20 From Newton to Hubble
In a universe expanding linearly according to Hubble’s law [Equation (1.12)], the
kinetic energy푇of the galaxy receding with velocity푣is
푇=1
2
푚푣^2 =
1
2
푚퐻 02 푟^2 , (1.29)
where푚is theinertial massof the galaxy. Although there is no theoretical reason for
the inertial mass to equal the gravitational mass (we shall come back to this question
later), careful tests have verified the equality to a precision better than a few parts
in 10^13. Let us therefore set푚G=푚. Thus the total energy is given by
퐸=푇+푈=^1
2푚퐻 02 푟^2 −^4
3
휋Gm휌푟^2 =푚푟^2(
1
2
퐻 02 −^4
3
휋퐺휌
)
. (1.30)
If the mass density휌of the Universe is large enough, the expansion will halt. The
condition for this to occur is퐸=0, or from Equation (1.32) thiscritical densityis
휌c=3 퐻 02
8 휋퐺
= 1. 0539 × 1010 ℎ^2 eV m−^3. (1.31)The valueℎ= 0 .696 from Equation (1.21) can be inserted here. A universe with density
휌>휌cis calledclosed;withdensity휌<휌cit is calledopen.
Expansion. Note that푟and휌are time dependent: they scale with the expansion.
Denoting their present values푟 0 and휌 0 , one has
푟(푡)=푟 0 푎(푡),휌(푡)=휌 0 푎−^3 (푡). (1.32)
The acceleration푟̈in Equation (1.27) can then be replaced by the acceleration of
the scale
푎̈=푟̈∕푟 0 =−4
3
휋퐺휌 0 푎−^2. (1.33)
Let us use the identity푎̈=^1
2d
d푎푎̇^2
in Equation (1.33) to obtain
d푎̇^2 =−8
3
휋퐺휌 0
d푎
푎^2.
This can be integrated from the present time푡 0 to an earlier time푡with the result
푎̇^2 (푡)−푎̇^2 (푡 0 )=8
3
휋퐺휌 0 (푎−^1 − 1 ). (1.34)
Let us now introduce the dimensionlessdensity parameter:훺 0 =휌 0
휌c=
8 휋퐺휌 0
3 퐻 02
. (1.35)
Substituting훺 0 into Equation (1.34) and making use of the relation in Equa-
tion (1.20),푎̇(푡 0 )=퐻 0 ,wefind
푎̇^2 =퐻 02 (훺 0 푎−^1 −훺 0 + 1 ). (1.36)