# Intuition of Geodesics

1. Nov 17, 2013

### sigma_

I'm taking an undergraduate level GR course, and from my text (Lambourne), the author describes a geodesic as a curve that "always goes in the same direction", and says that the tangent vector to the curve at some point u+du (where u is the parameter variable from which all the vector components of the space are defined) should be proportional to the parallel transported tangent vector on the same curve at the same point u+du.

I'm trying to picture this connection on a great circle in spherical geometry, and I can't see the following:

1. How does the tangent vector always point in the same direction at all points on a great circle of a sphere?

2. How can the parallel transported vector point in the same direction as the tangent vector at all points on a great circle of a sphere?

Could anyone show me with some sort of diagram? There is none in my text, and I'm really struggling with this.

P.S. please forgive me if this is not the correct subforum for this type of post.

Last edited: Nov 17, 2013
2. Nov 17, 2013

### PAllen

This parallel transport definition of geodesic is more normally stated:

- if you start with a tangent vector at one point on a curve, then parallel transport it along the curve, it will match (within scaling) the tangent vector at each point it is transported to. Any curve with this property is called a geodesic.

In the case of a great circle on a 2-sphere, especially as viewed from an embedding 3-space, the parallel transported tangent changes direction. But it always continues to match the tangent, wherever it is transported to. No other curve on the 2-sphere has this property.

3. Nov 17, 2013

### sigma_

Perhaps my understanding of parallel transport is incorrect then.

Lets say we are looking at the great circle of a 2-sphere from a perpendicular point of view, i.e. it looks like a regular circle, being viewed head-on

If we start with the tangent vector at the point which is at 12 oclock and say that that tangent vector is completely horizontal and points right, then the tangent vector at 3oclock would point directly downward and be completely vertical, then the tangent vector at 6 oclock would be completely horizontal and point left, and the tangent vector at 9oclock would be completely vertical and point upward etc etc etc.

If we parallel transported the tangent vector from 12 oclock to (lets say) the 3 oclock position, I dont see how it could be pointing in the same direction, being that the parallel transported vector, Vp, would be pointing right and the tangent vector, Vt, would be pointing downward.

I assume when you refer to scaling, you are referring to the magnitude of these vectors, not direction.

4. Nov 17, 2013

### PAllen

Yes, scaling refers to magnitude, not directions.

Yes, your understanding parallel transport is clearly incorrect. When the 12:00 vector is parallel transported to the 3:00 position it points downward (viewed from the embedding space, which confuses matters). You have to image the embedding space not being present. Parallel transport within the 2-sphere will be unaffected by whether you squash in a dimple on it as long as you dont stretch the surface in any way. Such an embedding change will affect how the transported vector looks in the embedding space, but that is irrelevant. In fact, anything you can change by varying the embedding without any stretching is by definition irrelevant to intrinsic geometry; parallel transport is a feature of intrinsic geometry.

5. Nov 17, 2013

### Staff: Mentor

It cannot point to the right, that would be pointing out of the manifold! Vectors in a manifold are always point somewhere in the tangent space, never perpendicular to it.

6. Nov 17, 2013

### sigma_

So then how can I visualize the notion of parallel transport of the tangent vector around the great circle on a spherical surface? If the whole idea of parallel transport is to keep a certain vector parallel to itself along an entire curve, how can a vector on a great circle ever be parallel to itself the entire way around the great circle without pointing out of the manifold?

If the definition of parallel is never intersecting, this then means that any two tangent vectors on a great sphere will never intersect, because they are essentially parallel transported copies of each other, yes?

Last edited: Nov 17, 2013
7. Nov 17, 2013

### PAllen

You have to try to ignore the embedding space. Your notion is to preserve parallelism in the embedding space. Instead, imagine you were a 2-d being in the 2-sphere. Vectors, and parallel transport are defined based on 2-geometry without reference to any embedding space. Try this: 3-spaces can can be curved. In fact, you live in one (the curvature is infinitesimal, but not zero). At every point, you have a set of directions available characterized by two angles. You have no option to consider a third angle representing a direction available if your 3-space was embedded in some 4-space.

Anyway, a parallel transport associates a vector in the tangent plane at one point, with a vector in the tangent plane at another point by a rule that (in general) depends on the path used between the points. Your book should have a formal definition of parallel transport (if I provide one, it is likely to used different terminology and rely on different background than used in your book - which I am totally unfamiliar with). If it doesn't, you should really get another book. You can't get very far in this relying on intuition.

8. Nov 17, 2013

### A.T.

It is easier to understand this with discrete steps, instead of continuous turning.

Imagine you walk around on a sphere. You can walk straight ahead or you can stop to turn and then walk straight again. You carry two arrows:
- tangent vector : is fixed to your current walking direction
- parallel transported vector : initially fixed to your current walking direction. But every time you turn by N degrees, you turn that vector by -N degrees relative to you. So it keeps its orientation relative to the ground during you are turning.

On a great circle you never stop to turn, so the arrows stay parallel.

9. Nov 17, 2013

### Staff: Mentor

Remember, a 2 sphere is a 2D manifold. There is only latitude and longitude, but no altitude. The vertical direction doesn't even exist.

So the only thing you are concerned about is not turning to the left or to the right. As long as you step straight ahead at each step without ever turning left or right then you have parallel transported the tangent vector.

10. Nov 17, 2013

### sigma_

Thank you all for the wonderful explanations! I think I've pieced it together now.

Since there is no up or down direction on this 2-D manifold, the only way that the tangent vectors could not be parallel transported copies of each other is if I turned and moved in some lateral direction, and not just continued in a "straight line" (great circle). Is this more or less correct?

11. Nov 18, 2013

### PAllen

Yes.

12. Nov 18, 2013

### pervect

Staff Emeritus
The question is slightly wrong - the tangent vector is parallel transported along itself, it doesn't quite mean that it's pointing in the same direction in any global sense.

There are at least a couple of ways to visualize parallel transport. One is

a geometrical construction based on parallelograms. Schild's ladder works only on manifolds without torsion, which is toe sort of manifolds we use in GR. You can find some remarks in Penrose's "Road to reality" about why it works - basically it follows from the fact that prallelograms are closed to third order on torsionless manifolds.

You migh find it helpful to use the Schild's ladder consruction on a sphere.

There's another way, which is more of a personal viewpoint. I believe it's sound, but because it's a personal viewpoint it may be less helpful than something you can find in a textbook that's documented.

Imagine a sphere, with some point P on it, and a plane that's a tangent to the sphere at P. Now imagine a great circle passing through P. Project it's image onto the tangent plane, in a way that's similar to the miller cylindrical projection, but onto the tangent plane rather than a cylinder. If you want something more formal on the details of the projection process, look up the term"exponential map". But hopefully saying that it's like the Miller projection process, but onto a plane rahter than a cylinder, suffices.

Imagine two more points on the great circle P-, one one side of P, and P+, on the other side, all close together. Parallel transport without torsion implies that the three points P-, P, and P+ are all collinear on the projection.

IF they were not parallel, we couldn't say that the segment P+ P was a continuation of the segment P- P. But they are collinear, so we can make this statement.

(The construction requires that the three points be "close", they can't be arbitrarily far apart).

A little thought should show you that given two points on a geodisic/ great circle, this requirement allows you to find the locations of nearby third points and hence "continue" the geodesic.