282 Gravitational instability in Newtonian theory
Problem 6.8Verify that the solution of (6.93) can be written in the following
parametric form:
a(R,η)=4 πG! 0
3 |F|
(1−cosη), t(R,η)=4 πG! 0
3 |F|^3 /^2(η−sinη)+t 0 , (6.94)forF<0, and
a(R,η)=
4 πG! 0
3 F(coshη−1), t(R,η)=
4 πG! 0
3 F^3 /^2(sinhη−η)+t 0 , (6.95)forF> 0 .Heret 0 ≡t 0 (R)is a further integration constant. Note that the same
“conformal time”ηgenerally corresponds to different values of physical timet
for differentR.Assuming that the initial singularity(a→ 0 )occurs at the same
moment of physical timet=0 everywhere in space, we can sett 0 = 0.
Let us consider the evolution of a spherically symmetric overdense region in a
flat, matter-dominated universe. Far away from the center of this region the mat-
ter remains undisturbed and hence! 0 (R→∞)→!∞=const. The condition of
flatness requiresF→0asR→∞.Taking the limit|F|→0 so that the ratio
η/
√
|F|remains fixed, we immediately obtain from (6.94)
a(R→∞,t)=( 6 πG!∞)^1 /^3 t^2 /^3. (6.96)The energy density is consequently
ε(R→∞,t)=!∞
a^3=
1
6 πGt^2, (6.97)
in complete agreement with what one would expect for a flat dust-dominated uni-
verse. Inside the overdense region,Fis negative and the energy density does not
continually decrease. Becauseε∝a−^3 , the density at some pointRtakes its min-
imal valueεmwhena(R,t)reaches its maximal value
am=8 πG! 0
3 |F|(6.98)
atη=π(see (6.94)). This happens at the moment of physical time
tm=
4 π^2 G! 0
3 |F|^3 /^2, (6.99)
when the energy density is equal to
εm(R)=! 0 (R)
am^3 (R)=
27 |F|^3
( 8 πG)^3!^20=
3 π
32 Gtm^2. (6.100)
Comparing this result with the averaged density att=tm, given by (6.97), we find
that in those places where the energy density exceeds the averaged density by a