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{{Other uses|Rotation operator (disambiguation){{!}}Rotation operator}}


This article derives the main properties of rotations in 3-dimensional space.


The three [[Euler angles|Euler rotations]] are one way to bring a [[rigid body]] to any desired orientation by sequentially making [[rotations]] about axis' fixed relative to the object. However, this can also be achieved with one single rotation ([[Euler's rotation theorem]]). Using the concepts of [[linear algebra]] it is shown how this single
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rotation can be performed.
 
==Mathematical formulation==
Let
:<math>\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3</math>
 
be a [[coordinate system]] fixed in the body that through a change in orientation is brought to the new directions
:<math>\mathbf{A}\hat e_1\ ,\ \mathbf{A}\hat e_2\ ,\ \mathbf{A}\hat e_3.</math>
 
Any [[euclidean vector|vector]]
:<math>\bar x\ =x_1\hat e_1+x_2\hat e_2+x_3\hat e_3</math>
 
rotating with the body is then brought to the new direction
:<math>\mathbf{A}\bar x\ =x_1\mathbf{A}\hat e_1+x_2\mathbf{A}\hat e_2+x_3\mathbf{A}\hat e_3</math>
 
i.e. this is a [[linear operator]]
 
The [[Matrix (mathematics)|matrix]] of this [[Operator (mathematics)|operator]] relative to the coordinate system
:<math>\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3</math>
 
is
:<math>
\begin{bmatrix}
  A_{11} & A_{12} & A_{13} \\
  A_{21} & A_{22} & A_{23} \\
  A_{31} & A_{32} & A_{33}
\end{bmatrix} =
\begin{bmatrix}
  \langle\hat e_1 | \mathbf{A}\hat e_1 \rangle & \langle\hat e_1 | \mathbf{A}\hat e_2 \rangle & \langle\hat e_1 | \mathbf{A}\hat e_3 \rangle \\
  \langle\hat e_2 | \mathbf{A}\hat e_1 \rangle & \langle\hat e_2 | \mathbf{A}\hat e_2 \rangle & \langle\hat e_2 | \mathbf{A}\hat e_3 \rangle \\
  \langle\hat e_3 | \mathbf{A}\hat e_1 \rangle & \langle\hat e_3 | \mathbf{A}\hat e_2 \rangle & \langle\hat e_3 | \mathbf{A}\hat e_3 \rangle
\end{bmatrix}
</math>
 
As
:<math> \sum_{k=1}^3 A_{ki}A_{kj}= \langle \mathbf{A}\hat e_i | \mathbf{A}\hat e_j \rangle
= \begin{cases}
0 & i\neq j, \\                                                                                1 & i = j,
\end{cases}
</math>
 
or equivalently in matrix notation
 
:<math>
\begin{bmatrix}
  A_{11} & A_{12} & A_{13} \\
  A_{21} & A_{22} & A_{23} \\
  A_{31} & A_{32} & A_{33}
\end{bmatrix}^T
\begin{bmatrix}
  A_{11} & A_{12} & A_{13} \\
  A_{21} & A_{22} & A_{23} \\
  A_{31} & A_{32} & A_{33}
\end{bmatrix} =
\begin{bmatrix}
  1 & 0 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1
\end{bmatrix}
</math>
the matrix is [[Orthogonal matrix|orthogonal]] and as a "right hand" base vector system is re-orientated into another "right hand" system the [[determinant]] of this matrix has the value 1.
 
===Rotation around an axis===
Let
 
:<math>\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3</math>
 
be an orthogonal positively oriented base vector system in <math>R^3</math>.
 
The linear operator
 
"Rotation with the angle <math>\theta</math> around the axis defined by <math>\hat e_3</math>"
 
has the matrix representation
 
:<math>
\begin{bmatrix}
  Y_1  \\
  Y_2  \\
  Y_3
\end{bmatrix}
=
\begin{bmatrix}
  \cos\theta & -\sin\theta & 0 \\
  \sin\theta &  \cos\theta & 0 \\
          0 &          0 & 1
\end{bmatrix}
\begin{bmatrix}
  X_1  \\
  X_2  \\
  X_3
\end{bmatrix}
</math>
 
relative to this basevector system.  
 
This then means that a vector
 
:<math>
\bar x=\begin{bmatrix}
  \hat e_1 & \hat e_2 & \hat e_3
\end{bmatrix}
\begin{bmatrix}
  X_1  \\
  X_2  \\
  X_3
\end{bmatrix}
</math>
 
is rotated to the vector
 
:<math>
\bar y=\begin{bmatrix}
  \hat e_1 & \hat e_2 & \hat e_3
\end{bmatrix}
\begin{bmatrix}
  Y_1  \\
  Y_2  \\
  Y_3
\end{bmatrix}
</math>
 
by the linear operator.  
 
The [[determinant]] of this matrix is
 
:<math>
\det
\begin{bmatrix}
  \cos\theta  & -\sin\theta & 0\\
  \sin\theta  &  \cos\theta & 0\\
          0  &          0 & 1
\end{bmatrix}=1
</math>
 
and the [[characteristic polynomial]] is
 
:<math>
\begin{align}
\det\begin{bmatrix}
  \cos\theta -\lambda  & -\sin\theta          & 0        \\
  \sin\theta          &  \cos\theta -\lambda & 0        \\
                    0 &                    0 & 1-\lambda
\end{bmatrix}
&=\big({(\cos\theta -\lambda)}^2 + {\sin\theta}^2 \big)(1-\lambda) \\
&=-\lambda^3+(2\ \cos\theta\ +\ 1)\ \lambda^2 - (2\ \cos\theta\ +\ 1)\ \lambda +1 \\
\end{align}
</math>
 
The matrix is symmetric if and only if <math>\sin\theta=0</math>, i.e. for <math>\theta=0</math>
and for <math>\theta=\pi</math>.
 
The case <math>\theta=0</math> is the trivial case of an identity operator.
 
For the case <math>\theta=\pi</math> the [[characteristic polynomial]] is
 
:<math>
-(\lambda-1){(\lambda +1)}^2
</math>
 
i.e. the rotation operator has the [[eigenvalue]]s
:<math>
\lambda=1 \quad \lambda=-1
</math>
 
The [[eigenspace]] corresponding to <math>\lambda=1</math> is all vectors on the rotation axis, i.e. all vectors
 
:<math>
\bar x  =\alpha \ \hat e_3 \quad -\infty <\alpha < \infty
</math>
 
The [[eigenspace]] corresponding to <math>\lambda=-1</math> consists of all vectors orthogonal to the rotation axis, i.e. all vectors
 
:<math>
\bar x  =\alpha \ \hat e_1 + \beta \ \hat e_2 \quad -\infty <\alpha < \infty \quad -\infty <\beta < \infty
</math>
 
For all other values of <math>\theta</math> the matrix is un-symmetric and as <math>{\sin\theta}^2 > 0</math> there is
only the eigenvalue <math>\lambda=1</math> with the one-dimensional [[eigenspace]] of the vectors on the rotation axis:
 
:<math>
\bar x  =\alpha \ \hat e_3 \quad -\infty <\alpha < \infty
</math>
 
The rotation matrix by angle <math>\theta</math> around a general axis of rotation
<math>
\mathbf{k} = \left[\begin{array}{ccc}
k_1  \\
k_2 \\
k_3
\end{array}\right]
</math>
is given by [[Rodrigues' rotation formula]].
: <math>
  R = I \cos\theta + [\mathbf{k}]_\times \sin\theta + (1 - \cos\theta) \mathbf{k} \mathbf{k}^\mathsf{T}
</math>,
 
where <math>I</math> is the [[identity matrix]] and <math>[\mathbf{k}]_\times</math> is the [[Hodge_star | dual 2-form]] of <math>\mathbf{k}</math> or
[[Cross_product | cross product matrix]],
 
: <math> [\mathbf{k}]_\times  =
\left[\begin{array}{ccc}
0 & -k_3 & k_2 \\
k_3 & 0 & -k_1 \\
-k_2 & k_1 & 0
\end{array}\right]
</math>.
 
Note that <math>[\mathbf{k}]_\times </math> satisfies <math>[\mathbf{k}]_\times \mathbf{v} = \mathbf{k}\times\mathbf{v} </math> for all  <math>\mathbf{v}</math>.
 
===The general case===
The operator
 
"Rotation with the angle <math>\theta</math> around a specified axis"
 
discussed above is an orthogonal mapping and its matrix relative to any base vector system is therefore an
[[orthogonal matrix]] . Furthermore its determinant  has the value 1.
A non-trivial fact is the opposite, i.e. that for any orthogonal linear mapping in <math>R^3</math> having  
determinant = 1 there exist base vectors
 
:<math>\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3</math>
 
such that the matrix takes the "canonical form"
 
:<math>
\begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &  \cos\theta & 0 \\
          0 &          0 & 1\end{bmatrix}
</math>
 
for some value of <math>\theta</math>.  
 
In fact, if a linear operator has the [[orthogonal matrix]]
 
:<math>
\begin{bmatrix}
  A_{11} & A_{12} & A_{13} \\
  A_{21} & A_{22} & A_{23} \\
  A_{31} & A_{32} & A_{33}
\end{bmatrix}
</math>
relative some base vector system
:<math>\hat f_1\ ,\ \hat f_2\ ,\ \hat f_3</math>
 
and this matrix is symmetric, the "Symmetric operator theorem" valid in <math>R^n</math> (any dimension) applies saying
 
that it has ''n'' orthogonal eigenvectors. This means for the 3-dimensional case that there exists a coordinate system
:<math>\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3</math>
 
such that the matrix takes the form
:<math>
\begin{bmatrix}
  B_{11} &    0 &    0 \\
      0  & B_{22} &    0 \\
      0  &    0 & B_{33}
\end{bmatrix}
</math>
As it is an orthogonal matrix these diagonal elements <math>B_{ii}</math> are either 1 or &minus;1. As the determinant is 1 these elements
are either all 1 or one of the elements is 1 and the other two are &minus;1.
 
In the first case it is the trivial identity operator corresponding
to  <math>\theta=0</math>.  
 
In the second case it has the form
 
:<math>
\begin{bmatrix}
    -1  &    0 &    0 \\
      0  &    -1 &    0 \\
      0  &    0 &    1
\end{bmatrix}
</math>
 
if the basevectors are numbered such that the one with eigenvalue 1 has index 3. This matrix is then of the desired form for <math>\theta=\pi</math>.
 
If the matrix is un-symmetric, the vector
:<math>
\bar E = \alpha_1\ \hat f_1 + \alpha_2\ \hat f_2 + \alpha_3\ \hat f_3
</math>
 
where
 
:<math>\alpha_1=\frac{A_{32}-A_{23} }{2} </math>
:<math>\alpha_2=\frac{A_{13}-A_{31}}{2}</math>
:<math>\alpha_3=\frac{A_{21}-A_{12}}{2}</math>
 
is non-zero. This vector is an eigenvector with eigenvalue
 
:<math>
\lambda=1
</math>
 
Setting
:<math>
\hat e_3=\frac{\bar E}{|\bar E|}
</math>
 
and selecting any two orthogonal unit vectors in the plane orthogonal to <math>\hat e_3</math>:
 
:<math>\hat e_1\ ,\ \hat e_2</math>
 
such that
 
:<math>\hat e_1\ ,\ \hat e_2,\ \hat e_3</math>
 
form a positively oriented triple, the operator takes the desired form with
:<math>\cos \theta=\frac{A_{11}+A_{22}+A_{33}-1}{2}</math>
:<math>\sin \theta=|\bar{E}|</math>
 
The expressions above are in fact valid also for the case of a symmetric
rotation operator corresponding to a rotation with <math>\theta = 0</math>
or <math>\theta = \pi</math>. But the difference is that for <math>\theta = \pi</math>
the vector
:<math>
\bar E = \alpha_1\ \hat f_1 + \alpha_2\ \hat f_2 + \alpha_3\ \hat f_3
</math>
 
is zero and of no use for finding the eigenspace of eigenvalue 1, i.e. the
rotation axis.
 
Defining <math>E_4</math> as <math>\cos \theta</math> the matrix for the
rotation operator is
 
:<math>
\frac{1-E_4}{{E_1}^2+{E_2}^2+{E_3}^2}
\begin{bmatrix}
E_1 E_1 & E_1 E_2 & E_1 E_3 \\
E_2 E_1 & E_2 E_2 & E_2 E_3 \\
E_3 E_1 & E_3 E_2 & E_3 E_3
\end{bmatrix}
+
\begin{bmatrix}
E_4 & -E_3 &  E_2 \\
E_3 &  E_4 & -E_1 \\
-E_2 &  E_1 &  E_4
\end{bmatrix}
</math>
 
provided that
 
:<math>
{E_1}^2+{E_2}^2+{E_3}^2 > 0
</math>
 
i.e. except for the cases <math>\theta=0</math> (the identity operator) and <math>\theta=\pi</math>
 
==Quaternions==
{{Main|Quaternions and spatial rotation}}
 
Quaternions are defined similar to <math>E_1\ ,\ E_2\ ,\ E_3\ ,\ E_4</math> with
the difference that the half angle  <math>\frac{\theta}{2}</math> is used
instead of the full angle <math>\theta</math>.
 
This means that the first 3 components <math>q_1\ ,\ q_2\ ,\ q_3\ </math> are components of a vector defined from
 
:<math>
q_1\ \hat{f_1}\ +\ q_2\ \hat{f_2}\ +\ \ q_3\ \hat{f_1}\ =\ \sin \frac{\theta}{2}\quad \hat{e_3}=\frac{\sin \frac{\theta}{2}}{\sin\theta}\quad \bar E
</math>
and that the fourth component is the scalar
:<math>
q_4=\cos \frac{\theta}{2}
</math>
 
As the angle <math>\theta</math> defined from the canonical form is in the interval
:<math>0 \le \theta \le \pi</math>
 
one would normally have that <math>q_4 \ge 0</math>. But a "dual" representation of a rotation with quaternions
is used, i.e.
:<math>q_1\ ,\ q_2\ ,\ q_3\ ,\ q_4\ </math>  
 
and
:<math>-q_1\ ,\ -q_2\ ,\ -q_3\ ,\ -q_4\ </math>
 
are two alternative representations of one and the same rotation.
 
The entities <math>E_k</math> are defined from the quaternions by
:<math> E_1=2 q_4 q_1</math>
:<math> E_2=2 q_4 q_2</math>
:<math> E_3=2 q_4 q_3</math>
:<math> E_4={q_4}^2 -({q_1}^2+{q_2}^2+{q_3}^2)</math>
 
Using quaternions the matrix of the rotation operator is
:<math>\begin{bmatrix}
2({q_1}^2+{q_4}^2)-1        &2({q_1}{q_2}-{q_3}{q_4})      &2({q_1}{q_3}+{q_2}{q_4})  \\
2({q_1}{q_2}+{q_3}{q_4})    &2({q_2}^2+{q_4}^2)-1          &2({q_2}{q_3}-{q_1}{q_4})  \\
2({q_1}{q_3}-{q_2}{q_4})    &2({q_2}{q_3}+{q_1}{q_4})      &2({q_3}^2+{q_4}^2)-1    \\
\end{bmatrix}</math>
 
==Numerical example==
 
Consider the reorientation corresponding to the [[Euler angle]]s
<math>
\alpha=10^\circ \quad \beta=20^\circ \quad \gamma=30^\circ \quad
</math>
relative a given base vector system
:<math>\hat f_1\ ,\ \hat f_2,\ \hat f_3</math>
 
Corresponding matrix relative to this base vector system is (see [[Euler angles#Matrix orientation]])
 
:<math>
\begin{bmatrix}
  0.771281 & -0.633718 &  0.059391 \\
  0.613092 &  0.714610 & -0.336824 \\
  0.171010 &  0.296198 &  0.939693
\end{bmatrix}
</math>
 
and the quaternion is
:<math>
(0.171010,\ -0.030154,\ 0.336824,\ 0.925417)
</math>
 
The canonical form of this operator
:<math>
\begin{bmatrix}
  \cos\theta  & -\sin\theta & 0\\
  \sin\theta  &  \cos\theta & 0\\
          0  &          0 & 1
\end{bmatrix}
</math>
with <math>\theta=44.537^\circ </math> is obtained with
:<math>\hat e_3=(0.451272,-0.079571,0.888832)</math>
 
The quaternion relative to this new system is then
:<math>
(0,\ 0,\ 0.378951,\ 0.925417) = (0,\ 0,\ \sin\frac{\theta}{2},\ \cos\frac{\theta}{2})
</math>
 
Instead of making the three Euler rotations
 
:<math>10^\circ,20^\circ,30^\circ</math>
 
the same orientation can be reached with one single rotation of size <math>44.537^\circ</math> around <math>\hat e_3</math>
 
==References==
* {{citation |title=An Introduction to the Theory of Linear Spaces|first=Georgi|last= Shilov|author-link =Georgii Evgen'evich Shilov|publisher= Prentice-Hall|year=1961|id=Library of Congress 61-13845}}.
 
[[Category:Linear algebra]]
[[Category:Kinematics]]

Revision as of 21:21, 5 February 2014


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