0
$\begingroup$

Let $\alpha(t)$ be a unit-speed curve in $\mathbb R^{3}$ with principal normal $ N(t)$ and binormal $B(t).$ We define a surface by $$ x(t,\theta) = \alpha(t) + a( N(t) \cos({\theta}) + B(t) \sin(\theta)),$$ where $a >0$ is a constant. I have to show that for any fixed $t_0$ the curve $\gamma(\theta) = x(t_0,\theta)$ is geodesic on $x.$

I have no idea from where to begin. Any help. Thank you.

$\endgroup$
1
$\begingroup$

Instead of your notation, I'll use $T_\alpha, N_\alpha$ and $B_\alpha$ for the Frenet trihedron of $\alpha$. Geometrically, the image of $x$ is a tube of radius $a$ around the image of $\alpha$. With this, you can see that $$N(x(t,\theta)) = N_\alpha(t)\cos \theta + B_\alpha(t)\sin \theta$$is a normal field along the surface. Now, if $\gamma(\theta) = x(t_0,\theta)$, then $$\gamma''(\theta) =- N_\alpha(t_0)\cos\theta - B_\alpha(t_0)\sin \theta$$is parallel to $N(\gamma(\theta))$, so $\gamma$ is a geodesic.

$\endgroup$
  • $\begingroup$ That's a great explanation. Thanks for your time. I have one query, will you please tell me how the surface look like ? If possible a link containing the picture of the surface. Thank you. @IvoTerek $\endgroup$ – hiren_garai Nov 5 '18 at 1:56
  • $\begingroup$ From Google, something like this. (Link for the image magically turned into a youtube link from the video where the figure was taken... go figure) $\endgroup$ – Ivo Terek Nov 5 '18 at 1:59
  • $\begingroup$ One more question, I've read the definition of a geodesic as, a curve $\gamma(t)$ is a geodesic to a surface if the vector field $\dot\gamma(t)$ is parallel along $\gamma(t)$,( dot denotes the derivative) is this the same definition that you're using ? Actually I'm facing some doubts there. @IvoTerek $\endgroup$ – hiren_garai Nov 5 '18 at 2:13
  • 1
    $\begingroup$ These definitions are equivalent. You can write $$\alpha''(t) = \frac{D\alpha'}{{\rm d}t}(t) + \alpha''_{\rm nor}(t),$$where $D\alpha'/{\rm d}t$ is the covariant derivative and $\alpha''_{\rm nor}(t)$ is normal to the surface. Then $D\alpha'/{\rm d}t =0$ if and only if $\alpha''(t) = \alpha''_{\rm nor}(t)$ is normal to the surface. $\endgroup$ – Ivo Terek Nov 5 '18 at 2:15
  • $\begingroup$ Now it's fine, but I've one more doubt, pardon my ignorance, how have you found the Normal field along the surface ? according to which formula , I'm unable to understand. @IvoTerek $\endgroup$ – hiren_garai Nov 5 '18 at 2:25

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.