Timeline for Rolling without slipping interpretation of torsion
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| S Jan 22 at 7:23 | history | suggested | The Amplitwist | CC BY-SA 4.0 |
fixed broken links to Wikipedia and Wordpress
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| Jan 22 at 4:57 | review | Suggested edits | |||
| S Jan 22 at 7:23 | |||||
| Jan 26, 2011 at 23:08 | comment | added | Willie Wong | I personally have a difficult time picturing, say, a 3-sphere rolling along Euclidean space all embedded in 4 dimensions. In 2D you'd have to given up the notion of easily telling which part of the transformation of the tangent space is due to torsion and which part due to "metric". | |
| Jan 26, 2011 at 23:04 | comment | added | Willie Wong | Also, a little bit about your second comment: the problem I have in my head is this: suppose you want two manifolds $(M,g,\nabla)$ and $(M,g,\tilde{\nabla})$ that are equivalent except for torsion (in particular, the geodesics agree and both connections are metric: one is Levi-Civita and the other is not), the easiest way is to "add" to the connection $\nabla$ a torsion tensor that is totally antisymmetric. But you can't do so in two dimensions, since there are no rank-3 totally antisymmetric tensors in 2D. The lowest dimension you can use is 3D. | |
| Jan 26, 2011 at 22:55 | comment | added | Willie Wong | I have a guess at what motivated their description: the Riemann curvature tensor has two pairs of antisymmetric indices, and you can treat it as the map taking as input a two-plane at a point and returns an infinitesimal rotation. In that sense it tells you that if you go in a small loop that is "in a certain plane" you'll pick up a rotation of the tangent space as such. The torsion tensor is type (1,2) and antisymmetric in the latter two indices. So you can think of it as mapping two-planes to vectors: hence the small loop and translation description. | |
| Jan 26, 2011 at 22:22 | comment | added | Willie Wong | The picture in my head is that the torsion is part of the parallel transport. Take the pure torsion example of the Cartan staircase. It is just like Euclidean 3-space (the geodesics are even the same). But when you parallel transport the unit $\partial_x$ vector along the $z$ direction, it spins around to become the $\partial_y$ vector, and then the $-\partial_x$ vector and so forth. | |
| Jan 26, 2011 at 22:10 | comment | added | David E Speyer | You write "In any case, since the surface geometry of physical objects tend to be Levi-Civita, I don't think it will be possible to actually demonstrate this using an analogous experiment". But the point of my worry (2) is that you should be able to demonstrate the lack of torsion physically. There should be some physical experiment you can perform on a surface, get 0 and say "yup, that's the torsion-free-ness". | |
| Jan 26, 2011 at 21:08 | comment | added | David E Speyer | Re your last paragraph: that's why I say that the $A$ part is the parallel transport. My understanding is that we are looking for an affine map whose linear term is the parallel transport, but whose constant term is more subtle. As for the rest, I'll have to think about it. | |
| Jan 26, 2011 at 20:35 | history | answered | Willie Wong | CC BY-SA 2.5 |