Doubt about our spacetime manifold

  1. I understand that accordingt to GR mass curves the spacetime (I'm not referring to spatial curvature k), so that the universe globally considered is a manifold with constant curvature, is this right?
    If so, is this curvature positive or negative in the current cosmological model?
     
  2. jcsd
  3. Good question, I'd actually just logged on to ask a similar one myself. As such I shall not directly answer you but I've just been reading this wiki page that seems pretty descriptive: http://en.wikipedia.org/wiki/Shape_of_the_Universe
     
  4. bapowell

    bapowell 1,850
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    The latest data from WMAP7 is that the universe is flat to within a few percent.
     
  5. nicksauce

    nicksauce 1,275
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    The spatial manifold (say M) has constant curvature. The spacetime manifold (RxM) has some non-trivial time-dependent curvature. Current data favours a universe with a slight negative spatial curvature, though the other scenarios are by no means ruled out.
     
  6. bapowell

    bapowell 1,850
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    Hi Nick,

    Could you provide a source for this? In reading through WMAP's most recent findings (arXiv:1001.4538), they report:

    WMAP+BAO+SN (95% CL): [tex]-0.0178 < \Omega_k < 0.0063[/tex]
    WMAP+BAO+H (95% CL): [tex]-0.0133 < \Omega_k < 0.0084[/tex]
     
  7. nicksauce

    nicksauce 1,275
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    Yes, you're right. I should have said, slight positive curvature, with slight negative curvature not ruled out.
     
    Last edited: Jun 11, 2010
  8. George Jones

    George Jones 6,396
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    But this refers to (lack of) curvature of spatial slices, not to (lack of) spacetime curvature, and the original poster wrote



    The spacetime curvature is non-zero, is constant on spatial slices, but is not constant in time. If curvature were not dynamical, einstein's equation wouldn't lead to a dynnaical Friedmann equation. For the components of the spacetime curvature tensor written in terms of the scale factor, see page 271 from

    http://books.google.com/books?id=IyJhCHAryuUC&printsec=frontcover&dq=gron&cd=3#v=onepage&q&f=false.
     
  9. bapowell

    bapowell 1,850
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    Indeed. Thanks for pointing this out.
     
  10. Thanks for the answers.
    I was thinking in terms of curvature R, as in this models from cosmoogy books: a de Sitter spacetime, and an Einstein spacetime have R>0, Anti de Sitter spacetime has R<0 , Minkowski spacetime has R=0. But of course all of these models are of static universes, I didn't realize that in our dynamical (expanding) universe the curvature is not so straightforward as is it is dynamical and I guess it can vary (noncostant and nonzero) as GeorgeJones pointed out.
    Am I on the right track?
     
  11. bapowell

    bapowell 1,850
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    Yes, you are on the right track. The scalar curvature, R, cannot be measured directly, but can be related to measurable dynamical quantities in a Friedmann-Robertson-Walker universe:

    [tex]R \propto \dot{H} + 2H^2 + \frac{k}{a^2}[/tex]

    where H is the Hubble parameter, k the curvature of spatial slices (the thing that WMAP constrains to be close to zero), and 'a' the scale factor. Using the Friedmann equations, this can be recast in terms of the energy content of the universe:

    [tex]R \propto \frac{1}{3}\rho - p = \rho\left(\frac{1}{3} - w\right)[/tex]

    where [tex]\rho[/tex] is the energy density of the universe and [tex]p[/tex] the pressure. The final equality is written in terms of observable parameters that are actively being constrained by current observations.
     
    Last edited: Jun 11, 2010
  12. bapowell, what's w stand for in the last equation?
    Thanks
     
  13. bapowell

    bapowell 1,850
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    Sorry. It's just a parameter that relates the energy density of the fluid to the pressure:

    [tex]p = w\rho[/tex]

    so nothing new...just a retooling of the previous equation. I write it this way because you frequently see [tex]w[/tex] constrained in experiments -- it is referred to as the equation of state parameter, or simply the equation of state. For reference, [tex]w = -1[/tex] is de Sitter expansion, [tex]w = 0[/tex] is pressureless dust, and [tex]w = 1/3[/tex] is radiation. You'll notice that a universe that is filled with radiation (uniformly) has R = 0.
     
  14. Chalnoth

    Chalnoth 5,305
    Science Advisor

    Well, no, that's not a correct interpretation of the data. The fact that zero is within the one-sigma limit means that curvature being zero is fully consistent with the data. Due to simple random statistical noise, we expect the experimental result to, on average, be about one sigma away from the true value anyway, so we cannot interpret any deviation from zero within one sigma as being evidence of a true value different from zero.
     
  15. Just one thing, I don't understand the case w=1/3. How is this a radiation filled universe? It would seem it correspnds to a universe with one third as much dust as radiation pressure if we follow that w=P/rho . What am I missing?
    Thanks
     
  16. Chalnoth

    Chalnoth 5,305
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    The pressure of radiation is one third its energy density, hence w = 1/3 for pure radiation (dust has no pressure).
     
  17. I tried to obtain this by myself but I didn't get the same, After adding the right terms of the Friedmann equations I got:

    [tex]R \propto - p [/tex]

    I know this is simple math but I'm not sure what I did wrong.
    Thanks in advance.
     
  18. Chalnoth

    Chalnoth 5,305
    Science Advisor

    Make sure you're using the second Friedmann equation that casts the time derivative of the Hubble parameter in terms of the density, as well as the first to simplify things.
     
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