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Why should the covariant derivative of the metric tensor be 0 ?

  1. Nov 15, 2006 #1
    That's a crucial point of GR !
    And I have always problems with that.

    Back to the basics, with your help.

    Thanks

    Michel
     
  2. jcsd
  3. Nov 15, 2006 #2

    dextercioby

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    The vanishing of the nonmetricity tensor in GR has to with the natural assumption that under parallel transport of vectors, neither the length of one vector, nor the angle between 2 such vectors would change.

    It's instructive to perform one of the 2 calculations referred to in the phrase above.

    Daniel.
     
  4. Nov 15, 2006 #3

    Demystifier

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    This assumption can probably be relaxed. But this gives you more freedom. I would say that this is merely the simplest (and geometrically the most natural) possibility.
     
  5. Nov 15, 2006 #4

    Hurkyl

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    A quick algebraic reason is that raising (lowering) indices then differentiating should be the same thing as differentiating then raising (lowering) indices.
     
  6. Nov 15, 2006 #5

    robphy

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    That condition is also needed for the "Fundamental Theorem of (Pseudo-) Riemannian Geometry".
     
  7. Nov 15, 2006 #6

    pervect

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    As far as I know, it's a defintional thing. If one adopts the defintion that two sides of a quadralateral are parallel when opposite sides have the same length, one can use a geometric construction based on this defintion of the parallelness of two different vectors to define parallel transport (also, the more elegant gometric construction known as "Schild's ladder") to perform covariant differentiation.

    (Note that given a metric, one can measure the length of the sides of a parallelogram.)

    Schild's ladder is discussed in MTW (but not many other textbooks include it).

    I've never quite understood torsion, though, or torsion-based theories of gravity.
     
  8. Nov 15, 2006 #7

    robphy

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    Here's another way to look at it.
    Given the metric tensor, if you don't choose the unique torsion-free metric-compatible derivative operator, you have to specify some more structure (say, the nonmetricity tensor or the torsion tensor or something equivalent) to pick out the derivative operator you prefer. From a physics point of view, you'd probably want to physically justify the choice.
     
  9. Nov 15, 2006 #8

    George Jones

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    I also don't know much know about torsion-based theories of gravity, but I think one story goes something like the following.

    Some theories relate torsion to particle spin. This is basically a coupling of spin to spacetime geometry, so that what constitutes a geodesic depends on a particle's spin.
     
  10. Nov 15, 2006 #9
    My understanding is that, if there is no force acting on your drawing compass, physically you'd expect the angle it subtends to remain constant. Mathematically: [itex]0 = \nabla_a (\boldsymbol{guv}) = \nabla_a \boldsymbol g[/itex] QED.

    To define the connection you also assume the covariant derivative is torsion free on scalar fields, and I'm not sure what that means physically.
     
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