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Energy-stress tenor = 0 => flat spacetime?

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  1. May 18, 2014 #1
    Mathematically,
    ##T_{\mu_\nu} = 0##
    ##\Rightarrow~R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = 0##

    Now multiply both sides by ##g_{\mu\nu}##
    with definition of ##R = g^{\mu\nu} R_{\mu\nu}##
    ##R - \frac{1}{2} R = 0##
    ##R = 0##

    Is that my imagination wrong? I thought 'empty space' might not be flat in general, like the Schwarzschild metric. And, I thought ##R = 0## meant 'the spacetime is flat.

    What is the point I missed? Thanks.
     
  2. jcsd
  3. May 18, 2014 #2

    Bill_K

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    In order to conclude that a space is flat, the entire Riemann curvature tensor must be zero, Rμνστ = 0. In four dimensions, the Riemann tensor has 20 independent components, while the Ricci tensor comprises 10 of them. In a vacuum region, Rμν = 0 ("Ricci flat") that leaves 10 independent components of curvature which may still be nonzero.
     
  4. May 18, 2014 #3
    Then what is the physical meaning of R? One could feel that of the Curvature tensor from parallel transport. But the contraction leaves me no clue to understand it by imagination.
     
  5. May 18, 2014 #4

    Bill_K

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    You don't really need to have an intuitive feel for it to use it! :wink: But in fact, this was discussed in a recent thread.
     
  6. May 18, 2014 #5

    pervect

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    Bill has already handled the math end, a zero Ricci tensor isn't sufficient to guarantee a zero Riemann tensor.

    An EM analogy might be helpful. Suppose you have no charges anywhere. Can you conclude there are no electric or magnetic fields?

    No - you can't conclude this, because an electromagnetic wave of any sort has fields and has no charge. (If you want a specific example, consider the plane electromagnetic wave).

    To ensure that you have no fields, in addition to not having charges, you must impose appropriate boundary conditions.

    The gravitational case is rather similar, asymptotic flatness (a boundary condition) is needed as well as a zero stress energy tensor to get a flat spacetime.
     
  7. May 18, 2014 #6

    PeterDonis

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    That's not a sufficient condition, because, for example, Schwarzschild spacetime is asymptotically flat and has zero SET everywhere, but it's not flat spacetime.
     
  8. May 18, 2014 #7

    pervect

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    I rather think of the Schwarzschld space-time as being a point mass in my analogy. I'm not sure how to make this idea rigorous, however. The objection is an important one and while I think it's fixable, I'm not sure what it takes to fix it.
     
  9. May 18, 2014 #8

    PeterDonis

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    How about geodesic completeness? That would rule out Schwarzschild spacetime. I'm not sure it would rule out all of the vacuum solutions that have nonzero curvature, though.

    Edit: Apparently it doesn't; according to Wikipedia, the Osvath-Schucking metric is globally defined and singularity-free, which should mean it's geodesically complete, but is not isometric to Minkowski spacetime.

    http://en.wikipedia.org/wiki/Ozváth–Schücking_metric
     
  10. May 18, 2014 #9
    We could also drag cosmological constant into this discussion.
     
  11. May 18, 2014 #10

    PeterDonis

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    As far as I can tell, this metric is *not* asymptotically flat. So it's possible that the combination of asymptotic flatness and geodesic completeness is sufficient to guarantee that the Riemann tensor is identically zero and thereby pin down Minkowski spacetime.
     
  12. May 18, 2014 #11

    Matterwave

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    Couldn't there be asymptotically flat spacetimes with gravitational waves and be geodesic-complete? Or how about standing gravitational waves? o.o
     
  13. May 18, 2014 #12
    Thank you everyone. Should have read recent thread before asking.
     
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