Confusion about parallel transport

[stevendaryl]

Could you explain this further? What is the source of the difference between the two answers?

Well, the best example is the one from A.T. about parallel transport on a globe. I'll use that example to illustrate the problem.
Suppose that we want to parallel-transport a vector around the following path:

1. Start at the North Pole and go straight south along the line of 0^o longitude until you get to the equator.
2. Go straight east along the equator until you get to the line of 90^o longitude.
3. Go straight north until you get to the North Pole.

Note that this is an "equilateral triangle" of sorts, because all 3 legs of your journey are the same length, 1/4 of the circumference of the globe. But unlike a triangle in flat 2D space, the angles are not 60^o but 90^o. In the figure below, on the left is shown the journey on a globe. I took a narrow strip centered on the path and "opened it up" to flatten it. The flattened version is shown on the right. Note that in order to flatten the path, I had to break one of the legs of the triangle. I chose to break the last leg, the journey north along the line of 90^o longitude.

Now, my claim is that parallel transport within the strip on the left is the same (at least approximately) as parallel transport within the flattened strip on the right. So within the strip, you can just use Cartesian coordinates to define parallel transport: just keep the components of the vector the same as you march around the strip.

However, one section of the strip is broken--the trip along the line of 90^o longitude. If you start off at the North Pole, there are two routes to get to this section: You can either travel south along 0^o longitude, then east along the equator, and then north along 90^o longitude. Or alternatively, you can just go south along 90^o longitude. The orientation of that section of the path if you go along one route is rotated 90^o with respect to the orientation of that section when you go along the other route.

So that's a way to think about path dependence of parallel transport in curved space. You could take a path and flatten it out, and just use Euclidean notion of parallel transport, but in order to flatten it out, you have to break some sections of the path. At those breaks, the orientation of that section of the path is different, depending on how you get to it. If the space is curved, then there is no way to flatten it without breaks, and the breaks imply path-dependence of orientation.

 

[TSny]

Wouldn't you have to normalize your vector at some point? If you keep projecting a vector on some surface that isn't in the plane of the vector, the projection length will be shorter than the original vector, correct?

To get some feel for why the length of the vector doesn't change during parallel transport, you can look at a simple example of transporting a vector around a circle.

Suppose you start with a vector of length $V_0$ that is tangent to the circle at point $p$. Slide the vector to point $p’$ such that it remains parallel to it’s original direction to produce the purple vector of length $V_0$. Then project this vector onto the tangent at $p’$ to produce the “transported" vector of length $V$. If $\epsilon$ is the angle subtended by arc $pp’$, then the length of the transported vector will be $V = V_0\cos \epsilon$.

If you were to let $\epsilon = \pi/4$, then you could get around the circle in 8 steps. Each step would reduce the length by a factor of $\cos \pi/4$ so that the length after returning to $p$ would be reduced overall by a factor of $(\cos \pi/4)^8 = 1/16$.

But, we want to consider very small steps. For small $\epsilon$ the number of steps to get once around the circle is about $2\pi/\epsilon$. So, the vector will be reduced overall by a factor of $(\cos \epsilon)^{2\pi / \epsilon}$. The limit of this expression as $\epsilon$ goes to zero is 1. So, the vector retains its length when you take "infinitesimal steps".

 

[WWGD]

From one of my science idols John Baez, (paraphrase).
Your Euclidean coordinates are only valid within a manifold chart. So think of a runner carrying a torch that s/he keeps at a fixed angle with respect to his body, while running along the loop. Maybe to simplify, say you keep your arm perpendicular to your body at all times, and then run along the loop (this would be the equivalent to the formal $\nabla^X_{\gamma'(t)} =0$, i.e., the vector field $X$ is parallel to the curve in the connection given by $\nabla$. Then doing a full loop around a space with non-zero curvature would not return you to the vector you started with. Maybe you can use a basketball with great circles drawn , and slide something like a tooth pick starting at one of the poles , go along a great circle, then when you hit the equator, travel around a segment of the equator, and then take a segment of a great circle back to the pole. The change between the initial and final positions is given by the holonomy group; it is a group of matrices (under composition).

 

[pervect]

I do know this superficially, but I am still unclear about how 'parallel' is defined on a curved surface.

Sorry for the late response - hope it still gets read.

I'd say that the simplest exact definition of defining parallel transport of a classical vector on a curved surface is the geometric construction via Schild's ladder, which I alluded to earlier with a wiki reference. It's worth a look if you haven't looked it up. Possibly, though, you might need a less simple definition, which includes some discussion of how a connection is defined. Unfortunately the less simple definition is - less simple. Hopefully it will help you, and you might need the more abstract definition of a connection to fully understand what's going on anyway. Unfortunately, you'll probably need a textbook and some study to really understand it, I suspect this short post won't do the job.

In GR, basically, we consider fully general connections only briefly, and then we look for the specific connection that has some desirable properties of preserving the metric, and use only that connection. In the Berry case, your connection is gauge dependent, so it doesn't really have any physical properties. Because of the gauge degree of freedom it's not even uniquely defined.

Imagine we have a sphere, and we have on the sphere two points p, and q, that are nearby. Point p has a tangent plane, and so does point q. They are planes because we are talking about a sphere as an example, if we were talking more generally we'd be talking about "tangent spaces" instead, where the space is of arbitrary dimension.

We also have some curve that connects p and q.

If we cheat a lot and imagine that when p and q are close and we can represent the displacement as a vector, then we can see that we have some linear map to the tangent space at Q from the tangent space at P and the displacement vector that is some sort of rank 3 tensor.

This tensor is called the connection, and is usually written in tensor notation something like $\Gamma^i{}_{jk}$

If we don't cheat a lot, we might introduce coordinates on the manifold, define the basis vectors at P and Q in terms of the coordinates (a so-called coordinate basis), and then write a differential equation that is satisfied along the curve, and that differential equation defines what it means by parallel transport along the curve terms of the specified connection.

That non-cheating approach is what is actuallly in my textbook, winds up as a differential equation for the components of the transported vector (the vector is x, with components $x^\mu$)

<br /> \frac{dx^\nu}{dt} + \sum_{\mu\lambda} t^\mu \Gamma^\nu{}_{\mu \lambda} x^\lambda = 0<br />

where t is the tangent vector to the curve, the curve should be imagined as being parmeterized so that it's tangent vector is a constant length, then $t^\mu = \frac{dr^\mu}{dt}$

Where does the connection come from? At this point we haven't said, it's just a general idea that we have one. In GR we demand that the connection have certain properties, that of preserving the length of vectors, and the dot product of vectors, and we go from a general arbitrary abstract idea to a more physical one. But as I mentioned, in the case of Berry phase, my references say the connection is gauge dependent, so it was never really physically significant in the first place.

[add]Since this got long, I'll recap. A connection defines a linear mapping from one tangent space (at P to Q on our example of a sphere) given a curve that goes from P to Q. Unlike Euclidean geometry, the mapping depends on the path you specify from P to Q, given the two points and a curve, you construct a map, in general a different curve connecting the same two points will generate a different mapping from the tangent spaces.

 

[emily1986]

So that's a way to think about path dependence of parallel transport in curved space. You could take a path and flatten it out, and just use Euclidean notion of parallel transport, but in order to flatten it out, you have to break some sections of the path. At those breaks, the orientation of that section of the path is different, depending on how you get to it. If the space is curved, then there is no way to flatten it without breaks, and the breaks imply path-dependence of orientation.

Thanks. This example really clicked with me. I believe I have some idea of what's going on now. You could take a flattened maps of the world, like one of these,

and then drew a series of parallel vectors across some path. When the map is flattened, the vectors are all parallel. When you fold up the map, the vectors are not longer parallel. I did this on a piece of paper and then made a cone. It became very obvious why the vector length stays the same and why the angle of the vector changes.

dxνdt+∑μλtμΓνμλxλ=0

So the first term in this DE keeps track of how the vector is changing with respect to our tangent plane, and the second term transport us from one tangent plane to the next? I'm assuming that the non-homogeneous version of this DE would be non-parallel transport then? Thanks.

 

[pervect]

So the first term in this DE keeps track of how the vector is changing with respect to our tangent plane, and the second term transport us from one tangent plane to the next?

Yes. In flat space with a cartesian coordinate system, for instance, the connection coefficients are all zero, therefore the condition of parallel transport is that the vector components don't vary at all.

I'm assuming that the non-homogeneous version of this DE would be non-parallel transport then? Thanks.

I would say that if the equation isn't satisfied, the vector is not being parallel transported by the connection. I'm not quite sure if this is equivalent to what you meant. For instance if we imagine a Cartesian coordinate system where all the connection coefficients are zero, but the vector components varied (were not constant), then the vector would not be being parallel transported by the chosen connection.

 

[emily1986]

I would say that if the equation isn't satisfied, the vector is not being parallel transported by the connection. I'm not quite sure if this is equivalent to what you meant.

After thinking about it, my question didn't really make sense. Thanks for your help, everything is much clearer now.