User sunayana ghosh - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:52:26Z http://mathoverflow.net/feeds/user/27446 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110304/generalization-of-vogts-theorem-for-curves-in-higher-dimension Generalization of Vogt's Theorem for curves in higher dimension Sunayana Ghosh 2012-10-22T07:27:36Z 2013-02-15T17:22:00Z <ol> <li><p>The Vogt's theorem for plane curves states that if A and B are endpoints of a spiral arc, the curvature nondecreasing from A to B. The angle $\beta$ of the tangent to the arc at B with the chord AB is not less than the angle $\alpha$ of the tangent at A with AB. $\alpha = \beta$ only if the curvature is constant.</p></li> <li><p>Does anyone know of a result which extend this theorem to space curves or curves in higher dimension. I have the following conjecture for space curves: Given a regular curve in space $\gamma : [0, l] \rightarrow \mathbb{R}^3$, parametrized by arc-length $s$, let $\kappa$ and $\tau$ denote the Euclidean curvature and torsion respectively. Let us assume that $\kappa$ is non-decreasing and $\tau$ is non-decreasing. Let $A = \gamma(0)$ and $B = \gamma(l)$ and let $\alpha$ be the angle between the tangent plane at $\gamma(0)$ and the chord $AB$ and let $\beta$ be the angle between the tangent plane at $\gamma(l)$ and the chord $AB$. We claim that $\alpha \leq \beta$ and equality holds only if $\gamma$ is a circular helix.</p></li> </ol> http://mathoverflow.net/questions/110304/generalization-of-vogts-theorem-for-curves-in-higher-dimension/119277#119277 Answer by Sunayana Ghosh for Generalization of Vogt's Theorem for curves in higher dimension Sunayana Ghosh 2013-01-18T15:54:29Z 2013-01-18T15:54:29Z <p>This is an attempt at an (as yet incomplete) proof of the above claim. I would be happy to receive any corrections and/or comments. </p> <p>Let us denote the curve $\gamma$ by the following parametrization $\gamma(s) = (x_1(s), x_2(s), x_3(s))$. Without loss of generality let us assume that $A = \gamma(0) = (0, 0, 0)$ and $B = \gamma(l) = (x_1(l), 0, 0)$. </p> <p>Let $\theta(s)$ denote the angle between tangent plane at $\gamma(s)$ and the chord $AB$, thus we have that: $\sin \theta(s) = \langle B(s), (1,0,0) \rangle$. Using Frenet-Serret formulae where :</p> <p>$T'(s) = \kappa(s) N(s)$ and $N'(s) = -\kappa(s) T(s) -\tau(s) B(s)$</p> <p>we have that : $\sin \theta(s) = \langle B(s), (1,0,0) \rangle = \frac{1}{\kappa(s)} (\gamma'(s) \times \gamma''(s)) = \frac{1}{\kappa(s)} (x_2'(s) x_3''(s) - x_3'(s) x_2''(s))$.</p> <p>Claim : $\alpha \leq \beta$ i.e., it is enough to prove that $\int_{\theta(0)}^{\theta(l)} \sin \theta d\theta \geq 0$.</p> <p>From the equation for $\sin \theta(s)$ we obtain that $d\theta(s) = \frac{\kappa(s) f'(s) - f(s) \kappa'(s)}{\kappa(s)\sqrt{\kappa(s)^2-f(s)^2}}$, where $f(s) := x_2'(s) x_3''(s)-x_2''(s) x_3'(s) = \kappa(s) \langle B(s), e_1 \rangle$ and $e_1:= (1,0,0)$.</p> <p>On further simplification using Frenet-Serret formulae we get:</p> <p>$\int_{\theta(0)}^{\theta(l)} \sin \theta(s) d\theta(s) = \int_0^l \frac{\tau(s)}{\kappa(s)} \frac{\langle B(s), e_1 \rangle \langle N(s), e_1 \rangle}{\sqrt{1-\langle B(s), e_1\rangle ^2}} ds$. </p> <p>From here it is not clear to me that product of the integrand is always positive, or using integration by parts the integrand is always positive. </p>