Show that the torsion satisfies $\tau(s)=\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}$

Question

Let $\alpha:I\rightarrow\mathbb{R}^{3}$ a curve parametrized by arc lenght $s$. Denote by $t(s)=\alpha'(s)$ the tangent vector, $\kappa(s)=|\alpha'(s)|$ the curvature, $n(s)=\alpha''(s)/\kappa(s)$ the normal vector, $b(s)=t(s)\wedge n(s)$ the binormal vector. The torsion is the number $\tau(s)$ such that $b'(s)=\tau(s)n(s).$

I need to show that $$\tau(s)=\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}$$.

I already showed that

$$\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}n(s)= \frac{-t(s)\wedge n(s)\cdot\alpha'''(s)}{\kappa(s)}n(s),$$

and, since $b'(s)=t'(s)\wedge n(s)+t(s)\wedge n'(s),$ I need to prove that

$$\frac{-t(s)\wedge n(s)\cdot\alpha'''(s)}{\kappa(s)}n(s)=t'(s)\wedge n(s)+t(s)\wedge n'(s). $$

How can I do that?

Answer 1

By definition, $$ \alpha'(s) = t(s) , $$ and differentiating twice, $$ \alpha''(s) = t'(s) = \kappa(s) n(s) \\ \alpha'''(s) = \kappa'(s) n(s) + \kappa(s) n'(s) . $$ We can find what $n'$ is: $$\begin{align} n \cdot n &= 1 \implies 0 = (n \cdot n)' = n \cdot n' \\ n \cdot t &= 0 \implies 0 = (n \cdot t)' = n' \cdot t + n \cdot t' = n' \cdot t + \kappa n \cdot n = n' \cdot t + \kappa \\ n \cdot b &= 0 \implies 0 = (n \cdot b)' = n' \cdot b + n \cdot b' = n' \cdot b + \tau \end{align}$$ so $ n' = -\kappa t - \tau b $, one of the Frenet–Serret formulae (note that Wikipedia uses a different, more common convention that $b' = -\tau n$). Thus, $$ \alpha'''(s) = \kappa'(s) n(s) - \kappa(s)^2 t(s) - \kappa(s)\tau(s)b(s) $$ If we now compute the triple product of the first three derivatives, we find $$ \alpha'(s) \times \alpha''(s) = \kappa(s) b(s), $$ so $$ (\alpha'(s) \times \alpha''(s)) \cdot \alpha'''(s) = -\kappa(s)^2 \tau(s), $$ whence the result. The key here is to work either completely in terms of $\alpha$, or completely in terms of $t,n,b$: mixing them tends to be confusing.

Answer 2

You can just calculate that $$\alpha'''=\left(\kappa N\right)'=\kappa N'+\kappa' N=\kappa \tau B+\cdots,$$ using the Frenet equations. We only care about the $B$ component due to the dot product we want to take. Hence, $$(\alpha'\times\alpha'')\cdot\alpha'''=\kappa^2\tau\implies \tau=\frac{(\alpha'\times\alpha'')\cdot\alpha'''}{\kappa^2}.$$

EDIT: Typically, one defines $\tau $ by $B'=-\tau N.$ I see now that you are using $B'=\tau N.$ In this case, the Frenet equations have some sign changes. Taking this into account, you get that $\alpha'''=-\kappa\tau B+\cdots,$ and taking the dot product gives $$(\alpha'\times\alpha'')\cdot\alpha'''=-\kappa^2\tau,$$ or $$\tau=-\frac{(\alpha'\times\alpha'')\cdot\alpha'''}{\kappa^2}.$$