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

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?

• Well, I know that $\dot{v}=n \kappa$, $\dot{n}=-v \kappa+b \tau$ and $\dot{b}=-n \tau$ by Frenet formulas. And $v=\dot{\gamma}$ @Sou燈馬想 Commented Nov 23, 2017 at 21:10
• Its better to include what've you tried and where are you stuck to your post. Commented Nov 23, 2017 at 21:18
• That's the problem that I don't even know where to start. :/ @Sou燈馬想 Commented Nov 23, 2017 at 21:19
• Well just try to compute $\dot{\gamma}$, $\ddot{\gamma}$, and $\dddot{\gamma}$ first. Commented Nov 23, 2017 at 21:21

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

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

• How come $\langle \dddot \gamma, \dddot \gamma \rangle = \kappa^4 + \kappa^2 \tau^2 + \dot \kappa^2$ Commented Nov 23, 2017 at 21:33
• Because $T, N, B$ are othonormal unit vectors. Does that help? Commented Nov 23, 2017 at 21:49
• Oh yes! Haha, I skipped that part by reading the whole proof :)) Sorry and thank you! Commented Nov 23, 2017 at 21:56
• No problemo, Senor! And thank you! Commented Nov 23, 2017 at 21:57