4.2.4 Rotations and Euclidean Motions

A general reflection at the plane $ w^\perp$ can be described by first rotating $ w$ into some vector $ v$, then reflect at $ v^\perp$ and now rotate back $ v$ to $ w$. In fact, let $ u:=v+w$. Then this first rotation is given by $ S_u\o S_w=R=S_v\o S_u$ and we have

$\displaystyle S_v\o R=S_v\o (S_v\o S_u) = (S_v)^2\o S_u=S_u=S_u\o (S_w)^2=R\o S_w.
$

How can rotations be described by matrices? Let us first consider 2-dimensional space. The rotation by $ 90^o$ maps any vector $ v$ to a normal vector $ v^\perp$ of the same length. There are only two possibilities for $ \langle a,b\rangle^\perp$ namely $ \pm \langle -b,a\rangle$, and the one with $ +$ is rotation in the positive direction, i.e. counterclockwise. The matrix corresponding to this rotation is given by

$\displaystyle R_{\pi/2} =\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}.
$

Now what about a rotation around 0 by an arbitrary angle $ \varphi $? By elementary geometric calculations we see, that the image of $ x$ is $ \cos{\varphi }\,x+\sin{\varphi }\,y$, and the image of $ y$ which is $ x$ turned $ 90^o$ counterclockwise is the vector $ -\sin{\varphi }\,x+\cos{\varphi }\,y$. Thus in coordinates this linear mapping is given by the unitarian matrix

$\displaystyle \begin{pmatrix}
\cos\varphi & -\sin\varphi \\
\sin\varphi & \cos\varphi
\end{pmatrix}$

Note that this can also be paraphrased by ``the reference frame is rotated in the opposite direction by $ \varphi $''. In three dimensions a rotation by $ \varphi $ around the $ z$-axes does not change the $ z$-coordinate and rotates the $ (x,y)$-coordinates as before and thus is given by the unitarian matrix

$\displaystyle \begin{pmatrix}
\cos\varphi & -\sin\varphi & 0 \\
\sin\varphi & \cos\varphi & 0 \\
0 & 0 & 1
\end{pmatrix}$

In Pov-Ray: rotate $ \varphi $*z.

Similarly rotations around the $ x$- and around the $ y$-axes are given by

$\displaystyle \begin{pmatrix}
1 & 0 & 0 \\
0 & \cos\varphi & -\sin\varphi \\
0 & \sin\varphi & \cos\varphi
\end{pmatrix}$    and $\displaystyle \begin{pmatrix}
\cos\varphi & 0 & -\sin\varphi \\
0 & 1 & 0 \\
\sin\varphi & 0 &\cos\varphi
\end{pmatrix}$

In Pov-Ray: rotate $ \varphi $*x and rotate $ \varphi $*y.

Note that the composition of two rotations by angles $ \varphi $ and $ \psi $ around the same center is the rotation by angle $ \varphi +\psi $. Expressing this via the corresponding matrices gives the addition laws for $ \sin$ and for $ \cos$.

All these rotation matrices $ M$ (as well as arbitrary compositions of such) satisfy $ \det(M)=1$ and $ M^t\cdot M=\operatorname{id}$, thus are special orthogonal matrices.

A general rotation around the axis spanned by the unit vector $ w$ by the angle $ \varphi $ is given by considering the orthogonal frame given by $ w$, $ v\times w$, $ v-\langle v\vert w\rangle w$. The length of these vectors are $ 1$, $ \sin(\angle(v,w))\Vert v\Vert$, $ \sin(\angle(v,w))\Vert v\Vert$. The vector $ v$ is given in this frame as

$\displaystyle v=\langle v\vert w\rangle\cdot w + 1\cdot(v-\langle v\vert w\rangle w) + 0\cdot v\times w,
$

so the rotation is

$\displaystyle v\mapsto \langle v\vert w\rangle\cdot w + \cos(\varphi )\cdot(v-\langle v\vert w\rangle w)
+ \sin(\varphi )\cdot v\times w
$

Figure: Orthogonal decomposition of a vector $ v$
\includegraphics[width=0.4\textwidth]{rot-3.eps}
Note that

$\displaystyle w\times (v \times u)=(u\times v)\times w$ $\displaystyle = \langle u\vert w\rangle v - \langle v\vert w\rangle u$    
and hence$\displaystyle \qquad v\times (v\times u)$ $\displaystyle = \langle u\vert v\rangle v - \langle v\vert v\rangle u = \langle u\vert v\rangle v - u.$    

See www.martinb.com/.../index.htm.

Let $ f$ be an arbitrary length preserving mapping (a so called EUCLIDEAN MOTION), i.e. $ \Vert f(v_1)-f(v_2)\Vert=\Vert v_1-v_2\Vert$ for all $ v_1,v_2$. Up to the translation by $ f(O)$ (i.e. replacing $ f$ by $ v\mapsto f(v)-f(0)$) it preserves also the origin $ O$ and hence $ \Vert f(v)\Vert=\Vert v\Vert$ for all $ v$. Furthermore by the polarization equality

$\displaystyle 2\langle v\vert w\rangle = \Vert v\Vert^2+\Vert w\Vert^2-\Vert v-w\Vert^2
$

we get

$\displaystyle 2\langle f(v)\vert f(w)\rangle$ $\displaystyle = \Vert f(v)\Vert^2+\Vert f(w)\Vert^2-\Vert f(v)-f(w)\Vert^2$    
  $\displaystyle = \Vert v\Vert^2+\Vert w\Vert^2-\Vert v-w\Vert^2 = 2\langle v\vert w\rangle,$    

i.e. $ f$ preserves the inner product and hence maps the standard orthogonal basis $ e_1=x$, $ e_2=y$ and $ e_3=z$ to some other orthonormal basis $ f_1:=f(e_1)$, $ f_2:=f(e_2)$ and $ f_3:=f(e_3)$. For a general vector $ v$ we have $ v=\sum_k v_k\,e_k$ with $ v_k=\langle v\vert e_k\rangle=\langle f(v)\vert f(e_k)\rangle=\langle f(v)\vert f_k\rangle$ and hence

$\displaystyle f(v)=\sum_k \langle f(v)\vert f_k\rangle f_k=\sum_k v_k\,f_k,
$

i.e. $ f$ is linear and, moreover, for the matrix $ [f]$ associated to $ f$ we have $ [f]^t\cdot [f]=\operatorname{id}$, i.e. $ f$ is given by an orthogonal matrix: In fact the $ jk^{th}$ entry in the matrix $ [f]^t\cdot [f]$ is given by

$\displaystyle ([f]^t\cdot [f])_{j,k}$ $\displaystyle = [f^*\o f]_{j,k} = \langle (f^* \o f)e_j\vert e_k\rangle = \langle fe_j\vert fe_k\rangle$    
  $\displaystyle = \langle e_j\vert e_k\rangle =\left\{\begin{array}{ll} 1 &\text{for}j=k, \\ 0 &\text{otherwise}.\end{array}\right.$    

Let now conversely an orthogonal $ 3\times 3$-matrix $ A$ be given. Then $ 1=\det(\operatorname{id})=\det(A^t\cdot A)=\det(A^t)\cdot \det(A)=\det(A)^2$, i.e. $ \det(A)\in\{-1,+1\}$. Let us assume first that $ \det(A)=1$.

Let $ F:=\operatorname{Fix}(A):=\{x\in E:Ax=x\}$ be the set of its fixed points, i.e. the eigenspace for the eigenvalue 1. This is a linear subspace.

We show next, that $ \dim(F)>0$. Let $ p(\lambda ):=\det(\lambda -A)$ be the characteristic polynomial of $ A$. We have

$\displaystyle p(\lambda )$ $\displaystyle = \det(\lambda -A) = \det((\lambda - A)^t) = \det(\lambda -A^t) = \det(\lambda -A^{-1})$    
  $\displaystyle = \det(-\lambda \left(\frac1{\lambda }-A\right)A^{-1}) = \det(-\lambda ) \cdot \det\left(\frac1{\lambda }-A\right)\cdot \det A^{-1}$    
  $\displaystyle = (-\lambda )^3 p(\frac1\lambda ),$    

thus we get for $ \lambda =1$ that $ 2p(1)=0$. Hence $ 1$ is an eigenvalue.

In case its dimension is 3, we have $ A=\operatorname{id}$.

In case its dimension is 2, we find a unit vector $ w$ such that $ F=w^\perp$. Since $ F$ is invariant under $ A$ and $ A$ is orthogonal the same is true for $ F^{\perp}$. Since $ w\in F^{\perp}$ and $ A$ is an isometry we have $ A(w)=-w$ and hence $ A$ is the reflection at the plane $ F=w^\perp$. In fact, $ x-\langle x\vert v\rangle v\in F$ for all $ x\in E$, since

$\displaystyle \langle x-\langle x\vert v\rangle v\vert v\rangle
= \langle x\vert v\rangle - \langle x\vert v\rangle \cdot \langle v\vert v\rangle = 0.
$

Thus

$\displaystyle Ax$ $\displaystyle = A\Bigl( ( x-\langle x\vert v\rangle v) + \langle x\vert v\rangle v \Bigr) = x-\langle x\vert v\rangle v + \langle x\vert v\rangle (-v)$    
  $\displaystyle = x-2 \langle x\vert v\rangle v.$    

Remains to consider the case, where $ \dim(F)=1$. Then there exists a unit vector $ w$ which spans $ F$. The orthogonal plane $ w^{\perp}=F^{\perp}$ is also $ A$ invariant, so $ A\vert _{F^{\perp}}$ is orthogonal on this plane and hence a reflection on a line in this plane or a rotation in this plane. Thus $ A$ is a reflection at the plane spanned by $ w$ and the reflection line of $ A\vert _{F^{\perp}}$ or a rotation with axis $ w$.

Therefore any orthogonal mapping on $ \protect\mathbb{R}^3$ is the composite of (at most three) reflections. It is a rotation around some axes $ v\ne 0$ by some angle $ \varphi $ iff it is a composite of two reflections.
In Pov-Ray: $ v\mapsto$vaxis_rotate $ (v,w,\varphi )$.
Finally the euclidean motions are exactly of the form $ v\mapsto R\cdot v + w$, where $ R$ is a rotation which is followed by the translation $ v\mapsto v+w$.

We show next that any rotation (i.e. special orthogonal matrix) can be obtained by composing 3 of the special rotations discussed above by the so called EULER ANGLES. Consider an airplane or an hang-glider: We have the basis given by the axes of airplane: the direction from the left to the right wing, the vertical direction, and the direction from back to front.

How do the corresponding matrices look like?
Figure:                 $ R_1$$ R_2$                
\includegraphics[width=0.45\textwidth]{drachen-1.eps}     \includegraphics[width=0.45\textwidth]{drachen-2.eps}
Figure:                 $ R_3$$ R_1$, $ R_2$, $ R_3$            
\includegraphics[width=0.45\textwidth]{drachen-3.eps}     \includegraphics[width=0.45\textwidth]{drachen.eps}

Another decomposition into 3 rotations is given via the following Euler-angles:
Let $ R$ be a rotation and $ f_j:= R(e_j)$ be the images of the standard-basis. We would like to express $ R$ as composition of 3 rotations around some coordinate-axes. It suffices to describe the images of these rotations on the first 2 vectors $ e_1$ and $ e_2$, since $ e_3=e_1\times e_2$ is the uniquely determined unit vector normal to $ e_1$ and $ e_2$ such that $ (e_1,e_2,e_3)$ is left oriented.

In order to rotate $ e_1$ to $ f_1$ we have to keep an axis $ k\in \{e_1\}^\perp \cap \{f_1\}^\perp
= \langle\{ e_2,e_3\}\rangle \cap \langle\{f_2,f_3\}\rangle
$ fixed. In order to rotate afterwards $ e_2$ to $ f_2$ without destroying the assignment $ e_1\mapsto f_1$, we could first rotate $ e_2$ to $ k$ around $ e_1$ and at the end rotate $ k$ to $ f_2$ around $ f_1$.

$\displaystyle \begin{matrix}
e_1 &\mapsto& e_1 &\mapsto& f_1 &\mapsto& f_1 \\
e_2 &\mapsto& k &\mapsto& k &\mapsto& f_2 \\
\end{matrix}$

Figure:                 Fixed axes $ k$$ R_1$                
\includegraphics[width=0.45\textwidth]{euler-ang-1}     \includegraphics[width=0.45\textwidth]{euler-ang-2}
Figure:                 $ R_2$$ R_3$                
\includegraphics[width=0.45\textwidth]{euler-ang-3}     \includegraphics[width=0.45\textwidth]{euler-ang-4}
Let $ \varphi _1,\varphi _2,\varphi _3$ be the so-called Euler-angles of $ R$, given by

$\displaystyle \varphi _1 := \measuredangle e_2 k;
\quad \varphi _2 := \measuredangle e_1 f_1;
\quad \varphi _3 := \measuredangle k f_2.
$

Thus the matrix-representation of the corresponding rotations $ R_1$, $ R_2$ and $ R_3$ are given by:

$\displaystyle [R_1]_{e_1,e_2}$ $\displaystyle = \begin{pmatrix}1 & 0 & 0 \\ 0 & \cos\varphi _1 & -\sin\varphi _1 \\ 0 & \sin\varphi _1 & \cos\varphi _1 \\ \end{pmatrix}$    
$\displaystyle [R_2]_{e_1,k}$ $\displaystyle = \begin{pmatrix}\cos\varphi _2 & 0 & -\sin\varphi _2 \\ 0 & 1 & 0 \\ \sin\varphi _2 & 0 & \cos\varphi _2 \\ \end{pmatrix}$    
$\displaystyle [R_1]_{f_1,k}$ $\displaystyle = \begin{pmatrix}1 & 0 & 0 \\ 0 & \cos\varphi _3 & -\sin\varphi _3 \\ 0 & \sin\varphi _3 & \cos\varphi _3 \\ \end{pmatrix}$    

Hence

\begin{multline*}[R_3\o R_2\o R_1]_{e_1,e_2,e_3} = \\
= \begin{pmatrix}
1 & 0 ...
...hi _3 \\
0 & \sin\varphi _3 & \cos\varphi _3 \\
\end{pmatrix}\end{multline*}

Andreas Kriegl 2003-07-23