Rigid Body and Rigid Motion
The motion of a particle moving in Euclidean space is described by giving the location of the particle at each instant of time. A trajectory of the particle is represented by the parameterized curve $p(t) = (x(t), y(t), z(t)) \in \mathbb{R}^3$.
However, in robotics, we are interested in collective motion of a set of particles. We define a rigid body as a collection of particles such that the distance between any two particles remains fixed. That is, for any point $p(t)$ and $q(t)$ on a rigid body at time $t$,
$$\| p(t) - q(t) \| = \| p(0) - q(0) \| = \text{constant}.$$
A rigid motion is then the continuous movement of the particles in the object that preserves the distance between any particles on the object.
Rigid Body Transformation
Preserving only distance may not preserve orientation (e.g., $(x, y, z) \to (x, y, -z)$). To remove this possibility, we require that the cross product between vectors in the body also be preserved. Formally, a mapping $g: \mathbb{R}^3 \to \mathbb{R}^3$ is a rigid body transformation if:
- Distance preservation. For all points $p, q \in \mathbb{R}^3$, $$\| g(p) - g(q) \| = \| p - q \|.$$
- Cross-product preservation. For all vectors $v, w \in \mathbb{R}^3$, $$g_*(v \times w) = g_*(v) \times g_*(w),$$ where $g_*(v) = g(p') - g(q')$ if $p', q'$ are some points (transformation acting on vectors).
The important thing here is that the cross product of vectors is preserved, not the cross product of points (this distinction will be dealt with later).
Interesting Property
Fact. From the above definition, the inner product of two vectors is also preserved under a rigid body transformation.
Proof. Using the polarization identity, $$v_1^\top v_2 = \tfrac{1}{4}\!\left(\| v_1 + v_2 \|^2 - \| v_1 - v_2 \|^2\right),$$ and the fact that $$\| v_1 + v_2 \| = \| g_*(v_1) + g_*(v_2) \|, \qquad \| v_1 - v_2 \| = \| g_*(v_1) - g_*(v_2) \|,$$ we conclude that for any two vectors $v_1, v_2$, $$v_1^\top v_2 = g_*(v_1)^\top g_*(v_2). \qquad\blacksquare$$