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History and geometry of flat universe

  1. Sep 18, 2009 #1
    1. The present universe is observed to be flat. Was it always flat, before it started its accelerated expansion?

    2. Is the Riemann tensor zero for this flat universe? Is its geometry that of special relativity?

    I'd appreciate if Marcus or Ich or any other science advisor weighed in on this. Thanks.
     
  2. jcsd
  3. Sep 18, 2009 #2

    nicksauce

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    If the universe is flat, is has always been flat and always will be. That is because the geometry of the universe is determined by a **constant**, k.

    The Riemann tensor is non-zero. According to my calculations...
    [tex]
    R^t_{xtx} = a\dot{\dot{a}}[/tex]
    [tex]
    R^x_{yxy} = (\dot{\dot{a}})^2
    [/tex]
    and all the other values come from symmetry between x<->y<->z.
    The geometry is not that of special relativity. In special relativity the metric is diag(-1,1,1,1). In a flat expanding universe the metric is diag(-1,a^2(t),a^2(t),a^2(t)), where a(t) is the scale factor of the universe.
     
  4. Sep 18, 2009 #3

    George Jones

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    For Friedmann-Walker-Robertson universes, space is flat but spacetime is curved.

    The Riemann tensor for Friedmann-Walker-Robertson universes is not zero because FRW spactimes are not flat. Three-dimensional spatial hypersurfaces orthogonal to cosmic time (i.e., space) are intrinsically flat, i.e., the Riemann tensor constructed from the spatial metric induced on these hypersurfaces by the spacetime metric is zero.
     
  5. Sep 19, 2009 #4

    Chalnoth

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    This is not expected to be the case. Rather, it is expected that our region of the universe started off with very significant curvature, but as it was dominated by an inflaton field that drove a very rapidly-accelerated expansion, it was driven to be almost perfectly flat in virtually no time.

    This happens because the effect of the curvature scales with the expansion as [tex]1/a^2[/tex], but during inflation, the dominant energy density was almost independent of expansion. As the scale factor increased by a factor of [tex]10^{30}[/tex] or more, the spatial curvature was driven to very near zero.

    Nicksauce has responded to this point well.
     
  6. Sep 19, 2009 #5
    Thank you all for the responses. Two more questions.

    1. Can the presently accelerating universe be flat without inflation?

    2. In a flat universe, omega = 1.
    Omega = matter(regular matter + dark matter) + dark energy.
    Can we have Omega = 1 without dark matter?
     
  7. Sep 19, 2009 #6

    Chalnoth

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    If there happens to be an alternative explanation for our current observations that point to inflation, I suppose it's possible.

    In principle it's a different issue. The flatness is related to the relationship between the average expansion rate and the total energy density. If the expansion rate is too fast compared to the energy density, then it's open. If it's too slow, then it's closed. If it's "just right", then it's flat.
     
  8. Sep 19, 2009 #7
    Is not the average expansion rate controlled by the total energy density, since dark energy that is driving the accelerated expansion is part of it?
     
  9. Sep 19, 2009 #8

    Chalnoth

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    Right, so, the first of the Friedmann equations is:

    [tex]H^2 = \frac{8\pi G}{3} \rho - \frac{k c^2}{a^2}[/tex]

    So basically if [tex]H^2 > \frac{8\pi G}{3} \rho[/tex], [tex]k < 0[/tex]. Likewise, if [tex]H^2 < \frac{8\pi G}{3} \rho[/tex], then [tex]k > 0[/tex].
     
    Last edited: Sep 19, 2009
  10. Sep 19, 2009 #9
    Both the inequalities are same, yet k changes. I'm confused.
     
  11. Sep 19, 2009 #10

    Chalnoth

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    Ack, sorry, typo. Fixed.
     
  12. Sep 19, 2009 #11
    Thanks
     
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