Chapter 3: THE FRIEDMANN MODELS

We may write the density of the Universe at any R in terms of the present density (i.e.
at R = R 0 ) which is parameterized by Ω 0 for the different components.
(3.32)
8πGρ
3H
2
0
⎛
⎜Ω
⎜
⎝
⎛ R ⎞
⎜
⎟
⎝ R0
⎠
−3
+ Ω
⎛ R ⎞
⎜
⎟
⎝ R0
⎠
−4
+ Ω
=
0, m
0, r
0, Λ
⎞
⎟
⎟
⎠
We can introduce a new time coordinate, η, called the conformal time though we
won’t encounter it much, such that
(3.33)
cdt
d η =
AR
so that, indicating differentiation w.r.t η with a prime, the Freidmann equation
becomes:
(3.34)
⎛ R′
⎞
⎜
⎟
⎝ R ⎠
2
8πGρ
2
= A R
2
c
0 3
2
⎛
⎜
⎝
R
R
0
⎞
⎟
⎠
2
⎛
− k
⎜
⎝
R
R
0
⎞
⎟
⎠
2
We can now substitute in the expression above for ρ(R)
⎛ R′
⎞
⎜
⎟
⎝ R0
⎠
2
=
H
c
2
0
2
⎛
⎜Ω
⎜
⎝
0, Λ
+ Ω
0, m
⎛
⎜
⎝
R
R
0
⎞
⎟
⎠
−3
+ Ω
0, r
⎛
⎜
⎝
R
R
0
⎞
⎟
⎠
−4
⎞
⎟ 2
A R
⎟
⎠
2
0
⎛
⎜
⎝
R
R
0
⎞
⎟
⎠
4
⎛
− k
⎜
⎝
R
R
0
⎞
⎟
⎠
2
and can clean this up with a = R/R 0 and substitute in the (3.27) relation:
k H
0
(3.35) = ( Ω0,
−1)
2 2 2 tot
R A c
and rearrange to get:
0
2
(3.36)
a′
2
=
k
( ) ( 2
4
Ω
)
0, r
+ Ω0,
ma
− ( Ω0,
tot
−1)
a + Ω0,
Λa
Ω −1
0
This is straightforward to solve provided that Ω 0,Λ = 0 (see 3.6 below).
Taking this equation and substituting in the relation for ρ(R) above (3.32) it is easy to
show that H(R) is given by:
2 2
−3
−4
−2
(3.37) H = H ( Ω + Ω a + Ω a − ( Ω −1)
a )
0
0, λ
0, m
0, r
0, tot
This tells us the value of Hubble’s parameter at all epochs.