PHYS 652: Astrophysics 20 5 Lecture 5: Solutions of Friedmann Equations “A man gazing at the stars is proverbially at the mercy of the puddles in the road.” Alexander Smith The Big Picture: Last time we derived Friedmann equations — a closed set of solutions of Einstein’s equations which relate the scale factor a(t), energy density ρ and the pressure P for flat, open and closed Universe (as denoted by curvature constant k = 0, 1, 1). Today we are going to − solve Friedmann equations for the matter-dominated and radiation-dominated Universe and obtain the form of the scale factor a(t). We will also estimate the age of the flat Friedmann Universe. From the definition of the Hubble rate H in eq. (72) a˙ H = (102) ≡ a ⇒ a¨ a¨ H˙ = H2 + = H2 1 H2 (1 + q) , (103) − a − − H2a ≡− we define a deceleration parameter q as a¨ q . (104) ≡−H2a Non-relativistic matter-dominated Universe is modeled by dust approximation: P = 0. Then, from eq. (95), we have a¨ 4πG + ρ = 0, (105) a 3 and, in terms of H 4πG H2q + ρ = 0. (106) − 3 Therefore 3H2 ρ = q. (107) 4πG Then the first Friedmann equation becomes a˙ 2 8πG k ρ = , a − 3 −a2 k H2 2H2q = , (108) − −a2 so k = a2H2(1 2q). (109) − − Since both a = 0 and H = 0, for flat Universe (k = 0), q = 1/2 (q > 1/2 for k = 1 and q < 1/2 for 6 6 k = 1). When combined with eq. (107), this yields critical density − 3H2 ρ = , (110) cr 8πG 20 PHYS 652: Astrophysics 21 the density needed to yield the flat Universe. Currently, it is (see eq. (73)) 2 2 h 1 year 2 3 × 10 × × 3H0 0.98 10 years 3600 24 365 sec −29 2 g −29 g ρcr = = = 1.87 10 h 10 . 8πG 8π (6.67 10−8cm3 g−1 s−2) × cm3 ≈ cm3 × (We used h 0.72 0.02.) ≈ ± It is important to note that the quantity q provides the relationship between the density of the Universe ρ and the critical density ρcr (after combining eqs. (107) and (109)): ρ q = . (111) 2ρcr The second Friedmann equation (eq. (101b)) for the matter-dominated Universe becomes a˙ ρ˙ + 3ρ = 0 a 3 2 d 3 3 3 a ρ˙ + 3ρaa˙ = 0 a ρ = 0 a ρ = a ρ0 = const. (112) ⇒ dt ⇒ 0 1 Radiation-dominated Universe is modeled by perfect fluid approximation with P = 3 ρ. The second Friedmann equation (eq. (101b)) becomes 1 a˙ a˙ ρ˙ + 3 ρ + ρ =ρ ˙ + 4ρ = 0 3 a a 4 3 d 4 4 4 a ρ˙ + 4ρaa˙ = 0 a ρ = 0 a ρ = a ρ0 = const. (113) ⇒ dt ⇒ 0 =0 = 1 Flat Universe (k , q0 2 ) Matter-dominated (dust approximation): P = 0, a3ρ = const. The first Friedmann equation (eq. (101a)) becomes a˙ 2 8πG a 3 = ρ 0 a2 3 0 a da 8πGρ a3 1 2 8πGρ a3 = 0 0 a1/2da = a3/2 + K = 0 0 t. (114) ⇒ dt 3 a1/2 ⇒ 3 3 r Z r At the Big Bang, t = 0, a = 0, so K = 0. Upon adopting convention a0 = 1, and the fact that the Universe is flat ρ0 = ρcr, we finally have 1/3 2/3 1/3 2/3 a = (6πGρ0) t = (6πGρcr) t 1 3 1 3 3H2 / 9H2 / 3H 2/3 = 6πG 0 t2/3 = 0 t2/3 = 0 t2/3. (115) 8πG 4 2 where we have used the eq. (110) in the second step. From here we compute the age of the Universe t0, which corresponds to the Hubble rate H0 and the scale factor a = a0 = 1 to be: 2 t0 = . (116) 3H0 h Taking H0 = 10 and h 72, we get 0.98×10 years ≈ 10 2 0.98 10 years 9 t0 = × × 9.1 10 years 9.1 (aeon). (117) 3 0.72 ≈ × ≡ A × 21 PHYS 652: Astrophysics 22 1 4 Radiation-dominated: P = 3 ρ, a ρ = const. The first Friedmann equation (eq. (101a)) becomes 2 a˙ 8πG a0 4 = ρ0 a2 3 a da 8πGρ a4 1 8πGρ a4 = 0 0 ada = 2a2 + K = 0 0 t. (118) ⇒ dt 3 a ⇒ 3 r Z r Again, at the Big Bang, t = 0, a = 0, so K = 0, and a0=1. Also ρ0 = ρcr. Therefore, 1 4 2 1/4 2 1/4 2 3H2 / H 1/2 a = πGρ t1/2 = πGρ t1/2 = πG 0 t1/2 = 0 t1/2. (119) 3 0 3 cr 3 8πG 2 Flat Friedmann Universe (k=0, q0=1/2) matter-dominated radiation-dominated a(t) t Figure 2: Evolution of the scale factor a(t) for the flat Friedmann Universe. =1 1 Closed Universe (k , q0 > 2 ) Matter-dominated (dust approximation): P = 0, a3ρ = const. The first Friedmann equation (eq. (101a)) becomes 2 a˙ 8πG a0 3 1 = ρ0 a2 3 a − a2 3 da 8πGρ0a0 da = 1 dt = 3 ⇒ dt 3a − ⇒ 8πGρ0a r Z Z 0 1 3a − q Rewrite the integral above in terms of conformal time given in eq. (83) (dη dt ): ≡ a da dη = 3 , (120) 8πGρ0a Z Z 0 a a2 3 − q 22 PHYS 652: Astrophysics 23 and define, after substituting a0 = 1 and using eqs. (107)-(109) 4πGρ0 2 q0 A = H0 q0 = . (121) ≡ 3 2q0 1 − Then a da˜ −1 a A 1 η η0 = = sin − + π. (122) 2 − 0 √2Aa˜ a˜ A 2 Z − But, the requirement η = 0 at a = 0 sets η0 = 0, so we have a A 1 − = sin η π = cos η a = A(1 cos η). (123) A − 2 − ⇒ − Now dt = adη, so t t0 = adη = A(1 cos η)dη = A (1 cos η) dη = A(η sin η). (124) − − − − Z Z Z But, the requirement η = 0 at t = 0 sets t0 = 0. Therefore, we finally have the dependence of the scale factor a in terms of the time t parametrized by the conformal time η as: q a = 0 (1 cos η), (125) 2q0 1 − q− t = 0 (η sin η). 2q0 1 − − 1 4 Radiation-dominated: P = 3 ρ, a ρ = const. The first Friedmann equation (eq. (101a)) becomes 2 a˙ 8πG a0 4 1 = ρ0 a2 3 a − a2 4 da 8πGρ0a0 da = 2 1 dt = 3 ⇒ dt 3a − ⇒ 8πGρ0a0 r Z Z 2 1 3a − q 8πGρ0 2q0 Again, rewrite the integral above in terms of conformal time and quantity A1 = 3 = 2q0−1 : a da˜ − a η η = = sin 1 . (126) 0 2 − 0 √A1 a˜ √A1 Z − Again, the requirement η = 0 at a = 0 sets η0 = 0, so we have a = A1 sin (η) , (127) and p t t0 = A1 cos (η) , (128) − The requirement η = 0 at t = 0 sets t0 = √Ap1, so we finally have 2q a = 0 sin η, (129) 2q0 1 r − 2q t = 0 (1 cos η) . 2q0 1 − r − 23 PHYS 652: Astrophysics 24 Closed Friedmann Universe (k=1, q0>1/2) matter-dominated radiation-dominated a(t) Big Crunch Big Crunch t Figure 3: Evolution of the scale factor a(t) for the closed Friedmann Universe. In both matter- and radiation-dominated closed Universes, the evolution is cycloidal — the scale factor grows at an ever-decreasing rate until it reaches a point at which the expansion is halted and reversed. The Universe then starts to compress and it finally collapses in the Big Crunch. 1 Open Universe (k = 1, q0 < ) − 2 Matter-dominated (dust approximation): P = 0, a3ρ = const. The first Friedmann equation (eq. (101a)) becomes a˙ 2 8πG a 3 1 = ρ 0 + a2 3 0 a a2 3 da 8πGρ0a0 da = + 1 dt = 3 ⇒ dt r 3a ⇒ 8πGρ0a0 Z Z 3a + 1 q Again, rewrite the integral above in terms of conformal time: da dη = 3 , (130) 8πGρ0a0 2 Z Z 3 a + a 4πGρ0 q0 q take a0 = 1, and define A˜ = . Then ≡ 3 2q0−1 a da˜ a + A˜ + a(2A˜ + a) a a a 2 η η0 = = ln = ln +1+ 2 + − 0 ˜ 2 qA˜ A˜ s A˜ A˜ Z 2Aa˜ +a ˜ p− a = cosh 1 + 1 . (131) ˜ A But, the requirement η = 0 at a = 0 sets η0 = 0, so we have a + A˜ = cosh η a = A˜(cosh η 1). (132) A˜ ⇒ − 24 PHYS 652: Astrophysics 25 Now dt = adη, so t t0 = adη = A˜(cosh η 1)dη = A˜ (cosh η 1) dη = A˜(sinh η η). (133) − − − − Z Z Z But, the requirement η = 0 at t = 0 sets t0 = 0. Therefore, we finally have the dependence of the scale factor a in terms of the time t parametrized by the conformal time η as: q a = 0 (cosh η 1), (134) 2q0 1 − q− t = 0 (sinh η η). 2q0 1 − − 1 4 Radiation-dominated: P = 3 ρ, a ρ = const. The first Friedmann equation (eq. (101a)) becomes a˙ 2 8πG a 4 1 = ρ 0 + a2 3 0 a a2 4 da 8πGρ0a0 da = 2 + 1 dt = 3 ⇒ dt r 3a ⇒ 8πGρ0a0 Z Z 3a2 + 1 q 8πGρ0 2q0 Again, rewrite the integral above in terms of conformal time and quantity A˜1 = : ≡ 3 2q0−1 a da˜ −1 a η η0 = = sinh (135) − ˜ 2 ˜ Z0 A1 +a ˜ A1 ! Again, the requirement η = 0 at a = 0 setsp η0 = 0, so we havep a = A˜1 sinh η, (136) q t t0 = A˜1 cosh η, (137) − q The requirement η = 0 at t = 0 sets t0 = A˜1, so we finally have p2q a = 0 sinh η, (138) 1 2q0 r − 2q t = 0 (cosh η 1) .