The Question is: Let $\gamma(t)$ be a regular curve lies on a sphere $S^2$ with center $(0, 0, 0)$ and radius $r$. Show that the curvature of $\gamma$ is non-zero, i.e., $κ \ne 0$.

My question:

We define $k$ as $k=\frac{(||\gamma'' \times \gamma'||)}{|| \gamma' || ^3} $

The equation of the sphere is: $x^2 + y^2 + z^2 = r$

So, we have $\gamma = (\sqrt{r} \cos u \sin v,\sqrt{r} \sin u \sin v, \sqrt{r} \cos v$)

How can I calculate $\gamma'$ and $\gamma''$ ? because $\gamma$ depends on two variables $u,v$

Thank you.

  • $\begingroup$ The sphere you wrote down has radius $\sqrt{r}$ rather than $r$ as stated. $\endgroup$ – Mikhail Katz May 30 '13 at 12:43

If you don't know much about Riemannian stuff, there is an easy and elementary proof of this fact.

Since $\gamma$ lies on the sphere of radius $r$, you get $\|\gamma(t)\|^2=r$. Taking the derivative of that expression, it follows that $$(1): \qquad \forall t, \langle \gamma(t),\gamma'(t)\rangle =0.$$

Finally, computing the derivative of the last expression will give you : $$(2):\qquad \langle \gamma(t),\gamma''(t)\rangle + \|\gamma'(t)\|^2=0.$$

Now, you want to show that the curvature does not vanish that is : for every $t$, $\|\gamma'(t)\times \gamma''(t)\|\neq 0$ or equivalently that the family $(\gamma'(t),\gamma''(t))$ is linearly independent at any time $t$.

Assume that there exists $t_0$ such that $\gamma''(t_0)=\lambda \gamma'(t_0)$ with $\lambda\in \mathbb R$. Then looking at $(1)$ and $(2)$ together you get a contradiction because $\gamma'$ is nonvanishing. So $\kappa(t)\neq 0$ for any $t$.


The quickest way of seeing this is by using the formula $k^2=k_g^2 + k_n^2$. Here $k_g$ is the geodesic curvature and $k_n$ is the normal curvature. In the case of the sphere the normal curvature is the inverse of the radius of the sphere. Hence curvature is positive.


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