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Here is a picture of a Möbius strip, made out of some thick green paper:

Möbius strip

I want to know either an explicit parametrization, or a description of a process to find the shape formed by this strip, as it appears in the picture. Now before you jump up and declare, "That's easy! It's just $\left(\left[1+u \cos \frac \theta2\right]\cos \theta,\left[1+u\cos\frac \theta2\right]\sin \theta,u\sin \frac \theta2\right)$ for $u\in\left[-\frac12\!,\frac12\right]$, $\theta\in[0,2\pi)$!" Understand that this parametrization misses some features of the picture; specifically, if you draw a line down the center of the strip, you get a circle, but the one in the picture is a kidney-bean shape, and non-planar. What equations would I need to solve to get a "minimum-energy" curve of a piece of planar paper which is being topologically constrained like this? Is it even true that the surface has zero curvature? (When I "reasonably" bend a piece of paper into a smooth shape, will it have zero curvature across the entire surface, or does some of paper's resistance correspond to my imparting non-zero curvature to the surface?)

This question is thus primarily concerned with the equilibrium shapes formed by paper and paper-like objects (analogous to minimal surface theory in relation to soap-bubble models). Anyone know references for this topic?

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I upvoted for the "before you jump up and declare…" part because that's exactly what I was on the point of doing. –  MJD Dec 17 '12 at 17:54
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+1 for the sheer aesthetic pleasure given by your picture. –  Georges Elencwajg Dec 17 '12 at 17:57
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So you don't want the exact curve represented here? Because you go out of your way to say you want the exact curve here. The set of constraints on your problem are a physics problem. The math problem inherent here is, given a set of constraints, what would the equation be? As such, you haven't given us a math problem, but a physics problem. –  Thomas Andrews Dec 17 '12 at 18:12
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This is an intriguing problem, but I agree with Thomas Andrews that we need input from physicists first. What integral gives us the total tension? (that we are supposed to minimize, I guess) Furthermore, an idealistic paper with an idealistic `join/glue' would still have a group of symmetries amounting to adding a constant to your parameter $\theta$ (modulo $2\pi$). But the solution manifestly does not have that symmetry! So some kind of spontaneous breaking of symmetry would have to happen, or may be the seam affects the equation after all? –  Jyrki Lahtonen Dec 17 '12 at 19:23
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@JyrkiLahtonen I am a physics ignoramus, but I imagine that there is an unstable symmetrical solution, which, slightly perturbed, becomes a configuration that collapses into a stable asymmetrical solution that is a member of a symmetrical family of configurations; exactly which one depends on the perturbation. I suppose the situation is analogous to that of a perfectly symmetrical pencil balanced on end. Which way does it fall? The pencil doesn't fall until it is perturbed, and it is the perturbation which breaks the symmetry. –  MJD Dec 17 '12 at 20:09
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1 Answer

up vote 19 down vote accepted
+50

The Möbius strip you show is a developable surface. No one, as far as I know, has been able to create a parametrization of it.

Since 1858, when the Möbius strip was discovered, mathematicians have been looking for a way to model it. The problem was finally solved in 2007 by E.L. Starostin and G.H.M. van der Heijden.

You might want to read their paper "The equilibrium shape of an elastic developable Möbius strip" by going to this site - http://www.ucl.ac.uk/~ucesgvd/pamm.pdf

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Wow! I can't believe someone's really looked into this in that much detail. +2! Also, it seems like they did a good job of a parametrization of the surface, since they have plots. –  Mario Carneiro Dec 18 '12 at 4:11
    
The standard parameterization I saw in university was something like the one given by Wikipedia, or by the original question in this post: $$x(u,v)= \left(1+\frac{v}{2} \cos \frac{u}{2}\right)\cos u\\ y(u,v)= \left(1+\frac{v}{2} \cos\frac{u}{2}\right)\sin u\\ z(u,v)= \frac{v}{2}\sin \frac{u}{2}.$$ It seems clear that this parametrization has everywhere-zero curvature. Why do you say “No one… has been able to create a parametrization of it.”? –  MJD Mar 31 at 15:01
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