1
$\begingroup$

I want to show that a geodesic with nowhere vanishing curvature is a plane curve if and only if it is a line of curvature.

I have done the following:

Let $\gamma$ be a geodesic with nowhere vanishing curvature and a plane curve.

Let $N$ be the unit normal of the surface.

We have that $\|N\|^2=1 \Rightarrow N\cdot N=1 \Rightarrow (N\cdot N)'=0\Rightarrow N'\cdot N=0 \Rightarrow N'\bot N$.

$\gamma'$ is the tangent vector of $\gamma$.

Since $N'$ is perpendicular to $N$, which is perpendicular to the curve $\gamma$ we have that $N'$ is parallel to $\gamma$, so it is also parallel to $\gamma'$. So $N'=-\lambda \gamma'$, for some $\lambda$, which means that $\gamma$ is a line of curvature.

Is this correct?

$$$$

For the other direction:

Let $\gamma$ be a geodesic with noehere curvature and a line of curvature.

Without loss of generality, we suppose that $\gamma$ is unit-speed.

Let $a=N\times\gamma$. Differentiating this we get $a'=N'\times\gamma'+N\times\gamma''$. Since $\gamma$ is a line of curvature we have that there is a $\lambda$ such that $N'= \lambda\gamma'$, i.e., $N'$ is paralle to $\gamma'$. So $N'\times\gamma'=0$. Since $\gamma$ is geodesic we have that $N\times\gamma''=0$.

How could we continue to conclude that the geodesic is a line of curvature.

$\endgroup$
5
  • 1
    $\begingroup$ "Since $N′$ is perpendicular to $N$, which is perpendicular to the curve $\gamma$ we have that $N′$ is parallel to $\gamma$". -- That reasoning is not correct. There are many unit vectors that are perpendicular to $N$. If you have two of them, you can't conclude that they are equal. $\endgroup$
    – bubba
    Commented Jan 14, 2016 at 1:06
  • $\begingroup$ I haven't really understood why we cannot conclude in that way that they are parallel... Could you explain it further to me? @bubba $\endgroup$
    – Mary Star
    Commented Jan 16, 2016 at 23:49
  • 1
    $\begingroup$ If A and B are perpendicular to C, then you can't conclude that A and B are parallel. Take three pencils, A, B, C. Stand C vertically. Put A and B on a horizontal table. Then A and B are both perpendicular to C. but you can turn A and B however you like -- they won't necessarily be parallel. $\endgroup$
    – bubba
    Commented Jan 17, 2016 at 1:58
  • $\begingroup$ Ah ok... I got it now... @bubba $\endgroup$
    – Mary Star
    Commented Jan 17, 2016 at 2:02
  • 2
    $\begingroup$ In general, differential geometry requires extensive use of pencils, pieces of cardboard, string, fingers, ping-pong balls, modeling clay, etc. Merely manipulating symbols won't get you very far.:-) $\endgroup$
    – bubba
    Commented Jan 17, 2016 at 2:07

1 Answer 1

3
$\begingroup$

Well, if $c\colon I\subset \mathbb{R}\longrightarrow S$ is a geodesic of a surface $S$ then we have : $$N(c(s))\parallel n(s),\ \forall s\in I,$$ where $N$ is the unit normal of $S$ and $n(s)$ the first normal of the curve $c$. Therefore, $$N(c(s))=\pm n(s),\ \forall s\in I.$$ By differentiating we obtain $$(N\circ c)'(s)=dN_{c(s)}(c'(s))=\pm n'(s),\ \forall s\in I.$$ By Frenet equations we have that $$n'(s)=-k(s)t(s)+\tau(s)b(s)$$ where $k(s)$ and $\tau(s)$ is the curvature and the torsion of $c$ respectively $(t(s):=c'(s)$ and $b(s)$ is the second normal of $c$. Thus, we obtain $$dN_{c(s)}(t(s))=-k(s)t(s)+\tau(s)b(s).$$ Now the result follows immediately:

  • $(\Rightarrow)$ If $c$ is a plane curve then the torsion $\tau=0$. Thus, $$dN_{c(s)}(t(s))=-k(s)t(s)$$ which implies that $c$ is a line of curvature.
  • $(\Leftarrow)$ If $c$ is a line of curvature then $dN_{c(s)}(t(s))=\lambda(s)t(s)$ and the equation above implies that $\tau=0$ and therefore $c$ is a plane curve.
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .