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Curved space?

  1. Sep 26, 2009 #1
    How do we distinguish (mathematically) between curved space and the choice of coordinates? For example, the flat space metric in spherical polar coordinates looks as if it is curved space. I can ask the same for gravitational waves - how do we know that it isn't the TT gauge which is wavelike, rather than spacetime itself?
     
  2. jcsd
  3. Sep 26, 2009 #2
    The space curvature R is the criterion. Whatever coordinate change you do in a flat space (R=0), the curvature remains to be zero.

    In A.A. Logunov's approach to gravity (RTG) the equations determining the harmonic coordinates are not the "coordinate conditions" but the field equations. In his construction the separation of the Minkowsky metric and the gravitational field is explicit due to involving the former in the field equations.
     
  4. Sep 26, 2009 #3
    Ok thanks that's all I needed to know. I'm actually working from Maggiore's text on gravitational waves. I'm having to learn GR in tandem with gravitional waves so I'm still trying to figure out a lot of conceptual issues.
     
  5. Sep 26, 2009 #4
    Sorry just another quick question. What do you mean by coordinates here? Presumably tranforming to a frame which is accelerating with respect to the flat frame would result in curvature, so I presume this is not what you mean by a coordinate change. At any point it is possible to find a frame which is locally flat, so I are you talking about globally flat here?
     
  6. Sep 26, 2009 #5
    No, no! Acceleration is different from gravity. Take simple Classical Mechanics and choose an accelerating RF. Such a change of variables introduces additional forces but the space remains flat, that's for sure. So by coordinate change I mean any coordinate change which is reversible (not singular). If it is done starting from a flat space, the curvature R=0 remains intact (it is an invariant anyway).
     
  7. Sep 26, 2009 #6

    Fredrik

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    A coordinate system is a function from an open set of spacetime into [itex]\mathbb R^4[/itex]. If the Riemann tensor is zero, its components are zero in all coordinate systems. Spacetime doesn't get curved just because you choose a coordinate system with the property that a hypersurface of constant time is a curved submanifold.
     
  8. Sep 26, 2009 #7
    Ok that's cleared some things up. But choosing an accelerating frame does introduce a non-zero affine connection right? Is it that the derivatives of the connection (which gives the Riemann tensor) are still zero, or that at least they combine in such a way as to let R=0?
     
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