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I may have some confusion about the twist and torsion of an elastic filament. The issue centers around this set-up.

I hold a cable in my hands, so that it forms a straight line. Keeping the endpoints clamped in my fingers, I move them closer together until they are right next to each other, so that the cable forms a circle. However, I can feel that there is significant twist-stress stored in the filament, so much so that for certain cables this is very hard to do. If I twist the endpoints as I move them together, though, I can feel that there is no stress, and the cable is clearly relaxed.

My question is, when is twist created in the first case? I have moved the endpoints in a straight line, and kept them clamped the whole time so they cannot rotate. It seems like the vanishing of the total twist should be topologically protected, but evidently not (or, the transformation I have performed on the filament somehow disturbed the topology). Where has my thinking gone wrong?

To be clear, my question is not about what, at the mechanical level, is causing the tension. I understand that as the material cross section rotates about the filament's tangent, stress is created. I'm more wondering how the cross section ends up rotating when I keep the endpoints clamped.

 

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So I managed to solve this question on my own through some reading and calculations. The crux of my issue seems to be my belief that the total twist of the filament is topologically protected.

Consider a helical filament subject to periodic boundary conditions. In this case, we know without doubt that there is no rotation occurring between the two endpoints, because of course there are no distinct endpoints. Let us assume that the filament's centerline forms a helix (since this minimizes $\int \kappa^2 \textrm{d}s$, the classical Euler elastic energy). What we want to measure when we talk about twist is the rate of rotation of the cross section of the filament. This is difficult because it is not obvious what an untwisted cross section really looks like in three dimensions. The first thing to do is therefore define the natural frame, an orthonormal vector frame $\{T, U, V \}$ where $T$ is the local tangent vector, defined at each point along the filament. This frame is notable in that it has no "twist" in the following sense: $U' \cdot V = -V' \cdot U = 0$, where prime denotes a derivative with respect to the filament's arclength. This shows that the frame does not rotate about the tangent vector, making it ideal as a reference point.

We can measure the orientation of the cross section relative to this frame. However, we know that from one end of the filament to the other, the cross section does not rotate, because the two ends are really the same point. So we can simply measure the total rotation of the natural frame over one period of the helix. A little work will show that for a helix of radius $r$ and pitch $2\pi c$, the natural frame rotates through an angle of $$ \Delta \psi = \frac{2 \pi c}{\sqrt{r^2+c^2}}.$$

I am amazed to see that this variable changes continuously with the pitch. The conclusion to draw here is that as we compress our helix with periodic boundary conditions, it will accumulate twist gradually. Once we have wrapped it into a circle and $c=0$, it will have gone through a full $2 \pi$ rotation. I guess this is why it is important for roadies to know how to wrap cables!

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  • $\begingroup$ See this link for a pdf of David Singer's lectures on elastic rods, which were enormously helpful for understanding this case.edu/artsci/math/singer/publish/elaslecs.pdf $\endgroup$ – ZachMcDargh Sep 28 '16 at 22:26
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    $\begingroup$ The next time I attempt an "answer" , I will absolutely look at the tags :). I am amazed, despite my naivety in answering, how much physics and math you can get out of a seemingly simple object. $\endgroup$ – user108787 Sep 28 '16 at 23:09
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    $\begingroup$ Very nice work - really enjoyed this Q and A. Indeed, I would probably have made the same "topological" mistake if I didn't know anything about this problem. Note that the "natural" frame you have discovered for yourself defines what are sometimes called "othogonal pipe co-ordinates". These also define how the electric and magnetic fields orthogonal to an optical fiber are "parallel transported" between different fiber cross sections, so that the field traversing a fiber loop acquires its "topological" phase - the rotation about the fiber optical axis. $\endgroup$ – WetSavannaAnimal Sep 29 '16 at 0:25

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