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?

  • $\begingroup$ I think you mos' likely want $\mathbf x = \lambda T + \mu N + \nu B$ in your second equation; am I right? Cheers! $\endgroup$ – 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.


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

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

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


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

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

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


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

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

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


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

we have by Cauchy-Schwarz that

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


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

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 $s$.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.