Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parallel transport on sphere

  1. Jul 14, 2011 #1
    If you walk at constant latitude with your arm always sticking towards the North pole, is that parallel transport of your arm?

    The equations don't seem to say it is.

    The vector field would be [itex]\vec{V}=V(\theta)e_\theta [/itex]. The component of the vector only depends on theta, but at constant latitude theta is constant so the component of V doesn't change as you circle the earth. [tex]e_\theta=(cos \theta cos \phi, cos \theta sin \phi, -sin \theta) [/tex]
    [tex]\partial_\phi e_\theta=cos \theta(-sin \phi, cos \phi, 0) [/tex]
    So [itex]D\vec{V}=V(\theta)cos(\theta) (-sin \phi, cos \phi, 0) d\phi[/itex]. This is only zero at the equator. So is the vector field [itex]\vec{V}=V(\theta)e_\theta [/itex] not parallel transported at constant latitude?

    I understand that you can integrate the above equation to get:
    [itex]\vec{V}(\phi_2,\theta)-\vec{V}(\phi_1,\theta)=V(\theta)cos(\theta) [(cos \phi_2, sin \phi_2, cos \theta)-(cos \phi_1, sin \phi_1, cos \theta)][/itex].

    Geometrically [itex]V(\theta)cos \theta[/itex] is the projection of your arm perpendicular to the z-axis (or parallel to the z=0 plane), and the difference [itex]\vec{V}(\phi_2,\theta)-\vec{V}(\phi_1,\theta)[/itex] can be attributed to the fact that the vector field rotates around in a circle about the z-axis.

    But wasn't parallel transport [itex]D\vec{V}=0 [/itex]
  2. jcsd
  3. Jul 14, 2011 #2
    Without going into the maths, the answer is no. If you are following a line of latitude (but not on the equator), your arm is constantly rotating relative to the path you are following or your heading. Bill K touched on this in post #21 in a similar thread here: https://www.physicsforums.com/showthread.php?t=514004&page=2

    At a latitude near the North pole the rotation rate of the vector is almost zero and at the equator it is exactly zero. Anyone care to work out at which latitude the rotation rate of the vector (per circumnavigation) is the greatest. It appears to be somewhere near the Equator.
    Last edited: Jul 14, 2011
  4. Jul 14, 2011 #3
    Then your 'vector' is gong to rotate, parallel in this case doesn't mean keep the same angle with geodesic, but really parallel in cartesian coordinates.
  5. Jul 14, 2011 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    It's not so much that your arm is rotating relative to the path, but that the path itself isn't straight. A geodesic parallel transports itself, but a line of constant lattitude isn't a geodesic.

    Thus, a vector pointing along the path of a line of constant lattitude isn't parallel transported by said path.
  6. Jul 15, 2011 #5
    If you are traveling in a circle on a circle of latitude, and always pointing your arm North, then doesn't your arm always make the same angle with respect to your path (90 degrees)? Isn't that the pictorial definition of parallel transport?

    The parallel transport equation is [itex]DV^i=0 [/itex] along a path for each coordinate i.

    This should be the same as saying [itex] u^k V^{i}_{;k}=0[/itex] all along the path, where u is the tangent vector of the path.

    As was pointed out in one of the links, there seems to be only one field that is parallel transported on circles of latitude based on the two more mathematical descriptions of parallel transport above, and it's a weird one, as it involves a field that rotates at a different azimuthal period from 2pi. In general, how many different types of vector fields are parallel transported by a given closed curve?
  7. Jul 15, 2011 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    I would agree that if your arm is pointing north, and you're traveling due east, your arm is maintaining a right angle to your path.

    It's also obvious that a circle of constant latitude is not a geodesic.

    Before I say more, I want to re-do some of the calculations for myself, not sure when I'll get the time...
  8. Jul 15, 2011 #7

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    As others have said, the key point is that a circle of constant latitude is not a geodesic. So, the path taken rotates with respect to a local orthonormal frame. Parallel transport really computes parallel transport, not "constant-angle-with-respect-to-my-heading" transport. So, since the tangent vector along the path is not itself being parallel transported, any vector that IS being parallel transported can't be expected to maintain a constant angle with respect to it.

    This is the same principle by which Foucault's pendulum works...the direction in which the pendulum swings is parallel transported around a line of constant latitude as the Earth rotates, causing the pendulum to slowly change direction.

    Incidentally, if you complete a closed loop, you will find that the parallel-transported vector does not return to itself, but ends up slightly rotated. The amount by which is it rotated turns out to be equal to the area, measured in steradians, enclosed by the loop. This is called "holonomy".
  9. Jul 15, 2011 #8
    When you say that a (nongeodesic) path rotates with respect to a local orthonormal frame, do you mean that you parallel transport the tangent vector 'u' of the path from point A of the old frame to the orthonormal frame of the new point B via adding: [itex] -\Gamma^{i}_{jk}u^j dx^k [/itex], and compare this result to actual tangent vector u' at point B, then the difference is a rotation, i.e., the quantity [tex]u'^{i}-(u^i-\Gamma^{i}_{jk}u^j dx^k)=DU^i[/tex] along the actual path between A and B is a measure of the rotation with respect to a local orthonormal frame (I'm assuming you mean the local frame at point B, after the value at A has been parallel transported to B for comparison)? So you're calling the covariant derivative a rotation?
    Last edited: Jul 15, 2011
  10. Jul 15, 2011 #9
    Keeping your arm or vector at the same angle to the path is not the pictorial definition of parallel transport (although it happens to be true only if you are following a great circle or geodesic). This was covered to some degree in this other thread: https://www.physicsforums.com/showthread.php?t=514004
    Although a vector is not parallel transported along a line of constant latitude by the path itself, would you agree that it is possible to parallel transport a vector along a line of constant latitude? If so, would you also agree that if a vector is parallel transported along such a path, that the vector would rotate relative to the path? For reference see section 2.1 (parallel transport can always be achieved) of this paper http://www.ocf.berkeley.edu/~adriand/classes/files/m140/17.pdf. i.e parallel transport of a vector along a path does not only apply to geodesics. However, a vector can only remain parallel to a given path if and only if the path is a geodesic.
    This is interesting but I am a little bit confused because the closed loop around the equator would enclose a larger area than a loop at any other constant latitude and yet the vector has zero rotation at the equator. How does that work out? Is it because at the equator the vector is considered to rotate through an angle that is a multiple of 360 degrees? Also, I would still be interested if anyone knows how to answer the question below that I posed in an earlier post:
    Last edited by a moderator: Apr 26, 2017
  11. Jul 15, 2011 #10

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    The angular area enclosed by the equator is [itex]2\pi[/itex].

    It depends what you mean by "largest angle", because of course any angle greater than [itex]\pi[/itex] is smaller than [itex]\pi[/itex] when measured in the other direction.
  12. Jul 15, 2011 #11


    User Avatar
    Staff Emeritus
    Science Advisor

    That's what I wanted to calculate for myself. And, I get the same answer everyone else did. Actually - I didn't check the magnitude for the amount of the rotation that others gavie, I just found that it was non-zero, as it was a very quick calculation.

    The covariant derivative of the metric preserves the dot product of two vectors, so if you have a set of orthogonal vectors, they remain orthogonal after the parallel transport.

    Obviously, "north" is orthogonal to "east".

    When you parallel transport these basis vectors along the non-geodesic path, what you get is that your new basis vectors are rotated. So you have some new basis vector that is no longer pointing along your curve, it started out pointing "east" but got rotated (you can imagine it picks up a bit of a South component, though this depends on which direction you're going.). And similarly, your "North" vector, after being parallel transported, gets an East component. But your new vector pair is guaranteed to be orthogonal after the parallel tranpsort if it was orthogonal before the parallel transport.

    So the simple method of defining parallel transport as "keeping the angle constant" which was popularized by Baez (as I recall), works if you are following a geodesic path (which is what he mentioned, I'm pretty sure), but doesn't really serve as a general definition of parallel transport.

    I tend to think of Parallel transport as being defined by Schild's ladder. But that demands that you use straight measuring rods (geodesics). The idea is rather similar to those drafting instruments one sees to draw parallel lines, it's based on the principle that if you draw a parallelogram where the opposite sides have equal length, the resulting lines are parallel. If you apply the same principle to a curved space-time, you'll also define parallel transport, where the "rigid rods" of your drafting instrument are mathematically defined by the length along a geodesic path.

    About the only reference I've ever seen that discusses Schild's ladder is a few usenet posts, and MTW"s gravitation. It's a shame, it's a rather interesting geometric construction. It actually doesn't need a metric space (though I"ve presented it that way by talking about the lengths of sides of parallelograms) - it will work in an affine geometry.

    YOu might ask - how does Schild's ladder define parallel transport along a non-geodesic path, like the example? The answer is pretty simple. You find a geodesic, that goes through the same point in the same direction, as your non-geodesic path. For the case at hand, you find a great circle that is locally the same as your line of constant latitude.

    The deviations of said geodesic from your path will be second order. So you just parallel transport over a small distance using the geodesic, ignoring the second order difference, and take the limit for small distances. The second order terms dissapear.

    So, the geodesic great-circle path will be close to the line of constant lattitude with only delta^2 discrepancies, but it will give you in the limit the "rate of rotation" defined relative to the n/s/e/w vector field that we normally use on a sphere for the non-geodesic path, which you take in the limiting case as a large number of different nearly-geodesic segments stitched together.
    Last edited: Jul 15, 2011
  13. Jul 15, 2011 #12


    Staff: Mentor

    So, the metric on the unit sphere is:
    [tex]ds^2 = d\theta^2 + \sin(\theta)^2 d\phi^2[/tex]

    The path, x, around a latitude line parameterized by longitude is given by
    [tex]\mathbf{x} = (\theta_0, \phi)[/tex]

    We will parallel transport an arbitrary vector, v, along the path, x
    [tex]\mathbf{v}(\phi) = (v^{\theta}(\phi), v^{\phi}(\phi))[/tex]

    Applying the parallel transport equation we get the following system of differential equations:
    [tex]v^{ \phi } ( \phi ) \sin ( \theta ) \cos ( \theta )= \partial_{\phi}v^{ \theta } ( \phi )[/tex]
    [tex]v^{ \theta } ( \phi ) \cot ( \theta )+ \partial_{\phi} v^{ \phi } ( \phi ) = 0[/tex]

    Solving those differential equations gives
    [tex]\mathbf{v}(\phi)=\left(c_1 \cos (\phi \cos (\theta ))+c_2 \sin (\theta ) \sin (\phi \cos (\theta
    )),c_2 \cos (\phi \cos (\theta ))-c_1 \csc (\theta ) \sin (\phi \cos (\theta

    For parallel transport completely around the line of latitude we start with phi=0 and end with phi = 2 pi
    [tex]\mathbf{v}(2\pi)=\left(c_1 \cos (2 \pi \cos (\theta ))+c_2 \sin (\theta ) \sin (2 \pi \cos (\theta
    )),c_2 \cos (2 \pi \cos (\theta ))-c_1 \csc (\theta ) \sin (2 \pi \cos (\theta

    So the cosine of the angle between v(0) and v(2pi) is given by
    [tex]\frac{g_{\mu\nu} v(0)^{\mu} v(2\pi)^{\nu}}{\sqrt{g_{\mu\nu} v(0)^{\mu} v(0)^{\nu}} \sqrt{g_{\mu\nu} v(2\pi)^{\mu} v(2\pi)^{\nu}}} = \cos (2 \pi \cos (\theta ))[/tex]

    Which is equal to -1 for [itex]\pi/3[/itex] and, as you say, equal to 1 for 0 and [itex]\pi/2[/itex]
    Last edited: Jul 15, 2011
  14. Jul 17, 2011 #13


    User Avatar
    Staff Emeritus
    Science Advisor

  15. Jul 17, 2011 #14
    Thanks for that excellent and skilful analysis Dalespam. Cheers :D
  16. Jul 17, 2011 #15
    As I understand it, Schild's ladder is an approximation of parallel transport that gets more accurate in the infinitesimal limit. I have a similar approximation that seems to work for a sphere and is perhaps more visual. First draw a point on a curve. Place a small disk that is tangential to the surface at that point and draw a vector on the disk. Now imagine a plane that passes through the centre of the sphere that the vector lies on. Draw a second point on the curve that you want to parallel transport the vector to. Another small tangential disk is centred on that second point. A copy of the original plane is moved so that it passes through the second point while remaining parallel to the original plane. This copied plane does not pass through the centre of the sphere. Where this plane intersects the second tangential disk is the position of the parallel transported vector. To continue the process the second plane is rotated using the second vector as an axis of rotation until it passes through the centre of the sphere. A further copy of this reorientated plane is then moved in such a way that it remains parallel to the first copy until it intersects the next point on the curve and so on. The smaller the steps the more accurate this approximation becomes.
  17. Jul 17, 2011 #16


    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, Schild's ladder is an approximation.

    One important difference between Schild's ladder and your construction, assuming I've understood it without a diagram, and assuming it works, is that your diagram makes reference to places that are not on the surface of the sphere (i.e. the center of the sphere).

    This is in my view an important difference, because when we are envisioning curved geometry, we usually want to take the role of a flatlander, exploring the geometry around himself and finding out that it's curved, without making any references to "extra-dimensional spaces" that it might or might not be embedded in.

    In short, we are exploring the geometry from a viewpoint within the geometry, without necessarily imaging there is something outside the geometry. (We can imagine such things if we like, but we'd prefer not to rely on them for calculations). So when we analyze curved space-time, we'd like to think about how to determine it's curvature with measurements we can actually make, measurements that don't require us to stand "outside" of space-time.

    Also, it's not clear how you would define "the center of the sphere" in the case of a general curved surface which might not be a sphere. Schild's ladder will serve as a suitable geometric construction on any curved surface, and it will be constructed entirely within said curved surface.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook