Prove curvature of sphere curve $\geq R^{-1}$

Question

Let $\bf x(s)$ be a sphere curve lying on the surface of a sphere with center $\bf{p}$ and radius $R$ satisfying $$|\bf{x}(s)-p|^2=R^2.$$ I want to prove that the curvature $\kappa \geq R^{-1}$.

I assumed the center is the origin as this does not affect the curvature. Differentiating I have

\begin{equation} \bf x'\cdot x=0, x''\cdot x=0. \end{equation}

I know I need to invoke the structure equations and that $\bf T,N,B$ form an orthonormal basis. I tried to write $$\bf x=\lambda T+\mu N+\nu. $$ Using the structure equations and $T=x'$ I found $$\mu'-\mu\kappa=1,\mu'-v\tau+\mu\kappa=0,v'=-\mu\tau$$

Using $\bf x' \cdot x=0$ I got $\bf\lambda^2+\mu^2+\nu^2=R^2$ and $\bf \lambda'\lambda+\mu\mu'+\nu\nu'=0$. So I want to establish $$\kappa^2\geq \frac{1}{\lambda^2+\mu^2+\nu^2}.$$ I have some relations between $\kappa,\tau,\lambda,\mu$ and $\nu$ and I get more if I look at $\bf x''\cdot x=0$. However, the equations become very long and I'm not sure how to show the inequality. Do I have the right approach?

Answer 1

OK, first of all I am inclined to say that though expanding $\mathbf x(s)$ in terms of the Frenet-Serret frame,

$\mathbf x = \lambda T + \mu N + \nu B, \tag 1$

as our OP user30523 has done, though at time-honored technique which often bears fruit--as in fact it does here, see my comments below--is a little more complicated than what is needed in the present context; so what I'll do is put forth my method for solving this problem--which incidentally only needs the Frenet-Serret equation $T = \kappa N$, as I will now demonstrate, and then offer a few comments on our OP's work.

I assume the curve $\mathbf x(s)$ is parametrized by its arc-length, which I take to be $s$; such usage is consistent with our OP's equation $\dot{\mathbf x}(s) = T(s)$, which only holds for unit-speed, arc-length parametrized curves.

From

$\mathbf x(s) \cdot \mathbf x(s) = R^2, \; \text{a constant}, \tag 2$

we obtain, upon differentiation with respect to $s$,

$T(s) \cdot \mathbf x(s) = \dot{\mathbf x}(s) \cdot \mathbf x(s) = 0, \tag 3$

and taking $d/ds$ once more we derive the equation:

$\dot T(s) \cdot \mathbf x(s) + 1 = \dot T(s) \cdot \mathbf x(s) + T(s) \cdot T(s) = \dot T(s) \cdot \mathbf x(s) + T(s) \cdot \dot {\mathbf x}(s) = 0. \tag 4$

The curvature $\kappa(s)$ is defined via the Frenet-Serret equation

$\dot T(s) = \kappa(s) N(s), \; \vert N(s) \vert = 1, \; \kappa(s) > 0; \tag 5$

also,

$\mathbf x(s) = R \mathbf n(s), \tag 6$

where $\mathbf n(s)$ is the unit normal to the sphere at point $\mathbf x(s)$; we substitute (5) and (6) into (4):

$\kappa(s) N(s) \cdot R \mathbf n(s) + 1 = 0, \tag 7$

or

$\kappa(s) N(s) \cdot \mathbf n(s) = -\dfrac{1}{R}; \tag 8$

we take absolute values, recalling $\kappa(s) > 0$:

$\kappa(s) \vert N(s) \cdot \mathbf n(s) \vert = \dfrac{1}{R}; \tag 9$

since

$\vert \mathbf n(s) \vert = \vert N(s) \vert = 1, \tag{10}$

we have by Cauchy-Schwarz that

$\vert \mathbf n(s) \cdot N(s) \vert \le \vert \mathbf n(s) \vert \vert N(s) \vert = 1, \tag{11}$

whence

$\kappa(s) = \dfrac{1}{\vert \mathbf n(s) \cdot N(s) \vert R} \ge \dfrac{1}{R}, \tag{12}$

as was to be proved. $\mathbf{OE\Delta}$.

A Few Remarks: Our OP's equation

$\ddot{\mathbf x}(s) \cdot \mathbf x(s) = 0 \tag{13}$

is erroneous, and does not follow from (3), as is shown in the preceding derivation; the correct consequence is (4), from which the desired result is readily reached.

Our OP's attempt, based upon

$\mathbf x = \lambda T + \mu N + \nu B \tag{14}$

bears further scrutiny. We have:

$T = \dot{\mathbf x} = \dot \lambda T + \lambda \dot T + \dot \mu N + \mu \dot N + \dot \nu B + \nu \dot B; \tag{15}$

using the complete set of Frenet-Serret equations,

$\dot T = \kappa N, \tag{16}$

$\dot N = -\kappa T + \tau B, \tag{17}$

$\dot B = -\tau N, \tag{18}$

we may re-write (15) in the form

$T = \dot{\mathbf x} = \dot \lambda T + \lambda \kappa N + \dot \mu N + \mu (-\kappa T + \tau B) + \dot \nu B -\nu \tau N$ $= (\dot \lambda - \mu \kappa)T + (\dot \mu + \lambda \kappa - \tau \nu)N + (\dot \nu + \mu \tau)B, \tag{19}$

from which I draw

$\dot \lambda - \mu \kappa = 1, \tag{20}$

$\dot \mu + \lambda \kappa - \tau \nu = 0, \tag{21}$

$\dot \nu + \mu \tau = 0; \tag{22}$

we further compute

$\lambda \dot \lambda + \mu \dot \mu + \nu \dot \nu = \lambda \dot \lambda + \mu \dot \mu + \nu \dot \nu - \lambda \mu \kappa + \mu \lambda \kappa - \mu \tau \nu + \nu \mu \tau$ $= \lambda (\dot \lambda - \mu \kappa) + \mu (\dot \mu + \lambda \kappa - \tau \nu) + \nu (\dot \nu + \mu \tau) = {\mathbf x} \cdot \dot {\mathbf x} = 0, \tag{23}$

which may also be derived by observing that, from (14),

$\lambda^2 + \mu^2 + \nu^2 = \mathbf x \cdot \mathbf x = R^2, \tag{23}$

and then differentiating with respect to $s$.

Answer 2

Decompose $\kappa\mathbf{N}=\frac{\mathrm{d}}{\mathrm{d}s}\mathbf{T}=\operatorname{I\!I}(\mathbf{T},\mathbf{T})+(\parallel\text{-component})$. Now use Pythagoras with the hypotenuse $|\kappa\mathbf{N}|=\kappa$ and the leg $\operatorname{I\!I}(\mathbf{T},\mathbf{T})$ has length $R^{-1}$.