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

Radius of the Universe

  1. Sep 5, 2010 #1
    "Radius" of the Universe

    Hi,

    I have a very simple question about GRT. I use a picture which I personally like. Let us have a shpere (2D, embedded in 3D). Physics do only work on this sphere in this theoretical universe, so the forces work only along geodatics (there is no "above" or "below"). The distance of objects depends on the radius of this sphere R(t) (which is in GRT determined by the energy-momentum tensor, but this not important here). Now we measure the distance to far objects by the redshift which means that we do not measure the radius but just the distance. For this theoretic universe we would need to follow that ds is always < c dt. The moving bodies are not allowed to be faster than c. If they would move with c we would never see them. This would hold for every part of this space since we have a sphere. Conclusion: The expansion allows to see objects, meaning it is always below c - but this is still not a radius we measure. Statistically it would be sufficient to measure some redshifts to have a measure for R(t). This seems to be different from the 4D space, embedded in intself. Nevertheless: The only think I can do on the sphere is to use a high resolution telescope to look as far as possible and potentially see myself if I look around.
    Why do we call it in the reality a radius since our measurement again is only along geodectics in space-time? Isn't it just the maximum on a geodetic we can look so far?

    Thanks!
     
  2. jcsd
  3. Sep 5, 2010 #2
    Re: "Radius" of the Universe

    I dont know who you think calls what a radius but I do know that the radius of the universe is unknown
     
    Last edited: Sep 5, 2010
  4. Sep 5, 2010 #3

    JesseM

    User Avatar
    Science Advisor

    Re: "Radius" of the Universe

    "Above" or "below" what, exactly? Are you taking the curved 2D surface of the sphere as an analogy for 3D space (as in the balloon analogy for an expanding universe with positive spatial curvature), so that we picture the beings living on the surface as 2D flatlanders who cannot travel "up" or "down" relative to the surface? Or is the sphere supposed to represent the edge of the observable universe for us, so we're at the center and can't see anything outside of it?
     
  5. Sep 5, 2010 #4
    Re: "Radius" of the Universe

    Exactly, I mean an 2D universe, a "flatlander" universe. This means there is no other dimension visible but time, so carefully I should call it a 3D hypersurface, since time is naturally present, unfortunately empedded in 4d. This is the tricky part, simply to accept that physics work only on the surface. SRT works on every part of this sphere, locally a flat space.
     
  6. Sep 5, 2010 #5
    Re: "Radius" of the Universe

    Thanks for the link, this is what I asked in another thread before. I like this 2D surface anology.
    The question is what this anology means for our measurements and for the meaning. Does a radius make sense if we have only geodetics? Isn´t the only thing we have measurements of curvature?
     
  7. Sep 5, 2010 #6

    Mentz114

    User Avatar
    Gold Member

    Re: "Radius" of the Universe

    In 2D the Riemann tensor has only one independent component, but I don't know about 2+1 D, or whether this is relevant to the question.
     
  8. Sep 5, 2010 #7

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: "Radius" of the Universe

    Actually that's not true: Davis and Lineweaver, "Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe," http://arxiv.org/abs/astro-ph/0310808

    Cosmologists don't typically talk about the radius of the universe, because that only works if it's a closed universe. Instead, they usually talk about the spatial curvature of the universe. This curvature can be positive (for a closed universe), zero (for a spatially flat universe), or negative (for an open universe). In theory, one way of determining this would be to construct a triangle out of laser beams, with all three vertices of the triangle moving with the Hubble flow, i.e., at rest with respect to the average motion of the nearby galaxies. You detect positive, zero, or negative curvature depending on whether the sum of the triangle's interior angles is greater than, equal to, or less than 180 degrees. In practice, that isn't what we actually do. The main sources of data are actually supernovae and the cosmic microwave background. See http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html#Section8.2 [Broken] , subsections 8.2.6 and 8.2.8.
     
    Last edited by a moderator: May 4, 2017
  9. Sep 6, 2010 #8
    Re: "Radius" of the Universe

    Interesting! I hope I will find the time to read it carefully (I am not a cosmologist). Since you are familiar with the paper (which is so far unpublished?), back to the 2D picture, how is exchange of information possible if v > c? Shouldn't "the light goes out" if this happens?

    That's what I expected.
    Exactly what I mean.
    This sounds impossible, isn't it? The biggest natural distance we could start this measurement from is 2 AE and I don't know where the reflector should be... please help me if I am wrong.
    Thanks again, I need to study this.
     
    Last edited by a moderator: May 4, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Radius of the Universe
  1. Schwarzschild radius (Replies: 6)

  2. Møller radius (Replies: 9)

Loading...