1
$\begingroup$

I am new to general relativity, and trying to understand parallel transport.

I want to transport a vector along a line of longitude, on a sphere, starting at the equator, and ending (just shy) of the North Pole. I want the vector to point south initially, in the $\theta$ direction.

If I understand Foster & Nightingale correctly, when parallel transporting a vector along a geodesic, the direction of the vector should not change.

But at the starting point (equator), the vector points "down/vertically," while at the ending point (North Pole) it is pointing horizontally. So the direction has changed?!

Is it because I am thinking "globally?" For example, if I look at the local coordinate system at the starting point, the vector is pointing in the $\theta$ direction. At the end point (just shy of the North Pole), although "horizontal" now, if I go to the local coordinate system, the vector is still pointing in the $\theta$ direction. In this sense, its direction has not changed. Is this the correct way to view things?

$\endgroup$
1
  • $\begingroup$ A vector pointing down isn’t a vector on the earth’s surface. $\endgroup$
    – Dale
    Nov 28, 2019 at 14:48

3 Answers 3

2
$\begingroup$

There are a couple of things that might help you with getting the intuition behind parallel transporting vectors in general relativity and also differential geometry in general.

First of all, when we have a vector on a manifold, e.g. a vector on a sphere as in your case, the vector has to belong to the tangent space of the manifold at the point where the vector resides. Less technically speaking, the vector cannot point "out of the manifold" so to speak, but it has to stay in the tangent plane of the surface at the point. That means, that at North Pole, there cannot be a vector pointing "down", if we have the sphere as our manifold where the vector resides. In other words, the problem is not so much in thinking "globally", but in thinking in the three dimensional space even though we have a two dimensional manifold.

The point you made last about the vector pointing in the $\theta$ direction is kind of closer to the right mind set, but it does not still actually work. Here's why:

In addition to the transport you specified, you can also first go a quarter round along the equator with parallel transporting your vector. After that, again head towards the Nort Pole and of course parallel transport your vector along it. Just shy of North Pole you are still facing the $\theta$ direction as in the first case. But the thing is, you can go all the way to the North Pole, and even move a little bit towards the starting point, so that you'll end up in the same point where you first parallel transported your vector, but now the parallel transported vector is orthogonal compared to the first case! This highlights an important point in transporting vectors in manifolds: The direction you'll end up with depends on the route of transportation. Note that we also used only geodesics on our second route, and so there is no point on the route where the vector has changed its direction.

It might be that the last thing I pointed out was already known to you and the first one (hopefully) already answered what you were seeking.

$\endgroup$
2
$\begingroup$

The parallel transport is defined completely by the space in question and does not require any information beyond it. In the case of sphere, you have information only about 2 directions that are tangent to the sphere and you don't need to use any information about the third direction pointing "out of the manifold" as @JustSaying explained.

Now, the parallel transport works like this:

You have some direction (vector) that you want to parallel transport along some curve. Therefore, you have also the direction (vector) of the curve at the place you are standing. Since the sphere is 2 dimensional, you can easily measure angle between these two directions (vectors).

Now parallel transporting means, then when you are moving in a very small (infinitesimal - so that the direction of the curve does not change as you are going) distance in the direction you have chosen, fix the direction of the vector you are transporting (that is, fix the angle between the vector and the direction of your movement) so that the angle does not change as you go and move the small (infinitesimal) distance. Congratulations, you are just parallel transported your first vector. Now iterate (infinitely) many times to actually go some longer distance. You can imagine this process to traverse the loop on the sphere that @JustSaying wrote about to intuitively understand what is going on. But don't forget that someone living in the sphere has no knowledge of the third direction and he can only measure one angle between vectors. So be careful not to mix the third direction in when imagining this process from the point of view of this "2-dimensional" person. Also note, that in the 3-dimensional space in which the sphere is embedded, the vector is not parallel transported, because you are changing the second angle, that you, as 3-dimensional being, has information about.

$\endgroup$
1
$\begingroup$

Locally the pendulum is always swinging relative to the local direction of gravity, in other words "down". That's true when you start at the equator and when you get to the pole. The real issue is if you take a closed path back to where you started. Start at (0,0) with pendulum swinging north-south and go north to the pole. Now head directly south to (0,90). Now head directly back to (0,0). When you arrive back you'll find that your pendulum is now going east-west. That's an effect of the curvature of the surface of the Earth.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.