Differential Geometry and Its Application - Exercise 1.3.12

Problem:

If a rigid body moves along a curve $\alpha(s)$ (which we suppose is unit speed), then the motion of the body consists of translation along $\alpha$ and rotation about $\alpha$. The rotation is determined by an angular velocity vector $\omega$ which satisfies $T' = \omega \times T$, $N' = \omega \times N$ and $B' = \omega \times B$. The vector $\omega$ is called the Darboux vector. Show that $\omega$, in terms of $T$, $N$ and $B$, is given by $\omega = \tau T + \kappa B$.

Solution:

$T$, $N$ and $B$ forms an orthonormal frame, so we can write:

$\omega = aT + bN + cB$.

To make things easier later, we document the cross product table here:

$\times$	$T$	$N$	$B$
$T$	$0$	$B$	$-N$
$N$	$-B$	$0$	$T$
$B$	$N$	$-T$	$0$

$\begin{eqnarray*} \kappa N = T' &=& \omega \times T &=& (aT + bN + cB) \times T &=& -bB + cN \\ -\kappa T + \tau B = N' &=& \omega \times N &=& (aT + bN + cB) \times N &=& aB - cT \\ -\tau N = B' &=& \omega \times B &=& (aT + bN + cB) \times B &=& -aN + bT \end{eqnarray*}$

Remember again $T$, $N$, $B$ is an orthonormal frame, so we proved $a = \tau$, $b = 0$ and $c = \kappa$, so $\omega = \tau T + \kappa B$.