Prove: $|\dddot{\gamma}|^2=\kappa^4+\kappa^2 \tau^2+\dot{\kappa}^2$, $\langle \dot{\gamma}, \dddot{\gamma} \rangle =-\kappa^2$

Question

We have $\gamma (t) $ (parametrised regulary) that has a curvature $\kappa$ and torsion $\tau$. Prove:

1. $|\dddot{\gamma}|^2=\kappa^4+\kappa^2 \tau^2+\dot{\kappa}^2$
2. $\langle \dot{\gamma}, \dddot{\gamma} \rangle =-\kappa^2$
3. $\langle \ddot{\gamma}, \dddot{\gamma} \rangle =\kappa \dot{\kappa}$.

Can anyone help me?

Answer 1

I assume $\gamma(s)$ is parametrized by arc-length $s$. Then

$\dot \gamma = T, \tag 1$

the unit tangent vector;

$\ddot \gamma = \dot T = \kappa N, \tag 2$

where $N$ a unit vector normal to $T$; thus

$\dddot \gamma = \dfrac{d(\kappa N)}{ds} = \dot \kappa N + \kappa \dot N; \tag 3$

by Frenet-Serret,

$\dot N = -\kappa T + \tau B, \tag 4$

so (3) becomes

$\dddot \gamma = \dot \kappa N - \kappa^2 T + \kappa \tau B; \tag 5$

since $T$, $N$, and $B = T \times N$ form an orthonormal triad, (5) yields

$\vert \dddot \gamma \vert^2 = \langle \dddot \gamma, \dddot \gamma \rangle = \kappa^4 + \kappa^2 \tau^2 + \dot \kappa^2; \tag 6$

using (1) and (5) we find

$\langle \dot \gamma, \dddot \gamma \rangle = \langle T, \dot \kappa N - \kappa^2 T + \kappa \tau B \rangle = -\kappa^2; \tag 7$

finally, using (2) and (5) we obtain

$\langle \ddot \gamma, \dddot \gamma \rangle = \langle \kappa N, \dot \kappa N - \kappa^2 T + \kappa \tau B \rangle = \kappa \dot \kappa. \tag 8$