yogi said:
Cepheid, Thanks for the info on the BAQs - I wasn't aware of these experiments
Also, I misspoke myself when typing "your post brings me back to my original question - I will re phrase it in the form of another question - why does the cosmological constant necessarily imply dark matter - why cannot space expand exponentially by some other mechanism. Perhaps we have been boxed in by models that may be too restrictive"
I meant to say "dark energy" rather than "dark matter" So I can understant your frustration
In your post #2 you state omega total must be one for flatness. My follow-up related to a pure de Sitter expansion - where Omega total is zero - Isn't the statement that "Omega must be one for flatness," a model dependent statement?
Thanks for your patience
Yogi
I'm not sure what you mean by model-dependent. First of all, let me address your lingering doubt that flatness requires Ω
tot = 1. To do that, let's take a look at the model (specifically the Friedmann world models). First we need to wrap our heads around the idea of the scale factor, 'a'. The scale factor is a dimensionless quantity, and it is a function of time: a(t). Consider any two objects in the universe. Basically, you can think of a(t) as the ratio: (separation of objects at time t)/(separation of objects now). So, right now, a = 1, and at any time in the past, a < 1, with it being smaller the farther back you go (because the universe has been expanding with time, so if you go back to earlier times, separations between objects decrease). So all of the information about the expansion history of the universe is encoded in this function a(t). This function is determined by a differential equation called the Friedmann equation, which is in turn derived from the Einstein Field Equations of General Relativity. The Friedmann equation is as follows:
\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}\rho_{\text{tot}} - c^2\kappa
where the overdot represents a derivative with respect to time i.e. \dot{a} \equiv da/dt. On the right hand side, \rho_{\text{tot}} is the total energy density of the universe (taking into account all constituents) and it is also a function of time \rho_{\text{tot}}(t). The second term has \kappa, which is the spatial curvature. It is the reciprocal of the radius of curvature. Basically, for κ > 0 (positive curvature) the geometry of the universe is closed, for κ = 0, the geometry of the universe is flat (i.e. Euclidean), and for κ < 0 (negative curvature), the geometry of the universe is open. Now, it can be shown that \dot{a}/a = H where H is the Hubble parameter (also a function of time). So, substituting that in, and rearranging the equation (and also switching to the c = 1 unit system that cosmologists use) we obtain:
\kappa = \frac{8\pi G}{3}\rho_{\text{tot}} - H^2
Now, let's consider the value of the spatial curvature today. In other words, consider this equation at time t = t
0, where t
0 is the present age of the universe. The value of the Hubble parameter today, H(t
0), is just H
0, what we call the Hubble constant. So we get:
\kappa = \frac{8\pi G}{3}\rho_{\text{tot}}(t_0) - H_0^2
So the condition for flatness (zero spatial curvature) is:
0 = \frac{8\pi G}{3}\rho_{\text{tot}}(t_0) - H_0^2
H_0^2 = \frac{8\pi G}{3}\rho_{\text{tot}}(t_0)
\rho_{\text{tot}}(t_0) = \frac{3H_0^2}{8\pi G}
The density required for flatness is therefore \rho_{\text{crit}} = 3H_0^2 / 8\pi G. Cosmologists call this value the
critical density. For the universe to be flat, the total energy density today must be equal to the critical density. I'm going to omit the t
0 argument to the density functions henceforth, and it will just be understood that all densities referred to are present values. Anyway,
by definition, the density parameter \Omega_i of the "i
th" constituent of the universe is given by:
\Omega_i \equiv \frac{\rho_i}{\rho_{\text{crit}}}
The total density parameter is just the sum of the density parameters for all the individual consituents (baryonic matter, dark matter, radiation (photons), and dark energy).
\Omega_{\text{tot}} = \frac{\sum_i \rho_i}{\rho_{\text{crit}}} = \sum_i \Omega_i
In light of the above, we can rewrite the flatness criterion that I derived above as:
\rho_{\text{tot}} = \rho_{\text{crit}}
\Rightarrow \frac{\rho_{\text{tot}}}{\rho_{\text{crit}}} = 1
\Rightarrow \Omega_{\text{tot}} = 1
In conclusion, that's why flatness implies that Ωtot = 1. It comes straight from the Friedmann equation.
Now, you posed a question about an empty universe (Ω
tot = 0). I can certainly understand your confusion. We all know that General Relativity says that mass/energy curves spacetime, so we'd expect an empty universe to have zero spatial curvature and to just reduce to the flat Minkowski spacetime of special relativity. However, if you look at the Friedmann equation, you can see that that's not what General Relativity actually says will happen. Instead, there is some intrinsic curvature to the spacetime of the Friedmann world model even with no energy content (ρ
tot = 0):
\kappa = \frac{8\pi G}{3}\rho_{\text{tot}}(t_0) - H_0^2
\Rightarrow \kappa = - H_0^2
Apparently, this model actually has maximally negative spatial curvature, and as a result its geometry is open (hyperbolic geometry). This particular Friedmann model, the empty universe, is known as the Milne model:
http://en.wikipedia.org/wiki/Milne_model
Note: a de Sitter universe (yet another special case of the Friedmann model) is not exactly an empty universe, because it assumes non-zero \Lambda and hence non-zero \Omega_\Lambda, which means that \Omega_{\text{tot}} \neq 0. The de Sitter space is empty of everything
except dark energy.
I hope that this addresses some of your concerns.
