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Problem understanding how universe is infinitely flat

  1. Oct 1, 2013 #1

    I can't visualize how the universe could be infinitely flat according to the big bang theory...

    The only way I can visualize this is like a cone surface, one dimension supressed and left with a circle line (the universe), the other dimension thus forming the cone, is time. Now, in a fourth dimension the ends of the circle never meet, thus forming a spiral line. This universe is flat (as cones are topologically flat surfaces), infinite and isotropically expanding.

    I am confused. I needed 4 dimensions to represent a universe which is only a line. How many dimensions are needed for the familiar 3D universe? 5 dimensions suffice?

    How is it done? Please help.

    Or maybe expansion is accelerated. Then we have a hyperboloid-like surface. Is my thinking any valid?
  2. jcsd
  3. Oct 1, 2013 #2
    I think your problem is wrapping the 1-D line back on itself into a circle. The 1 D line would stretch into infinity in both directions. Adding in time, this forms a 2-D sheet where the distance between points is found by sqrt(t^2 - x^2 ) or sqrt(x^2 -t^2 ) depending on which convention you like. Adding in another spatial dimension forms 3 D where we subtract or add y^2 like the x^2 in the previous two equations. Adding in a third spatial dimension forms a 4 D hypervolume where distances are found similarly: sqrt(+/- t^2 -/+ x^2 -/+ y^2 -/+ z^2 ).

    What is meant by it being flat is that parallel lines never converge or diverge, and triangles always have pi radians in interior angles, and the ratio of circumference to diameter is pi.

    Edit:NB, the universe is *presently* flat (to the best of our ability to measure). The shape has changed over time.
    Last edited: Oct 1, 2013
  4. Oct 1, 2013 #3


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    I don't think the big bang theory does say that it is infinitely flat. Can you provide a reference?
  5. Oct 1, 2013 #4
    Oh, never mind... I am a mathematician and not familiar with the theory.
    I think, I was trying to say infinite, flat and expanding. I was trying to visualize the thing macroscopically, not from the differential point of view. Sorry, if i have offended you.:redface:
  6. Oct 1, 2013 #5
    Well from a mathematics perspective, there's a scaling parameter that is a function of time: a(t). So distances (in any direction) is scaled by a(t). I'm not precisely sure how one would describe the overall space-time, but the "flatness" refers to a more-or-less Euclidean space, with time coupled like a Minkowski space.
  7. Oct 1, 2013 #6
    I was more thinking of a 3-torus, and trying to tile the infinite Minkowski plane with identical copies of this domain.
    The analogy to an infinite roll of paper folded to a cone for a constant-rate expanding universe is self-evident, but are cosmologists talking of the same thing here?
  8. Oct 1, 2013 #7
    There remains an open possibility that the universe could be a flat 3-torus, so long as the volume was larger than that of our observable universe. But it's an additional complication in the geometry unjustified by physical observation, so there's no particular push to drive acceptance of this model. The generally accepted one is that of an infinite Euclidean 3-D volume with a time-like dimension as well.
  9. Oct 1, 2013 #8
    I see... Because, all this time I have been thinking that all expansion-contraction was an issue of the curvature of space-time.
    But if the universe is truly flat, what is causing it to expand? More so, if the spatial universe boils down to an infinite stretching line, what could be the Minkowski spacetime for an isotropically, let's say, constant-rate expanding line?
    Why I can't visualize the thing without any curvature of the resulting spacetime ??!
  10. Oct 1, 2013 #9


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    The expansion is not related to the sectional curvature of the hypersurfaces in the space-like foliation determined by the CMB frame but rather to the spatial metric tensor relative to the CMB frame and the covariant derivative of the 4-velocity field of the CMB frame: ##\nabla_a u_b = \frac{1}{3}h_{ab}\theta## where ##\theta## is the expansion (defined by ##\theta = \nabla_a u^a##). By the way, the Friedman space-time is obviously not flat; when we say the universe is flat we just mean that the sectional curvature of the aforementioned hypersurfaces vanishes.
    Last edited: Oct 1, 2013
  11. Oct 1, 2013 #10
    what they said.

    Essentially a single "time slice" in a frame where the CMB is isotropic (ie at rest to the CMB frame, measured by the dipole moment of the measured anisotropy), and roughly around present time, is a 3D volume with no measured global curvature. (though of course there are many small pockets of local curvature, small being galactic clusters and smaller).

    What drives curvature is that measures of space and time are intimately coupled to the energy of "stuff" within that space-time. Some energy (like mass) pulls measures of space and time "closer together" (rulers are shorter and clocks run longer at the bottom of a gravitational well compared to the top). Other energy (like radiation pressure, dark energy) push measures of space and time "further apart" (rulers appear longer over time).

    Note also, that expansion does not occur in regions of space that have significant portions of mass (galactic clusters and smaller) where gravitation is the apparent behaviour of space-time. It is simply that our universe has a lot more space without massive stuff, so more expansion (on the whole) occurs than "contraction."
  12. Oct 2, 2013 #11


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    Well, it is. In fact, if you compute the Ricci curvature scalar in an FRW universe, you end up with two terms. One is directly related to the rate of expansion. The other is the spatial curvature. When people say, "the universe is flat," they're only talking about the spatial curvature.

    Of course, it's worth mentioning that the "spatial curvature" is coordinate-dependent, but the key point here is that it's possible to choose a set of coordinates where, at least to current measurement accuracy, there doesn't appear to be any spatial curvature.

    Either way, it's perfectly sensible to consider the expansion as being a manifestation of space-time curvature. It's just possible to select specific three-dimensional hypersurfaces that don't have any curvature (or, at least, that have a radius of curvature far larger than the observable universe).
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