# Torsion of tangent indicatrix curve

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.

• The post which looks the most like mine: (1)
• This one is also very similar, but has a more strict hypothesis (constant positive curvature and constant torsion), so the torsion amounts to zero. There are very detailed calculations there, but I couldn't use them for my case: (2)
• Your “useful” formula for torsion is wrong; you've lost the sign. – Ted Shifrin Oct 15 '20 at 21:05
• @TedShifrin Right below the problem: "Please mind that this book uses a definition of torsion which may have a different sign than other ones." – Wheepy Oct 15 '20 at 21:08
• @TedShifrin If you wish, you can use the usual definition, I don't mind :) – Wheepy Oct 15 '20 at 21:09
• I'm referring to your “formula for any parameter.” The absolute value is wrong. – Ted Shifrin Oct 16 '20 at 20:17
• @TedShifrin You're right, I hadn't even noticed! Too much copy-paste :| – Wheepy Oct 19 '20 at 23:14

## 1 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 $$\tau_1$$.

• You are right about the sign in the final formula for $\tau_1$. That was a typo, will edit. – Wheepy Oct 15 '20 at 22:30
• Your answer is really useful, thank you. But I'm missing a part. You say "adjust by the chain rule at the end to get the derivative of B with respect to the arclength of the new curve". I don't know exactly how to do that. I've tried reparametrizing $t(s)$ (the tangent indicatrix curve) by arc length, but I am not sure how to find the reparametrizing function, since $\alpha$ is a generic curve and I dont know how to compute the necessary arc length integral. Would you please give me a hint on that? – Wheepy Oct 15 '20 at 23:47
• Don't reparametrize. Use $\frac{dB}{d\sigma}=\frac{dB}{ds}\frac{ds}{d\sigma}$. You might want to download my differential geometry text, linked in my profile. – Ted Shifrin Oct 15 '20 at 23:52
• Thank you for letting me know, it is a valuable resource indeed! – Wheepy Oct 19 '20 at 23:19
• Glad to help, Wheepy! :) – Ted Shifrin Oct 19 '20 at 23:20