I Parallel transport

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In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist.



His task is to keep the arrow pointed in the same direction

How does he do this ? Does he use a reference point like the stars? (that only move very slowly)

If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in?

So ,although one refers to intrinsic curvature the embedding dimension is not discounted(at least in the way it is measured)
 
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geordief said:
How does he do this ? Does he use a reference point like the stars? (that only move very slowly)
No, definitely not. There should be no outside reference. If he walks straight forward, he keeps the arrow in the same direction relative to himself. If he walks in a curved path, he needs to adjust the arrow accordingly when he turns.
geordief said:
If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in?

No, see above.
geordief said:
So ,although one refers to intrinsic curvature the embedding dimension is not discounted(at least in the way it is measured)
No.
 
geordief said:
His task is to keep the arrow pointed in the same direction

How does he do this ?
It depends on what "in the same direction" means.

Parallel transport is one way of defining what "in the same direction" means. Unfortunately, that definition is most suited to transport along geodesics (which on a sphere are great circles), but the video shows transport along curves on a sphere that are not geodesics (lines of latitude other than the equator).

Along non-geodesic curves, there is Fermi-Walker transport, which in relativity is the way we generally define "in the same direction" locally. Along a geodesic, Fermi-Walker transport is the same as parallel transport, but it generalizes to non-geodesic curves in a way that parallel transport does not. The video does not seem to take this into account.
 
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To answer your question simply and directly, as a pilot, I would fly straight by keeping my wings level and applying and using the rudder only to avoid any yaw. Of course, in a real plane, this could not be done precisely and after an hour or two the plane would be flying in no particular direction.

But if the plane was ideal, and neglecting winds and the earths rotation, it would follow a great circle path.
That would be "keeping the arrow in the same direction".

In the video, he mentions the 3-D case where you could fix on a distant star. But for the 2D case, you need to deal with only the local geometry - as I did with the plane.
 
geordief said:
His task is to keep the arrow pointed in the same direction

How does he do this ?
Every time he himself turns by an angle x relative to the ground beneath his feet, he also turns the arrow by -x relative to himself, so the arrow retains its orientation relative to the local surface.

This is purely local and doesn't require extrinsic references.

Here is a mechanical analogy:

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.
 
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A.T. said:
Imagine a tank,
You beat me to it.

To reiterate, the tank can measure its rotation rate with a gyroscope internal to its hull and program its turret to counter-rotate at the same rate. No external references are needed.
 
If the turret of the tank is attached to the body by a mount that uses 100% non friction ball bearings ** and we send the tank around on a path that does not follow a great circle on the sphere does the inertia of the mass of the turret cause it to point in a different direction to the turning caterpillar wheels of the tank itself -so that the operator need not concern themselves with orienting the turret at all?

** ie it is basically in "free swing"
 
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PeterDonis said:
Along non-geodesic curves, there is Fermi-Walker transport, which in relativity is the way we generally define "in the same direction" locally. Along a geodesic, Fermi-Walker transport is the same as parallel transport, but it generalizes to non-geodesic curves in a way that parallel transport does not. The video does not seem to take this into account.
Parallel transport is perfectly well defined along non-geodesic curves. The manifold described in the video is a Riemannian manifold so there is no need to define (or use for) a Fermi-Walker transport.
 
geordief said:
If the turret of the tank is attached to the body by a mount that uses 100% non friction ball bearings ** and we send the tank around on a path that does not follow a great circle on the sphere does the inertia of the mass of the turret cause it to point in a different direction to the turning caterpillar wheels of the tank itself -so that the operator need not concern themselves with orienting the turret at all?

** ie it is basically in "free swing"
This analogy might work too, if the turret is perfectly balanced, so its center of mass is on the turret rotation axis. Otherwise linear accelerations of the tank would spin the turret. You have to ensure that any torques vectors applied to the turret around it's center of mass, never have any component vertical to the local ground.
 
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A.T. said:
This analogy might work too, if the turret is perfectly balanced, so its center of mass is on the turret rotation axis. Otherwise linear accelerations of the tank would spin the turret. You have to ensure that any torques vectors applied to the turret around it's center of mass, never have any component vertical to the local ground.
Very pleased if that is the case as I have been trying to understand this concept for the last 10 years.

I wonder if I can now try to get my head around the Levi -Civita(sp?) connection (which may hopefully be connected to non orthogonal bases. that I was inquiring after in a separate and recent thread.
 
  • #11
geordief said:
Very pleased if that is the case as I have been trying to understand this concept for the last 10 years.

I wonder if I can now try to get my head around the Levi -Civita(sp?) connection (which may hopefully be connected to non orthogonal bases. that I was inquiring after in a separate and recent thread.
The Levi-Civita connection is the unique metric compatible and torsion free connection on a Riemannian manifold (or pseudo-Riemannian). In general a connection has nothing to do with any basis, although its connection coefficients can be expressed in it.

The Levi-Civita connection is essentially what you think about when you hear this:
A.T. said:
This analogy might work too, if the turret is perfectly balanced, so its center of mass is on the turret rotation axis. Otherwise linear accelerations of the tank would spin the turret. You have to ensure that any torques vectors applied to the turret around it's center of mass, never have any component vertical to the local ground.

There are other connections that will not satisfy this and be different definitions of what it means to parallel transport a vector. For example, a metric compatible connection with torsion (on the sphere minus the poles) would be given by how a compass needle behaves when you transport it around the surface.

Edit: In relativity, the connection is generally the Levi-Civita connection.
 
  • #12
Orodruin said:
Parallel transport is perfectly well defined along non-geodesic curves.
I know it's well defined, I'm just not sure it's what the OP intended by "the same direction" along a non-geodesic curve.

Orodruin said:
The manifold described in the video is a Riemannian manifold so there is no need to define (or use for) a Fermi-Walker transport.
I wasn't aware of this; can you elaborate a little? Even in Riemannian manifolds, there is still a concept of path curvature (i.e., a curve being non-geodesic), and I would have expected that to affect the transport law along the curve.
 
  • #13
PeterDonis said:
It depends on what "in the same direction" means.

Parallel transport is one way of defining what "in the same direction" means. Unfortunately, that definition is most suited to transport along geodesics (which on a sphere are great circles), but the video shows transport along curves on a sphere that are not geodesics (lines of latitude other than the equator).

Along non-geodesic curves, there is Fermi-Walker transport, which in relativity is the way we generally define "in the same direction" locally. Along a geodesic, Fermi-Walker transport is the same as parallel transport, but it generalizes to non-geodesic curves in a way that parallel transport does not. The video does not seem to take this into account.
I disagree. The definition of a geodesics as a curve that parallel transports its tangent vector makes no sense without parallel transport also being defined for other curves. I have never seen a definition of parallel transport that assumes geodesics.

The connection is what defines which vectors at ‘nearby points’ are the same direction. It is thus the primary way to define what same direction means locally, thus forming the basis of the definition of parallel transport.
 
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  • #14
geordief said:
How does he do this ? Does he use a reference point like the stars? (that only move very slowly)
If you don't insist on the example in the video, consider this option:
With a couple of 2-meter ruler sticks, you can go forward by laying one stick fixed on the ground and then placing the second ahead of it so they both overlap for 1 meter. Then regard the second as the fixed one and repeat the process. This way you'll follow a 2D geodesic (assuming that the radius of curvature is much smaller than 1/meter). With a goniometer, you can carry a third 30cm ruler in such a way that its angle ##\alpha## relative to the long rulers is constant. This way, the short ruler is parallel-transported along the geodesic. If you decide to change the direction in which you advance by an angle ##\beta## , the short ruler should now be kept in a constant angle ##\alpha\pm\beta## relative to the long ones, so it is still parallel transported along the trajectory.

Much cheaper than a tank.
 
  • #15
PAllen said:
I have never seen a definition of parallel transport that assumes geodesics.
I didn't say the definition of parallel transport assumed geodesics. My concern was the difference between parallel transport and Fermi-Walker transport along non-geodesic curves. @Orodruin has said there is no such difference in Riemannian manifolds, and I've asked him to elaborate since I wasn't aware of that.

In a GR context, if we talk about using gyroscopes, for example, to define what "in the same direction means" along a worldline (and that has been used as an example in this thread), the proper transport law is Fermi-Walker transport. Along a geodesic that is the same as parallel transport, but along a non-geodesic curve it isn't.
 

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