2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The sphere you wrote down has radius $\sqrt{r}$ rather than $r$ as stated. $\endgroup$ – Mikhail Katz May 30 '13 at 12:43
1
$\begingroup$

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 $t$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

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.