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I couldn't prove that a regular curve, such that the torsion and curvature never equal zero, satisfies the relation $\frac{\tau}{\kappa}+(\frac{1}{\tau}(\frac{1}{\kappa})´)´=0$ iff it's spherical, i.e, it lies entirely on a sphere.

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    @MarianoSuárez-Alvarez where?2012-09-17

2 Answers 2

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Equivalently, let $\gamma(t)$ is a unit-speed curve with $\kappa(t) > 0$ and $\tau(t)$ never equal to zero. We claim $\gamma(t)$ is spherical iff

$ \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$

First assume that $\gamma(t)$ of spherical. Then there is some point $a$, the center of the sphere, such that

$||\gamma - a||^2 = r^{2}$

for the radius of the sphere $r$. Rewrite this as

$(*) \,\,\,\,\,\,(\gamma - a) \cdot (\gamma - a) = r^{2} $

To get the equality above, we are going to keep on taking messier and messier derivatives. C'est la vie. You should recall important equalities for unit-speed curves, like $\dot{t} = \kappa n$ and $\dot{b} = -\tau n$. I will use these implicitly in the calculation.

To start, take the derivative of $(*)$ to get

$ t \cdot (\gamma - a) = 0$

Now, take the derivative again

$t \cdot t + \kappa n \cdot (\gamma - a) = 0$

Note that $t \cdot t = 1$, so we can rewrite this as

$ n \cdot (\gamma - a) = -\frac{1}{\kappa}$

Take the derivative of both sides again, to get

$ n \cdot t + (-\kappa t + \tau b) \cdot (\gamma - a) = \frac{\dot{\kappa}}{\kappa^{2}}$

Since $t \cdot (\gamma - a) = 0$, this reduces to

$ b \cdot (\gamma - a) = \frac{\dot{\kappa}}{\tau\kappa^{2}}$

Take the derivative again to get

$b \cdot t - \tau n \cdot (\gamma - a) = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$

which is easily seen to be equivalent to

$ \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$

Now, we go the other direction, so assume

$(*) \,\,\,\,\,\, \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$

Let us define new quantities $\rho = 1/\kappa$ and $\sigma = 1/\tau$. Then $(*)$ implies that

$(\dagger) \,\,\,\,\, \rho = \frac{d}{ds}\big(-\sigma(\dot{\rho}\sigma)\big)$

Consider the quantity $\rho^2 + (\dot{\rho}\sigma)^2$. Let us take its derivative. By $(\dagger)$, we would have

$\frac{d}{ds} (\rho^2 + (\dot{\rho}\sigma)^2) = 2\rho\dot{\rho} + 2(\rho\sigma)\frac{d}{ds}(\rho\sigma) = 0$

Hence $\rho^2 + (\dot{\rho}\sigma)^2$ is some positive constant $r^2$. Let us go further and define the following "curve" $a$ (think about its definition a little)

$a = \gamma + \rho n + \dot{\rho}\sigma b$

Now, we take the derivative of $a(t)$, noting the equation following $(\dagger)$,

$\dot{a} = t + \dot{\rho}n + \rho(-\kappa t + \tau b) + \frac{d}{ds} (\dot{\rho}\sigma)b + (\dot{\rho}\sigma)(-\tau n) = 0$

and so $a$ is a constant, and is the center of our sphere by the following calculation

$\parallel \gamma - a \parallel^2 = ||\rho n + \dot{\rho}\sigma b||^2 = \rho^2 + (\dot{\rho}\sigma)^2 = r^2$

and the radius of this sphere is $r$. This demonstrates that $\gamma(t)$ lies on the surface of a sphere of radius $r$, and hence is spherical. QED.

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    Sorry, that step was incorrect. I've edited it accodingly.2012-09-18