Week 2 Part 2. The Friedmann Models: What are the constituents of the Universe?

Transcription

1 Week Part The Friedmann Models: What are the constituents of the Universe? We now need to look at the expansion of the Universe described by R(τ) and its derivatives, and their relation to curvature. For this we need gravity, i.e. General Relativity. 1

2 The Friedmann equation Solution of GR field equations for homogenous isotropic Universe gives the Friedmann equation linking the dynamics, density, and curvature of the Universe. Ṙ 8πG kc Λ = ρr + 3 A 3 R R R ρ kc A Λ - scale factor from the Robertson Walker metric - derivative w.r.t. cosmic time τ - density - curvature as in Robertson Walker metric - cosmological constant Friedmann equation can be written in several different ways (see later) Key role of ρ(r) in the solutions

3 The cosmological constant Λ Introduced by Einstein to allow static solutions (before 195) Subsequently described as biggest blunder etc. Non-zero Λ is equivalent to a term with ρ ~ constant, i.e. an energy density associated with the vacuum that does not dilute We will therefore ignore Λ but consider possible non-zero ρ Λ (but may loosely refer to lambda for a constant ρ component): Dark Energy is a component with ρ ~ constant. Is it exactly constant (i.e. Λ-like) or not? 3

4 Solutions to the Friedmann equation I We will assume the density composed of matter, radiation (+ relativistic particles) and/or false vacuum (Λ). Other possibilities easily derived. ρ tot = ρ + ρ + ρ m r Λ We now introduce the critical density ρ c and a density parameter Ω, both for individual density components and for the total Ω tot (often just written as Ω) We can rewrite the Friedmann equation to give the relationship (in various forms) between Ω tot and the physical curvature k/(ra) Important points Ω tot = 1 (i.e. ρ tot = ρ c ) is a flat Universe, Ω tot > 1 implies positive curvature (k=+1) and Ω tot < 1 implies negative curvature (k=-1) Ω tot -1 is a measure of the importance of the curvature, i.e. the ratio of the physical radius of curvature and the characteristic scale (c/h). Physical radius of curvature Characteristic length scale ~ O(cτ) ρ c = 3H 8πG Ω i = ρ i with Ω tot = Ω i ρ c i Ṙ = 8πG 3 ρr kc A H = H Ω tot kc A R k R A = H c ( RA ) = c H (Ω tot 1) k (Ω tot 1) 4

5 Which component dominates ρ tot and when? Let s examine each component density evolution with time and expansion factor; using the fact that expansion is adiabatic and the 1 st Law of Thermodynamics then de=-pdv; substituting the relativistic energy E=mc =ρvc, we have: c d(ρv ) = pdv dv V = 3 dr R dρ (ρ + p c ) = 3 dr R Then, for radiation and relativistic particles: ρc dρ dr p = ; = 3 ρ R 3 4 ρ R r R -4 3 For ordinary non-relativistic matter with low temperature: p=0 ρ m R -3 4 Why? Note: ρ T 4 For a hypothetical component (called Dark Energy ) with negative pressure: p=-ρc ρ Λ = constant What does negative pressure actually mean? 5

6 Which component dominates ρ tot and when? More generally, we can introduce an equation of state parameter w p = wρc dρ ρ = 3(1+ w) dr R ρ R 3(1+w) Matter w = 0 Radiation and relativistic particles w = +1/3 Dark Energy w ~ -1 Key question: Is it exactly -1, like a Λ- term, or just close to -1, as for example for an evolving scalar field Given the different density dependences on R, we would expect different components (if present) to dominate the total density at different times. Using the measured densities of the various components at current epoch in units of ρ c we have: Radiation ρ r R -4 ρ 0,r ~ 10-4 ρ c, so dominates at z > 300 Matter ρ m R -3 ρ 0,m ~ 0.5 ρ c, so dominates at 300 < z < 0.5 False vacuum ρ Λ ~ constant ρ 0,Λ ~ 0.75 ρ c, so dominates at z < 0.5 Overall equation of state will evolve as different components dominate. 6

7 Solutions to the Friedmann equation II Let s explore further the evolution of Ω tot : In a decelerating Universe, Ω tot diverges from unity. In an accelerating Universe, Ω tot is driven towards unity. k R A = H c (Ω tot 1) = c A 1 Ṙ (Ω tot 1) It can be useful to think of an Ω k coming from curvature. Then, the sum of all Ω i = unity. This is a check of the validity of the Friedmann equation, i.e. of GR itself. i.e. since we can neglect radiation at the present epoch, we would expect Ω Λ + Ω m + Ω k = 1 This is the basis of the cosmological triangle diagram as we will see later. Ω k = k R A = H c Ω tot + Ω k =1 kc (AR) H (Ω tot 1) 7

8 Geometry vs. density parameter k = 1 Ω k (τ) < 0 Ω tot (τ) > 1 k = 1 Ω k (τ) > 0 Ω tot (τ) < 1 k = 0 Ω k (τ) = 0 Ω tot (τ) = 1 8

9 Deceleration or acceleration? Second Friedmann equation (derived from the trace of Einstein s equation): R = 4π 3 Gρ 'R = 4π 3 G(ρ m + ρ r ρ Λ )R ρ ʹ = ρ + 3 p c R R Ṙ = 8πG 3H ρ m + ρ r ρ Λ q = Ω m + Ω Ω r Λ The Universe accelerates if Ω Λ > Ω m / + Ω r (q<0) NB: in our Universe, with Ω Λ,0 ~ 3 this occurred at z ~ 0.8 Deceleration parameter 9

10 Let s put all together: the cosmological triangle diagram

11 Let s put all together: the cosmological triangle diagram This diagram holds for any time τ. We are ignoring Ω r as it is always negligible, except at very high redshifts 11

12 Solving the Friedmann equation for R(t), ω(z) etc. Rewrite density ρ(r) in terms of the present-day densities, as parameterized with Ω 0,i, and using a as a short-hand for R/R 0 Substitute this into the Friedmann equation (dividing both sides by H 0 and a ) and using the relation between curvature and Ω o,tot today. ȧ a H = 8πGρ 0 3H 0 Ṙ = 8πG 3 ρr kc A 8πGρ 3H 0 = Ω 0,m a 3 + Ω 0,r a 4 + Ω 0,Λ ( ) kc R 0 A H 0 1 a H H = Ω m,0 a 3 + Ω r,0 a 4 kc ( + Ω Λ,0 ) 1 0 R 0 A H 0 a = ( a 3 + Ω r,0 a 4 + Ω Λ,0 (Ω tot,0 1)a ) To yield an expression for H(a), or H(z), and a differential equation for a, which must be evaluated numerically for general conditions H = H 0 ( Ω Λ,0 + a 3 + Ω r,0 a 4 (Ω tot,0 1)a ) ( ) = H 0 Ω Λ,0 + (1+ z) 3 +Ω r,0 (1+ z) 4 (Ω tot,0 1)(1+ z) ȧ = H 0 ( Ω Λ,0 a + a 1 +Ω r,0 a (Ω tot,0 1) ) 1

13 With H(z), we now now use RW metric to follow a photon s path: Rdω = cdτ = cdr R = cdr HR = cdz H ( 1+ z) R 0 dω = c H dz = c (1 Ω tot,0 )(1+ z) + Ω Λ,0 + (1+ z) 3 + Ω r,0 (1+ z) dz H 0 From this relation we can get ω(z) by integration. If we have a curved Universe, we know k/a in terms of Ω 0,tot and so we can also get D = R 0 S k (ω) etc. Unfortunately, simple analytic solutions only exist for Ω Λ = 0 Finally, it is easy to calculate Ω tot (z) in terms of present-day values by noticing that (Ω tot -1)(RH) is a constant (=k/a ) ( Ω tot (z) 1) = (Ω tot,0 1) H 0 (1+ z) H Ω = tot,0 1 1 Ω tot,0 + Ω Λ,0 (1+ z) + (1+ z)+ Ω r,0 (1+ z) ( ) Note: As expected, Ω tot (z) tends to unity at high z unless Ω m = Ω r = 0 13

14 ȧ = H 0 ( Ω Λ,0 a + a 1 + Ω r,0 a (Ω tot,0 1) ) Solve for simple cases where one of these terms dominates with Ω i ~ 1 Λ dominated ȧ = H 0 a a = e H (t t 0 ) Matter-dominated ȧ = H 0 a 1/ a = τ τ 0 /3 τ = 3 H 1 Radiation dominated ȧ = H 0 a 1 a = τ τ 0 1/ τ = 1 H 1 Curvature dominated (empty) ȧ = H 0 a = τ τ 0 τ = H 1 All non-accelerated Universes have R ~ 0 at some point in the finite past = Big Bang with τ = 0 H -1 is a characteristic expansion timescale, often loosely called the age of the Universe, and often more useful than any actual absolute age of the Universe 14

15 The time temperature relation in the early Universe: At z >> 3000, the Universe is radiation dominated and will have Ω r = 1, regardless of current conditions Ṙ = 8πG 3 ρ r R RṘ = 8πG ρ r R 4 3 R = 3πG ρ r R ρ r = τ 3πG There will be a precise relation between the density of radiation (plus any relativistic species present) and the cosmic epoch (again, independent of conditions today). The density of Black Body radiation is given uniquely by its temperature to the fourth power, so there must also be a precise time-temperature relation. We will see later that τ(sec) ~ T(10 10 K) - 1/ 1/4 τ 1/ 15

16 The Figures which follow compare various quantities for three cosmological models. Fifteen years ago, these would have been regarded as equally plausible possibilities. k = 0; Ω m = 1 (no Dark Energy) k = 0; Ω Λ,0 = 0.75; = 0.5 k = -1; = 0.5 (no Dark Energy) Einstein-de Sitter model Concordance model Open Universe 16

17 Ω m = 1 Einstein-de Sitter = 0.5; Ω Λ,0 = 0.75 Concordance = 0.5 Open Universe All models have R~0 at some finite time in the past = BIG BANG NB: Not true for pure ρ Λ model which expands exponentially or for some high ρ Λ models which would have bounced from an earlier contraction R(τ) and the Big Bang 17

18 Ω m = 1 Einstein-de Sitter = 0.5; Ω Λ,0 = 0.75 Concordance = 0.5 Open Universe All models approach Ω m = 1 in the past makes treatment of early Universe simple. Ω m as f(z) 18

19 Ω m = 1 Einstein-de Sitter = 0.5; Ω Λ,0 = 0.75 Concordance = 0.5 Open Universe Comoving distance to a galaxy with a given redshift varies greatly with models NOTE the change in the x-axis direction ω(z) 19

20 Ω m = 1 Einstein-de Sitter = 0.5; Ω Λ,0 = 0.75 Concordance = 0.5 Open Universe but all converge at some finite value of ω for very high z concept of horizon (see later). Convergence of ω 0

21 Ω m = 1 Einstein-de Sitter = 0.5; Ω Λ,0 = 0.75 Concordance = 0.5 Open Universe Note how two Ω = 1 models track each other at high z same geometry and same matter-dominated dynamics at high z Effective distance D(z) 1

22 Ω m = 1 Einstein-de Sitter = 0.5; Ω Λ,0 = 0.75 Concordance = 0.5 Open Universe Angular diameter distance does not decrease monotonically (even in spatially flat Universes!!).. Angular diameter distance D θ

23 Ω m = 1 Einstein-de Sitter = 0.5; Ω Λ,0 = 0.75 Concordance = 0.5 Open Universe so apparent angular size of objects of fixed physical size will actually start to increase beyond a certain redshift (typically z ~ 1). Angular size of 10 kpc rod 3

24 Friedmann equation with Newtonian gravity 4πρx x = G x 4π = Gρx 3 x = Rω x = R ω x = R ω OA OA OA O x A Ṙ 8πG kc Λ = ρr + R 3 A 3 Ṙ = 4 G ρ R 3π For pressure-less matter alone, this is all we need: i.e. the global dynamics of R(τ) are actually determined by local gravity this is the justification for the Newtonian approach RR R R 4πG R R 3 = ρ 3 R 8πG 1 = ρ R + ξ 3 R 8πG = ρr + ξ 3 3 Newtonian binding energy ξ solutions for recollapse (ξ < 0) or unbound expansion (ξ > 0) 4

25 Friedmann equation with Newtonian gravity Ṙ = 4 G ρ R 3π In the general case, we must use the active mass density in this equation to get the correct Friedmann equation ρ ʹ = ρ + 3 p c R = 4πG 3 ρr Ṙ R = 8πG 3 Ṙ = Ṙ = w = +1/3 radiation ρr 4 Ṙ R 3 8πG ρr R + ξ 8πG 3 ρr + ξ w = -1 Dark Energy R = + 4πG 3 ρr Ṙ R = + 8πG 3 Ṙ = Ṙ = [ ρ ] RṘ 8πG [ ρ ] R + ξ 3 8πG 3 ρr + ξ 5

26 Aside: Milne Model again We recovered the RW metric in the Milne model if we set the radius of curvature to A = c Ṙ This is exactly what we have in the Friedmann equation with ρ = 0. Ṙ = 8πG 3 ρr kc A 6

27 (Most) important points Friedmann equation links density, expansion rate and curvature. Key role played by the equation of state, relating pressure and density ρ(r) and determining R(τ). Matter and radiation produce deceleration, Dark Energy produces acceleration. The density as parameterized by Ω is intimately linked to the importance of curvature (i.e. physical curvature compared with the c/h length scale of Universe). Acceleration makes Ω tend to unity, deceleration makes Ω diverge from unity. With (observed) values of Ω i,0, we can calculate R(τ), H(z), Ω(z), ω(z), D(z) etc. Sometimes this must be done numerically. There must be a well defined relationship between time and temperature in the early Universe, regardless of the state of the Universe today. Dynamical effects in Friedmann equation are given also by Newtonian gravity (using active density). 7