How to intuitively understand parallel transport

Question

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.

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

Answer 1

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.).

Answer 2

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.

Answer 3

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.)

Answer 4

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.

Answer 5

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.

Answer 6

I also struggled with this for awhile, but here's a nice example on the sphere. First, you need to believe that I can make a $270^\circ$ triangle on the sphere, where this triangle is gotten by taking the arcs of three great circles. For example, let us take the standard $S^3:=\{(x,y,z)\in \mathbb{R}^3:x^2+y^2+z^2=1\}$, then the three arcs $\gamma_{x},\gamma_y,\gamma_z$ arcs lie respectively on the $x=0,y=0,z=0$ planes. Once you're convinced of this, suppose the north pole $N$ is the intersection of $\gamma_y,\gamma_z$. We parallel transport along this triangle from $N$ to $N$ (so topologically, this is a loop) the unit tangent vector of $\gamma_y$. Let's say we orient ourselves by first taking the $\gamma_y$ loop, then $\gamma_x$ then $\gamma_z$. Transporting along $\gamma_y$ preserves tangency. But since $\gamma_y$ is perpendicular to $\gamma_x$, now this vector which was previously tangent to $\gamma_y$ now becomes normal to $\gamma_x$. Transporting along $\gamma_x$ keeps the vector normal. Finally transporting along $\gamma_z$ preserves tangency. However, note that you've accumulated an extra $90^\circ$ once you're back at $N$.