The curvature of a tangent to a curve considered as a curve in its own right

Question

Suppose we have a curve $c:I \rightarrow \mathbb{R^3}$ that is parametrised with respect to arc length. We then consider $T,B,N$ (tangent, binormal and normal) as curve in there own right each defined at a time $t$. The task is to express the curvature of the curve $T$ in terms of the curvature and torsion of the orginal curve $c$. I am struggling to do this. So far I have:

$$ \begin{align*} \kappa_T &= \frac{\|T' \times T''\|}{\|T'\|^3} \\ &=\frac{\|T' \times T''\|}{\kappa^3}\\ &=\frac{\|c'' \times c'''\|}{\kappa^3}. \end{align*} $$

I need a expression for that top product, I am trying to link it to the alternative formula for torsion given by: $$ \tau = \frac{\det(c'(t),c''(t),c'''(t)}{\|c' \times c''\|^2}. $$

Any help welcome :)

Answer 1

I think I may have it now:

We can rewrite the numerator of the torsion formula in the following way $ \tau = \frac{det(c'(t),c''(t),c'''(t)}{||c' \times c''||^2} = \frac{c'\cdot (c'' \times c''')}{||c' \times c''||^2} $ Now we use that the denominator is actually $\kappa^2$, multiply up and take norms of both sides:

$ ||\tau \kappa^2|| = ||c'||||c'' \times c'''||\ = ||c'' \times c'''|| $

And substituting this into what I had earlier, I get:

$ \kappa_T = \frac{||\tau||}{\kappa} $

Answer 2

For clarity, denote the curve given by the tangent $T$ by $\gamma$. We will express $\gamma'$ and $\gamma''$ in term of the Frenet frame $T$, $N$, $B$ of $c$. We have $$ \begin{align*} \gamma' &= \kappa N \\ \gamma'' &= -\kappa^2 T + \kappa' N + \kappa \tau B\\ \gamma' \times \gamma'' &= \kappa (-\kappa^2 N \times T + \kappa \tau N\times B) \\ &= \kappa^2 (\tau T + \kappa B). \end{align*} $$

So the curvature $\kappa_\gamma$ of the curve $\gamma=T$ becomes $$ \begin{align*} \kappa_\gamma&= \frac{\|\gamma' \times \gamma''\|}{\|\gamma'\|^3} \\ &= \frac{\kappa^2 \sqrt{\tau^2 + \kappa^2}}{\kappa^3} \\ &= \sqrt{1 + \left(\frac{\tau}{\kappa}\right)^2}. \end{align*} $$