{\sim}{C} = \underset{\sim}{\delta}\) can then be expressed as:
\[\begin{aligned}
\underset{\sim}{C}^{T} \underset{\sim}{C} &=\left(\underset{\sim}{\delta}^{T}+\underset{\sim}{r}^{T}\right)(\underset{\sim}{Q}+\underset{\sim}{r}) \\
&=\underset{\sim}{\delta}^{T} \underset{\sim}{\delta}+\underset{\sim}{\delta}^{T} \underset{\sim}{r}+\underset{\sim}{\delta} \underset{\sim}{r}^{T}+\underset{\sim}{r}^{T} \underset{\sim}{r} \\
&=\underset{\sim}{\delta}+ \underset{\sim}{\delta} \underset{\sim}{r}+\underset{\sim}{\delta} \underset{\sim}{r}^{T}+\underset{\sim}{r}^{T} \underset{\sim}{r}=\underset{\sim}{\delta},
\end{aligned} \nonumber \]
(using the identities \(\underset{\sim}{\delta}^T\) = \(\underset{\sim}{\delta}\) and \(\underset{\sim}{\delta}\underset{\sim}{\delta}\) = \(\underset{\sim}{\delta}\)), and therefore
\[\underset{\sim}{r}+\underset{\sim}{r}^{T}+\underset{\sim}{r}^{T} \underset{\sim}{r}=0.\label{eqn:1} \]
Now suppose that the rotation is through a very small angle1. In this case, all components of \(\underset{\sim}{r}\) are \(\ll 1\), and the third term on the left-hand side of Equation \(\ref{eqn:1}\) is therefore negligible, leaving us with
\[\underset{\sim}{r}+\underset{\sim}{r}^{T}=0. \nonumber \]
So, for an infinitesimal rotation, the rotation matrix equals the identity plus an antisymmetric matrix whose elements are \(\ll 1\).
1More precisely, we take the limit as the angle goes to zero.