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Levi-cevita parallel propagation on a sphere

  1. May 21, 2012 #1
    Hello all,

    I have been trying to figure out a clear rule for parallel propagating vectors on spheres, such as in the wiki http://en.wikipedia.org/wiki/File:Parallel_transport.png. There seem to be lots of rules that are proposed in this forum and online, but they often don't work well to explain, for example, precession of a vector when parallel propagated around a line of constant latitude.

    Now, for some contenders:

    1. Bill K's solution: https://www.physicsforums.com/showpost.php?p=3402912&postcount=21 is nice, but it doesn't explain it intuitively.
    2. Matterwave's solution:
      This seems nice to make the most sense so far.
    3. Wikipedia solution:
      This makes sense in the `triangle' case, but on second thought, what does "axial rolling" mean? I seem to have a feeling that if your vector is fixed to the north pole axis (so rotates with it), and then rotate the sphere about the "vertical" so that the north pole axis traces out a line of constant latitude, then there will be some rotation about the north pole axis, i.e. precession of the vector. I can't justify that though.
    4. My professor's solution:
      Something along the lines of projecting the vector onto the xy, xz, yz, planes, and transporting them along the projection of the curve there, and then projecting them back onto the tangent plane of the sphere. I suspect this is similar to Matterwaves (and it also has to be infinitesimal), in that the cause of the precession is the rotation of the tangent plane.

    If someone could clarify which is correct, that would be great. I have a feeling all may be correct, but it'd be nice to link them all together. If one is up for a challenge, giving a geometrical explanation of Bill K's solution (in terms of projection onto rotating tangent planes) would have me most satisfied!

    Cheers
     
  2. jcsd
  3. May 21, 2012 #2

    lavinia

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    I think of the geodesic curvature as an instantaneous rotation and a parallel vector field as one that does not rotate. Thus with respect to the curve the parallel vector field must appear to rotate in the opposite direction. It will appear to precess with respect to an observer riding along the curve.

    So really to picture parallel translation, one needs to first picture geodesic curvature. Latitude lines on the sphere have constant geodesic curvature so a parallel vector field will rotate at a constant rate around the tangent. Geodesics do not rotate, so a parallel curve will not precess along a geodesic. Analytically, its angle to the tangent of the geodesic will be constant.

    An Example. Take a circle in the Euclidean plane. It's geodesic curvature is constant. A parallel vector field is just a constant vector field. Since the tangent to the circle is rotating, the constant field along it will appear to rotate in the the opposite direction.

    Generally, the total angular rotation is minus the integral of the geodesic curvature along the curve.
     
    Last edited: May 21, 2012
  4. May 21, 2012 #3
    Hmmm, interesting lavinia, though geodesic curvature is not what I am thinking of. At this stage, geodesics imply that you know the rule for parallel transport, so they can't really be used to define the rule.

    Cheers
     
  5. May 22, 2012 #4

    Ben Niehoff

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    For lines of constant latitude, there is an easy answer: Consider the cone that is tangent to the sphere at the line of constant latitude. Since the cone and the sphere are tangent, they share the same tangent spaces on the curve of tangency. Hence parallel transport along the curve of tangency must be the same on both surfaces.

    But we know how to do parallel translation on the cone, since it is a flat surface. You simply unwrap it onto a plane, and parallel transport along the same curve in the plane. It turns out the total angle of rotation after traveling around a line of constant latitude will be equal to the deficit angle of the tangent cone.

    For parallel transport along general curves on the sphere, consider the sphere to be rolling without slipping or twisting on a plane, such that the desired curve on the sphere is always in contact with the plane. Again, at each moment in time, the plane and the sphere share the same tangent space at the point of tangency. As the sphere moves, its point of tangency will trace out a curve in the plane.

    What remains is to relate the flat connection on the plane to the connection on the sphere. The "no twist" condition is what makes this the Levi-Civita connection. Hence parallel transport on the sphere can be achieved by rolling the sphere as described, and parallel transporting the vector along the curve within the plane according to the flat connection on the plane.

    There is another way to see this: Any small portion of a smooth curve on the sphere can be approximated by a line of constant latitude, and hence parallel transport along that portion can be obtained from the "osculating cone". Unrolling the osculating cone into the plane gives the same result as rolling the sphere on the plane without slipping or twisting.

    Note that this whole operation is equivalent to what Matterwave is describing: Isometrically embed the sphere into some Euclidean space (in this case R^3), parallel transport within the ambient space an infinitesimal distance, and project back onto the sphere (or alternatively, pull back the ambient connection to the embedded sphere).
     
  6. May 22, 2012 #5
    Thanks for the reply Ben, though I admit I'm missing something.

    Well I should have trusted my head, but I just actually cut out a cone to confirm it. Mine was simply a ninety degree cone (as in when laid flat was one quarter of a disk). If I propagate the vector along the edge of the cone when it is laid flat (such that the angle of the vector is the same with respect to the edge of the cone) then when I roll it up, hey presto, the vector is pointing in the same direction the whole way round, so no precession. You do get precession, however, if you keep the vector pointing in the same direction (e.g. North) when the cone is laid flat, so that the angle with respect to the cone edge changes. Then, when you wrap it up, you can see the precession of the vector.

    Is this what you mean?


    I like that description. Though, similarly to above, when the vectors are transported in the tangent plane, they are transported parallel with respect to the tangent plane coordinates, and not by keeping the same angle with the curve (i.e. like "North" above).

    Though this is a little beyond me, I can see where you're going, and I like that there is a mathematical way of formalising it.

    Cheers
     
  7. May 22, 2012 #6

    lavinia

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    If the surface is embedded in Euclidean space then the geodesic curvature is the projection of the ambient geodesic curvature onto the tangent plane.
     
    Last edited: May 22, 2012
  8. May 22, 2012 #7

    lavinia

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    Since parallel transport on the sphere and the cone is parallel transport in 3 space projected onto the tangent planes. But the tangent planes are the same along the latitude line.

    Unwrapping preserves angles and lengths so parallel translation is also the same.


    Can you explain this more?

    Sliding and twisting would create tangential forces on the sphere?No torques? I am thinking of a billiard ball bouncing of the table bumper. If there were a torque then the ball would skid rather than roll backwards.

    Or approximation by a piece of a great circle. Osculating cylinder?
     
    Last edited: May 22, 2012
  9. May 22, 2012 #8
    I also wondered about this twisting. I imagine the plane horizontally flat, and the sphere atop it being rolled along the plane so that he curve on the sphere's surface traces on out on the plane at the point of contact. I think "no twist" in this context means that the sphere is rolled such a way that a point infinitesimally far along the curve on the sphere is brought straight down. I think.
     
  10. May 22, 2012 #9

    lavinia

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    I am still not sure about this twisting but here is a thought.

    For an arbitrary curve on a surface in Euclidean space there is a surface whose tangent planes are the tangent planes to the surface at the points of the curve. This surface is flat since its tangent space depends on only one parameter, the parameter of the curve, and so may be flattened out without distortion. Ben's cone is a special case of this. Then the same argument as with the cone applies. Interestingly the geodesic curvature of the curve is just the regular curvature of its image in the flattened surface.

    This shows why the geodesic curvature is the rate at which the curve turns away from a parallel vector field and why its negative is the rate of precession of the parallel vector field.
    This can be shown by direct calculation as well.

    If t is the unit tangent to the curve, n the unit normal, and y is the parallel vector field - assume its length is 1 then the inner product,<t,y>, is the cosine of the angle,[itex]\theta[/itex], between t and y

    and its derivative with respect to t is k<n,y> = ksin([itex]\theta[/itex]) where k is the geoesic curvature.

    So -sin([itex]\theta[/itex])d[itex]\theta[/itex]/dt = ksin([itex]\theta[/itex])

    so d[itex]\theta[/itex]/dt = -k
     
    Last edited: May 22, 2012
  11. May 22, 2012 #10
    If I followed this right, I think I'd need a bit more evidence. Firstly, if the curve is on a sphere, then the same sphere is a surface whose tangent planes coincide with a sphere at all points of the curve. But the sphere is still curved. So there must be something special about this other surface you are looking for that I missed.

    Depending only on one parameter? I'm uneasy about this, as it depends in some way on the coordinates of the initial surface as well, and ignoring these ignores the original surface completely. Maybe that is your point? Anyway, the curve on the sphere depends only on one parameter, and the surface of the sphere is curved.

    OK, it turns out the surface of a cone is not intrinsically curved, which caught me off--guard. The reason would appear to be that a direct mapping can me made from the surface of the cone to the plane, and hence it is flat. (Or is it that the cone can actually be made from a sheet of paper, which is stronger than just "mapping"?) I gather the problem with a sphere is that it takes two charts to do so (due to the poles as such). (Aside: If I stuck two cones bases together and curved the join a little bit so it is smooth, am I right in saying this would not be flat, as it can be mapped to the sphere?)

    I think I now know what you mean lavinia, and I'll use the sphere as an example. For an arbitrary curve on the sphere, one takes the tangent space at each point as a little square plane tangent to the surface, and then sticks all these together to get a strip of paper. Since it is only a piece of paper, it is flat. Whether this can always be done, I am not sure.

    Lastly, I just found an image in Arnold of the cone/sphere situation, and it looks like the way I described it was alright:

    http://books.google.co.nz/books?id=...Qeq9ITODw&ved=0CFMQ6AEwBA#v=onepage&q&f=false

    (Hopefully that link works --- page 302.)
     
  12. May 22, 2012 #11

    Ben Niehoff

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    Yes, that's what I mean by "no twisting". The sphere should not rotate about the axis normal to the tangent plane.

    It has nothing to do with mapping the cone to the plane. It has to do with mapping the cone to the plane isometrically, which is the same as saying the cone "can be made from a sheet of paper".

    The cone also requires two charts to cover it.

    As I said, it has nothing to do specifically with mapping to the sphere. However, you know that if a surface has the same topology as a sphere, it must have curvature somewhere.

    The regions of your surface that are still exactly conical are going to be flat. The small region where you smoothed out the join between the cones is going to be curved. Also, the vertices of the cones can be considered to have infinite curvature.

    I am skeptical of Lavinia's assertion that a zero-curvature surface (or at least, a piece of a surface) can be found which is tangent to a sphere along any arbitrary curve in the sphere. Any proof of that?
     
  13. May 23, 2012 #12
    Thanks for clarifying things Ben. I'll look into isometric mappings, though I am guessing from the name this means lengths and angles are preserved.

    My concern was that if you tried to lay a strip of paper on an arbitrary curve you would have to bend it in the plane of the paper, causing it to crinkle as such. But no, you just cut your strip of paper with a bend, and it should (?) work. Curves crossing themselves could create problems etc.

    Taking a different approach, I suspect one could cut a strip of paper in any squiggly configuration one wanted and would be able to lay it on a sphere (?). While this reverse proof will not do, it is fun to think about. I now await a mathematical proof, though I suspect there may be a Feynman like hand waving one out there.
     
  14. May 23, 2012 #13
    Actually, I am more convinced one should be able to lay such a strip of paper for any arbitrary curve on an arbitrary surface. If you take two closely separated points on a sphere, the strip can either lie tangent to the sphere, or at an angle (i.e. so only one edge is touching it). If at an angle, then just shorten the upper edge a bit to bring it back down to tangent, by taking a wedge shaped slice out of the strip (fat bit of wedge at edge to be shortened). This is hard to explain without pictures, but in this way I think one could create a strip that would have intersecting tangent planes with the sphere at all points on an arbitrary curve.

    The question is whether the parallel transport can be modeled in the same way as for the cone example.
     
  15. May 23, 2012 #14

    lavinia

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    I said this intuitively generalizing the cone on the latitude line. At any point on a smooth curve on a surface there is a line perpendicular to the curve and tangent to the surface. These perpendiculars together with the curve itself seem to span a surface in an open segment of the curve around the point.

    This surface has a straight line through every point - just like the cone - and so is flat.

    Near a point where the geodesic curvature is not zero the surface is parameterized as c(t) + sn(t) where n is the unit normal to the curve on the surface.
    So what about points where the geodesic curvature is zero?

    Here is a try.

    If the curve is an embedded submanifold of the surface then the exponential map in the normal direction maps an open subset of the tangent space along the curve diffeomorphically onto a tubular neighborhood of the curve on the surface. Similarly, in Euclidean space it maps an open subset of the tangent space along the curve onto a solid tube around the curve. The intersection of these two sets in the tangent space along the curve is mapped diffeomorphically onto a surface by the exponential map in Euclidean space - which is just moving along the straight lines normal to the curve.

    If the curve is piecewise smooth then this surface will have a crease. This doesn't seem to prevent the same argument from working.

    I still don't see why no twisting and sliding is the condition that makes the connection a Levi-Civita connection.
     
    Last edited: May 23, 2012
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