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 $.