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Flat Universe

  1. Aug 26, 2010 #1
    Now I don't quite understand what a flat universe means. Clearly we are not talking about dimensions right? Obviously universe has more than 2 dimensions. So what does flat stand for ? Don't start with Euclidian Geometry please.
  2. jcsd
  3. Aug 26, 2010 #2
    I'm not sure what you mean by "don't start with Euclidean Geometry" because a flat Universe is completely defined by the aspects of Euclidean Geometry, essentially it follows Euclid's postulates: http://mathworld.wolfram.com/EuclidsPostulates.html. By studying WMAP's representation of the CMB (http://en.wikipedia.org/wiki/WMAP/http://en.wikipedia.org/wiki/Cosmic_microwave_background_radiation [Broken]) Cosmologists have concluded with a 2% margin of error that the Universe is spatially flat hence the constant curvature (http://en.wikipedia.org/wiki/Constant_curvature) is [tex]\Omega[/tex]0 = 0. Other forms of geometry for the Universe include Hyperbolic Geometry (http://en.wikipedia.org/wiki/Hyperbolic_geometry) and Spherical/Elliptical Geometry (http://en.wikipedia.org/wiki/Spherical_geometry). The Constant of curvature in these geometries are:
    Hyperbolic = [tex]\Omega[/tex]0 = -1
    Spherical/Elliptical = [tex]\Omega[/tex]0 = 1
    Last edited by a moderator: May 4, 2017
  4. Aug 26, 2010 #3
    As far as I understand it, it is like this: Imagine a strictly fictional 4D Euclidean space (nothing to do with space-time). Now, let us consider hyper-surfaces (i.e. 3D surfaces in 4D space given by some relation [itex]f(x_{1}, x_{2}, x_{3}, x_{4}) = 0[/itex]) that are rotationally invariant, i.e. the functional dependence above is of the form:

    F(r, x_{4}) = 0, \; r = \sqrt{x^{2}_{1} + x^{2}_{2} + x^{2}_{3}}

    Here, we may use an analogy with 3D space and 2D (hyper)surfaces. If they are rotationally invariant around [itex]x_{3}[/itex], then they are of the form [itex]F(\rho, x_{3}) = 0, \; \rho = \sqrt{x^{2}_{1} + x^{2}_{2}}[/itex].

    One hypersurface is spherical if:

    r^{2} + x^{2}_{4} = a^{2}

    in analogy to a 3D sphere [itex]\rho^{2} + x^{2}_{3} = a^{2}[/itex].

    Another surface is hyperbolic (and open along the [itex]x_{4}[/itex] direction) if:

    x^{2}_{4} - r^{2} = a^{2}

    in analogy to the 3D hyperboloid [itex]x^{2}_{3} - \rho^{2} = a^{2}[/itex].

    Finally, a flat hyper-surface perpendicular to the [itex]x_{4}[/itex] direction may be written as:

    x_{4} = a \Rightarrow x^{2}_{4} = a^{2}

    in analogy to a 3D plane perpendicular to the [itex]x_{3}[/itex] direction [itex]x_{3} = a[/itex].

    A clever observation is that all of these cases correspond to an implicit functional relationship of the form:

    x^{2}_{4} + \Omega \, r^{2} = a^{2}

    and the different cases correspond to the values:

    [itex]\Omega = 1[/itex] - spherical

    [itex]\Omega = 0[/itex] - flat

    [itex]\Omega = -1[/itex] - hyperbolic

    Now, let us return to the question of the metric along this surface. The squared distance between to infinitesimally distant points is still given by the famous Euclidean form:

    ds^{2} = dx^{2}_{1} + dx^{2}_{2} + dx^{2}_{3} + dx^{2}_{4}

    but, not all 4 coordinates are independent because of the above functional relationship (we are fixed on the 3D hyper surface and cannot wander off of it). We can use that relationship to eliminate one of the coordinates, namely [itex]x_{4}[/itex]. Let us differentiate the above relation:

    2 \, x_{4} \, dx_{4} + \Omega \, 2 \, r \, dr = 0 \Rightarrow dx_{4} = -\frac{\Omega \, r \, dr}{x_{4}}

    Furthermore, we can still use spherical coordinates for the remaining three Cartesian coordinates and write:

    dx^{2}_{1} + dx^{2}_{2} + dx^{2}_{3} = dr^{2} + r^{2} \, \left( d\theta^{2} + \sin^{2}{\theta} \, d\phi^{2} \right)

    everywhere. The squared differential of the fourth Cartesian coordinate is:

    dx^{2}_{4} = \left( -\frac{\Omega \, r \, dr}{x_{4}} \right) = \frac{\Omega^{2} \, r^{2} \, dr^{2}}{x^{2}_{4}} = \frac{\Omega^{2} \, r^{2} \, dr^{2}}{a^{2} - \Omega \, r^{2}}

    where we have used the functional relation once more to eliminate the remaining [itex]x^{2}_{4}[/itex]. Combining everything together and simplifying the coefficient in front of [itex]dr^{2}[/itex], we get:

    ds^{2} = \frac{a^{2} + \Omega \, (\Omega - 1) \, r^{2}}{a^{2} - \Omega \, r^{2}} \, dr^{2} + r^{2} \, \left( d\theta^{2} + \sin^{2}{\theta} \, d\phi^{2} \right)

    This metric corresponds to specific classes of non-Euclidean 3D spaces.
  5. Sep 2, 2010 #4
    i think serpens asks a good question. it marks a chasm in the understanding of the layscientist (like myself) from the understanding of those with a heavy and appropriate mathematical background.

    i bet he's littlerally trying to picture "flat" space in 3d terms and going "what the hell."

    i wish that elementary schools and highschools would do a better job of teaching how our model of the big bang has evolved in the last 10 years....
  6. Sep 2, 2010 #5


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    Calling it a "flat" universe is JARGON. When scientists say "flat", they mean that the principles of Euclidean geometry are applicable (parallel lines never converge, three angles of a triangle add up to 180 degrees etc.). Again for emphasis: Euclidean is what flat means in this context. It does not mean two-dimensional. The OP seems to be wracking his/her brain over a contradiction that just stems from terminology/nomenclature.
  7. Sep 2, 2010 #6
    are you a science teacher or something? i wish that when i was in secondary school that notions of time and space, etc, were defined and taught with more of their relativistic signifigance....

    a lot gets lost - again for the layperson - in JARGON conflict!
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