19
$\begingroup$

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

$\endgroup$
  • $\begingroup$ You have merrily changed the vector direction. Please consult more books under Levi-Civita. $\endgroup$ – Narasimham Nov 9 '17 at 17:52
12
$\begingroup$

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.).

$\endgroup$
  • $\begingroup$ The picture is not accessible as the link is broken $\endgroup$ – binaryfunt Nov 7 '16 at 17:37
  • 2
    $\begingroup$ @binaryfunt: Thanks for pointing it out! I found the new address and have updated the link. $\endgroup$ – Hans Lundmark Nov 7 '16 at 18:34
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Arturo don Juan Sep 27 '15 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 '15 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? $\endgroup$ – Arturo don Juan Sep 27 '15 at 20:35
  • 2
    $\begingroup$ 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. $\endgroup$ – Ted Shifrin Sep 28 '15 at 0:37
4
$\begingroup$

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.)

$\endgroup$

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.