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Physical intuition behind geodesics and parallel transport

  1. Mar 16, 2010 #1
    Hi all,

    Sorry if this is a dumb question, but what exactly do we mean by the term parallel transport? Is it just the physicist's way of saying isometry?

    Also, in my class we have just defined geodesics, and we're told that having a geodesic curve cis equivalent to demanding that the unit tangent vector be parallel transported along c. This confuses me. I can read the proof of this, and it's fine, but I don't feel any physical intuition developing here. Can someone explain, as physically as possible, why these two conditions are equivalent?

  2. jcsd
  3. Mar 16, 2010 #2


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    I don't think it is related to isometry.

    In a curved spacetime, there is no obvious notion of "parallel". But you need one, so you just define it, by defining the notion of a derivative, a rate of change, an acceleration. No acceleration = no change in velocity = no change in tangent vector = parallelly transported (by definition, but hopefully it will seem like a graceful generalization of the terms we use for flat spacetime). There are many possible dervatives, and hence many possible notions of parallel. In GR, the derivative is chosen by specifying that the connection be the Levi-Civita connection http://en.wikipedia.org/wiki/Connection_(mathematics)
    Last edited: Mar 16, 2010
  4. Mar 17, 2010 #3


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    Put a gyroscope in a box. Transport it along some path. Observe the orientation of the gyroscope at the end.
  5. Mar 17, 2010 #4


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  6. Mar 17, 2010 #5
    1) Geodesic is free fall.
    2) Parallel transport reveals curvature.
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