Test for arbitrary-speed curve lying on a sphere

Question

My textbook, ONeill's Elementary Differential Geometry, gives as an exercise the following test function to determine if a unit-speed curve lies on a sphere. If the test function is constant then the curve lies on a sphere. For a later exercise, we are invited to extend the test function so that it applies to arbitrary-speed curves. The prescription given is to multiply the derivative in the original formula by the speed of the curve. I tried that with a test curve that I am pretty confident lies on a sphere. But the results do not check out; the test function is not constant. I would like to know the above-stated prescription for taking into account arbitrary speed is as simple as I have suggested.

The test function for unit-speed curves is $\rho^2+(\rho^\prime\sigma)^2$, where $\rho$ is the inverse of the curvature, $\sigma$ is the inverse of the torsion, $N$ is the principal normal and $B$ is the binormal.

For the speed, I use $\nu=||\frac{d\alpha}{dt}||$. So for the modified test function I get $\rho^2+(\nu\rho^\prime\sigma)^2$

My curve that I believe lies on a sphere is $\alpha(t)=t/\sqrt{t^2+1}, cos(t)/\sqrt{t^2+1}, sin(t)/\sqrt{t^2+1}$

Answer 1

Well, after going through the derivation and extending it for arbitrary speed, it becomes clear that rather than multiplying by the speed, one has to divide by the speed where the derivative appears. Perhaps this should be obvious from the equation $\frac{d\rho}{ds}=\frac{\frac{d\rho}{dt}}{\frac{ds}{dt}}$

Answer 2

There’s no need for unit speed: any Frenet curve $c$ with non-vanishing torsion $\tau$ and curvature $\kappa$ lies on a sphere with center $p$ and radius $r$, that is satisfying $\|c-p\|^2=r^2$, iff $$ \left(\frac{1}{\kappa}\right)^2+ \left(\left(\frac{1}{\kappa}\right)'\right)^2\cdot\left(\frac{1}{\tau}\right)^2=r^2.$$

Answer 3

Assuming it lies on a sphere, $||\alpha-C||^2$ is constant. $\rho=1/\kappa$, $\sigma=1/\tau$.

$\nu$ is the speed.

$(\alpha-C)\cdot T=0$

$\alpha^\prime\cdot T+(\alpha-c)\cdot T^\prime=0$

$\nu=||T||^2=-(\alpha-C)\cdot T^\prime=-(\alpha-c)\cdot\kappa\nu N$

$\kappa(\alpha-C)\cdot N=-1$

So the component of $\alpha-C$ along N is $-\frac{1}{\kappa}=-\rho$

$\frac{d}{dt}||\alpha-C||^2=2(\alpha-C)\cdot T=0$

$\alpha$ is a linear combination of $B$ and $N$.

$\kappa^\prime(\alpha-C)\cdot N=-\kappa(\alpha-C)N^\prime=-\kappa(\alpha-C)\cdot(-\kappa \nu T+\tau B)=-\kappa(\alpha-C)\cdot \tau \nu B$

$(\alpha-C)\cdot B=-\frac{\kappa^\prime(\alpha-C)\cdot N}{\kappa\nu\tau}=\frac{\kappa^\prime}{\kappa^2\nu\tau}$

$(\alpha-C)\cdot B=-\frac{\kappa^\prime}{\kappa^2\nu\tau}=-\frac{\rho^\prime}{\nu\tau}=-\frac{\rho^\prime\sigma}{\nu}$

This is the component of $\alpha-C$ along $B$

$\alpha-c=-\rho N-\frac{\rho^\prime\sigma}{\nu} B$

So for the curve to lie on a sphere, $||\alpha-c||^2=\rho^2+(\frac{\rho^\prime\sigma}{\nu})^2$ is constant and vice versa.