Chapter 3: THE FRIEDMANN MODELS

Here S k (u) and C k (u) are analogous to S k (ω) and are sin, sinh, cos, cosh etc.
depending on the value of k. Then
2
Sk
( u)
C
dη
= 2
1−
kS
= 2S
2
k
( u)
du
= k(1
− C
k
⎛ S
η = k⎜u
−
⎝
2
k
( u)
du
( u)
(2u))
du
k
k
(2u)
⎞
⎟
2 ⎠
Now making a final substitution of Θ = 2u, we get parametric solutions
k
η = ( Θ−Sk
( Θ))
2
k
ζ = ( −Ck
2 1 ( Θ))
So, we finally have for R and τ:
(3.46)
kGM
R =
2
Ac
kGM
τ =
3
c
{ 1−
C ( Θ)
}
k
{ Θ − S ( Θ)
}
k
Note that dτ = AR/c dΘ. The parameter Θ is (as noted above) called the "conformal
time" (although we will not make great use of it), and dω is proportional to dΘ for a
light ray.
3.6.2 Determining S k (ω) in the matter-dominated case
We saw in the previous chapter how the effective distance D enters into almost all the
relations between the intrinsic and observed properties of objects. Recall that D = R 0
S k (ω). Thus we need to know S k (ω) or rather S k (z) since the redshift z is the only real
observable related to distance.
We want to get ω(z), which we can get by integrating dω/dz, which in turn can be
obtained by recognizing that:
dω
dz
dω
dt
= × ×
dt dR
We had from equation (3.21):
dR
dz
(3.47)
R&
R
2
2
= H ( 1+Ω
z)
2 0
0
0