Screw theory

Remarks

On the axis, the velocity is parallel to $\omega$: for $p = \omega \times v / \abs{\omega}^2$ one finds $\omega \times p = \omega (\omega \cdot v)/\abs{\omega}^2 - v$, so $u(p) = h \omega$. So the body turns about the axis and slides along it with speed $h \abs{\omega}$, which is the infinitesimal form of Chasles' theorem. Example: for $\omega = (0, 0, 1)$ and $v = (0, -1, 2)$, the pitch is $h = 2$ and $p = \omega \times v = (1, 0, 0)$. Check: $u(p) = (0, -1, 2) + (0, 0, 1) \times (1, 0, 0) = (0, -1, 2) + (0, 1, 0) = (0, 0, 2) = 2\omega$; the axis is the vertical line through $(1, 0, 0)$. If $\omega = 0$, the twist is a pure translation with velocity $v$, a screw of infinite pitch. Non-example: the field $u(x) = A x$ with $A = \operatorname{diag}(1, 0, 0)$ is not a twist, since $A$ is not skew-symmetric; it stretches the body along the first axis instead of moving it rigidly. The twists form the Lie algebra of the Euclidean group, with bracket $[(\omega_1, v_1), (\omega_2, v_2)] = (\omega_1 \times \omega_2, \omega_1 \times v_2 - \omega_2 \times v_1)$.