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In class the teacher was talking about normal vectors. $r = \langle x,y\rangle$ then the normal vector is

$$ N\left(t\right) = \frac{T^{\prime}\left(t\right)}{||T^{\prime}\left(t\right)||} $$

where $T\left(t\right) = \frac{r^{\prime}}{||r^{\prime}||}$

Now, one of the naughty students asked the question in the title.

I thought it was a cool question so I am asking here.

ps. I have no idea, still trying to figure out how the unit norm vector needs the 2nd derivative of the tangent vector T.

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The question makes no sense. You gave the definition of the normal vector for a parametrised curve in $\mathbb{R}^2$; this is not the same formula (but it is similar) as that which describes the normal vector for a surface. The problem with the Moebius strip is that depending on parametrisation (in your case $t\mapsto -t$) there are two distinct normal vectors, and for the Moebius strip there is no consistent choice of a global parametrisation that chooses exactly one of the two normal vectors. I don't see what this has to do with "the normal vector intersecting." –  Willie Wong Oct 31 '12 at 8:50
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Also, what do you want the normal vector to intersect? Usually intersection requires two objects. And a vector by itself (thinking in terms of a directed line segment) quite clearly cannot self intersect (unlike a curve). –  Willie Wong Oct 31 '12 at 8:51
    
@WillieWong Thanks, I didn't realize that, you have a nice and informative response. –  yiyi Oct 31 '12 at 12:21

1 Answer 1

up vote 1 down vote accepted

Note that your definition of the normal vector along a curve $\gamma$ does not work at points where the curvature is zero. In stretches of $\gamma$ where the curvature is nonzero your normal vector points inward, which might be to the right or to the left vs. the forward direction.

If $\gamma$ is given as $$\gamma:\ t\mapsto{\bf r}(t)=\bigl(x(t),y(t)\bigr)$$ one might also define the normal vector by $${\bf n}(t):={\bigl(-\dot y(t),\dot x(t)\bigr)\over\bigl|\dot{\bf r}(t)\bigr|}\ .$$ In this way the normal vector is defined at all points of $\gamma$ and points to the left. Now the curvature, when nonzero, has a sign.

On a Moebius band there is no curve going around once which has everywhere positive absolute curvature. You might say that the normal vector flips at one of the points where the curvature vanishes.

Using the other definition of normal vector, a normal vector "pointing to the left" becomes a normal vector "pointing to the right" after one orbit.

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