Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why isn't everything expanding in an expanding universe?

  1. May 11, 2015 #1
    Viewpoint 1:
    Because gravity or other forces that are holding the thing concerned (be it a galaxy, a ruler or an atom) together are way stronger than the "force" caused by the expansion of space. So strictly speaking, space does in fact expand everywhere, including the space inside an atom between its nucleus and its electrons. But gravity or other forces keep the size of the object constant.

    Viewpoint 2:
    The space within the thing concerned (such as the space inside an atom) is of a different nature from the space between two galaxies (the space that "occupies" the vast region of space). Let's call the former type-1 space and the latter type-2 space. Only type-2 space expands for some reason; eg., type-2 space is filled with (or littered with) dark energy, which drives the expansion, but type-1 space is void of dark energy.

    I think the correct or the widely accepted viewpoint is the first one. But I don't understand how exactly the size of an object is being kept constant.
     
    Last edited: May 11, 2015
  2. jcsd
  3. May 11, 2015 #2

    phinds

    User Avatar
    Gold Member
    2016 Award

    Viewpoint 2 sounds like nonsense to me. Where did you hear this? Do you have reputable citations?
     
  4. May 11, 2015 #3

    Chalnoth

    User Avatar
    Science Advisor

    Everything isn't expanding because the universe doesn't have perfectly uniform density. When you work through the equations of what gravity predicts in the face of differences in density from place to place, you get an average, large-scale expansion, but small-scale bound systems that are reasonably stable over time (e.g. galaxy clusters and smaller).
     
  5. May 11, 2015 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    There is no such thing; the expansion of the universe does not give rise to any force.
     
  6. May 11, 2015 #5
    Yes, that's why it was in quotation marks. iirc, Prof Susskind mentioned in one of his lectures that the expansion can be modelled using a small force.
     
  7. May 11, 2015 #6

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Do you have a reference? I suspect the use of the word "force" in this connection was ill-advised, since it leads to the erroneous inference you have made. Expansion does not "try" to push bound objects apart, which is what the word "force" implies.
     
  8. May 11, 2015 #7

    wabbit

    User Avatar
    Gold Member

    I suspect he may have been discussing accelerated expansion in the presence of a cosmological constant, where something like that is possible (within limits) but I don't think it makes sense at all for unaccelerated expansion.
     
  9. May 11, 2015 #8

    Chronos

    User Avatar
    Science Advisor
    Gold Member

    We can safely conclude the historical earth sun recession rate is much smaller than that predicted by the expansion rate of the universe based on a variety of unrelated lines of evidence as discussed in this paper;http://arxiv.org/abs/1306.3166, A Closer Earth and the Faint Young Sun Paradox: Modification of the Laws of Gravitation, or Sun/Earth Mass Losses? Given the existing expansion rate measured at 1% per 140,000,000 years the earth would have been about 27% closer to the sun at the beginning of the Archean era 3.8 billion years ago. This is inconsistent with a vast body of climate and paleontology evidence.
     
  10. May 11, 2015 #9
    As far as I know, the universe at the smallest level does expand. A possible fate of the universe is called the big rip, where the expansion of space becomes so fast that atoms aren't able to keep themselves together.

    Chronos, that's interesting, I never thought of that before, so why has the solar system not expanded? A gravitational orbit in an expanding space shouldn't remain stable should it?
     
  11. May 11, 2015 #10

    wabbit

    User Avatar
    Gold Member

    I don't understand how expansion per se can affect orbits at all. The earth is not comoving and the geometry is locally Schwarzchild. Can it be affected by a distant symmetric distribution of galaxies, whether these are receding or approaching?
    The cosmological constant should have an effect though, I recall seeing a paper analysing that, will try to dig it up.
     
  12. May 11, 2015 #11

    Chalnoth

    User Avatar
    Science Advisor

    In some equations, the expansion acts as a sort of friction, tending to damp differences in velocity between things in the universe over time. That *might* be what he was talking about, but I'm not sure of the context.
     
  13. May 11, 2015 #12

    Chalnoth

    User Avatar
    Science Advisor

    No, it doesn't. Bound objects are stable in an expanding universe without a cosmological constant. This drops right out of the linearized equations using perturbation theory to describe a universe that isn't homogeneous (linearized equations only describe really big objects accurately....smaller objects are going to be even less impacted by the expansion).
     
  14. May 11, 2015 #13

    wabbit

    User Avatar
    Gold Member

    I was saying there might be an effect with a CC, not without it - not sure though.
    Couldn't find the paper I had in mind but this one seems relevant:
    http://arxiv.org/abs/0810.2712
    Influence of global cosmological expansion on local dynamics and kinematics
    Matteo Carrera, Domenico Giulini
     
  15. May 11, 2015 #14

    Chalnoth

    User Avatar
    Science Advisor

    Right. I was a little ambiguous in my reply. I was agreeing with you (my first sentence responds to your first sentence).
     
  16. May 11, 2015 #15

    wabbit

    User Avatar
    Gold Member

    Ah OK, got it now thanks:)

    For a spherically symmetric distribution, distant stars should have zero effect - this is correct in Newtonian gravity but is it always true in GR? Looking at these papers the issue doesn't seem so simple so presumably this may be only correct up to a Newtonian approximation, or otherwise up to second order effects?
     
    Last edited: May 11, 2015
  17. May 11, 2015 #16

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Yes; the "shell theorem" holds in GR as well as Newtonian gravity.
     
  18. May 11, 2015 #17

    wabbit

    User Avatar
    Gold Member

    Great. Can we also say that if we add one mass at the center of this spherical vacuum, we get a Schwarzchild geometry in place of flat spacetime?
    Newton-wise we would just add forces, but in GR unless the sphere is large enough there should some correction?
    Otherwise the orbits around a comoving mass surrounded by a spherically symmetric distribution of matter would be unaffected even by a cosmological constant - but this isn't quite true, in the vacuum case we get Schwarzchild-de Sitter geometry with some corrections to orbits?

    Edit: but how does the shell theorem work with a CC? It must be saying something like, not the vacuum is flat, but it has de Sitter geometry? Otherwise a sphere cut out in de Sitter space would seem to provide a counterexample.
     
    Last edited: May 11, 2015
  19. May 11, 2015 #18

    Chalnoth

    User Avatar
    Science Advisor

    Found a proof here:
    http://arxiv.org/abs/0908.4110

    They say that locally, the vacuum solution in a spherically-symmetric space-time in the presence of a cosmological constant is necessarily equivalent to either de Sitter space-time, or Schwarzschild-de Sitter space-time (Theorem 1 in the paper). However, they also say that the predicted space-time is not necessarily static, and I don't entirely understand what that means.
     
  20. May 11, 2015 #19

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This is a proof of Birkhoff's Theorem, not the shell theorem; they're two different things.

    Birkhoff's Theorem, in the generalized form proven in this paper, says that any vacuum, spherically symmetric solution to the EFE with cosmological constant must be the Schwarzschild-de Sitter geometry.

    The shell theorem says that, if we have a distribution of stress-energy that is spherically symmetric outside some region, the geometry inside that region is unaffected by the stress-energy distribution outside it.

    We can combine the two results to say that, for example, if the stress-energy distribution is spherically symmetric outside some region, and inside the region we have a spherically symmetric vacuum, then the geometry inside the region must be Schwarzschild-de Sitter.

    It means that the Schwarzschild-de Sitter geometry is only static between the two horizons--outside the black hole horizon and inside the cosmological horizon. In the other regions it is not static; there is still an extra Killing vector field (in addition to the ones implied by spherical symmetry), but it is not timelike (it is null on the two horizons and spacelike inside the black hole or outside the cosmological horizon).
     
  21. May 12, 2015 #20

    Chalnoth

    User Avatar
    Science Advisor

    I don't think so. One consequence of Birkhoff's theorem is that the spacetime inside a spherical shell is Minkowski.

    Right, which reduces to de Sitter geometry when m=0. And as with Birkhoff's original theorem, this is also valid for vacuum inside a spherical shell.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why isn't everything expanding in an expanding universe?
Loading...