I Question about 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.
 
  • #10
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.
 
  • #16
PeterDonis said:
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.
I've generally seen FW transport only defined along timelike curves, so that definition would simply not apply to Riemannian manifolds.

For Lorentzian manifolds, for a vector tangent at some point to a timelike curve, the FW transport of the vector remains tangent to the curve, while the parallel transported vector deviates from the tangent precisely to the degree the curve deviates from being geodesic. Thus, the difference between the two is a measure of geodesic deviation.

Thus, parallel transport locally defines a vector being the same at different points, while FW transport preserves 'orientation' relative to the curve (even as the curve may deviate from a geodesic).
 
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  • #17
PAllen said:
the difference between the two is a measure of geodesic deviation.
While I know what you mean by this, I would point out as a caution that the term "geodesic deviation", at least in a GR context means something different: the amount by which nearby geodesics deviate from each other due to the curvature of the manifold.
 
  • #18
PAllen said:
I've generally seen FW transport only defined along timelike curves
Generally yes, since it's generally used in a relativity context along worldlines.

However, the definition only makes use of path curvature and the unit tangent vector, so I don't see why it couldn't be generalized to spacelike curves (and therefore to Riemannian manifolds). That's why I asked @Orodruin to elaborate on his statement.
 
  • #19
PeterDonis said:
While I know what you mean by this, I would point out as a caution that the term "geodesic deviation", at least in a GR context means something different: the amount by which nearby geodesics deviate from each other due to the curvature of the manifold.
Yes, not the best choice of words.
 
  • #20
PeterDonis said:
Generally yes, since it's generally used in a relativity context along worldlines.

However, the definition only makes use of path curvature and the unit tangent vector, so I don't see why it couldn't be generalized to spacelike curves (and therefore to Riemannian manifolds). That's why I asked @Orodruin to elaborate on his statement.
Ok, yes, I don't see a reason you couldn't apply the definition in a Riemannian manifold - books just assume a timelike curve. For the Riemannian context, the elements allowing the definition of FW transport are sometimes called Frenet-Serret formulas, but authors don't seem to define FW transport in the Riemannian case.
 
  • #21
PeterDonis said:
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.


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.
The entire point of FW transport is to keep a spacelike vector, such as a polarization vector, orthogonal to the timelike tangent of the world line (and the timelike tangent is also FW transported). The corresponding definition in a Riemannian manifold would do the same, it would keep the tangent of the curve FW transported even when you turn, which is not really what you want - you want to keep pointing in the same direction (as defined by the connection).

Edit: The difference is: Parallel transport keeps the vector pointed in the same direction (as defined by the connection, think gyroscope), FW transport rotates the vector along with the path without twisting (think a rigid frame fixed to your spacecraft turning without twisting).
 
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  • #22
PeterDonis said:
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.
The FW transport in GR is the correct thing to use for gyroscopes precisely because we are talking about its frame describing spacelike directions orthogonal to the timelike tangent of the world line. In the Riemannian setting, consider for example a gyroscope in regular Euclidean space, which will maintain its orientation regardless of how you move - which is parallel transport and not FW transport. Same thing with a turret rotating freely on a moving tank.
 
  • #23
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 connection defines how vector components at one point in a space are related to vector components at a nearby point.

Consider a flat plane with polar coordinates. I stand somewhere on it and point in the ##\vec r## direction. If I take a small step in some other direction while continuing to point in the same direction, I will no longer be pointing in the ##\vec r## direction because the radial direction has changed. The connection coefficients are the things that give me the change in my vector components. Naturally, the connection coefficients are zero in Cartesian coordinates on a flat plane.

The Levi-Civita connection turns out to be the particular definition of connection coefficients that yields our natural sense of "this vector hasn't changed even though its components have done". You can define other ones, and some theories of gravity consider them. GR does not.

The connection coefficients come into the concept of covariant differentiation. Going back to me standing on the plane, as I moved the ##r## and ##\phi## components of my vector changed - that is ##\partial x^a/\partial \tau\neq 0##. But the vector didn't really change (I'm still pointing in the same direction) even if its components did. It would be really useful if we could define a sort of derivative that measures how much a vector really changed (not at all) even when its components do, since all the things like forces would depend on that. That is the covariant derivative, which is basically the familiar partial derivative plus a correction term for how the coordinates changed - i.e. the connection coefficient. Thus if the covariant derivative of a vector is zero as you carry it along a line then it is pointing in the same direction. That is parallel transport.

There is one vector that is special to a curve, which is its tangent vector, which points in the direction you are moving if you follow that curve. A curve that parallel transports its tangent vector, then, is a curve where you don't change direction as you follow it. In a flat space that would be a straight line, but in general we call it a geodesic.
 
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  • #24
geordief said:
I wonder if I can now try to get my head around the Levi -Civita(sp?) connection
I think the video you posted explains it quite well, starting at 7:25: When you approximate the curved surface with flat facets, the connection tells you what happens when you cross from one facet to another.
 
  • #25
Ibix said:
... the tank can measure its rotation rate with a gyroscope internal to its hull ...
For the analogy I prefer to use the difference in left and right track / wheel distance. Or the inertia of a perfectly balanced turret, as proposed by the OP in post #7.

A gyroscope is tricky within this analogy (2D curved surface embodied in flat 3D), as it tends to keep its orientation within the 3D embedding space, not within the 2D space intrinsic to the curved surface. You might even end up with the gyroscope axis pointing vertically up, and then you cannot detect the turning of the tank. A gyroscope makes sense when you go to higher dimensions, where the 3D space curved.
 
  • #26
A.T. said:
A gyroscope is tricky within this analogy (2D curved surface embodied in flat 3D), as it tends to keep its orientation within the 3D embedding space
Ah yes, good point.
 
  • #27
A.T. said:
For the analogy I prefer to use the difference in left and right track / wheel distance. Or the inertia of a perfectly balanced turret, as proposed by the OP in post #7.

A gyroscope is tricky within this analogy (2D curved surface embodied in flat 3D), as it tends to keep its orientation within the 3D embedding space, not within the 2D space intrinsic to the curved surface. You might even end up with the gyroscope axis pointing vertically up, and then you cannot detect the turning of the tank. A gyroscope makes sense when you go to higher dimensions, where the 3D space curved.
I mean, it is a two-dimensional gyroscope, i.e., one where the gyroscope is only free to rotate around one axis ... :wink:

Ibix said:
The connection coefficients come into the concept of covariant differentiation. Going back to me standing on the plane, as I moved the ##r## and ##\phi## components of my vector changed - that is ##\partial x^a/\partial \tau\neq 0##. But the vector didn't really change (I'm still pointing in the same direction) even if its components did. It would be really useful if we could define a sort of derivative that measures how much a vector really changed (not at all) even when its components do, since all the things like forces would depend on that. That is the covariant derivative, which is basically the familiar partial derivative plus a correction term for how the coordinates changed - i.e. the connection coefficient. Thus if the covariant derivative of a vector is zero as you carry it along a line then it is pointing in the same direction. That is parallel transport.
I'd like to add some additional structure to this: In a Euclidean space - such as the flat plane - it is clear what "the same direction" means between points (precisely because the space is flat). Taking any set of basis vector fields ##\{\vec E_a\}## (let us take ##\{\hat e_r, \hat e_\phi\}## of the polar coordinates for concreteness) that generally depend on the position, a vector field can be written as
$$
\vec v = v^r \hat e_r + v^\phi \hat e_\phi
$$
Along a curve with curve parameter ##s##, we can then consider how ##\vec v## varies along the curve by taking the derivative
$$
\frac{d\vec v}{ds} = \dot v^r \hat e_r + v^r \dot{\hat e}_r + \dot v^\phi \hat e_\phi + v^\phi \dot{\hat e}_\phi
$$
where the dot represents differentiation wrt ##s## and we have simply applied the chain rule. For constant basis vectors, such as the cartesian ones, the derivatives of the basis vectors vanish, but this is not the case in general. Instead, we can expand
$$
\dot{\hat e}_a = \frac{d\hat e_a}{ds} = \sum_b \dot x^b \partial_b \hat e_a
$$
where we know that ##\partial_b \hat e_a## is a vector and may be written as a linear combination of ##\hat e_r## and ##\hat e_\phi## with some coefficients ##\Gamma_{ab}^c##:
$$
\partial_b \hat e_a = \Gamma_{ba}^r \hat e_r + \Gamma_{ba}^\phi \hat e_\phi
$$
The ##\Gamma## coefficients describe precisely how the basis changes when you change position and are the Christoffel symbols, i.e., the connection coefficients of the Levi-Civita connection.

The change in the vector field ##\vec v## along the curve is then described by two separate parts, one describing how the components change along the curve and another that describes how the basis changes:
$$
\frac{d\vec v}{ds} = [\dot v^r \hat e_r + \dot v^\phi \hat e_\phi] + v^r (\dot r [\Gamma_{rr}^r \hat e_r + \Gamma_{rr}^\phi \hat e_\phi]+ \dot\phi [\Gamma_{\phi r}^r \hat e_r + \Gamma_{\phi r}^\phi \hat e_\phi]) + \ v^\phi \dot{\hat e}_\phi (\dot r [\Gamma_{r\phi}^r \hat e_r + \Gamma_{r\phi}^\phi \hat e_\phi]+ \dot\phi [\Gamma_{\phi \phi}^r \hat e_r + \Gamma_{\phi \phi}^\phi \hat e_\phi])
$$

Now that looks like quite a mouthful, but is significantly more compact with Einstein summation notation, where repeated indices are summed over their ranges:
$$
\frac{d\vec v}{ds} = \dot v^a \hat e_a + \dot x^a v^b \Gamma_{ab}^c \hat e_c
$$
 
  • #28
Orodruin said:
I mean, it is a two-dimensional gyroscope, i.e., one where the gyroscope is only free to rotate around one axis ... :wink:
I think if you would constrain the spin-axis of a regular spinning gyro to stay tangential the local spherical surface (allow the gyro-axis to rotate around the local vertical axis only), then it would precess, even when transported along a geodesic along the surface.

Maybe a Sagnac gyroscope oriented in the local tangent plane would work. But I fear that would make the practical analogy more difficult to understand than the math it is trying to represent.
 
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