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Critical density formula hides fundamental relationship?

  1. Jun 22, 2011 #1
    Observations show that the current density of the Universe is close to the critical density:

    rho = 3 H_0^2 / 8 Pi G

    Assuming a spherical observable Universe of constant density, together with the relation c = H_0 * R, one can rearrange the above equation to give the relation:

    G * M / R = c^2 / 2

    where M is the mass and R is the radius of the observable universe at the current cosmological time.

    Perhaps this is a fundamental relationship that holds at all times. It seems to embody Mach's principle in that it says that half the mass/energy of any particle comes from the mutual gravitational potential energy between the particle and the rest of the observable Universe. (To see this multiply both sides by the particle mass m). In fact this relationship seems to provide a natural definition of the "observable" universe as the "gravitationally interacting" Universe.

    If you plug the above relationship into the Friedmann equations one gets a linearly expanding model that behaves exactly like the massless Milne model!
     
    Last edited: Jun 22, 2011
  2. jcsd
  3. Jun 23, 2011 #2

    Chalnoth

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    If you do it properly, everything will cancel, because the equation you wrote down is the equivalent of the Friedmann equations.
     
  4. Jun 23, 2011 #3
    The equation defines a model of the Universe that is spatially flat and linearly expanding that obeys the Friedmann equations.

    The first Friedmann equation says:

    [(a' / a)^2 - 8/3 G Pi rho] a^2 = -kc^2

    where a is the scale of the Universe, rho the density and k is the spatial curvature.

    If I use the relationship G M / R = c^2 / 2 for a spherical observable universe of Mass M and radius R then I can get a formula for the density rho as

    rho = M / V = (c^2 * R / 2 G) * 1 / (4/3 Pi R^3)

    rho = 3 c^2 / 8 Pi G R^2

    Now for the Hubble constant we have the formula:

    c = H R

    Therefore the term 8 /3 G Pi rho is actually H^2.

    We also know the term (a' / a)^2 = H^2 by the definition of the Hubble constant.

    Thus we find that this solution implies

    -k c^2 / a^2 = H^2 - H^2 = 0

    therefore the spatial curvature is zero. Notice we haven't set it to zero beforehand it came out naturally from the model.
     
  5. Jun 23, 2011 #4

    bapowell

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    This is a linear approximation valid only below the curvature scale; so you are implicitly assuming a flat universe by using the Hubble relation.
     
  6. Jun 23, 2011 #5

    bapowell

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    And this is a spherical mass distribution embedded in flat space...
     
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