• flat universe
• W_m ~ 0.3
• mass of galaxy M_g ~ 10¹²M_sun
• present horizon radius d ~ 2/H₀

Let M(t) = ⁴/₃pd³(t) r_m(t) be the mass contained within the horizon radius d. Using (18.47), the constancy of r_ma³ and the present value r_m(t₀) = W_mr_crit = 3W_mH₀²/8p 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)

æ è   dâ dt ö ø 2   = Wv â2 +  Wm ä

This can be solved by separation of variables (the substitution y = â^3/2 helps),

â(t) = æ è  Wm Wv ö ø 1/3   sinh2/3(3/2 Wv½ H0t).

The present moment occurs when â(t₀) = 1, namely

t0 = æ è  2 3H0 Wv½ ö ø sinh-1 é ë æ è  wv Wm ö ø 1/2   ù û .

18-24
a) Look for the maximum of U_eff of (18.78): this occurs when   -(W_m/â²) + 2W_vâ = 0   or   r_m = 2 r_v = L/4p.

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

c) The plot of U_eff in Fig 18.9 shows that a small change in r_m 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 k_BT ~ 10 Mev; for smaller a it should rise beyond the value it has a a_max 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

b) Exchange sinh and cosh in the above, and the equation does have a solution namely

r(u) = cosh u     z(u) = ó õ u 0  (1 - sinh2u¢)1/2 du¢

which defines an embedded surface but only up to u = sinh^-1(1) = 0.88.