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Inflation and WMAP

  1. Mar 4, 2015 #1
    I read the following passage from a book that does not look correct to me. I understood that the geometry of the universe as far as we can tell is flat to within the error bars of our measurements and that is therefefore either infinite or at least much much larger than our observable patch. However the passage reads as this:
    "Current observational data indicate a closed-geometry universe, contrary to the predictions of inflationary multiverse models. Additionally, the low power seen in the low spherical harmonics from WMAP indicate that the total universe may not be much larger than the observable universe."
    I was pretty sure WMAP and Planck results were considered inflation friendly. Have i got something wrong or is this passage wrong?
     
  2. jcsd
  3. Mar 4, 2015 #2

    bapowell

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    What is the reference?
     
  4. Mar 4, 2015 #3

    Chalnoth

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    Yeah, it's not correct.

    There are some inflation models which are made unlikely by the latest Planck (and other CMB) data, but there are many other models which fit the data just fine. There is no evidence (yet) of any deviation from flatness.
     
  5. Mar 4, 2015 #4
    Thanks, thought so.
     
  6. Mar 4, 2015 #5

    Chronos

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    It is interesting to note that Einstein strongly favored a closed universe. His equations become rather awkward in an open universe
     
  7. Mar 4, 2015 #6

    Chalnoth

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    I'm not sure that's important. But it is true that most theorists who consider models for the birth of our universe tend to find it easier to work with topologically closed universes.

    But just because a universe is closed topologically doesn't mean we should always expect to measure positive curvature. To take a simple example, consider a torus. A two-dimensional torus embedded in three dimensions (basically a doughnut) has positive curvature along the outside and negative curvature along the inside. If our universe as a whole had a similar topology, it would be possible for the observable patch to fall along the interior side that has negative curvature, despite the closed topology.
     
  8. Mar 4, 2015 #7

    mfb

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    Really? I can draw a square everywhere and the angles are always 90°, indicating zero curvature.
     
  9. Mar 4, 2015 #8

    Chalnoth

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    That's true for an idealized torus. It's not true for the 2D torus embedded in three dimensions.
     
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