Problem #1 [Structure Formation I: Radiation Era]

For the present horizon size we must set z = 0 into the above expressions. For Ω m = 0.3 we use
[
]
c
d H (z = 0) = √ cosh −1 2(1 − 0.3)
1 + = 8.7 Gpch −1
H 0 1 − 0.3 0.3
For Ω m = 1 we use
For Ω m = 3 we use
d H (z = 0) =
d H (z = 0) = 2c
H 0
= 6.0 Gpch −1
[
c
√ cos −1 1 −
H 0 3 − 1
]
2(3 − 1)
= 4.0 Gpch −1
3
(c). Show that at high redshifts 1 + z ≫ Ω −1
0,m
, a good approximation for all three cases is given
by
2c
d H (z) ≈ √ (1 + z) −3/2
Ω0,m H 0
(you may find the identity cosh −1 x = ln(x + √ x 2 − 1) useful.)
For Ω 0,m < 1 we will use
d H (z) =
c
H 0
√
1 − Ω0,m
(1 + z) −1 cosh −1 (1 + α)
where α = 2(1 − Ω 0,m)
(1 + z)Ω 0,m
using the above relationship for cosh −1 we find
√
cosh −1 (1 + α) = ln[1 + α + (1 + α) 2 − 1] = ln[1 + α + √ 2α]
but we know that
√
2α > α
and we can just write
and we know that the Taylor expension for
so we now get
so now the distance horizon becomes
d H (z) ≈
cosh −1 (1 + α) = ln[1 + √ 2α]
ln(1 + x) ≈ x
cosh −1 (1 + α) ≈ √ 2α
c
H 0
√
1 − Ω0,m
(1 + z) −1√ 2α ≈
6
2c
√
Ω0,m H 0
(1 + z) −3/2