Why is the hypersphere not seriously considered by cosmologists as the best model for the overall shape of the universe?

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Why is the hypersphere not seriously considered by cosmologists as a model for the overall shape of the universe?

This is simply not true.

FLRW models are the most important models used by cosmologists, and they come in two flavors, open and closed (plus a borderline case, which is flat). The closed type has spatial cross-sections that are hyperspheres. If we fit parameters to a variety of cosmological observations, the universe is constrained to be very nearly spatially flat, but the error bars on the spatial curvature are big enough to allow both open and closed cosmologies. Therefore a hypersphere is currently a perfectly reasonable candidate for the spatial geometry of our universe, and cosmologists do take it seriously.

If this is the geometry, then models that fit the data constrain the radius of the hypersphere to be very large (IIRC orders of magnitude greater than the radius of the observable universe).

And if one assumes the radius of the hypersphere is 13,820 Million Light-Years, and is increasing at the speed of light then one can calculate the expansion rate of the universe (which we all know as Hubble's constant). It would be the speed that the circumference of the hypersphere is increasing (2 pi c) divided by the length of the circumference of the hypersphere (2 pi R).

Here you seem to be mixing up the observable universe with the entire universe. Even if we're talking about the observable universe, it doesn't expand at c.

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Updated on August 01, 2022

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  • DG123
    DG123 7 days

    Cosmologists seem to not seriously consider the hypersphere as the best model for the universe even though they mention it as a candidate from time to time. If you look closely, it seems to be a very good fit.

    The surface of a hypersphere is 3D and unbounded, just like our universe. And if one assumes the radius of the hypersphere is 13,820 Million Light-Years, and is increasing at the speed of light then one can calculate the expansion rate of the universe (which we all know as Hubble's constant). It would be the speed that the circumference of the hypersphere is increasing (2 pi c) divided by the length of the circumference of the hypersphere (2 pi R). So it would be (2 pi c)/(2 pi R) or (c/R).

    Plugging in the numbers one gets 21,693 m/s/Mly, and after multiplying by 3.2615 to convert to mega-parsecs and dividing by 1,000 to convert to kilometers one gets 70.75 km/s/Mpc (which are the units for Hubble's constant astronomers like to use). The currently accepted value of Hubble's constant is around 71 km/s/Mpc.

    So the hypersphere model fits the facts very well, but cosmologists do not embrace the model. What are the objections of the cosmologists to the hypersphere model of the universe? What facts or observations does the hypersphere model not fit?

    • rob
      rob over 4 years
      A link-only answer: This was Einstein's opinion (though it would take me some time to dig up a citation), but it's not consistent with the observed flatness of the universe.
    • DG123
      DG123 over 4 years
      The surface of the hypersphere is curving in the fourth dimension not in our three dimensions. Assuming light follows a curved path in 4D space along the surface of the hypersphere it would appear to be moving in a straight line to us, wouldn't it? I don't know how scientists have measured flatness, but it seems that if they use light, they might not be able to detect curvature in the fourth dimension. So until scientists show they can measure curvature in the fourth dimension, the flatness results cannot be used to refute the hypersphere model.
    • Admin
      Admin over 4 years
      @DG123: In your comment, you seem to be misunderstanding how general relativity works. It fundamentally uses spacetime curvature in 3+1 dimensions, it doesn't assume that anything is embedded in a higher-dimensional space, and the curvature it talks about is intrinsic, not extrinsic, so there is not part of it that's physically undetectable from within the spacetime.
    • PhillS
      PhillS over 4 years
      Curvature can be measured for example by counting the number of galaxies within a certain distance of your position. In a flat universe it goes up as $r^3$. In s closed universe it rises more slowly with distance. In a universe with negative curvature it rises faster. (Not that galaxy counts are a great way of measuring curvature because of look back time: at very large distances you will see no galaxies because you are looking at a time before they formed, to give a crude idea). The point is there are lots of effects of curvature that can be measured in principle
    • DG123
      DG123 over 4 years
      All three of the dimensions available to us on the surface of the hypersphere are parallel to the surface of the hypersphere. In order to measure the curvature of the surface of the hypersphere we would need to be able to measure distance in the fourth dimensional direction, but that direction is unavailable to us, so to believe one can measure curvature in the fourth dimension by making measurements parallel to its surface seems impossible to me. I would be interested in hearing (in detail) how you would accomplish it.
    • DG123
      DG123 over 4 years
      @PhillS: Counting galaxies may be a viable way to determine if our universe has the shape of a hypersphere, but the current circumference is 86.8 billion light years (assuming a radius of 13.82 Bly) and one would have to count galaxies beyond a quarter of the length of the circumference (21.7 Bly) in order to see a deviation from a flat 3D universe. It would be difficult but maybe possible.
  • DG123
    DG123 over 4 years
    Actually, the radius of the observable universe is expanding at c. Multiply 21,693 m/s/Mly (Hubble's constant) by 13,820 Mly (radius of the observable universe) and see what you get. But, no, I'm not mixing up the radius of the observable universe with the radius of the hypersphere. The center of the hypersphere is in 4D space. The center of our observable universe and its entire volume is on the surface of the hypersphere. Every point in our world is a point on the surface of the hypersphere and is the same distance from the center of the hypersphere. Just saying, I think it needs more study.
  • Admin
    Admin over 4 years
    @DG123: Actually, the radius of the observable universe is expanding at c. You're mistaken. physics.stackexchange.com/questions/26549/… The center of the hypersphere is in 4D space. No, here you're showing further confusion about GR. You might want to post this as a separate question.
  • PM 2Ring
    PM 2Ring over 4 years
    @DG123 IF the universe is closed then yes space is like a 3-sphere (a 4D hypersphere), but that doesn't necessarily mean that there's an extra spatial dimension: GR works fine with spacetime that's intrinsically curved, it doesn't need spacetime to be embedded in a higher dimensional manifold.
  • DG123
    DG123 over 4 years
    @Ben Crowell: I'm sorry to have to disagree with you again but the fact that objects 13.82 billion light years from us are moving away from us at the speed of light would be readily affirmed by any astrophysicist. The farther from us an object is, the faster it is moving, and its speed can be found by Hubble's law, v = Hd. At a distance of 13.82 Bly, v=c. It's a well established fact of astrophysics.
  • bapowell
    bapowell over 4 years
    @DG123: your mistake is that the Hubble scale does not set the size of the observable universe.
  • Admin
    Admin over 4 years
    @DG123: You're falling prey to a couple of common misconceptions about cosmological expansion. See this answer physics.stackexchange.com/a/141219/4552 and/or the Scientific American article by Lineweaver that I linked to from there.
  • DG123
    DG123 over 4 years
    I looked at the answer you gave in the link you provided. You refer to the expansion rate of the universe as a velocity. It's not a velocity, its a velocity divided by a length. Hubble's constant is usually stated in units of km/s/Mpc, which is a velocity divided by a length. You state in your link, "it doesn't make sense to ask for "the" velocity of expansion of the universe. There is not one velocity but many." It sounds like you're saying there is not one Hubble's constant but many. If that's what you're saying then you are just wrong. The universe is expanding at the same rate everywhere.
  • Admin
    Admin over 4 years
    @DG123: Re your most recent comment, you seem to be stating a mistaken characterization of the content of my answer.
  • DG123
    DG123 over 4 years
    Your statement, "it doesn't make sense to ask for "the" velocity of expansion of the universe. There is not one velocity but many." confused me. I misunderstood what you meant. I thought maybe you meant 'rate of expansion' when you said 'velocity of expansion', although, I see now what you said is true, it's just a strange way to talk about the expansion rate of the universe - from a velocity perspective - which, yes, the velocity of expansion is different for different distances. I've just never heard anyone talk about the expansion rate from that perspective.
  • PyRulez
    PyRulez over 3 years
    "If we fit parameters to a variety of cosmological observations, the universe is constrained to be very nearly spatially flat, but the error bars on the spatial curvature are big enough to allow both open and closed cosmologies." It should be noted that if the universe is flat, the error bars will always include the open and closed cosmologies.