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I Can a finite universe end in heat death?

  1. Feb 17, 2017 #1
    Hello guys,

    I was reading some models about the topology and size of the universe (always a controversial topic), then a question came to my mind.

    It is predicted that our universe will expand until it reaches heat death. Can a closed, finite universe also reach heat death and be described by the standard de Sitter cosmology?

    Or in other words, can a closed, finite universe reproduce the standard flat lambda model which prioritizes dark energy as the source of accelerated expansion?

    Thanks in advance
     
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  3. Feb 17, 2017 #2

    phinds

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    Speaking from a base of almost pure ignorance, I'd say that it MUST be possible because "finite" does not constrain the size, really, so things could get far enough apart that it would amount to heat death.
     
  4. Feb 17, 2017 #3

    Jorrie

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    Yes, even if our universe was mildly closed today (e.g. Omega = 1.5), but with the presently observed cosmological constant, it would eventually have evolved towards a de Sitter universe with Omega =1 and heat death would still follow. This is because the Omega_Lambda would eventually dominate anyway.
     
  5. Feb 17, 2017 #4

    Chronos

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    A finite universe would imply it is a closed system. Thermodynamics push closed systems towards thermal equilibrium [i.e., heat death]. Thus, heat death is inevitable in a finite universe. Of course, as usual, there is a caveat. Under the rules of thermodynamics a closed system must have a boundary and we already know that boundary conditions and infinity do not play well together. What kind of boundary conditions logically apply to all that exists [the universe]? What is outside/separate from everything that exists, nothing? Nothing cannot possess any definable properties without forfeiting its claim to be nothing, and would make a terrible excuse for a boundary between all that exists and all that does not exist. Boundary condition get very messy where the universe is involved.
     
  6. Feb 17, 2017 #5

    Chalnoth

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    I don't think this is true. A toroidal-shaped universe would be finite but without any boundary in space. I don't think thermodynamics has anything to say about the potential existence of that possibility.
     
  7. Feb 17, 2017 #6

    Chronos

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    You can make that argument so long as you are willing to relax the notion of boundaries on a closed system. A system without boundaries is like a definite integral without limits: Difficult to solve and not necessarily continuous across all values. I would agree thermodynamics has little to say about such a possibility because it is no more meaningful than any other singularity that arises in a mathematical function.
     
  8. Feb 18, 2017 #7
    How about a closed, spherical universe?

    How can a closed universe evolve to de Sitter space? Can de Sitter have curvature?
     
  9. Feb 18, 2017 #8

    Chronos

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    A closed spherical universe has a boundary, hence is spared thermodynamic paradoxes. Being finite, however, imposes a heavy enough burden on any closed universe model [what lays beyond everything in existence?]. A de Sitter space, by definition has positive curvature, and positive curvature strongly implies a closed universe- although there are ways closure can be avoided, as Chalnoth noted.
     
  10. Feb 18, 2017 #9
    Just for curiosity, what paradoxes did you have in mind while typing the first sentence?

    I thought that de Sitter space by definition has zero curvature, hence it is what will happen in an expanding flat universe. Can you expand why de Sitter has positive curvature and does that in someway imply that the whole universe should be finite (and closed)? And what happens to the cosmological horizon in the case of de Sitter space which has curvature? Objects go away from us on curved trajectories?

    Sorry for the accumulation of questions, but I think they are all pretty much connected. Thanks for the answers :wink:
     
  11. Feb 18, 2017 #10

    Bandersnatch

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    This post doesn't sound right. Admittedly, I'm operating beyond the edge of what I understand, so feel free to educate me.
    How does a sphere (or hypersphere) have a boundary if it's not embedded in a higher-dimensional space? Your response to Chalnoth's counterexample of toroidal space didn't clarify what you meant.
    Why would there be a 'heavy burden' of the type you describe? Finite and closed means that it makes no sense to ask about anything beyond.
    Why does positive space-time curvature (i.e. de Sitter space) imply spatially closed universe? It can be sliced into spatial slices of positive, negative or flat curvatures. Using FLRW metric the slicing of de Sitter space along constant t gives flat 3D space, or so I'm told.
    I thought Chalnoth's toroidal space (which is flat) example shows that a closed space doesn't need a boundary - you seem to be saying that it shows that a positively curved space doesn't need to be closed. Can you explain what you mean here?
     
  12. Feb 18, 2017 #11

    PeterDonis

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    It does? A spatial slice of such a universe has topology ##S^3##, which is a compact manifold without boundary.
     
  13. Feb 18, 2017 #12

    PeterDonis

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    That is not correct. It has positive curvature, as Chronos said. More precisely, it has positive spacetime curvature (the Ricci scalar is positive everywhere--it's just a multiple of the positive cosmological constant). There are slicings of de Sitter spacetime with positive, zero, and negative spatial curvature.
     
  14. Feb 18, 2017 #13

    Jorrie

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    There may be some confusion between the meanings of "de Sitter space" and a "de Sitter universe". AFAIK, a de Sitter universe is spatially flat, but is expanding exponentially, while de Sitter space is spatially static (and closed). Or do I misunderstand?
     
  15. Feb 18, 2017 #14

    PeterDonis

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    You misunderstand. A de Sitter universe is just de Sitter space by a different name. One possible coordinate chart on de Sitter space corresponds to a spatially flat, exponentially expanding universe. But there is another chart on the same space (same manifold, same geometry) which is spatially static and closed. (These two charts do not cover the same portion of the entire manifold.)
     
  16. Feb 18, 2017 #15

    Can it be closed and exponentially expanding? Which in some sense highlights the main question in my thread.
     
  17. Feb 18, 2017 #16

    PeterDonis

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    These are properties of slicings, not of the spacetime. The spacetime admits a closed slicing (which is static) and an exponentially expanding slicing (which is spatially flat). But it's the same spacetime in both cases.
     
  18. Feb 18, 2017 #17
    I am afraid that I don't understand the thesis presented in the current way.

    If the whole universe is not infinite in size (space that you mentioned), but finite - it curves back on itself, therefore we have a closed and finite universe. Right?


    Now in de Sitter case, space is exponentially expanding. You said that it implies that the topology is flat. Can the space be curved (so that it closes on itself) and still exponentially expanding?
     
  19. Feb 18, 2017 #18

    Jorrie

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    Thanks, I think I understand. Is it then correct to say that the slicing used for the LCDM cosmological model will evolve towards the latter case, as matter density is vanishing? Or must one be more exact than this?
     
  20. Feb 18, 2017 #19

    PeterDonis

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    In the LCDM model, the flat slicing is the only one that is homogeneous and isotropic; the presence of matter density breaks the symmetry of exact de Sitter spacetime. As the matter density gets smaller and smaller compared to the cosmological constant, other possible slicings on de Sitter spacetime will get closer and closer to being homogeneous and isotropic slicings on the actual spacetime of the universe (as they are on exact de Sitter spacetime).
     
  21. Feb 18, 2017 #20

    Chalnoth

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    Typically it's actually simpler. You just use periodic boundary conditions, with the periodicity dependent upon precisely how the universe is connected.
     
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