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- Thread starter Kidphysics
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Chestermiller

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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

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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 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

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WannabeNewton

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A geodesic parallel transports its

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Yes, but since parallel transport also preserves the dot product I think that you could probably generalize it to arbitrary vectors.A geodesic parallel transports itsown tangent vector, [itex]\triangledown _{\mathbf{U}}\mathbf{U} = 0[/itex]; nothing is being said about arbitrary vectors being transported along the curve.

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WannabeNewton

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

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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.

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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?Yes indeed it does but generalize what sorry?

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

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A.T.

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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.

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

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WannabeNewton

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

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

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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.

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WannabeNewton

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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.

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A.T.

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If the opening angle is less than 60° it has closed geodesics around the apex.Is there such a geodesic on a cone? I didn't think that a cone had any closed geodesics

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

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Cool, I didn't know that.If the opening angle is less than 60° it has closed geodesics around the apex.

- #18

A.T.

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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:Cool, I didn't know that.

2*pi*sin(opening_angle / 2)

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Bill_K

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A.T.

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

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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 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.

- #22

A.T.

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

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Yes, but I think it is intersecting itself at an angle.It doesn't have a corner, it is intersecting itself.

- #24

A.T.

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

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

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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

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