Parallel transport of a vector on a sphere

[bres gres]

TL;DR Summary

i get stuck when i watch the tensor video in this link:
https://www.youtube.com/watch?v=Af9JUiQtV1k&list=PLJHszsWbB6hpk5h8lSfBkVrpjsqvUGTCx&index=21
hope someone can help me as it is in my head for several days
thank you

1.how can i estimate the effect of the "straight component" in different small circles on the sphere and plot them on the graph

2.there is contradiction between the graph and the definition?

question1 :

if you draw a small circle around the north pole (it should be the same at every points because of the symmetry of the sphere),then it is approximately a flat space ,then we can translate the vector on sphere just like what we have done in flat space(which translate the vector "passively" because we just move our feet but the vector is hold in the same direction)(case1) .

There is no straight line on sphere,however the straightest line is the path lie on the great circle,therefore we can bend the vector actively to change the direction such that it seems like the vector is unchanged when we move forward in a "straight line" therefore we should subtract the normal component(and there is no change in tangential component since the vector seems to be unchanged in the observers frame). (case2)

Between these two extreme case, the person moving on the Earth will inevitably bend his fit into some direction in his frame(because the path on the great circle is straight one in his frame and other paths will be non-straight one even in his frame),at the same time there is "straight component" as well just like as case (2) mentioned,if we combine the effect in case1 and case 2 in a miraculous way,we will gradually bend the vector into some direction.here is what i understand the only thing i get stuck is i cannot imagine how two things combined in a way that i can plot the vectors on the graph,it seems hard to me. how can i estimate the effect of the "straight component" in different small circles.
question2:
according to the definition of the covariant derivative (which is the ordinary vector derivative in R^3 space subtract the normal component),it seems the graph in 3:40 fits the definition as well since there is no rate of change of the vector in R^3 space,then if we use x-y-z coordinate to calculate the rate of change of the vector,then it should be zero. The normal component should be zero as well(since the normal component of the rate of change of this vector is zero),therefore the final value will be zero. because the author uses extrinsic perspective in the graph,therefore we can use normal xyz coordinate to treat the vector as normal,if every point on the sphere assigned with a vector that is pointed into x direction,then the "ordinary part" would be zero(there is no normal part as well),therefore this fits the definition as well. However this cannot be true since in 3:40 ,the author said that the vector will be pointed into the sky geometrically(after the vector is moved after a quarter of great circle) ,but algebraically we can get zero which fits the definition. there is contradiction between his explanation and the definition?thank for help!
this is my first time to ask physics question

 

[Dale]

Your video is >33 min long. Can you make your question self-contained so that watching such a long video is not needed in order to understand your question?

it seems the graph in 3:40 fits the definition as well since there is no rate of change of the vector in R^3

Yes, the red vector would be parallel transported to the blue vector in R^3. However, note that the blue vector is not a vector in S^2 (or more precisely the tangent space at that point on the equator in S^2). In fact, in R^3 the different tangent spaces at each point in S^2 are oriented differently. So to parallel transport a vector in S^2 will necessarily be different from parallel transport in R^3. Parallel transport in S^2 will require rotation in R^3 in order to keep the vector in the tangent space of S^2.

 

[bres gres]

Your video is >33 min long. Can you make your question self-contained so that watching such a long video is not needed in order to understand your question?Yes, the red vector would be parallel transported to the blue vector in R^3. However, note that the blue vector is not a vector in S^2 (or more precisely the tangent space at that point on the equator in S^2). In fact, in R^3 the different tangent spaces at each point in S^2 are oriented differently. So to parallel transport a vector in S^2 will necessarily be different from parallel transport in R^3. Parallel transport in S^2 will require rotation in R^3 in order to keep the vector in the tangent space of S^2.

thank for your help
the first part is related to the example in 31:32 where u2 = pi/4 (not on great circle)
the first question is [ that the vector moving along the latitude in 31:32 leaned toward to left] is unclear
i try to understand intuitively what "degree of straight" will the walker on this curve feel when he is deriated from the "straight line" as he moved along this curve on latitude.

 

[bres gres]

Your video is >33 min long. Can you make your question self-contained so that watching such a long video is not needed in order to understand your question? Yes, the red vector would be parallel transported to the blue vector in R^3. However, note that the blue vector is not a vector in S^2 (or more precisely the tangent space at that point on the equator in S^2). In fact, in R^3 the different tangent spaces at each point in S^2 are oriented differently. So to parallel transport a vector in S^2 will necessarily be different from parallel transport in R^3. Parallel transport in S^2 will require rotation in R^3 in order to keep the vector in the tangent space of S^2.

i get trouble when i try to understand a vector which is parallel transport in constant latitude."However, note that the blue vector is not a vector in S^2"
why? i think it is "zero vector" in S^2"in R^3 the different tangent spaces at each point in S^2 are oriented differently. So to parallel transport a vector in S^2 will necessarily be different from parallel transport in R^3. "

according to the definition,it is "the ordinary derivative minus the normal component"
this definition doesn't tell us "the parallel transport in R2 will be different from parallel transport in R^3) ?

 

[WWGD]

i get trouble when i try to understand a vector which is parallel transport in constant latitude."However, note that the blue vector is not a vector in S^2"
why? i think it is "zero vector" in S^2"in R^3 the different tangent spaces at each point in S^2 are oriented differently. So to parallel transport a vector in S^2 will necessarily be different from parallel transport in R^3. "

according to the definition,it is "the ordinary derivative minus the normal component"
this definition doesn't tell us "the parallel transport in R2 will be different from parallel transport in R^3) ?

But $\mathbb R^2$ is flat , so parallel transport is trivial with identity connection . Maybe you can consider circle embedded in $\mathbb R^2$ to have a meaningful problem.

 

[pervect]

A curve of constant latitude on the sphere isn't a great circle, unless the circle is at the equator. Therfore it isn't a geodesic, it isn't a "straight line".

Therefore, to be able to parallel transport a vector along a curve of constant latitude, you need to be able to parallel transport a vector along something that isn't a geodesic, i.e. the curved space version of a "straight line".

I suspect that this is where your confusion is. I would suggest considering the following flat space example. Parallel transport the vertical vector along the two dotted lines, labelled "A" and "B" in the plane. The result is another vertical line, independent of the path because we are on a plane.

The description in the video of how to perform parallel transport is a bit vague, and I suspect that there are communication difficulty here between what the author meant and how you interpreted it.

There are a couple of alternatives here.

You can approximate the circle of constant latitude with a number of great circles, then use the method you are familiar with from the lecture to parallel transport the vector along the great circle segments.

You're probably used to seeing maps where curves of constant latitude appear to be straight lines. But they're not geodesics, they are not curves of shortest distance between two points that lie entirely on the sphere.

Approximating the curve of constant latitude via series of great circles will give a curve that looks something like this on the usual map (where curves of constant latitude appear to be straight lines, even though they are not geodesics):

As another alternative, you can also try reading a less-vague description of how to do parallel transport geometrically, namely "Schild's ladder"

To be able to utilize this construction, all you'll be able to need to do is to draw great circles on a sphere, and find their midpoints.

The wiki link is here. There is also a write up in MTW's text "Gravitation", if you have access to it.
https://en.wikipedia.org/w/index.php?title=Schild's_ladder&oldid=910236915

I hope that's everything - I am going to rush this out, there may be typos or things that could be clearer, as the holidays are calling me away.

 

[Dale]

why? i think it is "zero vector" in S^2

The blue vector is clearly not the zero vector. There are no basis vectors in the tangent space that can form the blue vector.

according to the definition,it is "the ordinary derivative minus the normal component"
this definition doesn't tell us "the parallel transport in R2 will be different from parallel transport in R^3) ?

The “normal component” is what makes it different (assuming you meant S2 and not R2)

 

[pervect]

I'm not sure if the OP is still with us, but, having more time, I wanted to expand on something I said earlier. Possibly this is redundant, and the OP has already figured out the answer to their question.

In the following diagram, we are parallel transporting a vector via two different paths on a plane (the paths are represented by dotted lines, labelled A and B) from a point on the left to the point on the right.

Because we are parallel transporting on a plane, we know that the parallel-transported vector always points upwards. This won't be true in general, but by focusing on the simple case of parallel transport on the plane, we can understand the rules better.

When we parallel transport a vector, represented by a spear in the video that the OP mentioned in his original post, the angle between this vector and the path always remains constant. In the diagram, this is curve A, and the angle is always 90 degrees.

There is a formal way of saying this. The angle between the tangent vector to the curve, and the vector being parallel transported, always remains constant as we parallel transport a vector along a geodesic.

This is always true, in flat or in curved space-time. When you parallel transport a vector along a geodesic, the angle between the vector and the curve it's being transported along (technically, we say the angle between the vector and the tangent to the curve) remains constant.

Parallel transporting a vector along a geodesic using this rule is one of the easiest cases to describe, and I think the video described parallel transport in this manner, i.e. it tells you how to parallel transport a vector along a "straight" path, aka a geodesic. But it skips over the describe how to parallel transport a vector along a curved path, such as path B. THis is logical, the video doesn't need to consider any other case to make their point.

We can note, instantly, that the angle between the vector (modeled as a carried spear in the video) and the path does NOT stay constant in case B. So, we need a different rule to define parallel transport a vector along a curve that is not geodesic than the "constant angle" rule.

So, what is the rule for parallel transporting a vector along a curved path? Well, we can approximate the curved path by a number of straight line segments.

Then the angle between the vector and each of the segments remains constant as long as we are on that segment. But when we change from one segment to the next, the angle changes. By considering the limiting process where we approximate the curve as a large number of straight segments, we can carry out the process.

The video, though, doesn't consider this case. It doesn't need to to make the point the video makes - which is to say that when we parallel a transport a vector along a spherical triangle made up of three great circles, it comes back rotated.

The case of parallel transporting a vector a long a path that is not a great circle is not covered by the video, as it's not relevant to the point the video is trying to make. But we can make a good informal argument as to how to carry out parallel transport along a general curve by approximating the general curve by an ever-increasing number of segements, and taking the limit as the number of segments approaches infinity.

As an alternative, as I mentioned previously, we can use the geometric construction known as Schild's ladder to do the parallel transport. The end goal of both argumetns is to demonstrate the existence and unqiueness of this process we call "parallel transport", and to build up some intuition.

 

[A.T.]

So, what is the rule for parallel transporting a vector along a curved path? Well, we can approximate the curved path by a number of straight line segments.

Then the angle between the vector and each of the segments remains constant as long as we are on that segment. But when we change from one segment to the next, the angle changes.

@bres gres Here a very practical analogy of the above rule:

Imagine a tank, with the gun turret rotation inversely coupled to the tank steering: When the tank hull turns X degree relative to the local ground, the turret turns -X degree relative to the hull, so the gun keeps its orientation relative to the local ground. This is how you parallel transport a gun.

On the sphere, the only way to prevent the turret from rotating relative to the hull, is to move on great circles (geodesics). To move on a circle of constant latitude off the equator, the tracks of the tank must be running at different speeds, so the tank is turning, and the turret is rotating relative the hull. When the tank arrives at the starting position, the gun will have a different orientation, then it started with.

 

[Dale]

the tracks of the tank must be running at different speeds

This is a key point. This is how the tank knows that it’s path is not a geodesic and therefore how much to turn the turret relative to the body in order to parallel transport it.

 

[bres gres]

thank you
the explanations are clear and it seems i understand