Lecture 2: Cosmological Background

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1 Lecture 2: Cosmological Background Houjun Mo January 27, 2004 Goal: To establish the space-time frame within which cosmic events are to be described.

2 The development of spacetime concept Absolute flat space and absolute time Flat space and absolute time Flat space and relative time (Special Relativity) Curved spacetime (General Relativity)

3 Spacetime metric The spacetime structure in cosmology is very different from our common sense (space is Euclidean and time flows uniformly). The space may be curved and time is not uniformly flowing. The geometric properties are described by spacetime metric: ds 2 = g µν dx µ dx ν g µν dx µ dx ν µ,ν In Special Relativity: ds 2 = c 2 dt 2 dl 2, dl 2 = dx 2 + dy 2 + dz 2

4 Two-dimensional analog For a general 2-dimensional surface, the distance between two infinitesimally close points is: dl 2 = g i j (x)dx i dx j. i, j In general, the form of g i j (x) can be complicated. Special cases: isotropic and homogeneous 2-d surfaces (only 3 kinds): ( ) dr dl 2 = a Kr + 2 r2 dθ 2 K = 0: dl 2 = a 2 (dr 2 + r 2 dθ 2 ): 2-d plane, (r,θ) as polar coordinates K = +1: dl 2 = a 2 (dχ 2 + sin 2 χdθ 2 ) K = 1: dl 2 = a 2 (dχ 2 + sinh 2 χdθ 2 ) (r = sinχ): 2-d sphere (r = sinhχ): 2-d hyperbolic surface

5 The standard cosmology (or Big Bang Cosmology) Cosmological Principle (boundary condition) + General Relativity (dynamics) Cosmological Principle: sufficiently large scales The Universe is homogeneous and isotripic on The proposal of this principle was based on philosophical considerations, it is only a hypothesis but has observational support.

6 The consequence of Cosmological Principle The spacetime properties is determined by the metric g µν (x). Like in twodimensional cases, g µν can be complicated in general, but only very special forms are allowed by the the symmetry imposed by the Cosmological Principle. Robertson-Walker metric: [ ] dr ds 2 = c 2 dt 2 a 2 2 (t) 1 Kr + 2 r2 (dθ 2 + sin 2 θdφ 2 ) ] g µν = diag [c 2, a2 r 2, a 2 r 2 sin 2 θ 1 Kr 2, a2 where K = +1, 0, 1 are the curvature signature; a(t) is the scale factor relateing coordinates to physical distance. In this case, the spacetime is completely described by the value of K and the form of a(t).

7 Coordinates and distances The other form of R-W metric ds 2 = c 2 dt 2 a 2 (t) [ dχ 2 + f 2 K(χ)(dθ 2 + sin 2 θdφ 2 ) ] f K (χ) = r = sinχ (K = +1) χ (K = 0) sinhχ (K = 1) Proper distance: distance measured with physical scale Comoving distance: distance measured with scale a.

8 The properties of the R-W metric ds 2 = c 2 dt 2 a 2 (t)dχ 2 (radial) ds = 0 ds 2 = c 2 dt 2 a 2 (t) f 2 K(χ)dθ 2 (transverse) (for photons) The motion of free photons: redshift The motion of free non-relativistic particles: Thermodynamics and equation of state Distances: luminosity distance and angular-diameter distance Hubble s constant and deceleration parameter: H 0 ȧ a (t 0); q 0 äa ȧ 2 (t 0)

9 Cosmic redshift λ 0 /λ e = ν e /ν 0 = a(t 0 )/a(t e ), z (λ 0 λ e )/λ e, 1 + z = a(t 0 )/a(t e ).

10 The motion of free photons Suppose δt = t 2 t 1 t 1. Then v 2 v 1 = O(δt). To the 1st order of δt, δl v 1 δt. Relative to O 1, the velocity of the (free) particle at t 2 is v 2 + H(t 2 )δl = v 1 Thus, the change of peculiar velocity δv = v 2 v 1 = H(t)v 1 δt = ȧ(t) a(t) v 1δt, which gives v(t) 1/a(t)

11 Thermodynamics and the equation of state Consider gas in a small volume which expands with the universe V a 3 (t). Since there should not be any net heat flow across the boundary of V in a homogeneous universe, the First Law of thermodynamics gives d[(p + ε)v ] = V dp, where ε ρc 2 = nmc 2 + u. Setting V a 3, we have ( ) dε P + ε da + 3 = 0. a For nonrelativistic gas P 2u/3, ρ a 3 For relativistic particles P = ε/3, ρ P a 4 ; T a 1 For vacuum energy (cosmological constant) P = ε, ρ = const..

12 Relativistic Cosmology Goal: To obtain the value of K and to derive the dynamical equation for a from Einstein s field equations: R µν = 8πG ( T c 4 µν 1 ) 2 g λ µνt λ. The Ricci tensor is defined by R µν R σ µσν, R µ ναβ = α Γ µ νβ β Γ µ να + Γ µ σαγ σ νβ Γ µ σβγ σ να, Γ µ αβ = g µσ ( β g σα + α g σβ σ g αβ ) /2, where µ / x µ, and the energy-momentum tensor is T µν = (ρc 2 + P)U µ U ν + g µν P, where ρ and P are the density and pressure, U µ = dx µ /ds is the four-velocity.

13 Friedmann equation Inserting R-W metric into Einstein s field equations, and using the fact that U µ = (1,0,0,0) for a homogeneous and isotropic fluid, we have (0, 0) component : ( ä a = 4πG ρ + 3P ) 3 c 2 ä (space space) : (ρ a + 2ȧ2 a + 2 2Kc2 a = 4πG Pc ) 2 2 These two equations combined give the Friedmann equation: (ȧ ) 2 = 8πGρ a 3 K a 2.

14 From Friedmann equation, we see that K is given by where and Kc 2 = 8πGρ 0 a 2 0 ȧ 2 0 = H 2 3 0a 2 0(Ω 0 1), Ω 0 = ρ 0 /ρ crit,0, ρ crit,0 = 3H2 0 8πG is the critical density, and subscript 0 denote quantities at the present time.

15 Friedmann equation from Newtonian physics Under the assumption implied by the cosmological principle, a global quantity should have the same value as the corresponding local quantity, e.g. the mean density of the universe at a give time should be the same as that at any location of the universe. Since Newtonian theory applies on small scale, we can apply it to a small spherical region: ä = 4πG ρa 3 3 a 2 = 4πG 3 ρa. Using ä = dȧ/dt = (1/2)dȧ 2 /da, and integrating the above equation once gives ȧ 2 = 8πGρ 3 a2 + const. Thus, if we choose the integration constant to be Kc 2, we obtain the Friedmann equation.

16 The Friedmann equation to be solved Suppose the universe contains a nonrelativistic component, a relativistic component, and a cosmological constant. The Friedmann equation can be written: (ȧ a ) 2 = Kc2 a + 8πGρ m,0 2 3 ( a0 a ) 3 + 8πGρ r,0 3 ( a0 a ) 4 8πGρ Λ, (Note that this equation is valid only when there is no energy exchange among these energy components). In order to specify this equation, we must know the densities at the present time: ρ m,0, ρ r,0, ρ Λ,0, and K (or H 0 ). In principle, all these quantities should be determined by the physical processes in the early universe, but the theory is incomplete, and so one has to rely on observations to determine their values.

17 The present values of cosmic densities and H 0 h H 0 /(100kms 1 Mpc 1 : 0.7 and so ρ crit, h 2 kgm 3 Ω m,0 : Ω r,0 : dominated by the 2.7K cosmic microwave background: ε = 4σT 4 /c (σ: Stefan-Boltzmann constant) Ω γ,0 = h 2 Ω Λ,0 : 0 1 The epoch of mass-radiation equality: Since ρ m /ρ γ Ω m,0 h 2 (1 + z) 1, ρ m = ρ γ at 1 + z eq Ω m,0 h 2.

18 The solutions of the Friedmann equation The goal is to find the form of a(t). We consider several interesting cases. 1. The radiation-dominated epoch (z z eq ): Since T (1 + z) we have a a 0 = T K kt 1MeV ( ) 1/4 32πGργ,0 t 1/2 3 ρ kg m 1 + z ( t ) 1/ sec Hot Big Bang!!

19 2. The matter-dominated epoch (z z eq ): Ω 0 = 1 (K = 0, flat universe): a a 0 = ( ) 2/3 3 2 H 0t Ω 0 < 1 (K = 1, open universe): Ω m,0 Ω m,0 a = 1 (coshθ 1); H 0 t = 1 a 0 21 Ω m,0 2(1 Ω m,0 ) 3/2(sinhθ θ). a t 2/3 (for t 0); a t (for t ).

20 Ω 0 > 1 (K = +1, close universe): a = 1 Ω m,0 a 0 2Ω m,0 1 (1 cosθ); H 0t = 1 Ω m,0 2(Ω m,0 1) 3/2(θ sinθ). a t (for t ). Maximum expansion: a max = Ω m,0 a 0 Ω m,0 1 ; H 0t max = π Ω m,0 2 (Ω m,0 1) 3/2.

21

22 3. Flat universe with cosmological constant (z z eq, Ω m,0 + Ω Lambda,0 = 1) In this case, the Friedmann equation is (ȧ a ) 2 = H 2 0 [ ( a0 ) ] 3 Ω m,0 + ΩΛ,0 a For Ω Λ,0 > 0, the solution is a a 0 = ( Ωm,0 Ω Λ,0 ) 1/3 [ ( )] 2/3 3 sinh 2 Ω1/2 Λ,0 H 0t a t 2/3 (for t 0) and a exp( Ω Λ,0 H 0 t) (for t ).

23 Cosmological tests Since our present knowledge of the matter content of the universe is limited, we must design observations to test cosmological models. The age of the universe where t(z) = dȧ a = 1 H 0 z dz (1 + z)e(z), E(z) = 1 H 0 ȧ a = [ Ω Λ,0 + (1 Ω 0 )(1 + z) 2 + Ω m,0 (1 + z) 3 + Ω r,0 (1 + z) 4] 1/2.

24 Radiation-dominated era t(z) = [(1 + z)/10 10 ] 2 sec Matter-dominated era and Ω 0 = 1: t(z) = (2/3H 0 )(1 + z) 3/2 (2/3h)(1 + z) 3/ yr. Matter dominated era and Ω 0 < 1 Matter dominated era and Ω 0 > 1 Flat universe with cosmological constant: Ω 0 = Ω Λ,0 + Ω m,0 = 1 t 0 t(0) > 13Gyr, because it must be larger than the age of the oldest objests (globular clusters) in the universe.

25 Cosmological distances and volumes 1. Angular-diameter distance and luminosity distance Consider a terrestrial object of size D and intrinsic luminosity L, put at some distance d from us. The value of d can be estimated by either measuring the angular size θ subtended by the object, d = D/θ, or by measuring the flux F from the object, d = (L/4πF) 1/2. Thus, we define the angular-diameter distance, d A, and the luminosity distance, d L, of a cosmological object as d A = D/θ; d L = (L/4πF) 1/2. For a Robertson-Walker metric, it can be shown that d A = a 0r e 1 + z e ; d L = a 0 r e (1 + z e ). Thus, we need to find the expression for r e = r(z e ).

26 For R-W metric, r = f K (χ), ds = 0 χ(z) = cdt a = c da aȧ = c H 0 a 0 z 0 dz E(z ). For example, for a matter dominated universe with Ω Λ,0 = 0: a 0 r = 2c H 0 Ω 0 z + (2 Ω 0 )[1 (Ω 0 z + 1) 1/2 ] Ω 2 0 (1 + z).

27 Stabdard candle (objects with known L) and standard ruler (objects with known D) can be used to obtain d L and d A by measuring F and θ. By measuring the redshifts of these objects, one can obtain cosmological parameters. An example is type Ia supernovae, whose peak luminosity is found to be almost a constant. m-m (mag) (m-m) (mag) MLCS Ω M =0.28, Ω Λ =0.72 Ω M =0.20, Ω Λ =0.00 Ω M =1.00, Ω Λ = z

28 2. Number of intersections per unit redshift along a sightline dn(z) = n(z)a(z) dl dz dz, where dl is the proper distance corresponding to dz Applications: dl = a(z)dχ dl dz = c H 0 1 (1 + z)e(z). The number of QSO absorption lines systems per unit redshift. The number of gravitational lensing events per unit redshift.

29 3. Cosmological volume Proper volume element: dv = a 3 (t)r 2 dχdω, where a(t)dχ = dl, a(t)rdθ is the the proper distance corresponding to dθ at radial coordinate r. Thus dv = c H 0 dz [a 0 r(z)] 2 dω. (1 + z)e(z) (1 + z) 2 Application: if the number density of a population of cosmological objects is n(z) = n 0 (1 + z) 3, then d 2 N dzdω dv n(z) = dzdω = cn 0 [a 0 r(z)] 2. H 0 E(z)

30 The production and annihilation of particles The hot Big Bang provided the condition to produce elementary particles, nuclei, and atoms. Basic principle: i + j a + b dn i dt + 3ȧ a n i = α(t )n a n b β(t )n i n j, where α(t ) and β(t ) are the reaction rates. Example: γ + γ e + e +. At kt > m e c MeV, or T > K, a lot of electrons and positrons can be produced. Other particles can also be produced through chanels like e + e ν e + ν e, etc.

31 If the reaction is sufficiently fast, i.e. nσv is high enough, then the number density of particles is given by the equilibrium distribution function (Planckian distribution, Bose distribution or Fermi distribution). The final number of a particle species is determined by nσv/h(t), if this ratio is much smaller than 1, then the reactions are no longer effective, and so the the number of particles is conserved. The freeze-out time is given by n(t) = H(t)/σv(t), while the final density is given by n(t freeze out ). Important applications: The mass density of dark matter Cosmic nucleosynthesis Recombination and decoupling of photons

32 The difficulties in the standard cosmology Consider the quantity: t0 cdt χ(r e ) = t e a(t), which has the meaning that a light signal emitted from (r e,t e ) reaches the origin χ = 0 (or r = 0) at the time t 0. If χ(r e ) is finite for t e 0, i.e. χ h = t 0 0 cdt/a(t) is finite, then there must be points with χ(r) > χ h. Light signals emitted even at t = 0 from these points will not be able to reach r = 0 even at the present time t 0, i.e. these points cannot have any causal connection with the orgin r = 0. Note that χ h = t0 t e cdt a(t) = a0 a e cda aȧ = a0 Thus, if ρa 2 as a 0, then χ h is finite. a e da a [ 8πGρa 2 3c 2 ] 1/2 K

33 It can be shown that χ h (z) 6000 Ωm,0 (1 + z) h 1 Mpc. Thus, χ h 100h 1 Mpc at z The problem: If no causal connections for different parts separated more than 100 Mpc apart, why the universe is so homogeneous on large scales? One solution: Assuming that the energy content of the universe is such that ρa 2 is finite as a 0, then χ h, and all points can have causal connection. One example of such energy content is ρ = const.. In this case a exp(ht), and we have rapid expansion of the universe, i.e. inflation! The very rapid expansion implies that the present observable universe was very small before the inflation and casual connection could have been established then.

The Friedmann Equation R = GM R 2. R(t) R R = GM R GM R. d dt. = d dt 1 2 R 2 = GM R + K. Kinetic + potential energy per unit mass = constant

The Friedmann Equation R = GM R 2. R(t) R R = GM R GM R. d dt. = d dt 1 2 R 2 = GM R + K. Kinetic + potential energy per unit mass = constant The Friedmann Equation R = GM R R R = GM R R R(t) d dt 1 R = d dt GM R M 1 R = GM R + K Kinetic + potential energy per unit mass = constant The Friedmann Equation 1 R = GM R + K M = ρ 4 3 π R3 1 R = 4πGρR

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