Configuration manifold of a rigid body

Question

As I know, a rigid body is a set of $N$ particles in three-dimensional space subject to the following constraint: if $b_1,\dots,b_N\in \mathbb{R}^3$ are the initial positions of the particles and if $c_1,\dots,c_N: \mathbb{R}\to \mathbb{R}^3$ are their time evolutions, then

$$|c_{i}(t)-c_{j}(t)|=|b_i-b_j| , \quad \forall t \in \mathbb{R}.$$

In other words, the distance between each pair of points is maintained constant throughout the time evolution. With this in mind, it is usual to consider that the configuration manifold of a rigid body is in truth

$$Q = \mathbb{R}^3\times SO(3).$$

In Spivak's Mechanics book it is said:

This all can be expressed in a more familiar, geometric, way by considering the "configuration space" of $\mathbf{b}$, which is the subset $\mathscr{M}\subset (\mathbb{R}^3)^N$ of all points that can be reached from $\mathbf{b}$ at the end of a rigid motion. In other words,

$$\mathscr{M}=\{(A(b_1),\dots,A(b_N)) : A \ \text{an orientation preserving isometry of } \ \mathbb{R}^3\}.$$

When $\mathbf{b}$ is non-planar, $\mathscr{M}$ is a $6$-dimensional manifold diffeomorphic to the set of all orientation preserving isometries $A$ of $\mathbb{R}^3$, and thus to $\mathbb{R}^3\times SO(3)$. With this picture, a rigid motion of $\mathbf{b}$is simply a curve in $\mathscr{M}$, so a virtual infinitesimal displacement $\mathbf{v}$ of $\mathbf{b}$ is simply a tangent vector to $\mathscr{M}$ at $\mathbf{b}$.

Although this seems really simple, I'm having a hard time to understand that the configuration manifold of a rigid body is $\mathbb{R}^3\times SO(3)$. I see that in truth if $A$ is a curve of isometries, then really $t\mapsto (A(t)(b_1),\dots,A(t)(b_N))$ is a curve that satisfies the constraints.

But why this says the configuration manifold is the set of all orientation preserving isometries? Also, where the orientation preserving came from? In short, how to see that the configuration manifold of a rigid body is $\mathbb{R}^3\times SO(3)$?

Answer 1

This isn't too hard.

1. Translations clearly are isometries: if $\vec a' = \vec a + \vec c$ and $\vec b' = \vec b + \vec c$ then $|\vec a' - \vec b'| = |\vec a - \vec b|.$
2. Consider any isometry $f$; consider the isometry $g(x) = f(x) - f(0)$ which preserves the origin. It's not too hard to see that this has to be linear and is therefore described by a matrix $$M = \begin{bmatrix}\hat u_1&\dots&\hat u_n\end{bmatrix}.$$These have to be unit vectors because they are the transformations of the unit vector$\begin{bmatrix}0\dots&1&\dots 0\end{bmatrix}^T,$ and they must be orthogonal for a similar reason, which means $M^T M = I.$ So it has to be a subset of $O(n).$
3. The restriction $\det M = +1$ is only needed to make the family of isometry transforms continuous with the identity.

Answer 2

The $\mathbb{R}^{3}$ and the $SO(3)$ respectively parametrize the center-of-mass trajectory and the relative motion (i.e., rotation) about the center of mass.

To see why the latter is $SO(3)$, imagine three mutually orthogonal unit vectors that are fixed to the rigid body and whose base points all coincide with the center of mass. One can completely describe the relative rotational motion by specifying which directions the three unit vectors point to. Furthermore, the rotational motion preserves the relative orientation of these vectors. This is exactly the defining properties of $SO(3)$.