18-10.
We need some assumptions:
• flat universe
• Wm ~ 0.3
• mass of galaxy Mg ~ 1012Msun
• present horizon radius d ~ 2/H0

Let M(t) = 4/3pd3(t) rm(t) be the mass contained within the horizon radius d. Using (18.47), the constancy of rma3 and the present value rm(t0) = Wmrcrit = 3WmH02/8p we find
 dM dt = 4prm(t) d2(t)        so at present dM dt ~ 6Wm ~ 1.8.
The required time Dt is therefore
 Dt = Mg dM/dt ~ 1012 Msun×(1.5 km/Msun) 1.8 ×(3×105 km/s) ~ 106 s ~ 1 month.

18-20
The Friedmann equation (18.77) becomes (with â(t) = a(t)/a0)
 ćč dâ dt öř 2 = Wv â2 + Wm ä
This can be solved by separation of variables (the substitution y = â3/2 helps),
 â(t) = ćč Wm Wv öř 1/3 sinh2/3(3/2 Wv½ H0t).
The present moment occurs when â(t0) = 1, namely
 t0 = ćč 2 3H0 Wv½ öř sinh-1 éë ćč wvWm öř 1/2 ůű .

18-24
a) Look for the maximum of Ueff of (18.78): this occurs when   -(Wm2) + 2Wvâ = 0   or   rm = 2 rv = L/4p.

b) From the Friedman equation (18.77) we have   Ueff(â) = 1/2 Wc   or

-8p/3 (rm +rvac ) a2 = -1     hence     a = L-½

so the volume is   V = 2p2 a3 = 2p2 L-3/2.

c) The plot of Ueff in Fig 18.9 shows that a small change in rm will either cause the universe to expand to infinite volume, or to collapse to a singularity.

18-27
a) From (18.63) and (18.77) we have
 r(a) = - 3 4pa2 Ueff(a)
The first law of thermo (18.20) gives
 p(a) = - d(ra3) d(a3) = 1 4pa2 d[aUeff(a)] da
That's your parametric representation.

b) The potential should be like (18.78) down to a universe radius when kBT ~ 10 Mev; for smaller a it should rise beyond the value it has a amax in order to cause a bounce.

c) From the above equations,
 r+ 3p = 3 4pa2 dUeff da
So r+ 3p must be negative in the 'smaller a" region of b).

18-30
a) We need r(c) and z(c) such that the geometry to be embedded can be written

dS2 = dc2 + sinh2cdf2 = dr2 + r2 df2 + dz2
Evidently r(c) = sinh(c), but then
 ćč d\z dc öř 2 = 1 - cosh2c < 0
does not work.

b) Exchange sinh and cosh in the above, and the equation does have a solution namely
 r(u) = cosh u     z(u) = óő u 0 (1 - sinh2u˘)1/2 du˘
which defines an embedded surface but only up to u = sinh-1(1) = 0.88.