Torsion formula derivation

I am trying to derive the formula for torsion of a curve. In many example proofs I've seen the final step is that $$\kappa=\|\dot{\lambda}\times\ddot{\lambda}\|$$. However I thought curvature was defined to be $$\kappa=\frac{\|\dot{\lambda}\times\ddot{\lambda}\|}{\|\dot{\lambda}\|^3}$$ or $$\kappa=\|\ddot{\lambda}\|$$ if $$\lambda$$ is unit-speed. However, I would like to note, I have only assumed that $$\lambda$$ is a regular curve with non-zero curvature.

Edit: essentially how does one derive the formula in the screenshot.

• Excuse me. Could you tell me the name of the book you use? Commented Dec 21, 2019 at 23:23
• This is not from a book, but from my lecturers notes.
– kam
Commented Dec 22, 2019 at 12:16
• Okay. Thanks! ... Commented Dec 22, 2019 at 14:18

We must use the Frenet formulas for regular curves: \begin{align*} T' &= v\kappa N\\ N' &= -v\kappa T + v \tau B\\ B' &= -v\tau N. \end{align*} Here $$T, N, B$$ is the Frenet frame along the curve $$\gamma$$, $$v = \|\gamma'\|$$ is the speed and $$\kappa$$ and $$\tau$$ are of course the curvature and torsion.
By definition of $$v$$ we have $$\gamma' = vT.$$ The second derivative of $$\gamma$$ is $$\gamma'' = v'T + vT' = v' T + v^2\kappa N.$$ The cross product $$\gamma'\times \gamma''$$ becomes $$\gamma' \times \gamma'' = \kappa v^3 B.$$ If we take the norm of both sides, we obtain $$\kappa = \frac{\|\gamma'\times\gamma''\|}{v^3} = \frac{\|\gamma'\times \gamma''\|}{\|\gamma'\|^3}. \tag{1}$$ The binormal vector $$B$$ is a unit vector perpendicular to $$T$$ and $$N$$. From the expressions for $$\gamma'$$ and $$\gamma''$$, it follows that $$B$$ points in the same direction as $$\gamma'\times \gamma''$$, so $$B = \frac{\gamma'\times\gamma''}{\|\gamma'\times\gamma''\|}. \tag{2}$$
Deriving one more time gives $$\gamma'''$$: \begin{align*} \gamma''' &= v'' T + v'T' + (v^2\kappa)'N + v^2\kappa N' \\ &= v'' T + (v'v\kappa + (v^2\kappa)')N + v^3\kappa (-\kappa T + \tau B). \end{align*} Take the inner product with $$B$$ and obtain $$\gamma'''\cdot B = v^3\kappa \tau$$. Combined with $$(2)$$ this gives $$\tau = \frac{\gamma''' \cdot (\gamma'\times\gamma'')}{\|\gamma'\times\gamma''\|v^3\kappa}.$$ Finally, using $$(1)$$, we may conclude that the torsion is given by $$\tau = \frac{(\gamma'\times\gamma'')\cdot\gamma'''}{\|\gamma'\times\gamma''\|^2}.$$
• Yes, that's a typo; forgot to add the ${}^2$. Thanks for pointing out. Commented Dec 23, 2019 at 15:44