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I Does the Universe's Expansion lead to Critical Mass < 1?

  1. Feb 14, 2016 #1
    We understand that for the Universe to be flat, the critical mass of everything that has mass must total an omega of 1.0.

    At first, with our lack of knowledge regarding dark energy, we thought that the universe that we detected (ordinary matter and dark matter) did not even add up to a third of this critical amount. Yet, the observable universe was found to be flat.

    Enter dark energy. Its corresponding mass (I'm making the assumption that this is due to mass-energy equivalence) is thought to bring up the universe's mass up to the critical density needed for a flat universe.

    As the universe expands, there is more total dark energy as the amount of volume increases and the density remains constant. Yet, the amount of ordinary matter and dark matter remain the same.

    Therefore, as the universe expands, the total amount of mass per unit volume must be decreasing.

    So wouldn't that imply that over time, the density of the universe is decreasing to below a density of 1?

    Below is a mathematical interpretation of what I'm getting at:

    • Ordinary Matter: 0.04 of Critical Density
    • Dark Matter: 0.24 of Critical Density
    • Dark Energy: 0.72 of Critical Density
    In a flat universe:

    (0.04 + 0.24 + 0.72) / 1 = 1.00 Flat Universe

    Now, let's say that the universe expands to twice its size:

    Ordinary matter (0.04) and dark matter (0.24) still have the same amount of mass, but spread over twice the volume. Only dark energy's density is constant, so its mass increases by the corresponding amount (0.74 * 2):

    (0.04 + 0.24 + 0.72 * 2) / 2 = 0.02 + 0.12 + 0.72 = 0.86 Saddle Shaped Universe

    Is it possible that as the volume of the universe increases, its density (and hence shape) changes so that it's no longer flat, given that no new ordinary matter or dark matter is added?
  2. jcsd
  3. Feb 14, 2016 #2


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    A constant ##\Lambda## does not imply a constant ##\Omega_\Lambda##. The critical density changes with time as it depends on the Hubble rate.
  4. Mar 31, 2016 #3
    We should know where the Ω function comes.
    [tex]\Omega_i=\frac{\rho_i}{\rho_c}[/tex] where ρc is the critical density (those who gives a flat universe) and ρi is the density of what we're considering (matter, dark energy...).
    If the unverse is flat, the critical density and the density is the same, so Ω=1.
    If we write [tex]\Omega=\sum{\Omega_i}=\sum{\rho_i}=\rho[/tex] and we multiply both sides by the Volume, we get that [tex]E(t)=\sum{E(t)_i}[/tex] where E(t) is the energy (equivalent to the matter). So the volume doesn't modify the equation.
    Finally we can ask for the energy to be constant. That's the same as do [tex]\frac{dE(t)}{dt}=0=\frac{d}{dt}\sum{E(t)_i}=0→\sum{E(t)_i}=0[/tex] so, if the sum of the energies of each part of the universe must be zero (and then is needed to exist some negative energy, that's the dark energy). If this happens, he universe is flat.
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