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)

Move matrix material to other page(s)?

I think, the matrix material would better fit into the general context of rigid motion, isometries of Euclidean spaces, orthogonal matrices, and linear operators.
Comments? Peter Schmitt 22:34, 7 June 2009 (UTC)

Introduction

I have rewritten the introduction in the attempt to make the statement of the theorem simpler (the fixed point need not be in the body), and to describe the modern mathematical view. Peter Schmitt 22:53, 7 June 2009 (UTC)