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Rotation Matrices

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Introduction

A rotation matrix, \({\bf R}\), describes the rotation of an object in 3-D space. It was introduced on the previous two pages covering deformation gradients and polar decompositions.

The rotation matrix is closely related to, though different from, coordinate system transformation matrices, \({\bf Q}\), discussed on this coordinate transformation page and on this transformation matrix page.

A transformation matrix describes the rotation of a coordinate system while an object remains fixed. In contrast, a rotation matrix describes the rotation of an object in a fixed coordinate system. The amazing fact, and often a confusing one, is that each matrix is the transpose of the other.

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Summary

The rotation matrix, \({\bf R}\), is used in the rotation of vectors and tensors while the coordinate system remains fixed. The vector or tensor is usually related to some object that is actually undergoing the rotation, and the vector and/or tensor is along for the ride.

The general rules for applying the rotation matrix are the same as for the coordinate transformation matrix.

Vectors		\({\bf v'}\)	\( = \)	\({\bf R} \cdot {\bf v}\)
2nd Rank Tensors		\(\boldsymbol{\sigma'}\)	\( = \)	\({\bf R} \cdot \boldsymbol{\sigma} \cdot {\bf R}^T\)
4th Rank Tensors		\({\bf C'}\)	\( = \)	\({\bf R} \cdot {\bf R} \cdot {\bf C} \cdot {\bf R}^T \cdot {\bf R}^T\)

and in tensor notation...

Vectors		\(v'_i\)	\( = \)	\(R_{ij} v_j\)
2nd Rank Tensors		\(\sigma'_{mn}\)	\( = \)	\(R_{mi} R_{nj} \sigma_{ij}\)
4th Rank Tensors		\(C'_{mnop}\)	\( = \)	\(R_{mi} R_{nj} R_{ok} R_{pl} C_{ijkl} \)

In two dimensions, \({\bf R}\) is

\[ {\bf R} = \left[ \matrix {\cos \theta & -\sin \theta \\ \sin \theta & \;\;\;\cos \theta} \right] \]
where \(\theta\) is the rotation angle. Note that \({\bf R}\) is just the transpose of \({\bf Q}\), i.e., \({\bf R} = {\bf Q}^T\) and \({\bf Q} = {\bf R}^T\).
The general definition of \({\bf R}\), in 3-D, is

\[ {\bf R} = \left[ \matrix { \cos(x',x) & \cos(y',x) & \cos(z',x) \\ \cos(x',y) & \cos(y',y) & \cos(z',y) \\ \cos(x',z) & \cos(y',z) & \cos(z',z) } \right] \]
where \((x',x)\) represents the angle between the \(x'\) and \(x\) axes, \((x',y)\) is the angle between the \(x'\) and \(y\) axes, etc. Yes, this is indeed the transpose of \({\bf Q}\).

An alternative way of interpreting and generating the rotation matrix is as follows.

\[ {\bf R} = \left[ \matrix { \left( \matrix{\text{x-comp} \\ \text{of } {\bf i'}} \right) & \left( \matrix{\text{x-comp} \\ \text{of } {\bf j'}} \right) & \left( \matrix{\text{x-comp} \\ \text{of } {\bf k'}} \right) \\ \left( \matrix{\text{y-comp} \\ \text{of } {\bf i'}} \right) & \left( \matrix{\text{y-comp} \\ \text{of } {\bf j'}} \right) & \left( \matrix{\text{y-comp} \\ \text{of } {\bf k'}} \right) \\ \left( \matrix{\text{z-comp} \\ \text{of } {\bf i'}} \right) & \left( \matrix{\text{z-comp} \\ \text{of } {\bf j'}} \right) & \left( \matrix{\text{z-comp} \\ \text{of } {\bf k'}} \right) } \right] \]
where "x-comp of \({\bf i'}\)" means the x-component of the \({\bf i'}\) unit vector. Pay close attention here to which terms have primes on them and which don't. Don't confuse this with "the x'-component" because "the x'-comp of \({\bf i'}\)" is simply 1. Yet another way to say this is, "the first component of the \({\bf i'}\) unit vector in the non-primed reference \(x, y, z\) coordinate system." Once again, this is just the transpose of the \({\bf Q}\) matrix.

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Successive Rotations - Roe Convention

The 3-D rotation matrix can be viewed as a series of three successive rotations about coordinate axes. There must be dozens of variations of this since any combination of axes can be chosen in any order to rotate about. One popular choice is the so-called Roe convention.

As shown in the figure, it consists of (i) a rotation through angle \(\psi\) about the z-axis, then (ii) a rotation of angle \(\theta\) about the new y-axis (which has already rotated itself due to the first rotation about z), and finally, (iii) a second rotation of \(\phi\) about the now-tilted z-axis.

The Roe convention is very popular despite one key challenge it contains. That is... if \(\theta = 0\), then \(\psi\) and \(\phi\) become indistinguishable and only the sum of the two is what matters.



\[ \begin{eqnarray} {\bf R} & = & \left[ \matrix { \cos \psi & \! -\sin \psi & 0 \\ \sin \psi & \;\;\; \cos \psi & 0 \\ 0 & 0 & 1 } \right] \left[ \matrix { \;\;\; \cos \theta & 0 & \sin \theta \\ \;\; 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta } \right] \left[ \matrix { \cos \phi & \! -\sin \phi & 0 \\ \sin \phi & \;\;\; \cos \phi & 0 \\ 0 & 0 & 1 } \right] \\ \\ \\ & = & \left[ \matrix { \cos \psi \cos \theta \cos \phi - \sin \psi \sin \phi & -\cos \psi \cos \theta \sin \phi - \sin \psi \cos \phi & \cos \psi \sin \theta \\ \sin \psi \cos \theta \cos \phi + \cos \psi \sin \phi & -\sin \psi \cos \theta \sin \phi + \cos \psi \cos \phi & \sin \psi \sin \theta \\ -\sin \theta \cos \phi & \;\;\; \sin \theta \sin \phi & \;\; \cos \theta } \right] \end{eqnarray} \]
This time, the rotation matrices are indeed written in the same order: \(\psi, \theta, \phi\).

Going In Reverse - Determining Roe Angles from a Rotation Matrix

Suppose you have a rotation matrix populated with values as shown here and would like to know what Roe transformation angles were responsible for producing the matrix.

\[ {\bf R} = \left[ \matrix { R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} } \right] \]
The first and easiest step is to look at the \(R_{33}\) component. It is simply the cosine of \(\theta\), so \(\theta = \text{Cos}^{-1}(R_{33})\).

The second step is to determine \(\psi\) as follows

\[ {R_{23} \over R_{13}} = { \sin \psi \sin \theta \over \cos \psi \sin \theta } = { \sin \psi \over \cos \psi } = \tan \psi \] So
\[ \psi = \text{Tan}^{-1} \left( {R_{23} \over R_{13}} \right) \] The same process leads to
\[ \phi = \text{Tan}^{-1} \left({R_{32} \over -R_{31}} \right) \] So in summary
\[ \psi = \text{Tan}^{-1} \left( {R_{23} \over R_{13}} \right) \qquad \qquad \theta = \text{Cos}^{-1} \left( R_{33} \right) \qquad \qquad \phi = \text{Tan}^{-1} \left( {R_{32} \over -R_{31}} \right) \]
UNLESS!!!....... \(\theta\) proves to be very small! In this case, \(\sin \theta\) will be very small and therefore, so will \(R_{31}, R_{32}, R_{13}\), and \(R_{23}\) because all these terms contain \(\sin \theta\). This can lead to major round-off errors in any such calculations.

Fortunately, the solution for this (\(\theta \rightarrow 0\)) is to recall that \(\psi\) and \(\phi\) become indistinguishable from each other. This permits \(\phi\) to be set to zero, and \(\psi\) to be computed according to \(\psi = \text{Sin}^{-1}(R_{21})\).

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Rotation About An Axis

Yet another way of specifying the rotation matrix is through a rotation axis vector, \({\bf p}\), and a rotation angle, \(\alpha\), about the \({\bf p}\) axis.

In this case, the rotation matrix is written as

\[ {\bf R} = \cos \alpha \; {\bf I} + (1 - \cos \alpha) {\bf p} \otimes {\bf p} - \sin \alpha \; {\bf P} \]
with
\[ {\bf P} = \left[ \matrix { \;\;\; 0 & \;\;\; p_3 & -p_2 \\ -p_3 & \;\; 0 & \;\;\; p_1 \\ \;\;\; p_2 & -p_1 & \;\;\; 0 } \right] \]
Writing the matrix out gives

\[ {\bf R} = \left[ \matrix { \cos \alpha + (1-\cos \alpha) p^2_1 & (1 - \cos \alpha) p_1 p_2 - \sin \alpha \; p_3 & (1 - \cos \alpha) p_1 p_3 + \sin \alpha \; p_2 \\ (1 - \cos \alpha) p_2 p_1 + \sin \alpha \; p_3 & \cos \alpha + (1-\cos \alpha) p^2_2 & (1 - \cos \alpha) p_2 p_3 - \sin \alpha \; p_1 \\ (1 - \cos \alpha) p_3 p_1 - \sin \alpha \; p_2 & (1 - \cos \alpha) p_3 p_2 + \sin \alpha \; p_1 & \cos \alpha + (1-\cos \alpha) p^2_3 } \right] \]
As usual, this can be written quite concisely in tensor notation

\[ R_{ij} = \cos \alpha \; \delta_{ij} + (1 - \cos \alpha) p_i p_j - \sin \alpha \; \epsilon_{ijk} \; p_k \]
Note the similarity to the tensor equation of \({\bf Q}\). The only difference is the + or - in front of \(\sin \alpha\).

\[ Q_{ij} = \cos \alpha \; \delta_{ij} + (1 - \cos \alpha) p_i p_j + \sin \alpha \; \epsilon_{ijk} \; p_k \] As with the Roe angles, it is possible to back-out \({\bf p}\) and \(\alpha\) given a rotation matrix, \({\bf R}\). The first step is to determine \(\alpha\). This is done by taking the trace of the matrix.

\[ \alpha = \text{Cos}^{-1} \left\{ {1 \over 2} \Big[ \text{tr}({\bf R}) - 1 \Big] \right\} \]
Once \(\alpha\) is determined, the components of \({\bf p}\) are computed by

\[ p_1 = { R_{32} - R_{23} \over 2 \sin \alpha } \qquad \qquad p_2 = { R_{13} - R_{31} \over 2 \sin \alpha } \qquad \qquad p_3 = { R_{21} - R_{12} \over 2 \sin \alpha } \]
The three equations can be conveniently summarized in tensor notation as

\[ p_i = { - \epsilon_{ijk} R_{jk} \over 2 \sin \alpha } \]
Note that \({\bf p}\) becomes undefined when \(\alpha = 0\). This means, "the axis you rotate about doesn't matter if you don't rotate in the first place." The above example closes the loop on converting rotations from one system, Roe, to a second system, about an axis. In this case, and any other, the transformation matrix \({\bf R}\) is the go-between permitting the conversion. The procedure is always to calculate the \({\bf R}\) matrix in the first system and then back-out the angles in the second system from \({\bf R}\).

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Inverses and Transposes

All rotation matrices possess a rather remarkable property - their transpose is their inverse. So it becomes trivial to compute the inverse of such a matrix.

\[ {\bf R}^T = {\bf R}^{-1} \qquad \text{so} \qquad {\bf R}^T \cdot {\bf R} = {\bf I} \]