Introduction to Cosmology

Friedmann–Lemaitre Cosmologies 83

At our present time푡 0 when the mass density is휌 0 , the cosmic scale is 1, the Hub-
ble parameter is퐻 0 and the density parameter훺 0 is given by Equation (1.35), Fried-
mann’s equation (5.4) takes the form

푎̇^20 =

8

3

휋퐺휌 0 −푘푐^2 =퐻 02 훺 0 −푘푐^2 , (5.7)

which can be rearranged as

푘푐^2 =퐻 02 (훺 0 − 1 ). (5.8)

It is interesting to note that this reduces to the Newtonian relation (1.35). Thus the
relation between the Robertson–Walker curvature parameter푘and the present density
parameter훺 0 emerges: to the푘values+1, 0 and−1 correspond an overcritical density
훺 0 >1, a critical density훺 0 =1 and an undercritical density 0<훺 0 <1, respectively.
The spatially flat case with푘=0 is called theEinstein–de Sitter universe.

General Solution. When we generalized from the present퐻 0 to the time-dependent
Hubble parameter퐻(푡)=푎̇∕푎in Equation (2.47), this also implied that the critical den-
sity [Equation (1.31)] and the density parameter [Equation (1.35)] became functions
of time:

휌c(푡)=

3

8 휋퐺

퐻^2 (푡), (5.9)

훺(푡)=휌(푡)∕휌c(푡). (5.10)

Correspondingly, Equation (5.8) can be generalized to

푘푐^2 =퐻^2 푎^2 (훺− 1 ). (5.11)

If푘≠0, we can eliminate푘푐^2 between Equations (5.8) and (5.11) to obtain

퐻^2 푎^2 (훺− 1 )=퐻^20 (훺 0 − 1 ), (5.12)

which we shall make use of later.
It is straightforward to derive a general expression for the solution of Friedmann’s
Equation (5.4). Inserting푘푐^2 from Equation (5.8), and replacing( 8 휋퐺∕ 3 )휌by훺(푎)퐻 02 ,
Equation (5.4) furnishes a solution for퐻(푎):

퐻(푎)≡푎̇∕푎=퐻 0

√

( 1 −훺 0 )푎−^2 +훺(푎). (5.13)

Here we have left the푎dependence of훺(푎)unspecified. As we shall see later, various
types of energy densities with different푎dependences contribute.
Equation (5.13) can be used to solve for thelookback time푡(푧)∕푡 0 or푡(푎)∕푡 0 (nor-
malized to the age푡 0 ) since a photon with redshift푧was emitted by writing it as an
integral equation:

∫

푡(푎)

0

d푡=

∫

푎

1

d푎

aH(푎)

. (5.14)

The age of the Universe at a given redshift is then 1−푡(푧)∕푡 0. We shall specify this in
more detail later.