Introduction to Cosmology

(Marvins-Underground-K-12) #1
Model Testing and Parameter Estimation 195

occurred when훺m푎−^3 =훺r푎−^4. Using the value of훺mfrom (8.50) and훺rfrom (8.54)
would give푎−^1 = 1 +푧≈5773. This is not correct, however, because the푎dependence
of훺rshould actually be given by


훺r(푎)=

푔∗(푎)


2


푎S푇^4


휌c

=


푔∗(푎)


2


푎S


휌c

(


2. 725



) 4


, (8.55)


using the function푔∗discussed in Chapter 5. In the region of 1+푧≳1000 neutrinos
will be relativistic and푔∗= 3 .36 instead of 2 [the contribution to the integral in Equa-
tion (5.55) from large푧≳ 108 ,where푔∗(푎)> 3 .36 is negligible]. Temperature scales
inversely with푎, thus at푧eq=3280 from WMAP 9-year data [12].
Inserting the values of the energy densities in Equations (8.50) and (8.53) into
Equation (5.55), one finds the age of the Universe to a remarkable precision,


푡 0 = 13. 82 ± 0 .05 Gyr, (8.56)

is in excellent agreement with the independent determinations of lesser precision in
Section 1.5. The WMAP team have also derived the redshift and age of the Universe
at last scattering and the thickness of the last scattering shell (noting that the WMAP
team use the termdecouplingwherewehaveusedlast scattering surface—see our def-
initions in Section 6.3):


푡LSS= 0. 379 +−^00 .. 007008 Myr, 훥푡LSS= 0. 118 +−^00 ..^003002 Myr,

1 +푧LSS= 1089 ± 1 ,훥푧LSS= 195 ± 2.

}


(8.57)


Deceleration Parameter. Recalling the definitions


퐻 0 =

푎̇ 0


푎 0


,훺m=

8 휋퐺휌m
3 퐻 02

,훺휆=



3 퐻 02


,푞 0 =−


푎̈ 0


푎 0 퐻^20


,


and ignoring훺r, since it is so small, we can find relations between the dynamical
parameters훺휆,훺m,퐻 0 , and the deceleration parameter푞 0. Substitution of these
parameters into Equations (5.17) and (5.18) at present time푡 0 , gives


퐻 02 +


푘푐^2


푎^20


−휆


3


=훺m퐻 02 , (8.58)

− 2 푞 0 퐻 02 +퐻 02 +


푘푐^2


푎^20


−휆=− 3 훺m퐻 02 푤, (8.59)

where푤denotes the equation of state푝m∕휌m푐^2 of matter.
We can then obtain two useful relations by eliminating either푘or휆.Inthefirst
case we find


훺m( 1 + 3 푤)= 2 푞 0 + 2 훺휆, (8.60)

and, in the second case,


3
2

훺m( 1 +푤)−푞 0 − 1 = 푘푐

2
푆 02 퐻 02

. (8.61)

Free download pdf