2.4 Einstein’s static model and Lemaître’s model

So far we have shown that solutions of the Einstein equation are dynamical in general, i.e., the scale factor a is time-dependent. As a digression, let us examine why Einstein once introduced the Λ-term to obtain a static cosmological solution. This is mainly important for historical reasons, but is also interesting to observe how the operationally identical parameter (the Λ-term, the cosmological constant, the vacuum energy, the dark energy) shows up in completely different contexts in the course of the development of cosmological physics.

Consider first the case of Λ = 0 in Equations (4View Equation) and (5View Equation). Clearly the necessary and sufficient condition that the equations admit the solution of a = const. is given by

3K ρ = − 3p = ------2. (26 ) 8πGa
Namely, any static model requires that the Universe is dominated by matter with either negative pressure or negative density. This is physically unacceptable as long as one considers normal matter in the standard model of particle physics. If Λ ⁄= 0 on the other hand, the condition for the static solution is
K = 4 πG (ρ + p )a2, Λ = 4πG (ρ + 3p), (27 )
1 ( 3K ) 1 ( K ) ρ = ----- -2--− Λ , p = ----- Λ − --2 . (28 ) 8πG a 8 πG a
Thus both ρ and p can be positive if
K 3K -2-≤ Λ ≤ --2-. (29 ) a a
In particular, if p = 0,
-Λ--- 2 ρ = 4πG , K = Λa . (30 )
This represents the closed Universe (with positive spatial curvature), and corresponds to Einstein’s static model.

The above static model is a special case of Lemaître’s Universe model with Λ > 0 and K > 0. For simplicity, let us assume that the Universe is dominated by non-relativistic matter with negligible pressure, and consider the behavior of Lemaître’s model. First we define the values of the density and the scale factor corresponding to Einstein’s static model:

Λ ρE = ----, K = Λa2E. (31 ) 4πG
In order to study the stability of the model around the static model, consider a model in which the density at a = aE is a factor of α(> 1) larger than ρE. Then
( ) ( ) aE- 3 α-Λ-- aE- 3 ρ = αρE a = 4πG a , (32 )
and Equations (4View Equation) and (5View Equation) reduce to
Λ ( 2αa3 ) a˙2 = -- ---E-− 3a2E + a2 , (33 ) 3 [ a ] ¨a Λ (aE )3 --= -- 1 − α --- . (34 ) a 3 a
For the period of a ≪ aE, Equation (33View Equation) indicates that a ∝ t2∕3 and the Universe is decelerating (¨a < 0). When a reaches α1 ∕3aE, ˙a2 takes the minimum value Λa2(α2 ∕3 − 1) E and the Universe becomes accelerating (¨a > 0). Finally the Universe approaches the exponential expansion or de Sitter model: ∘ ---- a ∝ exp(t Λ ∕3). If α becomes closer to unity, the minimum value reaches zero and the expansion of the Universe is effectively frozen. This phase is called the coasting period, and the case with α = 1 corresponds to Einstein’s static model in which the coasting period continues forever. A similar consideration for α < 1 indicates that the Universe starts collapsing (˙a2 = 0) before ¨a = 0. Thus the behavior of Lemaître’s model is crucially different if α is larger or smaller than unity. This suggests that Einstein’s static model (α = 1) is unstable.
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