Parallel transport and geodecics

A.T.

Quote by PeterDonis

Is it constant curvature that defines the special cases? More precisely, is it that the surface enclosed by the closed geodesic has to have constant curvature? That would explain why a sphere is a special case but an ellipsoid of revolution is not.

On the ellipsoid of revolution there are some self intersecting geodesics which preserve the direction. On the sphere all of them do. This seems to be due to the higher degree of symmetry on the sphere, which is of course connected to the globally constant curvature.

DaleSpam

Quote by A.T.

Based on this reasoning there is trivially (per definition) no example of "other manifold" that you mention in post #9 (as a counter example to the sphere). Is there?

Not that I can think of, but I am concerned that is due to a lack of imagination on my part.

DaleSpam

Yay, I just thought of an example: a Moibus strip.

So, you can take a Moibus strip and draw a geodesic around the strip. The tangent vector is parallel transported along the geodesic, and so obviously it parallel transports to itself.

A vector orthogonal to the tangent vector is parallel transported such that its dot product with the tangent vector remains the same (0), but it winds up on the opposite side of the tangent vector. So it does not map to itself despite the tangent vector doing so and the dot product remaining the same.

phinds

Quote by yuiop

I think kidphysics is thinking of the often quoted example of transporting a vector along a great circle around a sphere which is an analogy to a geodesic.

I don't understand why you say it is an ANALOG to a geodesic. My understanding of the word agrees w/ the first definition I found on-line just now:

Of, relating to, or denoting the shortest possible line between two points on a sphere or other curved surface.

So how is a great circle not a geodesic but just an "analog" of a geodesic???? What am I missing?

A.T.

Quote by DaleSpam

Yay, I just thought of an example: a Moibus strip.

So, you can take a Moibus strip and draw a geodesic around the strip. The tangent vector is parallel transported along the geodesic, and so obviously it parallel transports to itself.

A vector orthogonal to the tangent vector is parallel transported such that its dot product with the tangent vector remains the same (0), but it winds up on the opposite side of the tangent vector.

Are you sure about the orthogonal vector being flipped? On the standard Moibus strip?

PeterDonis

Quote by Austin0

from what you said is it correct to infer that closed geodesics only refers to static manifolds

I don't think so; being static is a very restrictive condition on manifolds. The paper doesn't seem to talk about this, as far as I can tell.

Quote by Austin0

and that the transport means repositioning vectors as a mathematical operation, with no implication of actual translation through spacetime?

No. Parallel transport does imply "moving" vectors from one event to another; that's what it's for, to allow you to "compare" vectors at different events by moving one of them from one event to the other.

atyy

Shifrin comments on this in his notes, Corollary 1.5 and the paragraphs following on p79-80.
http://math.berkeley.edu/~reshetik/1...rinDiffGeo.pdf

DaleSpam

Quote by A.T.

Are you sure about the orthogonal vector being flipped? On the standard Moibus strip?

Yes. The Moibus strip is non-orientable.

A.T.

Quote by DaleSpam

Yes. The Moibus strip is non-orientable.

I see what you mean. You treat the strip as a "single layer" which represents the manifold.

I treated the two sides of the strip as "two layers" which represent different regions of the same manifold: Walking on the strip will bring you to your original position (on the same side) without flipping the orthogonal vector.

DaleSpam

Quote by A.T.

I see what you mean. You treat the strip as a "single layer" which represents the manifold.

Yes, that is what I meant, sorry about not being clear.

Austin0

Quote by DaleSpam

Yay, I just thought of an example: a Moibus strip.

So, you can take a Moibus strip and draw a geodesic around the strip. The tangent vector is parallel transported along the geodesic, and so obviously it parallel transports to itself.

A vector orthogonal to the tangent vector is parallel transported such that its dot product with the tangent vector remains the same (0), but it winds up on the opposite side of the tangent vector. So it does not map to itself despite the tangent vector doing so and the dot product remaining the same.

How does this work that it ends up in the opposite orientation?

. it seems that a complete circuit returning to the starting point does retain an orthogonal vector on the same side as the original.

DaleSpam

No, you have to go two complete circuits to get an orthogonal vector back to the same side. I don't know how to describe how it works, but you can see for yourself. Use a piece of transparent material so that you don't have to worry about the fact that a piece of paper is a 3D object and has two flat sides whereas a flat 2D manifold does not.

Austin0

Quote by A.T.

If the opening angle is less than 60° it has closed geodesics around the apex.

Replace the pointy tip with a small spherical dome. Or simply look at the geodesics on an ellipsoid of revolution.

Why wouldn't any conic section that was not hyperbolic be a closed geodesic?
i am just trying to get the picture here.

A.T.

Quote by Austin0

How does this work that it ends up in the opposite orientation?

. it seems that a complete circuit returning to the starting point does retain an orthogonal vector on the same side as the original.

See my comments in post #43. You are thinking (like did) about walking on the strip, and thus being on one or the other side of it.

DaleSpam is thinking about moving within the strip. The loop you travel here is just half the length of the loop "on the surface", and you arrive flipped.

A.T.

Quote by Austin0

Why wouldn't any conic section that was not hyperbolic be a closed geodesic?

Conic sections are not geodesics in general. See post #19 for an explanation on how geodesics on a cone are constructed.

Austin0

Quote by DaleSpam

No, you have to go two complete circuits to get an orthogonal vector back to the same side. I don't know how to describe how it works, but you can see for yourself. Use a piece of transparent material so that you don't have to worry about the fact that a piece of paper is a 3D object and has two flat sides whereas a flat 2D manifold does not.

I understand that. But the fact that it makes two closed circuits as viewed from a 3 d space doesn't seem relevant when considering it as a 2d manifold where it makes only a single circuit.
But then i'm not sure what you talking about as there seems to be a difference of view here.
AN early post mentioned a geodesic as a helical world line which I get. The path of an inertial particle.
Then PeterDonis seemed to indicate you were not talking about this but some other concept in the context of geometric topology . SO ???
When you talk about geodesics on a sphere aren't you talking about the surface as a 2 d topology?
Thanks

Austin0

Quote by PeterDonis

I don't think so; being static is a very restrictive condition on manifolds. The paper doesn't seem to talk about this, as far as I can tell.

No. Parallel transport does imply "moving" vectors from one event to another; that's what it's for, to allow you to "compare" vectors at different events by moving one of them from one event to the other.

OK I'll take another try. A particle can travel on a closed geodesic but it's world line will of course be unbounded.
Then wrt closed geodesic:
a) it's world line would be a set of point worldlines together comprising a tube.
b) it is a single entitiy whose world line is a tube.
c) It is an abstraction that can't really be said to have a worldline.

so does parallel transport include a velocity term?

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atyy Recognitions: Science Advisor	Shifrin comments on this in his notes, Corollary 1.5 and the paragraphs following on p79-80. http://math.berkeley.edu/~reshetik/1...rinDiffGeo.pdf

DaleSpam Mentor	No, you have to go two complete circuits to get an orthogonal vector back to the same side. I don't know how to describe how it works, but you can see for yourself. Use a piece of transparent material so that you don't have to worry about the fact that a piece of paper is a 3D object and has two flat sides whereas a flat 2D manifold does not.