Definitions

Rotation operator (vector space)

The three Euler rotations is an obvious way to bring a rigid body into any desired orientation by sequentially making rotations about axis fixed relative the body. But it is a non-trivial fact is that this also can be achieved with one single rotation. Using the concepts of linear algebra it is shown how this single rotation can be found.

Mathematical formulation

Let

$hat e_1 , hat e_2 , hat e_3$

be a coordinate system fixed in the body that through a change in orientation is brought to the new directions

$mathbf\left\{A\right\}hat e_1 , mathbf\left\{A\right\}hat e_2 , mathbf\left\{A\right\}hat e_3.$

Any vector

$bar x =x_1hat e_1+x_2hat e_2+x_3hat e_3$

of the body is then brought to the new direction

$mathbf\left\{A\right\}bar x =x_1mathbf\left\{A\right\}hat e_1+x_2mathbf\left\{A\right\}hat e_2+x_3mathbf\left\{A\right\}hat e_3$

i.e. this is a linear operator

The matrix of this operator relative the coordinate system

$hat e_1 , hat e_2 , hat e_3$

is


begin{bmatrix} A_{11} & A_{12} & A_{13} A_{21} & A_{22} & A_{23} A_{31} & A_{32} & A_{33} end{bmatrix} = begin{bmatrix} langlehat e_1 | mathbf{A}hat e_1 rangle & langlehat e_1 | mathbf{A}hat e_2 rangle & langlehat e_1 | mathbf{A}hat e_3 rangle langlehat e_2 | mathbf{A}hat e_1 rangle & langlehat e_2 | mathbf{A}hat e_2 rangle & langlehat e_2 | mathbf{A}hat e_3 rangle langlehat e_3 | mathbf{A}hat e_1 rangle & langlehat e_3 | mathbf{A}hat e_2 rangle & langlehat e_3 | mathbf{A}hat e_3 rangle end{bmatrix}

As

$sum_\left\{k=1\right\}^3 A_\left\{ki\right\}A_\left\{kj\right\}= langle mathbf\left\{A\right\}hat e_i | mathbf\left\{A\right\}hat e_j rangle$
= begin{cases} 0 & ineq j, 1 & i = j, end{cases}

or equivalently in matrix notation


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} the matrix is 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

$hat e_1 , hat e_2 , hat e_3$

be an orthogonal positively oriented base vector system in $R^3$

The linear operator

"Rotation with the angle $theta$ around the axis defined by $hat e_3$"

has the matrix representation


begin{bmatrix}
` Y_1  `
` Y_2  `
` Y_3`
end{bmatrix} = begin{bmatrix}
` costheta & -sintheta & 0 `
` sintheta &  costheta & 0 `
`          0 &           0 & 1`
end{bmatrix} begin{bmatrix}
` X_1  `
` X_2  `
` X_3`
end{bmatrix}

relative this basevector system

This then means that a vector


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}

is rotated to the vector


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}

by the linear operator

The determinant of this matrix is


det begin{bmatrix}
` costheta  & -sintheta & 0`
` sintheta  &  costheta & 0`
`          0  &           0 & 1`
end{bmatrix}=1

and the characteristic polynomial is


begin{align} detbegin{bmatrix}
` costheta -lambda  & -sintheta          & 0        `
` sintheta           &  costheta -lambda & 0        `
`                    0 &                    0 & 1-lambda`
end{bmatrix} &=big({(costheta -lambda)}^2 + {sintheta}^2 big)(1-lambda) &=-lambda^3+(2 costheta + 1) lambda^2 - (2 costheta + 1) lambda +1 end{align}

The matrix is symmetric if and only if $sintheta=0$, i.e. for $theta=0$ and for $theta=pi$

The case $theta=0$ is the trivial case of an identity operator

For the case $theta=pi$ the characteristic polynomial is


-(lambda-1){(lambda +1)}^2

i.e. the rotation operator has the eigenvalues



The eigenspace corresponding to $lambda=1$ is all vectors on the rotation axis, i.e. all vectors


bar x =alpha hat e_3 quad -infty

The eigenspace corresponding to $lambda=-1$ consists of all vectors orthogonal to the rotation axis, i.e. all vectors


bar x =alpha hat e_1 + beta hat e_2 quad -infty

For all other values of $theta$ the matrix is un-symmetric and as $\left\{sintheta\right\}^2 > 0$ there is only the eigenvalue $lambda=1$ with the one-dimensional eigenspace of the vectors on the rotation axis:


bar x =alpha hat e_3 quad -infty

The general case

The operator

"Rotation with the angle $theta$ around a specified axis"

discussed above is an orthogonal mapping and its matrix relative any base vector system is therefore an orthogonal matrix . Further more its determinant has the value 1. A non-trivial fact is the opposite, i.e. that for any orthogonal linear mapping in $R^3$ having determinant = 1 there exist base vectors

$hat e_1 , hat e_2 , hat e_3$

such that the matrix takes the "canonical form"


begin{bmatrix} costheta & -sintheta & 0 sintheta & costheta & 0 0 & 0 & 1end{bmatrix}

for some value of $theta$.

In fact, if a linear operator has the orthogonal matrix


begin{bmatrix} A_{11} & A_{12} & A_{13} A_{21} & A_{22} & A_{23} A_{31} & A_{32} & A_{33} end{bmatrix} relative some base vector system
$hat f_1 , hat f_2 , hat f_3$

and this matrix is symmetric the "Symmetric operator theorem" valid in $R^n$ (any dimension) applies saying

that it has n orthogonal eigenvectors. This means for the 3-dimensional case that there exists a coordinate system

$hat e_1 , hat e_2 , hat e_3$

such that the matrix takes the form


begin{bmatrix} B_{11} & 0 & 0 0 & B_{22} & 0 0 & 0 & B_{33} end{bmatrix} As it is an orthogonal matrix these diagonal elements $B_\left\{ii\right\}$ are either 1 or −1. As the determinant is 1 these elements are either all 1 or one of the elements is 1 and the other two are −1.

In the first case it is the trivial identity operator corresponding to $theta=0$.

In the second case it has the form


begin{bmatrix}
`    -1  &     0 &    0 `
`     0  &    -1 &    0 `
`     0  &     0 &    1`
end{bmatrix}

if the basevectors are numbered such that the one with eigenvalue 1 has index 3. This matrix is then of the desired form for $theta=pi$.

If the matrix is un-symmetric the vector


bar E = alpha_1 hat f_1 + alpha_2 hat f_2 + alpha_3 hat f_3

where

$alpha_1=frac\left\{A_\left\{32\right\}-A_\left\{23\right\} \right\}\left\{2\right\}$
$alpha_2=frac\left\{A_\left\{13\right\}-A_\left\{31\right\}\right\}\left\{2\right\}$
$alpha_3=frac\left\{A_\left\{21\right\}-A_\left\{12\right\}\right\}\left\{2\right\}$

is non-zero. This vector is an eigenvector with eigenvalue


lambda=1

Setting


hat e_3=frac{bar E}
>

and selecting any two orthogonal unit vectors in the plane orthogonal to $hat e_3$:

$hat e_1 , hat e_2$

such that

$hat e_1 , hat e_2, hat e_3$

form a positively oriented trippel the operator takes the desired form with

$cos theta=frac\left\{A_\left\{11\right\}+A_\left\{22\right\}+A_\left\{33\right\}-1\right\}\left\{2\right\}$
$sin theta=|bar\left\{E\right\}|$

The expressions above are in fact valid also for the case of a symmetric rotation operator corresponding to a rotation with $theta = 0$ or $theta = pi$. But the difference is that for $theta = pi$ the vector


bar E = alpha_1 hat f_1 + alpha_2 hat f_2 + alpha_3 hat f_3

is zero and of no use for finding the eigenspace of eigenvalue 1, i.e. the rotation axis.

Defining $E_4$ as $cos theta$ the matrix for the rotation operator is


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}

provided that


{E_1}^2+{E_2}^2+{E_3}^2 > 0

i.e. except for the cases $theta=0$ (the identity operator) and $theta=pi$

Quaternions

Quaternions are defined similar to $E_1 , E_2 , E_3 , E_4$ with the difference that the half angle $frac\left\{theta\right\}\left\{2\right\}$ is used in stead of the full angle $theta$.

This means that the first 3 components $q_1 , q_2 , q_3$ are components of a vector defined from


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}}{sintheta}quad bar E and that the forth component is the scalar

q_4=cos frac{theta}{2}

As the angle $theta$ defined from the canonical form is in the interval

$0 le theta le pi$

one would normally have that $q_4 ge 0$. But a "dual" representation of a rotation with quaternions is used, i.e.

$q_1 , q_2 , q_3 , q_4$

and

$-q_1 , -q_2 , -q_3 , -q_4$

are two alternative representations of one and the same rotation.

The entities $E_k$ are defined from the quaternions by

$E_1=2 q_4 q_1$
$E_2=2 q_4 q_2$
$E_3=2 q_4 q_3$
$E_4=\left\{q_4\right\}^2 -\left(\left\{q_1\right\}^2+\left\{q_2\right\}^2+\left\{q_3\right\}^2\right)$

Using quaternions the matrix of the rotation operatator is

$begin\left\{bmatrix\right\}$
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}

Numerical example

Consider the reorientation corresponding to the Euler angles $alpha=10deg quad beta=20deg quad gamma=30deg quad$ relative a given base vector system

$hat f_1 , hat f_2, hat f_3$

Corresponding matrix relative this base vector system is (see Euler angles#Matrix notation)


begin{bmatrix}
` 0.771281 & -0.633718 &  0.059391 `
` 0.613092 &  0.714610 & -0.336824 `
` 0.171010 &  0.296198 &  0.939693`
end{bmatrix}

and the quaternion is


(0.171010, -0.030154, 0.336824, 0.925417)

The canonical form of this operator


begin{bmatrix}
` costheta  & -sintheta & 0`
` sintheta  &  costheta & 0`
`          0  &           0 & 1`
end{bmatrix} with $theta=44.537deg$ is obtained with
$hat e_3=\left(0.451272,-0.079571,0.888832\right)$

The quaternion relative this new system is then


(0, 0, 0.378951, 0.925417) = (0, 0, sinfrac{theta}{2}, cosfrac{theta}{2})

Instead of making the three Euler rotations

$10 deg + 20 deg + 30 deg$

the same orientation can be reached with one single rotation of size $44.537 deg$ around $hat e_3$

Reference

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