User talk:Paul Wormer/scratchbook1

Rotations in $\mathbb {R} ^{3}$

Consider a real 3×3 matrix R with columns r₁, r₂, r₃, i.e.,

$\mathbf {R} =\left(\mathbf {r} _{1},\mathbf {r} _{2},\mathbf {r} _{3}\right)$.

The matrix R is orthogonal if

$\mathbf {r} _{i}\cdot \mathbf {r} _{j}=\delta _{ij},\quad i,j=1,2,3.$

The matrix R is a proper rotation matrix, if it is orthogonal and if r₁, r₂, r₃ form a right-handed set, i.e.,

$\mathbf {r} _{i}\times \mathbf {r} _{j}=\sum _{k=1}^{3}\,\varepsilon _{ijk}\mathbf {r} _{k}.$

Here the symbol × indicates a cross product and $\varepsilon _{ijk}$ is the antisymmetric Levi-Civita symbol,

${\begin{aligned}\varepsilon _{123}=&\;\varepsilon _{312}=\varepsilon _{231}=1\\\varepsilon _{213}=&\;\varepsilon _{321}=\varepsilon _{132}=-1\end{aligned}}$

and $\varepsilon _{ijk}=0$ if two or more indices are equal.

The matrix R is an improper rotation matrix if its column vectors form a left-handed set, i.e.,

$\mathbf {r} _{i}\times \mathbf {r} _{j}=-\sum _{k}\,\varepsilon _{ijk}\mathbf {r} _{k}\;.$

The last two equations can be condensed into one equation

$\mathbf {r} _{i}\times \mathbf {r} _{j}=\det(\mathbf {R} )\sum _{k=1}^{3}\;\varepsilon _{ijk}\mathbf {r} _{k}$

by virtue of the the fact that the determinant of a proper rotation matrix is 1 and of an improper rotation −1. This can be proved as follows: The determinant of a 3×3 matrix with column vectors a, b, and c can be written as

$\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )$.

Remember that for a proper rotation the columns of R are orthonormal and satisfy,

$\mathbf {r} _{1}\cdot (\mathbf {r} _{2}\times \mathbf {r} _{3})=\sum _{k}\,\varepsilon _{23k}\,\mathbf {r} _{1}\cdot \mathbf {r} _{k}=\varepsilon _{231}=1.$

Likewise the determinant is −1 for an improper rotation, which ends the proof.

Theorem

A proper rotation matrix R can be factorized thus

$\mathbf {R} =\mathbf {R} _{z}(\omega _{3})\;\mathbf {R} _{y}(\omega _{2})\;\mathbf {R} _{x}(\omega _{1})$

which is referred to as the z-y-x parametrization, or also as

$\mathbf {R} =\mathbf {R} _{z}(\alpha )\;\mathbf {R} _{y}(\beta )\;\mathbf {R} _{z}(\gamma )\quad$

the z-y-z Euler parametrization.

Here

$\mathbf {R} _{z}(\varphi )\equiv {\begin{pmatrix}\cos \varphi &-\sin \varphi &0\\\sin \varphi &\cos \varphi &0\\0&0&1\\\end{pmatrix}},\quad \mathbf {R} _{y}(\varphi )\equiv {\begin{pmatrix}\cos \varphi &0&\sin \varphi \\0&1&0\\-\sin \varphi &0&\cos \varphi \\\end{pmatrix}},\quad \mathbf {R} _{x}(\varphi )\equiv {\begin{pmatrix}1&0&0\\0&\cos \varphi &-\sin \varphi \\0&\sin \varphi &\cos \varphi \\\end{pmatrix}}.$

Proof

First the z-y-x-parametrization will be proved by describing an algorithm for the factorization of R. Consider to that end

$\mathbf {R} _{z}(\omega _{3})\,\mathbf {R} _{y}(\omega _{2})={\begin{pmatrix}\cos \omega _{3}\cos \omega _{2}&-\sin \omega _{3}&\cos \omega _{3}\sin \omega _{2}\\\sin \omega _{3}\cos \omega _{2}&\cos \omega _{3}&\sin \omega _{3}\sin \omega _{2}\\-\sin \omega _{2}&0&\cos \omega _{2}\\\end{pmatrix}}\equiv (\mathbf {a} _{1},\mathbf {a} _{2},\mathbf {a} _{3}).$

Note that the multiplication by R_x(ω₁) on the right does not affect the first column, so that r₁ = a₁. Solve $\omega _{2}$ and $\omega _{3}$ from the first column of R,

$\mathbf {r} _{1}={\begin{pmatrix}\cos \omega _{3}\;\cos \omega _{2}\\\sin \omega _{3}\;\cos \omega _{2}\\-\sin \omega _{2}\\\end{pmatrix}}.$

This is possible. First solve $\omega _{2}$ for $-\pi /2\leq \omega _{2}\leq \pi /2$ from

$\sin \omega _{2}=-R_{31}\equiv -(\mathbf {r} _{1})_{3}.$

Then solve $\omega _{3}$ for $0\leq \omega _{3}\leq 2\pi$ from

${\begin{aligned}\cos \omega _{3}=&{R_{11} \over \cos \omega _{2}}\\\sin \omega _{3}=&{R_{21} \over \cos \omega _{2}}.\end{aligned}}$

This determines the vectors a₂ and a₃.

Since a₁, a₂ and a₃ are the columns of a proper rotation matrix they form an orthonormal right-handed system. The plane spanned by a₂ and a₃ is orthogonal to $\mathbf {a} _{1}\equiv \mathbf {r} _{1}$ and hence contains $\mathbf {r} _{2}$ and $\mathbf {r} _{3}$. Thus,

$(\mathbf {r} _{2},\mathbf {r} _{3})=(\mathbf {a} _{2},\mathbf {a} _{3}){\begin{pmatrix}\cos \omega _{1}&-\sin \omega _{1}\\\sin \omega _{1}&\cos \omega _{1}\\\end{pmatrix}}.$

Since $\mathbf {r} _{2},\mathbf {a} _{2},\mathbf {a} _{3}$ are known unit vectors we can compute

${\begin{aligned}\mathbf {a} _{2}\cdot \mathbf {r} _{2}=&\cos \omega _{1}\\\mathbf {a} _{3}\cdot \mathbf {r} _{2}=&\sin \omega _{1}.\end{aligned}}$

These equations give $\omega _{1}$ with $0\leq \omega _{1}\leq 2\pi$. Augment the matrix to $\mathbf {r} _{x}(\omega _{1})$, then

${\begin{aligned}\mathbf {R} \equiv &(\mathbf {r} _{1},\mathbf {r} _{2},\mathbf {r} _{3})=(\mathbf {r} _{1},\mathbf {a} _{2},\mathbf {a} _{3})\mathbf {R} _{x}(\omega _{1})\\=&(\mathbf {a} _{1},\mathbf {a} _{2},\mathbf {a} _{3})\mathbf {R} _{x}(\omega _{1})=\mathbf {R} _{z}(\omega _{3})\,\mathbf {R} _{y}(\omega _{2})\,\mathbf {R} _{x}(\omega _{1}).\end{aligned}}$

This concludes the proof of the z-y-x parametrization.