The simplest rotation

To motivate many of our later developments, we start with the simplest case of a rotation: a rotation about a fixed axis {\bf p}_3 through an angle \theta = \theta(t). This example should be familiar to you from many different venues. To describe this rotation, we consider the action of this rotation on the set of fixed, orthonormal, right-handed basis vectors \left\{ {\bf p}_1, \, {\bf p}_2, \, {\bf p}_3\right\}. As depicted in Figure 1, we suppose that these vectors are transformed to the set \left\{ {\bf t}_1, \, {\bf t}_2, \, {\bf t}_3\right\} by the rotation.

Figure 1. The transformation of basis vectors  \left\{ {\bf p}_1, \, {\bf p}_2, \, {\bf p}_3\right\} into  \left\{ {\bf t}_1, \, {\bf t}_2, \, {\bf t}_3\right\} by a rotation about  {\bf p}_3 through an angle  \theta.

A matrix representation of the rotation

Using a matrix notation, we can represent the transformation from the set of basis vectors \left\{ {\bf p}_1, \, {\bf p}_2, \, {\bf p}_3\right\} to \left\{ {\bf t}_1, \, {\bf t}_2, \, {\bf t}_3\right\} as

(1)   \begin{equation*} \left[ \begin{array}{c} {\bf t}_1 \\ {\bf t}_2 \\ {\bf t}_3 \end{array} \right] = \underbrace{\left[ \begin{array}{c c c } \cos(\theta)  &  \sin(\theta)  & 0 \\ - \sin(\theta)  & \cos(\theta)  & 0  \\ 0 & 0 & 1 \end{array} \right]}_{\mathsf{R}} \left[\begin{array}{c} {\bf p}_1 \\ {\bf p}_2 \\ {\bf p}_3 \end{array} \right] . \end{equation*}

It is easy to see that the matrix \mathsf{R} in (1) has a unit determinant, and its inverse is its transpose: \mathsf{R}^{-1} = \mathsf{R}^T. That is, the matrix \mathsf{R} is proper-orthogonal. By differentiating (1) with respect to time, we find that

(2)   \begin{equation*} \left[ \begin{array}{c} \dot{\bf t}_1 \\ \dot{\bf t}_2 \\ \dot{\bf t}_3 \end{array} \right] = \underbrace{ \dot{\theta} \left[ \begin{array}{c c c } - \sin(\theta)  &  \cos(\theta)  & 0 \\ - \cos(\theta)  & - \sin(\theta)  & 0  \\ 0 & 0 & 0 \end{array} \right]}_{ \dot{\mathsf{R}} } \left[\begin{array}{c} {\bf p}_1 \\ {\bf p}_2 \\ {\bf p}_3 \end{array} \right]. \end{equation*}

Using the fact that \mathsf{R}^{-1} = \mathsf{R}^T, we can easily replace {\bf p}_i (i = 1, \, 2, \, 3) in (2) with {\bf t}_i to obtain

(3)   \begin{eqnarray*} \left[ \begin{array}{c} \dot{\bf t}_1 \\ \dot{\bf t}_2 \\ \dot{\bf t}_3 \end{array} \right] \!\!\!\!\! &=& \!\!\!\!\! \underbrace{ \dot{\theta} \left[ \begin{array}{c c c } - \sin(\theta)  &  \cos(\theta)  & 0 \\ - \cos(\theta)  & - \sin(\theta)  & 0  \\ 0 & 0 & 0 \end{array} \right] }_{ \dot{\mathsf{R}} } \underbrace{ \left[ \begin{array}{c c c } \cos(\theta)  &  - \sin(\theta)  & 0 \\ \sin(\theta)  & \cos(\theta)  & 0  \\ 0 & 0 & 1 \end{array} \right] }_{ \mathsf{R}^T } \left[\begin{array}{c} {\bf t}_1 \\ {\bf t}_2 \\ {\bf t}_3 \end{array} \right] \hspace{1in} \scalebox{0.001}{\textrm{\textcolor{white}{.}}} \\ \\[0.05in] &=& \!\!\!\!\! \underbrace{\dot{\theta} \left[ \begin{array}{c c c } 0  &  1  & 0 \\ - 1  & 0  & 0  \\ 0 & 0 & 0 \end{array} \right]}_{\dot{\mathsf{R}}\mathsf{R}^T} \left[\begin{array}{c} {\bf t}_1 \\ {\bf t}_2 \\ {\bf t}_3 \end{array} \right]. \end{eqnarray*}

Notice that this is equivalent to the familiar results \dot{\bf t}_1 = \dot{\theta}{\bf t}_2 and \dot{\bf t}_2 = - \dot{\theta}{\bf t}_1 from cylindrical polar coordinates. It should also be clear from (3) that \dot{\mathsf{R}}\mathsf{R}^T is a skew-symmetric matrix. A vector \dot{\theta}{\bf t}_3 = \dot{\theta}{\bf p}_3 can be introduced that has the useful property that

(4)   \begin{equation*} \dot{\bf t}_k = \dot{\theta}{\bf t}_3 \times {\bf t}_k \quad (k = 1, \, 2, \, 3). \end{equation*}

You should notice how the vector \dot{\theta}{\bf t}_3 = \dot{\theta}{\bf p}_3 can be inferred from the components of \dot{\mathsf{R}}\mathsf{R}^T.

A tensor representation of the rotation

It is convenient to use a tensor notation1 to describe the rotation we have been discussing. In particular, we can write (1) in the form

(5)   \begin{equation*} {\bf t}_k = {\bf R}{\bf p}_k , \end{equation*}

where {\bf R} is the tensor

(6)   \begin{equation*} {\bf R} = \cos(\theta)\left( {\bf p}_1\otimes{\bf p}_1  + {\bf p}_2\otimes{\bf p}_2\right) - \sin(\theta)\left( {\bf p}_1\otimes{\bf p}_2  - {\bf p}_2\otimes{\bf p}_1\right) + {\bf p}_3\otimes{\bf p}_3. \hspace{1in} \scalebox{0.001}{\textrm{\textcolor{white}{.}}} \end{equation*}

It can be shown that this representation of the rotation is equivalent to (1). Furthermore, because

(7)   \begin{eqnarray*} && {\bf I} - {\bf p}_3\otimes{\bf p}_3 = {\bf p}_1\otimes{\bf p}_1  + {\bf p}_2\otimes{\bf p}_2, \\ \\ && {\bf I} - {\bf t}_3\otimes{\bf t}_3 = {\bf t}_1\otimes{\bf t}_1 + {\bf t}_2\otimes{\bf t}_2, \\ \\ && \bepsilon{\bf p}_3 = {\bf p}_1\otimes{\bf p}_2  - {\bf p}_2\otimes{\bf p}_1, \\ \\ && \bepsilon{\bf t}_3 = {\bf t}_1\otimes{\bf t}_2 - {\bf t}_2\otimes{\bf t}_1, \end{eqnarray*}

we can express the tensor {\bf R} entirely in terms of the axis of rotation {\bf p}_3 = {\bf t}_3 and the angle of rotation \theta:

(8)   \begin{equation*} {\bf R} = \cos(\theta)\left( {\bf I} - {\bf p}_3\otimes{\bf p}_3 \right) - \sin(\theta) \left( \bepsilon{\bf p}_3 \right) + {\bf p}_3\otimes{\bf p}_3, \end{equation*}

or, equivalently,

(9)   \begin{equation*} {\bf R} = \cos(\theta)\left( {\bf I} - {\bf t}_3\otimes{\bf t}_3 \right) - \sin(\theta) \left( \bepsilon{\bf t}_3 \right) + {\bf t}_3\otimes{\bf t}_3. \end{equation*}

As we shall see later, these equivalent representations naturally lead to the general form of a tensor that represents a rotation about an arbitrary axis through an arbitrary angle of rotation. As with the matrix representation of this simple rotation, the rotation tensor {\bf R} is a proper-orthogonal tensor because it has a determinant of one and its inverse is its transpose: \mbox{det}\left({\bf R}\right) = 1 and {\bf R}^{-1} = {\bf R}^T. Differentiating (5), we find that

(10)   \begin{equation*} \dot{\bf t}_k = \dot{\bf R} \underbrace{ {\bf p}_k }_{ {\bf R}^T{\bf t}_k } + \ {\bf R} \underbrace{ {\dot{\bf p}_k} }_{ \mathbf{0} } = \left(\dot{\bf R} {\bf R}^T\right){\bf t}_k , \end{equation*}

where we have used the fact that {\bf p}_k are constant vectors. Some straightforward computations involving (6) reveal that

(11)   \begin{equation*} \dot{\bf R} = - \dot{\theta}\sin(\theta)\left( {\bf p}_1\otimes{\bf p}_1  + {\bf p}_2\otimes{\bf p}_2\right) - \dot{\theta}\cos(\theta)\left( {\bf p}_1\otimes{\bf p}_2  - {\bf p}_2\otimes{\bf p}_1\right) \hspace{1in} \scalebox{0.001}{\textrm{\textcolor{white}{.}}} \end{equation*}

and

(12)   \begin{eqnarray*} \dot{\bf R} {\bf R}^T \!\!\!\!\! &=& \!\!\!\!\!  \dot{\theta} \left( - {\bf p}_1\otimes{\bf p}_2  + {\bf p}_2\otimes{\bf p}_1\right) \\[0.10in] &=& \!\!\!\!\! \dot{\theta} \left( - {\bf t}_1\otimes{\bf t}_2 + {\bf t}_2\otimes{\bf t}_1\right). \end{eqnarray*}

With the help of these results, we conclude from (10) that

(13)   \begin{equation*} \dot{\bf t}_k = \left(\dot{\bf R} {\bf R}^T\right){\bf t}_k = \dot{\theta} {\bf t}_3 \times{\bf t}_k . \end{equation*}

As expected, this result is in agreement with (4). Expression (13) implies that the vector \dot{\theta} {\bf t}_3 = \dot{\theta} {\bf p}_3 is the axial vector of the skew-symmetric tensor \dot{\bf R} {\bf R}^T.

Notes

  1. For those unfamiliar with tensor notation, a review of tensors is available in the background section on linear algebra.