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A cosmology question

  1. Apr 19, 2009 #1
    Hi All

    Let's assume the Universe's topology is a standard hypersphere (i.e. it's finite) and it's only a bit over-dense (Ωo = 1.02 - 1.01.) Say it has a mass-energy break-down like our own, 27% matter and 73% cosmological constant. In that case how do we work out its current radius? Can it have a compact topology even with so much cosmological constant?

    From what I've read an "open" Universe could still be finite - or have I misunderstood something?
  2. jcsd
  3. Apr 19, 2009 #2
    What do you mean by "current radius"????
  4. Apr 19, 2009 #3


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    They do this in the 2008 WMAP report (5th year data, implications for cosmology) by Komatsu et al.

    The authors include Ned Wright, Joanne Dunkley, David Spergel.

    They use a simple formula for the radius of curvature
    and get a lower bound for the RoC of about 100 billion lightyears.

    The idea is suppose you think, like you proposed hypothetically, that
    Ωo = 1.015

    The key number here is Ωo - 1 = 0.015

    You just take the Hubble distance c/H and divide by the square root of 0.015!

    Really simple. they do this in the notes to Table 2 on page 4 of the Komatsu et al report.

    Let's actually work it out, say the Hubble distance is 13.8 billion LY so we just do
    13.8/sqrt(0.015) = 112.7 billion LY.

    Call it 100, don't want to seem too precise,

    the current volume of a hypersphere with RoC = R is V = 2 pi2 R3

    so if you plug in R = 100 billion LY you can get the current volume in cubic lightyears.
    Last edited: Apr 19, 2009
  5. Apr 19, 2009 #4
    Present radius then? Not talking about anything electromagnetic, just the present hyper-radius of the cosmos, from which you can get the antipodal distance. In a matter only closed Universe the present radius is:

    Ro = LHo - 1)

    (LH is the Hubble Distance, c/Ho which is 20.1 billion light-years in our Universe's case. In a closed Universe with Ωo = 2, then Ro = LH.)

    the maximum radius is:

    Rm = Roo - 1)/Ωo = LHΩoo - 1)-3/2

    and the antipodal distance is:

    LA = πRo

    The 3-Volume of the hypersphere is:

    3-V = 2π2Ro3

    Current age of the Universe involves a substitution...

    cosη = (2 - Ωo)/Ωo

    so to = (Rm/2)(η - sinη)

    where Rm is in light-years, so the age is in years. In metres just divide by c to get to in seconds.

    So that's all very easy when it's a matter dominated Universe. It's the Cosmological Constant, Λ, that makes things trickier.
  6. Apr 19, 2009 #5


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    graal, I understand most of what you say here, it's rather along the same lines as my post. But I don't know where you get the number for the Hubble distance. You say 20.1 billion LY.
    Do you have a source?

    I think a more usual estimate is around 13.8 billion LY.
  7. Apr 19, 2009 #6


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    The 'time now' thing unnecessarily complicates matters, IMO. I prefer that which predicts the current size of the observable universe - 13.7 billion light years.
  8. Apr 20, 2009 #7
    Doesn't this issue go to the measured value of H as opposed to the value which may exist at present - for example if the measured z redshift is physically a stretching of space then the present scale divided by an earlier scale is independent of the velocity profile (i.e., whether the universe is accelerating or decelerating)- but if the z redshift is actually due to a Doppler effect - then the age of the universe will depend on whether the expansion rate is changing. Also, the third possibility is that space is not stretching, but rather the observed redshift is due to the addition of space - in which case the nebula are actually not moving with respect to the spatial background (as is the case with convention Doppler) - but rather wafted apart by the growing distance - in which case we have a Doppler like sensitivity to the history of the expansion
    Last edited: Apr 20, 2009
  9. Apr 20, 2009 #8
    Follow up to Post 7...So I guess my query jells down to whether the effective Hubble radius could be different than than c/Rp where Rp is what we measure to arrive at H approximately equal to 70
  10. Apr 20, 2009 #9
    Hi Marcus

    Hmmm... I think I remultiplied something I shouldn't have. Let's see, if Ho = 70 km/s, then c/Ho is 300,000/70 = 4,300 Mpc = 14 Gly. If Ωo is close to 1, then to is 2/3 of LH - I had multiplied that by 13.4 Gly... my mistake being that in our accelerated Universe to = LH approximately. Thanks for spotting my gaffe.

    I read that paper by Komatsu et. al. (skim-read it, I mean) and found the computations for radius of curvature. Doesn't quite resolve my alternative topology question, but it does explain how to derive the radii involved. I wonder if we'll ever be certain if Ωo deviates at all from 1 and whether our Cosmos is open or closed, finite or infinite.

    Can a Cosmos with hyperbolic topology - an Open Universe - be finite and unbounded? Does it have to have a boundary if it is finite? I know there's no spatial boundary in the 3-Volume of a hypersphere - it's the usual example of a finite, but unbounded Cosmos. Can such a finite Universe still expand forever?
  11. Apr 20, 2009 #10


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    Certainly can! BTW after 1998 the terms "closed" and "open" changed meaning or became ambiguous. So be careful. Better to say spatially closed or spatially finite, if that is what you mean. All universe cases can expand forever, given adequate Lambda.

    Before 1998 if a universe was spatially closed then that implied it would have to recollapse (because Lambda was assumed zero). So people would say closed when they meant "destined to collapse". But the semantics changed.

    Statements like this rely on a model. The mainstream consensus model now is LCDM (lambda cold dark matter) and it works pretty well. In terms of that model we currently have a 95% confidence interval for Ωo which is [0.992, 1.018]. (Table 2, Komatsu again)

    It is quite possible that we will in future get a 95% interval like [1.002, 1.016].
    That could happen with the Planck observatory that was just launched. Or with the big surveys planned for the next decade. It could happen.

    Using LCDM we would then be 95% certain that the universe is closed and destined to expand forever.

    (I mean spatially closed, of course. The finite volume boundaryless case that you mentioned, the hypersphere.)

    It is also possible that future confidence intervals will just shrink down closer and closer around 1.00000 like the next generation might say [0.999, 1.002]. Could go either way.

    In reading Table 2 of the Komatsu paper, keep in mind that Ωk = 1 - Ωo
    or in other words
    Ωo = 1 - Ωk

    It is just a convention they use.

    So if they say Ωk = - 0.018
    that means the same as
    Ωo = 1.018
    Last edited: Apr 20, 2009
  12. Apr 20, 2009 #11
    Well that makes sense. Did seem a bit confusing.
    What do you think? Does the infinite Universe make sense or is it just convenient to assume? Max Tegmark's discussion of copies of Earth at super-exponential distances does make the idea seem a tad pointless. But then the Everett Many-Worlds implies an infinite number of quantum copies as well. Seems equally anti-parsimonious in some way.
  13. Apr 20, 2009 #12


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    Have to be really really careful not to let one's own prejudices or preferences influence one's expectations about the course of future research.

    To be honest, I feel instinctively more comfortable with the idea of a spatially closed finite volume boundaryless universe----say topologically equivalent to the 3D hypersphere.

    But at the moment I can see no rational grounds for expecting that nature will turn out to be the way I am more comfortable with.

    Other people around here may be able to shed more light on this.


    In any case, as you surely know, the standard cosmo picture is homogeneous and isotropic. That is, whatever space is (finite or infinite volume) it is boundaryless and it is uniformly filled with matter---matter is approximately evenly distributed throughout all space.
    Last edited: Apr 20, 2009
  14. Apr 20, 2009 #13
    Thanks for the reminder - Nature is under no obligation to fulfill expectations.
  15. Apr 21, 2009 #14
    It's a while since I looked at the Lambda-CDM model but I thought it was predicted that the universe was leaning towards slightly open rather than slightly closed (or has slightly closed always been the case?). If Ωo was negative, what would the implications be?
  16. Apr 21, 2009 #15
    I think most investigators believe flatness is the most parsimonious choice since it's neither positive or negative curvature. Some versions of inflation produce perfectly flat universes reinforcing the perception it's parsimonious.
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