Astroparticle Physics

368 17 Solutions17.11 Chapter 111. This probability can be worked out fromφ = σ Th Nd,where σ Th is the Thomson cross section (665 mb), N the number of target atoms percm 3 ,andd the distance traveled.Since the universe is flat, one has ϱ = ϱ crit . Because of ϱ crit = 9.47 × 10 −30 g/cm 3 andm H = 1.67 × 10 −24 g, and the fact that only 4% of the total matter density is in the formof baryonic matter, one has N = 2.27 × 10 −7 cm −3 . The distance travelled from thesurface of last scattering corresponds to the age of the universe (14 billion years),d = 14 × 10 9 × 3.156 × 10 7 s × 2.998 × 10 10 cm/s = 1.32 × 10 28 cmresulting inφ = 1.99 × 10 −3 ≈ 0.2% .2. The critical density ϱ c = 3H 2 /8πG as obtained from the Friedmann equation has to bemodified if the effect of the cosmological constant is taken into account:ϱ c,Λ = 3H 2 − Λc 2.8πGSince ϱ c,Λ cannot be negative, one can derive a limit from this equation:3H 2 − Λc 2 > 0 , Λ < 3H 2c 2 = 1.766 × 10 −56 cm −2 .This leads to an energy densityϱ ≤3. p 1 =− dE classdV∼ddVc48πG Λ = 8.51 × 10−10 J/m 3 = 5.3GeV/m 3 = 5.3keV/cm 3 .=− ddV(− GmM )= GmM dRdV( 1V 1/3 )=− 1 3 V −4/3 ∼− 1 R 4 ,( 1V 1/3 )( 4π3) 1/3p 2 =− dE ΛdV∼− ddV (−ΛR2 ) ∼ddV V 2/3 = 2 3 V −1/3 ∼+ 1 R .p 1 is an inward pressure due to the normal gravitational pull, whereas p 2 is an outwardpressure representing a repulsive gravity.