# Concept of parallel transport and geodesics

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## Summary:

Physical intuition and conditions of parallel transport

## Main Question or Discussion Point

I am currently reading Foster and Nightingale and when it comes to the concept of parallel transport, the authors don't go very deep in explaining it except just stating that if a vector is subject to parallel transport along a parameterized curve, there is no change in its length or direction and hence its derivative with respect to the parameter is equal to zero.
Later, there is an example of a vector being parallelly transported on a latitude of a sphere and its final direction is different from the initial direction at the same point! Then how can we say that it was parallelly transported?
I am unable to get the concept of parallel transport.

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vanhees71
Gold Member
2019 Award
The specific parallel transport depends on the definition of a connection on a manifold.

If you learn GR, there it's by assumption the uniquely defined torsion-free connection compatible with the metric. A connection is metric compatible, if the length of parallel transported vectors locally don't change, and it's torsion free if the corresponding Christoffel symbols are symmetric in their lower indices. From these assumptions you can express the Christoffel symbols defining the connection in terms of partial derivatives of the components of the pseudometric:
$${\Gamma^{\alpha}}_{\beta \gamma} = \frac{1}{2} g^{\alpha \mu} (\partial_{\beta} g_{\mu \gamma} + \partial_{\gamma} g_{\beta \mu}-\partial_{\mu} g_{\beta \gamma}).$$
The parallel transport of a vector $V$ along a curve $q^{\mu} = q^{\mu}(\lambda)$ is defined by the differential equation [EDIT: corrected in view of #3]
$$\dot{V}^{\mu} + {\Gamma^{\mu}}_{\rho \sigma} \dot{q}^{\rho} V^{\sigma}=0.$$

Last edited:
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Just to add a bit of intuition to the reply of @vanhees71 :

In Euclidean space (or a general affine space) a straight line is a curve whose tangent vector is constant (or can be reparametrised such that it is). A geodesic is a generalisation of this concept to an arbitrary manifold.

Similarly, a vector field in Euclidean space that is parallel is constant throughout that space. In a general manifold, there is no notion of what a ”constant vector field” would be because the tangent spaces at different points are different. Instead, we need to introduce what is called a ”connection” that has essentially all of the properties you would expect from a directional derivative. You may see this connection as a definition of what it means for a field to be ”constant” in a direction. A connection can be fully specified by its connection coefficients $\Gamma^a_{bc}$ in some coordinate system.

When you say that a field is parallel along a curve, what you mean is that the directional derivative in the tangent direction along that curve is zero, i.e., it is ”constant” along that curve.

The parallel transport of a vector $V$ along a curve $q^{\mu} = q^{\mu}(\lambda)$ is defined by the differential equation
$$\dot{V}^{\mu} + {\Gamma^{\mu}}_{\rho \sigma} \dot{q}^{\rho} V^{\sigma}.$$
$=0$ ;)

pervect
Staff Emeritus
Summary:: Physical intuition and conditions of parallel transport

I am currently reading Foster and Nightingale and when it comes to the concept of parallel transport, the authors don't go very deep in explaining it except just stating that if a vector is subject to parallel transport along a parameterized curve, there is no change in its length or direction and hence its derivative with respect to the parameter is equal to zero.
Later, there is an example of a vector being parallelly transported on a latitude of a sphere and its final direction is different from the initial direction at the same point! Then how can we say that it was parallelly transported?
I am unable to get the concept of parallel transport.
That doesn't sound like a good definition to me. It works fine on the plane, I would assume that your text books comments about parallel transport "not changing the direction" were intended to apply only to flat surfaces such as the plane, but I can't really say for sure exactly what they wrote, as I don't have the text in question.

If one is able to draw geodesics between two points, there is a geometric construction known as "Schild's ladder" that constructs the parallel transported vectors that may be helpful to developing some intuition.

In the usual approach, drawing geodesics is defined in terms of parallel transport, but in the special case of the Levi-civita connection, we can use a more intuitive definition of a geodesic as the curve of shortest distance between two points. This definition really applies only to Riemannian manifold (so it needs some modification to work on the pseudo-Riemannian manioflds in General Relativity), but it is good enough to be able to work out parallel transport on the sphere, and see how it differs from parallel transport on the plane. Note that when I talk about the curve of shortest distance, it is implied that all the points lie on the manifold (surface) of interest. So in the case of the sphere, which is the surface of a ball, all the points of the curve have to lie on the sphere, curves that leave the surface of the ball to go into its interior are not allowed.

This approach is rather oversimplified, but since it will work for both the flat plane and the surface of the sphere, it seems to me to be a cut above what you're getting out of your current textbook.

Defining geodesics in this manner also assumes the use of a particular connection, the important Levi-Civita connection. But that's the only connection we need to use in GR, and it's a known property of geodesics in the Levi-Civita connection of Riemannian manifolds that they are curves of shortest distance. While it's probably a good idea to someday learn about more general connections, learning how to do parallel transport in the connection that GR actually uses is at least a good first step, and may be all that you ever really need.

On the sphere, all we really need to know is that geodesics are "great circles". This allows us to do parallel transport on the sphere using this goemtric construction and some fairly basic spherical trig.

The name of the construction that does the parallel transport is "Schild's ladder".

The full details are given in for instance https://en.wikipedia.org/w/index.php?title=Schild's_ladder&oldid=910236915, but we can simplify the somewhat long procedure as defined in the wiki even further. The simplification will be less general, but we don't need full generality, we're just trying to develop some basic intuitions that give the correct answers for a plane, and for the surface of a sphere at this point. The full technique is actrually more general than just working on a sphere, but it's more work to describe when it works and when it doesn't than to just describe the technique.

There's also a desription of Schild's ladder in the textbook "Gravitation", by Misner, Thorne, and Wheeler, if you happen to have access.

The simplified versions of Schild's ladder is that if we draw a quadrilateral (of geodesics) whose opposing sides have the same length, then in the limit of a small quadrilateral, the opposite sides are parallel.

The following notes assume that you look at the figures in the wiki article I cited above.

In the wiki example, the idea is that the vector representing by the geodesic segement A_0 - X_0 is parallel transported along the curve to become the vector A_1 - X_1.

The purpose of the geometric construction is basically to construct a some quadrilateral whose opposing sides have the same length, so that A_0 - X_0 has the same length as A_1- X_1, and A_0 - A_1 has the same length as X_0 - X_1.

We've played a few games by representing vectors as arrows, but that's a common graphical technique.

A.T.
Summary:: Physical intuition and conditions of parallel transport

I am currently reading Foster and Nightingale and when it comes to the concept of parallel transport, the authors don't go very deep in explaining it except just stating that if a vector is subject to parallel transport along a parameterized curve, there is no change in its length or direction and hence its derivative with respect to the parameter is equal to zero.
Later, there is an example of a vector being parallelly transported on a latitude of a sphere and its final direction is different from the initial direction at the same point! Then how can we say that it was parallelly transported?
I am unable to get the concept of parallel transport.
Parallel transport preserves the direction relative to the local tangential geodesic, but a path of constant latitude is not a geodesic (except of the equator). If the path diverges from a geodesic, the direction of the transported vector diverges from the path tangent vector.

To get an intuitive grasp on parallel transport consider this mechanical model from a previous thread:
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.

When the tank arrives at the starting position, the gun will have a different orientation, then it started with.
Just to add to this, when we’re talking about lengths of paths a tank might typically take in relation to the curvature of the earth, this deviance might not even be noticeable. But if you imagine the tank going on a great journey, starting at the equator, traveling along it a quarter of the way around the earth, then turning due north, going all the way to the north pole, then making a 90 degree turn to go back south to where it started, it’s easy to work out that the gun will end up oriented 90 degrees off from where it started.

A.T.
Just to add to this, when we’re talking about lengths of paths a tank might typically take in relation to the curvature of the earth, this deviance might not even be noticeable. But if you imagine the tank going on a great journey, starting at the equator, traveling along it a quarter of the way around the earth, then turning due north, going all the way to the north pole, then making a 90 degree turn to go back south to where it started, it’s easy to work out that the gun will end up oriented 90 degrees off from where it started.
Yes, in this example the tank moves on geodesic segments (where the turret doesn't rotate) and turns twice by 90° (where the turret rotates). This is somewhat easier to understand.

In the constant latitude case the tank turns towards the pole continuously, as the track closer to the equator runs faster, and so the turret rotates all the time. This is easier to visualize closer to the pole.

If the tank was on a perfectly flat plane, the gun would never change direction in global space, no matter what path the tank drives along. It would always arrive back parallel to how it started, hence "parallel transport".

I forgot to ask one thing: For any path other than a geodesic, does the tangent vectors not form a parallel field,i.e., if I take the tangent vector at a point and parallel transport it to the next point, does it coincide with the tangent vector at the next point only for geodesics?

PeterDonis
Mentor
2019 Award
if I take the tangent vector at a point and parallel transport it to the next point, does it coincide with the tangent vector at the next point only for geodesics?
Yes. In fact, one way to define geodesics is as curves that parallel transport their tangent vectors along themselves.

vanhees71
Indeed, you get the geodesics equation by just setting $V^{\mu}=\dot{q}^{\mu}$ in the equaion in #2. That exactly describes what @PeterDonis states: the parallel transfport of the tangent vectors along themselves.