# Prove that a curve is spherical iff it satisfies the relation

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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This was already asked here. –  Mariano Suárez-Alvarez Sep 17 '12 at 17:39
@MarianoSuárez-Alvarez where? –  user40276 Sep 17 '12 at 17:43

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

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

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

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

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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In the step where you say "Taking the dot product with $b$, we get", with what are you taking the dot product? –  robjohn Sep 17 '12 at 21:23
Sorry, that step was incorrect. I've edited it accodingly. –  Isaac Solomon Sep 18 '12 at 17:50