34
$\begingroup$

In the article I've referenced below, and many other articles for that matter, the notion of parallel transport along a line of latitude $\theta=\theta_0$ on the unit 2-sphere is spoken about. What I can't understand is, intuitively, how there is any "rotation" after circuiting the full line-of-latitude path.

enter image description here

The above illustration is my understanding of what is meant by "parallel transport of a vector along the $\theta$ or $\phi$ direction". Clearly, after completing a full path, the rotation of the vectors is always exactly $2\pi$, and hence the parallely transported vectors are left unchanged. What have I misunderstood? My understanding doesn't seem to match with the results derived in the article below.

Reference: http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/Geodesics.pdf

$\endgroup$
1
  • $\begingroup$ You have merrily changed the vector direction. Please consult more books under Levi-Civita. $\endgroup$
    – Narasimham
    Nov 9, 2017 at 17:52

5 Answers 5

22
$\begingroup$

I was also having trouble with this for a long time. The explanation which finally worked for me was the following:

For the purposes of parallel transport along a particular circle of latitude, the sphere can be replaced by the cone which is tangent to the sphere along that circle, since a “flatlander” living on the surface and travelling along the circle would experience the same “twisting of the tangent plane in the ambient space” regardless of whether the surface is a sphere or a cone.

And for the cone, there's an easy way of seeing that there is indeed a rotation of the transported vector with respect to the tangent vector of the curve: just cut the cone open and lay it flat on the table, so that parallel transport becomes simply ordinary parallel transport on the plane.

A picture says more than a thousand words, and I found a good one here: A simple discussion of the Berry Phase (N. P. Ong, Physics, Princeton Univ.).

$\endgroup$
7
  • $\begingroup$ The picture is not accessible as the link is broken $\endgroup$
    – binaryfunt
    Nov 7, 2016 at 17:37
  • 6
    $\begingroup$ @binaryfunt: Thanks for pointing it out! I found the new address and have updated the link. $\endgroup$ Nov 7, 2016 at 18:34
  • $\begingroup$ In the example of the cone laid out flat, parallel transport is defined as transport in which $\vec v$ never rotates about the normal-axis $\hat n$. In the 2-dimensional plane, $\hat n$ always points in the upward direction, so $\vec v$ literally never changes direction. However if we go back to the original 3-D trajectory where $\hat n$ is changing, parallel transport cannot be defined as "not rotating about $\hat n$" because then $\vec v$ couldn't have changed direction after going all the way around. In light of this, how does transport along the cone relate to the sphere? $\endgroup$ Nov 24, 2021 at 20:44
  • $\begingroup$ Also, in relating the laid-out cone to the line-of-latitude on the sphere, the angle between the vector and tangent at the end of the trajectory is identified as the ultimate "change in angle". But clearly, at any intermediate point along the path, we cannot say the same. For example, along a great circle a vector initially parallel to the trajectory will remain parallel throughout, however in the cone analogy the vector will rotate $2\pi$ with respect to the tangent. How do we resolve this? I suspect it has something to do with the fact that the beginning and end points are identified. $\endgroup$ Nov 24, 2021 at 20:53
  • 1
    $\begingroup$ @ArturodonJuan: First comment: I don't understand what you're talking about. What does $\hat{n}$ have to do with this? Second comment: For a great circle, you don't get a cone, but a cylinder. $\endgroup$ Nov 25, 2021 at 7:29
11
$\begingroup$

The red and blue vector fields in your picture are not parallel along the pink curve. One way to see this is to note that you can compute the covariant derivative of a vector field along a curve in the sphere by computing its ordinary derivative in $\mathbb R^3$, and then orthogonally projecting that onto the tangent plane. At any point on the pink circle, the ordinary derivative of the blue vector field points toward the center of the pink circle. Since that is not orthogonal to the tangent plane, its orthogonal projection onto the tangent plane is nonzero.

$\endgroup$
4
  • $\begingroup$ Wait but doesn't the ordinary derivative of the red vector field (pointing along line of latitude) point towards the center of the sphere, which is orthogonal to the tangent plane at any point along the pink curve? $\endgroup$ Sep 27, 2015 at 20:18
  • $\begingroup$ @ArturodonJuan No. The ordinary derivative of the red vector field is a horizontal vector for every point. $\endgroup$
    – Aloizio Macedo
    Sep 27, 2015 at 20:26
  • $\begingroup$ Hmmm... I really don't understand how to visualize this parallel transport along a line of latitude on a 2-sphere. How can I change my picture to illustrate this correctly? $\endgroup$ Sep 27, 2015 at 20:35
  • 3
    $\begingroup$ Note that the projection of the derivative of the blue vector field onto the tangent plane of the sphere heads "uphill" (or to the left), and so to parallel translate we must rotate it to the right to compensate. Feel free to look at my differential geometry text for a discussion of this and various other concrete examples. $\endgroup$ Sep 28, 2015 at 0:37
7
$\begingroup$

One way to see that the blue vector field is not parallel is by noting that it is the vector field corresponding to velocity. If it were parallel, the red circle would be a geodesic. But geodesics in the sphere are the great circles.

(Actually, this is the same thing that Jack Lee is saying, but phrased differently.)

$\endgroup$
4
$\begingroup$

Hilbert and Cohn-Vossen in their “Geometry and Imagination” give the example of driving a small car on the globe. You never have to turn the front wheel from the straight position only if you’re driving on a great circle of the globe. The blue vectors need to turn to compensate for the wheel turning required on the latitudinal drive.

$\endgroup$
1
$\begingroup$

Let's try to understand this intuitively . Let's say you are sitting in a car somewhere very close to the north pole, may be just 50 meters from the pole. That means you are at very high latitude, probably around 89.9 degrees north. The pole is marked by a flag that you can clearly see.

Now if I ask you to circle the pole in your car, first you will need to turn at right angles from the pole. Then , as you start circling the pole keeping the flag in sight, you can clearly feel that you will need to constantly turn the steering wheel towards the pole to be able to drive in a circle. If you try to drive straight, you will drive away from the pole.

Now I ask you to repeat the same experiment when you are sitting in your car at the equator, which is 0 degree north. If I ask you to circle the north pole, you can intuitively feel that all you need to do is to keep the steering wheel straight, and you will be able to circle the pole even when driving perfectly straight.

So what is different between these two situations, apart from the latitude which is just a number. If you think closely about these two situations, you will arrive at the solution.

When you are walking along the smaller red circle, you are NOT walking in a straight line. OTOH, when you walk along the equator, you are walking in a straight line.

One way to understand this is following. Assume the sphere in your picture is earth and the smaller red circle is a latitude line. You are walking along this latitude. At any point on this latitude line you can imagine a tangent plane. The normal to the tangent place is line from the center to the earth to the point where you are standing. In fact your upright posture IS the normal to the tangent plane, because you feet point directly to the center of the earth.

Now, when you walk/run/drive along this latitude line, you experience a centrifugal acceleration which will be along a horizontal direction, i.e. from the center of the latitude circle to the point where you are standing. So, as you can see, there is a non-zero angle between the centrifugal acceleration vector and the tangent plane's normal vector. It is this non-zero angle that determines that you are not walking in a straight line.

When you walk along the equator, (or any great circle on earth for that matter), this aforementioned angle vanishes. That is why walking on any great circle on a spherical surface is considered walking in a straight line on a curved 2-D surface embedded in an ambient 3-D surface.

When you drive along a high latitude, the tangent plane of the instantaneous location constantly tilts towards the north pole.

$\endgroup$

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.