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Steady state state model consistent with Big Bang

  1. Aug 17, 2011 #1
    The model I have is very simple it is:

    a(t) = A * (exp(H t) - 1)

    where H^2 = (8 Pi G /3 ) rho_min

    and A is just a factor to make a = 1 at the present time t_0:

    A = 1 / exp(H t_0 - 1)

    and rho_min is the limiting density of matter in the Universe for large t.

    The model starts off linear with q = 0, like Milne's massless model, and then quickly tends to Fred Hoyle's steady state Universe with an accelerating expansion q = -1, a constant Hubble radius and continuous matter generation. The equation for the deceleration parameter q = -1 + exp(-H t). It has no cosmological constant but it has an equation of state:


    Pressure = - density c^2 (3 - 2 exp(-H t)) / 3

    As t -> 0: Pressure = - density c^2 / 3

    As t -> infinity: Pressure = - density c^2

    For large t we have the second equation of state. As the pressure is multiplied by three in the Friedman acceleration equation:

    a'' / a = - 4 Pi G / 3 (rho + 3 p / c^2),

    this means that this model predicts that the "dark energy" density should be 3 times the matter density which is consistent with the current consensus.

    The problem with Hoyle's steady state was that it was not consistent with the cosmic microwave background and the Big Bang.

    However my model does describe a point of infinite mass density as t approaches zero. Interestingly as t->0 the equation of state becomes:

    pressure = - density c^2 / 3

    This imples the space-time curvature a'' / a -> 0 so there is no singularity at t = 0. The model behaves like the Milne model as t -> 0! I think this is better than the standard big bang model. Having no singularity implies the entropy at t = 0 is zero. This is consistent with the 2nd law of thermodynamics and the arrow of time.
     
    Last edited: Aug 17, 2011
  2. jcsd
  3. Aug 17, 2011 #2
    Just wondering if you could show the basic Friedmann equation and how you modeled the terms to arrive at this solution of the scale factor?
     
  4. Aug 18, 2011 #3
    I start from the Friedmann equations with spatial curvature k=0 and cosmological constant=0:

    (a' / a)^2 = 8 Pi G rho / 3

    2 a'' / a + (a' / a)^2 = -8 Pi G p / c^2

    where rho is the matter density and p is the pressure.

    If you substitute my scale factor equation:

    a(t) = A (exp(H t) - 1)

    where H and A are constants

    into the Friedmann equations then you get a solution to them provided the following equation of state holds:

    p = - rho c^2 (3 - 2 exp(-H t)) / 3.
     
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