Lecture 21 Dynamics of the Universe Objectives: • The Friedmann equations Reading: Schutz 12; Hobson 14; Rindler 16 21.1 Friedmann’s equation The evolution of the Universe in GR is determined as follows: 1. The FRW interval =⇒ the metric, e.g. grr = −R2 /(1 − kr2 ) 2. The metric =⇒ Γα βγ , the connection. 3. The metric and connection =⇒ Rαβ , the Ricci tensor. 4. The Ricci tensor and field equations =⇒ differential equations for the size factor R and the fluid density ρ. Jumping straight in at step 4, consider R̈ Rtt = 3 . R Use field equations in the form Rαβ ¶ 1 = k Tαβ − T gαβ . 2 µ Assume perfect fluid: ³ p´ Tαβ = ρ + 2 Uα Uβ − pgαβ . c 85 See handout 7 86 LECTURE 21. DYNAMICS OF THE UNIVERSE Fluid is static in co-moving coordinates of FRW metric so U i = 0 and gαβ U α U β = gtt U t U t = c2 , so since gtt = c2 , U t = 1. Hence Ut = gtt U t = c2 , and ³ p´ Ttt = ρ + 2 c4 − pc2 = ρc4 , c while ³ ³ p´ p´ T = gαβ T αβ = ρ + 2 gαβ U α U β − pgαβ g αβ = ρ + 2 c2 − 4p = ρc2 − 3p. c c Therefore ¶ µ 1 2 R̈ 4 2 3 = k ρc − (ρc − 3p)c . R 2 Putting k = −8πG/c4 we obtain 4πG R̈ = − 3 µ 3p ρ+ 2 c ¶ R. (21.1) This is the acceleration equation. Similarly the rr, θθ and φφ components all lead to: Ṙ2 + kc2 = 8πG 2 ρR . 3 (21.2) This is the Friedmann equation. Finally, taking the time derivative of the Friedmann equation and substitutProve this ing for R̈ from the acceleration equation it is simple to show: ρ̇ + 3Ṙ ³ p´ ρ + 2 = 0. R c (21.3) which is the fluid equation. Alternatively this comes from T αβ ;α = 0. 21.1.1 Newtonian interpretation Each of Eqs 21.1, 21.2 and 21.3 has an approximate Newtonian interpretation. If one considers an expanding uniform density sphere then R̈ = − 4πG ρR. 3 LECTURE 21. DYNAMICS OF THE UNIVERSE 87 There is no Newtonian explanation for the pressure term in the acceleration equation. Conserving energy for a particle on the edge of such a sphere gives: 1 2 4πG 2 E ρR = . Ṙ − 2 3 m Newtonian equivalent for curvature term kc2 is total energy per unit mass. Finally the fluid equation follows directly from T dS = dU + p dV, setting dS = 0 (reversible adiabatic, no temperature gradients) and using mass–energy equivalence. Such Newtonian interpretations are a fudge: Eqs 21.1, 21.2 and 21.3 are relativistic. 21.2 The cosmological constant In 1917 Einstein modified the field equations to read 1 Rαβ − Rg αβ + Λg αβ = kT αβ , 2 where Λ is the cosmological constant. Still satisifes T αβ ;α = 0 since g αβ ;γ = 0. Nowadays, it is usual to place the new term on the right as the stress–energy tensor of the vacuum. ¶ µ Λ αβ 1 αβ αβ αβ . R − Rg = k T − g 2 k Second term in brackets on the right has the form of a perfect fluid ³ pΛ ´ α β ρΛ + 2 U U − pΛ g αβ , c if ρΛ + and pΛ = pΛ = 0, c2 Λ Λc4 =− , k 8πG and thus Λc2 . 8πG i.e. a fluid of constant density and negative pressure. ρΛ = This is “dark energy”, perhaps the most puzzling problem in modern physics. LECTURE 21. DYNAMICS OF THE UNIVERSE 21.2.1 88 Einstein’s static universe Negative pressure allows a static Universe. From µ ¶ 3p 4πG ρ + 2 R, R̈ = − 3 c R̈ can be zero if 3p = 0. c2 Here ρ and p are the sums of contributions from all components. Considering matter and Λ only, for matter pM ¿ ρM c2 so ρ+ ρ+ 3pΛ 3p ≈ ρM + ρΛ + 2 = ρM − 2ρΛ . 2 c c Thus 4πG (ρM − 2ρΛ ) R, 3 which is zero if ρM = 2ρΛ . This is Einstein’s static universe. Unfortunately it would not be static for long since it is unstable. Consider a perturbation ρM = 2ρΛ + ρ′ , R = R0 + R′ . To first order R̈ = − R̈′ = − 4πG ′ ρ R0 . 3 If R′ > 0 we expect ρ′ < 0 since matter is diluted as the universe expands, hence R̈′ > 0 and the perturbation will grow =⇒ instability. The universe either contracts or expands away from R = R0 . Λ therefore can give a static but not a stable universe. Had Einstein realised this, he could have predicted an expanding or contracting universe. Perhaps this was why he once referred to the cosmological constant as “my greatest blunder” (as quoted by Gamow, 1970).