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Vector parallel to our instantaneous direction of travel

  1. Jun 22, 2005 #1
    "If we trace out a close path on a sphere, requiring that we always hold some vector parallel to our instantaneous direction of travel, at the end of our trip, the vector will no longer point in the same direction as it did at the time of departure."

    I did a small test with what he claims. But it wasn't true at all.
    Did I miss something?
     
  2. jcsd
  3. Jun 22, 2005 #2

    matt grime

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    start at the north pole and start walking down the lne iof 0 degrees longitude, surely you can see that is is easy to end upi walking back to the north pole with the vector pointing in some other direction? eg loop round and approach from the line of 90 degrees longitude.
     
  4. Jun 22, 2005 #3
    He should have said "may" instead of "will", as you can find closed paths that do return the vector to its original orientation.
     
    Last edited: Jun 22, 2005
  5. Jun 22, 2005 #4

    matt grime

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    why is this impossible on euclidean space? move the vector round the edge of a (smoothed off if we want to not worry about sharp undifferentiable bends) square it comes back pointing at 90 degrees. of course if the closed paths are required to be smooth at all points, including where it joins up (ie a smooth non intersecting map of S^1 into the space) then it is impossible in any space. it is impossible to parallel transport a vector to a different on in R^n, and possible on S^2, but that isn't what the question was after.
     
    Last edited: Jun 22, 2005
  6. Jun 22, 2005 #5
    Ah, good catch. I was thinking of parallel transport when I added that.
     
  7. Jun 22, 2005 #6
    But then, if I were to loop around and come back via the 90 deg longitude, I would have a sharp bend at the point where 90 deg is connected to 0 deg longitude. So, if I were to "smooth out" this sharp bend, I would finally end up with a vector pointing at the same direction as the initial direction.
     
  8. Jun 23, 2005 #7

    matt grime

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    you don't have to smooth out the final join though. it doesn't say the path is smooth and closed, just closed. as i said, if the path has to closed and smooth (homeomorphic and infinitely differnetiable image of S^1) then it is wrong.

    i don't know what the author is getting at though, as this is true in any space. it sounds to me like they're getting parallel transport messed up.
     
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