Parallel transport and geodecics

[Kidphysics]

So from what I understand if you pass a vector (using parallel transport) through a closed curve where there is curvature in the interior, the vector will come back not to it's original vector but with a changed sense. However if the vector is on a geodesic it will not change its sense after it gets parallel transported but isn't this a contradiction (since it's a closed curve with curvature in the interior) or is this just the definition of a geodesic?

 

[Chestermiller]

So from what I understand if you pass a vector (using parallel transport) through a closed curve where there is curvature in the interior, the vector will come back not to it's original vector but with a changed sense. However if the vector is on a geodesic it will not change its sense after it gets parallel transported but isn't this a contradiction (since it's a closed curve with curvature in the interior) or is this just the definition of a geodesic?

I don't think it is possible to have a closed curve geodesic, since a geodesic is a world line for a particle in free fall. You can form a closed curve comprised of segments of geodesics, and if you parallel transport around one of these, the vector will change its sense.

Chet

 

[yuiop]

I don't think it is possible to have a closed curve geodesic, since a geodesic is a world line for a particle in free fall. You can form a closed curve comprised of segments of geodesics, and if you parallel transport around one of these, the vector will change its sense.

Chet

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.

 

[WannabeNewton]

So from what I understand if you pass a vector (using parallel transport) through a closed curve where there is curvature in the interior, the vector will come back not to it's original vector but with a changed sense. However if the vector is on a geodesic it will not change its sense after it gets parallel transported but isn't this a contradiction (since it's a closed curve with curvature in the interior) or is this just the definition of a geodesic?

A geodesic parallel transports its own tangent vector, \triangledown _{\mathbf{U}}\mathbf{U} = 0; nothing is being said about arbitrary vectors being transported along the curve.

 

[Dale]

A geodesic parallel transports its own tangent vector, \triangledown _{\mathbf{U}}\mathbf{U} = 0; nothing is being said about arbitrary vectors being transported along the curve.

Yes, but since parallel transport also preserves the dot product I think that you could probably generalize it to arbitrary vectors.

 

[WannabeNewton]

Yes, but since parallel transport also preserves the dot product I think that you could probably generalize it to arbitrary vectors.

Yes indeed it does but generalize what sorry?

 

[Kidphysics]

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.

yes this is what I was referencing sorry for not including this. So this was during a susskind lecture and he was saying that if you parallel transport a vector along the geodesic then it returns full circle and is not displaced by any angle. I assume this is just a property of geodesics they are the only closed paths which has curvature in the interior such that when a vector is parallel transported along that closed path it is invariant to changing orientation. I was just looking for an affirmative.

 

[Dale]

Yes indeed it does but generalize what sorry?

If a closed geodesic parallel transports its own tangent vector then it must parallel transport all other vectors also since it preserves the dot product also, right?

 

[Dale]

yes this is what I was referencing sorry for not including this. So this was during a susskind lecture and he was saying that if you parallel transport a vector along the geodesic then it returns full circle and is not displaced by any angle. I assume this is just a property of geodesics they are the only closed paths which has curvature in the interior such that when a vector is parallel transported along that closed path it is invariant to changing orientation. I was just looking for an affirmative.

It is true for geodesics on a sphere, but I am not sure that it is true in other manifolds.

 

[A.T.]

yes this is what I was referencing sorry for not including this. So this was during a susskind lecture and he was saying that if you parallel transport a vector along the geodesic then it returns full circle and is not displaced by any angle. I assume this is just a property of geodesics they are the only closed paths which has curvature in the interior such that when a vector is parallel transported along that closed path it is invariant to changing orientation. I was just looking for an affirmative.

It is true for geodesics on a sphere, but I am not sure that it is true in other manifolds.

I don't think so. Try a closed geodesic path on a cone that encloses the apex.

 

[Dale]

a closed geodesic path on a cone that encloses the apex.

Is there such a geodesic on a cone? I didn't think that a cone had any closed geodesics, but I must admit that I am having trouble visualizing it for sure. I am definitely less confident about this than most of my posts.

 

[WannabeNewton]

If a closed geodesic parallel transports its own tangent vector then it must parallel transport all other vectors also since it preserves the dot product also, right?

Well the geodesic will maintain \triangledown _{\mathbf{U}}g(U,U) = 0, with \mathbf{U} being the tangent to the geodesic, but I'm not seeing why this or the definition of the geodesic would imply \triangledown _{\mathbf{U}}\mathbf{V} = 0 for some arbitrary vector \mathbf{V}. I'm probably missing something so I apologize in advance.

 

[Dale]

No need to apologize, I am not confident that I am right either.

 

[PeterDonis]

I don't think it is possible to have a closed curve geodesic, since a geodesic is a world line for a particle in free fall.

A *timelike* geodesic is, yes. And a null geodesic is a world line for a light ray in "free fall" (i.e., no waveguides or other stuff present). But there are also spacelike geodesics. Those can be closed if the manifold is curved. For example, yuiop gave the example of a great circle on a sphere.

(There are spacetimes, such as the Godel universe, where there are closed timelike curves; but I'm not sure whether those curves can be geodesics. I think they can, but I'm not positive.)

I don't think so. Try a closed geodesic path on a cone that encloses the apex.

The apex is a singularity of the manifold; I believe there is a theorem that says parallel transport around a closed geodesic brings all vectors back to the same vectors, but only if the closed geodesic doesn't enclose a singularity.

 

[WannabeNewton]

No need to apologize, I am not confident that I am right either.

Sorry, I was thinking of segments of geodesics being connected together to form a closed loop. But yeah for a single closed geodesic I agree with what you said before.

 

[A.T.]

Is there such a geodesic on a cone? I didn't think that a cone had any closed geodesics

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

The apex is a singularity of the manifold

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

 

[Dale]

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

Cool, I didn't know that.

 

[A.T.]

Cool, I didn't know that.

Me neither. I figured it out after you questioned it. So better check it. It seems that the direction change after a loop around the apex is:

2*pi*sin(opening_angle / 2)

 

[Bill_K]

The surface of a cone is metrically equivalent to flat space. A cone is flat space with a sector removed and the two "edges" glued together. The geodesics on it are straight lines. It will be possible to have a geodesic that hits both "edges" if and only if the sector you removed is greater than 180^o, and if so, it will go all the way around.

 

[A.T.]

The surface of a cone is metrically equivalent to flat space. A cone is flat space with a sector removed and the two "edges" glued together. The geodesics on it are straight lines. It will be possible to have a geodesic that hits both "edges" if and only if the sector you removed is greater than 180^o, and if so, it will go all the way around.

That was my idea too. And if you remove more than 270° it will go all the way around twice, right?

 

[Dale]

It will be possible to have a geodesic that hits both "edges" if and only if the sector you removed is greater than 180^o, and if so, it will go all the way around.

Yes, but I don't think that it will be a geodesic. I think it will be straight everywhere except at the place where the edges join where it will have a corner.

 

[A.T.]

Yes, but I don't think that it will be a geodesic. I think it will be straight everywhere except at the place where the edges join where it will have a corner.

It doesn't have a corner, it is intersecting itself.

 

[Dale]

It doesn't have a corner, it is intersecting itself.

Yes, but I think it is intersecting itself at an angle.

 

[A.T.]

Yes, but I think it is intersecting itself at an angle.

Yes, that is the point. The direction of a vector changes after being parallel transported along a closed loop on a single geodesic. So this:

I assume this is just a property of geodesics they are the only closed paths which has curvature in the interior such that when a vector is parallel transported along that closed path it is invariant to changing orientation.

is wrong. The orientation is unchanged only in special cases, like the sphere.

 

[Kidphysics]

Yes, that is the point. The direction of a vector changes after being parallel transported along a closed loop on a single geodesic. So this:

is wrong. The orientation is unchanged only in special cases, like the sphere.

I know it is wrong and the example of the cone is perfect. I first thank you and another contributor for helping me.

Now if I can ask why does the definition seem to say that the sense of the vectors remains unchanged??

In the presence of an affine connection, geodesics are defined to be curves whose tangent vectors remain parallel if they are transported along it.

from wikipedia: geodesics

 

[Kidphysics]

The apex is a singularity of the manifold; I believe there is a theorem that says parallel transport around a closed geodesic brings all vectors back to the same vectors, but only if the closed geodesic doesn't enclose a singularity.

this theorem would be great to find! I will try to find it

 

[PeterDonis]

The orientation is unchanged only in special cases, like the sphere.

So what defines the special cases? I see what you mean about the presence of a "singularity" not being the right thing to look for, since you can alter the manifold slightly to eliminate it without eliminating the change in a vector parallel transported around a closed loop, e.g. putting a small spherical dome at the tip of the cone.

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.

 

[PeterDonis]

this theorem would be great to find! I will try to find it

After seeing A.T.'s follow up post, I think I was mistaken; the theorem, if there is one, won't be that general. It will only cover the "special cases" like a sphere, not all manifolds without singularities.

 

[Dale]

The direction of a vector changes after being parallel transported along a closed loop on a single geodesic.

But then it isn't a geodesic. A geodesic parallel transports it own tangent vector. The path you describe doesn't parallel transport its own tangent vector since it has a bend. Therefore it isn't a geodesic.

 

[PeterDonis]

Found an interesting paper titled "Closed Geodesics on Positively Curved Manifolds" which bears on this topic:

http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/Annals82_BTZ.pdf

 

[Austin0]

Found an interesting paper titled "Closed Geodesics on Positively Curved Manifolds" which bears on this topic:

http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/Annals82_BTZ.pdf

In this paper there was reference to closed hyperbolic geodesics.

this seems a contradiction ,could you explain the geometric meaning of this??

I understood that in a gravitational context hyperbolic orbits were necessarily escape trajectories.

 

[PeterDonis]

I understood that in a gravitational context hyperbolic orbits were necessarily escape trajectories.

"Hyperbolic" in the paper is a technical term; I believe it refers to the type of differential equation describing the curve, as here:

http://en.wikipedia.org/wiki/Partial_differential_equation#Classification

But I am not really familiar with much of the technical terminology in this paper.

Also, a point of clarification about orbits: yes, hyperbolic orbits are not bound orbits in GR, but bound orbits are not closed geodesics of spacetime either; they are "helical" when the time dimension is included. Their projections into spacelike slices will be closed spacelike curves, but those curves probably won't be geodesics of the spacelike slice.

 

[Austin0]

"Hyperbolic" in the paper is a technical term; I believe it refers to the type of differential equation describing the curve, as here:

http://en.wikipedia.org/wiki/Partial_differential_equation#Classification

But I am not really familiar with much of the technical terminology in this paper.

Also, a point of clarification about orbits: yes, hyperbolic orbits are not bound orbits in GR, but bound orbits are not closed geodesics of spacetime either; they are "helical" when the time dimension is included. Their projections into spacelike slices will be closed spacelike curves, but those curves probably won't be geodesics of the spacelike slice.

thanks for the link, although mostly over my head.
from what you said is it correct to infer that closed geodesics only refers to static manifolds and that the transport means repositioning vectors as a mathematical operation, with no implication of actual translation through spacetime?

 

[A.T.]

But then it isn't a geodesic. A geodesic parallel transports it own tangent vector. The path you describe doesn't parallel transport its own tangent vector since it has a bend. Therefore it isn't a geodesic.

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?

The entire line extending to infinity is a geodesic which intersects itself. The loop, that the geodesic encloses, is of course not a geodesic if you jump to the other part at the intersection. But it is a closed loop made from a single geodesic segment. The fact the total change in direction at the corner(s) is different from 2pi implies curvature inside the loop. A full great circle on a sphere has no corners at all, so the total corner change in direction is zero which also implies curvature.

 

[A.T.]

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.

 

[Dale]

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.

 

[Dale]

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]

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.]

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]

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.

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/140/ShifrinDiffGeo.pdf

 

[Dale]

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

Yes. The Moibus strip is non-orientable.

 

[A.T.]

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.

 

[Dale]

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]

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.

 

[Dale]

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]

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.]

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.]

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]

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