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I'm studying the book "elementary differential geometry" by o'neil. There is a collorary which states that if two curves a(t), b(t) which is defined in the same real line interval has the same speed, curvature, torsion(torsion may differ by sign), then they are congruent.

The preceding theorem says that two unit speed curve defined on the same real line interval are congruent if their curvature and torsion are equal(torsion may differ by sign)

So i would like to prove the collorary as follows: since the speeds of the two curves are the same, the arc length parametrizations are the same( with the same lower limit of integration), say t=t(s)

So the unit speed parametrization of the two curves has the same curvature and torsion since the original curve has the same curvature and torsion and arc length paramerization. So the two unit speed parametrization of two curves are congruent. And substituting s=s(t), the original two curves are congruent.

Is there any fallacy in my reasoning? Any help will be greatly appreciated. Thank you for reading

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The reasoning is fine. I would write something more concrete, though. Since $|a'(t)|\equiv |b'(t)|\equiv C$, the curves $\tilde a(t)=a(C^{-1}t)$ and $\tilde b(t)=b(C^{-1}t)$ have unit speed. The chain rule implies that the curvature and torsion functions satisfy
$$\kappa_{\tilde a}(t)=C\kappa_a(Ct),\quad \tau_{\tilde a}(t)=C\tau_a(Ct)$$ and similarly for $\tilde b$. It follows that $\tilde a$ and $\tilde b$ are congruent, and therefore the original curves are congruent.

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  • $\begingroup$ Thanks! Appreciate it very much! $\endgroup$ – Mathcho Jun 11 '15 at 3:16
  • $\begingroup$ When i look through your argument, i see you assumed the speed of curves are constant. But this is not the case in the book. But thanks anyway! $\endgroup$ – Mathcho Jun 11 '15 at 4:27
  • $\begingroup$ Yes, this is how I interpreted "the same speed". Anyway, there's not much to say other than: your argument is correct. $\endgroup$ – user147263 Jun 11 '15 at 4:29

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