Steady state state model consistent with Big Bang

In summary, the model I have is very simple it is: a(t) = A * (exp(H t) - 1) where H and A are constants and A is just a factor to make a = 1 at the present time t_0: A = 1 / exp(H t_0 - 1). 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:
  • #1
johne1618
371
0
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.
 
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  • #2
johne1618 said:
The model I have is very simple it is:

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

...
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?
 
  • #3
edgepflow said:
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?

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.
 

1. What is the Steady State Model?

The Steady State Model is a cosmological theory that suggests the universe is both expanding and in a constant state, meaning the overall density of matter and energy remains the same over time. This theory is in contrast to the Big Bang Theory, which proposes that the universe began with a single, explosive event.

2. How does the Steady State Model explain the expansion of the universe?

The Steady State Model proposes that new matter and energy are constantly being created to fill the gaps left by the expansion of the universe. This process is known as the "continuous creation" of matter. This theory also suggests that the universe has no beginning or end, and has always existed in a steady state.

3. Is the Steady State Model consistent with the Big Bang Theory?

No, the Steady State Model is in direct contradiction to the Big Bang Theory. While the Big Bang Theory proposes that the universe began with a single event, the Steady State Model suggests that the universe has always existed in a constant state.

4. What evidence supports the Steady State Model?

One piece of evidence that was once thought to support the Steady State Model was the observation of distant galaxies. These galaxies were thought to be evenly distributed throughout the universe, which would be consistent with a constant state. However, further research and observations have since disproven this theory.

5. Is the Steady State Model widely accepted in the scientific community?

No, the Steady State Model has largely been rejected by the scientific community in favor of the Big Bang Theory. This is due to the overwhelming amount of evidence that supports the Big Bang Theory, including the cosmic microwave background radiation, the abundance of light elements in the universe, and the observed redshift of distant galaxies.

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