Out of the Ashes of the Friedmann Model: Is the Universe Condemned?
Updated: Oct 18, 2019
The Friedmann model presumably combines a space curvature, k, with a "radius of curvature" of the universe, a(t). Neglecting numerical factors, and the Einstein cosmological constant, the Friedmann equation of cosmology is
where rho is the constant density of matter. Taking the time derivative of this equation leads to
where the asterisk denote time derivatives. From this we conclude that the Friedmann model treats an accelerating universe. From the first equation, this permits the space curvature to be of indeterminate sign.
Now, one can lodge a complaint that in order to derive the sectional curvature, the "effective material density of the universe" must be constant. Then, there would be no critical density that occurs at a critical time when the space curvature, k, vanishes. Yes, that is true, for, otherwise the eigenvalues of the Ricci tensor would not all turn out to be 3 rho, but contain time derivatives of the density, and would be unequal. However, this would destroy the isotropy of the universe. It would look different in different directions to our chagrin. Consequently, the sectional curvatures must all be equal (homogeneity), and constant (isotropy).
This implies that the density, rho, must be constant, and, moreover, there is no difference between the eigenvalues of the Ricci tensor. That the universe is decelerating implies that the spatial curvature, k, must be positive. (In the case of an accelerating universe, it can be negative; but, it must always be bounded from above by the square of the velocity, a*.)
So setting them equal to different components of the energy-stress tensor is simply wrong. As Robertson and Noonan have pointed out, it would make the density equal to a negative pressure!
At a constant radial coordinate that supposedly is comoving with the galactic gas as the universe accelerates, the metric is
where dS is an element of a 2-sphere. Since the radius of curvature, a(t) is a sole function of time, we can swap time dt for da by writing da=a*dt, so that the metric becomes
ds^2=-da^2/(k+rho a^2)+a^2 dS.
The geodesic equation of motion is
where A=1/(k+rho a^2) and the prime denotes differentiation with respect to a.. The coefficient of the second term is the pertinent Christoffel symbol. It is a simple matter to check that it is the same as the sectional curvature above.
If we go from constant mass density to the gravitational potential, the coefficient A changes into A=1/(k-2m/a^3), where m is the mass of the star. It is apparent that it has the wrong sign since k, the total energy, would necessarily be positive But, this is precisely what is required in order to get an accelerating universe:
with positive sectional curvature. Again, it is easy to check that the sectional curvatures all add up to zero, just as in the Schwarzschild outer solution.
Now, it is anti-gravity that is driving the expansion of the universe. Hence. the opposite must be true, which brings to mind Newton's "island universe" that would collapse into itself. Transforming from potential to constant density, Newton believed that it was more realistic to believe that matter was originally smeared out uniformly throughout an infinite universe. That being the case, Bentley queried Newton that there would be no reason to move and matter wouldn't clump together in the form of stars and galaxies. Even appealing to the time reversibility of mechanics, Newton was not able to overcome the contradictions expressed by Poisson's equation when applied to the universe.
What is sure however, is that the Friedmann model does not predict any critical density at a specific instant in time when the so-called spatial curvature vanishes. The density is constant like in Poisson's equation. All reference to things like dark matter, and dark energy, are figments of an incorrect model being applied to understand the evolution of the universe. In this regards, science has not progressed since Newton's time!