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

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

$$$$\bf x'\cdot x=0, x''\cdot x=0.$$$$

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?

• I think you mos' likely want $\mathbf x = \lambda T + \mu N + \nu B$ in your second equation; am I right? Cheers! – Robert Lewis Sep 30 '18 at 20:48

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 $$s$$.

• But $x'' \cdot x+1=0$ should be correct? – user30523 Sep 30 '18 at 20:24
• @user30523: yes, This is equation (4) of my answer. And thanks for the "acceptance". Cheers! – Robert Lewis Sep 30 '18 at 20:34

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 $$R^{-1}$$.

• What does II stand for? – user30523 Sep 30 '18 at 19:05
• The second fundamental form of the sphere $\{\mathbf{x}\in\mathbb{E}^3:|\mathbf{x}-\mathbf{p}|=R\}$. – user10354138 Sep 30 '18 at 19:07