Talk:Euler's theorem (rotation)

What is a rotation?

As I understand the first sentence, a rotation is defined to be "a motion of the rigid body that leaves at least one point of the body in place", but what is a rigid body motion? I think SE(3), i.e., all transformations of the form

${\mathbf {x}}\mapsto {\mathbf {Rx}}+{\mathbf {b}}$

with R in SO(3), however that does not seem to be what is meant in the article. -- Jitse Niesen 10:50, 14 May 2009 (UTC)

Yes, when b = 0 it is a rotation, provided R is an orthogonal matrix. When R = E it is a pure translation. I thought that "rigid body motion" would not have to be defined. See also Rotation matrix where I wrote the same (I'm still working on the latter). --Paul Wormer 11:23, 14 May 2009 (UTC)

But there are combinations of rotations and translations that leave points of the body in place. For instance, take

${\mathbf {x}}\mapsto {\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\end{bmatrix}}{\mathbf {x}}+{\begin{bmatrix}1\\0\\0\end{bmatrix}}.$

This transformation leaves the point (1/2, 1/2, 0) in place, but it's not a rotation. So I think it's wrong to define a rotation as "a motion of the rigid body that leaves at least one point of the body in place". -- Jitse Niesen 16:47, 14 May 2009 (UTC)

The vector (1/2, 1/2, 0) is an element of the difference space of the 3D real affine space. It represents the equivalence class consisting of pairs of ordered points P→Q related to each other by parallel translation (class of parallel oriented line segments). For instance O→P ∼ O′→P′ are represented by the same triplet with

${\begin{aligned}O\rightarrow &{\begin{bmatrix}0\\0\\0\end{bmatrix}}\quad P\rightarrow {\begin{bmatrix}1/2\\1/2\\0\end{bmatrix}}\\O'\rightarrow &{\begin{bmatrix}-1\\0\\0\end{bmatrix}}\quad P'\rightarrow {\begin{bmatrix}-1/2\\1/2\\0\end{bmatrix}}.\\\end{aligned}}$

Namely,

${\begin{bmatrix}1/2\\1/2\\0\end{bmatrix}}=\left({\begin{bmatrix}1/2\\1/2\\0\end{bmatrix}}-{\begin{bmatrix}0\\0\\0\end{bmatrix}}\right)=\left({\begin{bmatrix}1/2\\1/2\\0\end{bmatrix}}-{\begin{bmatrix}1\\0\\0\end{bmatrix}}\right)-\left({\begin{bmatrix}0\\0\\0\end{bmatrix}}-{\begin{bmatrix}1\\0\\0\end{bmatrix}}\right).$

A rotation is a motion of affine space that leaves invariant one point of affine space. To map affine space on the difference space (consisting of triplets of real numbers) we take a system of axes with origin in the invariant point. Hence an orthogonal map of difference space is a rotation of affine space if and only if it leaves the triplet (0, 0, 0) invariant, or, in other words, iff b = 0 and R orthogonal.

--Paul Wormer 08:23, 16 May 2009 (UTC)

To be precise: A rotation (at least, as used here) is not a motion in affine space (which has no metric), but in Euclidean affine space. Peter Schmitt 22:25, 7 June 2009 (UTC)

Slight change of title?

I think "Euler's theorem on rotation" (or similar) is a better title since "(rotation)" points to disambiguation. Peter Schmitt 22:29, 7 June 2009 (UTC)