Disclaimer: there are many posts about this problem here, but none with exactly what I'm looking for. I'll state the problem and then link some references.
This is the problem:
Let $\alpha(s)$ be a regular curve parametrized with respect to arc lenght with nonzero curvature.
Let $t(s)$ be the curve defined by the tangent vectors to $\alpha$. This curve is called the spherical tangent indicatrix of $\alpha$.
Prove the following relations for the curvature and torsion of $t(s)$:
$k_1^2 = \dfrac{k^2 + \tau^2}{k^2} \quad $
(I've already proved this one)
$\blacktriangleright\blacktriangleright\blacktriangleright\quad$ $\tau_1 = \dfrac{k\tau' – k'\tau}{k (k^2 + \tau^2)}, \quad $
(this is where I'm stuck) $\quad\blacktriangleleft\blacktriangleleft\blacktriangleleft$
where $k_1$ and $\tau_1$ are the curvature and torsion of $t(s)$, respectively, and $k$ and $\tau$ are the curvature and torsion of $\alpha(s)$.
Please mind that this book uses a definition of torsion which may have a different sign than other ones. To clarify, here are the definitions/formulas I think may be different for some people and/or might be useful (but do feel free to use the version of the definition/formulas you prefer):
$k = \dfrac{\lvert \alpha' \times \alpha'' \rvert}{\lvert \alpha'\rvert^3}\quad$ (formula for any parameter)
$\tau = \dfrac{ \langle \alpha' \times \alpha''', \alpha'' \rangle}{\lvert \alpha' \times \alpha'' \rvert^2}\quad$ (formula for any parameter)Frenet formulas:
$t'(s) = k(s) n(s)$
$ n'(s) = -k(s) t(s) – \tau(s) b(s) $
$b'(s) = \tau(s) n(s) \quad\quad$ (thus $\quad \tau(s) = \langle b'(s), n(s) \rangle$ )
where $\{t,n,b\}$ is the Frenet frame of $\alpha(s)$.
I've lost so much time in these calculations that I don't even know which one of them (if any) will lead me to an answer, so any help is appreciated. Thanks in advance.
Best Answer
My recommendation is to figure out the Frenet frame of the tangent indicatrix and use the chain rule to adjust for the fact that it is not arclength parametrized (i.e., its speed is $k$). I will use capital letters for the Frenet frame of the tangent indicatrix and little letters, as you did, for the original curve. You should find that \begin{align*} T &= n; \\ N &= -\frac{kt+\tau b}{\sqrt{k^2+\tau^2}}; \\ B &= T\times N = \frac{kb-\tau t}{\sqrt{k^2+\tau^2}}. \end{align*} Now calculate $\langle B',N\rangle$. (Here I differentiate with respect to $s$, and adjust by the chain rule at the end to get the derivative of $B$ with respect to the arclength of the new curve.) Note that you can see that a priori the terms that come from differentiating $t$, $b$, and $\sqrt{k^2+\tau^2}$ will all disappear. Warning: With your sign convention on the definition of $\tau$, there may be a minus sign that appears in the final formula for $\tau_1$.