Cosmology. Lecture Scripts. Daniel Baumann concave convex

Cosmology Lecture Scripts 0.25 0.20 0.15 concave convex 0.10 small-field large-field 0.05 0.00 natural 0.94 0.96 0.98 1.00 Daniel Baumann dbaumann@damtp.cam.ac.uk

Chapter 1. GEOMETRY and DYNAMICS 1.1. GEOMETRY The geometry of spacetime is encoded in the metric, which turns observer-dependent coordinates X µ (t, x i ) into the invariant line element ds 2 = 3 µ,ν=0 g µν dx µ dx ν g µν dx µ dx ν In SR, the metric is constant: g µν = diag(1, 1, 1, 1). In GR, the metric depends on the spacetime position: g µν (t, x). What is the metric of the universe? When averaged over large scales, the universe looks isotropic = the same in all directions. gravity = geometry! If we don t live at a special place, then the universe is also homogeneous = the same everywhere. These symmetries fix the form of the metric: symmetric 3-space ds 2 = dt 2 a 2 (t) dl 2 scale factor 1

There are only 3 maximally symmetric 3-spaces: I. zero curvature = E 3 (Euclidean) dl 2 = dx 2 = δ ij dx i dx j II. positive curvature = S 3 (3-sphere) embedded in E 4 x 2 + u 2 = a 2, dl 2 = dx 2 + du 2 III. negative curvature = H 3 (3-hyperboloid) embedded in R 1,3 x 2 u 2 = a 2, dl 2 = dx 2 du 2 Let us manipulate II and III: If x ax, then dl 2 = a 2 [dx 2 ± du 2 ] ( ), with x 2 ± u 2 = ±1 ( ) u au Take the differential of ( ), udu = x dx, and substitute into ( ): dl 2 = a 2 [ dx 2 ± ] (x dx)2 1 x 2 This can be unified with case I: [ ] dl 2 = a 2 dx 2 (x 0 Euclidean E 3 dx)2 + k, where k = 1 kx 2 +1 spherical S 3 1 hyperbolic H 3 It is nice to write this in polar coordinates (r, θ, φ): Using dx 2 = dr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) dr 2 + r 2 dω 2 we get x dx = rdr [ ] dr dl 2 = a 2 2 1 kr + 2 r2 dω 2.. 2

Finally, let a be a function of time a(t) and incorporate dl 2 into ds 2 : [ ] dr ds 2 = dt 2 a 2 2 (t) 1 kr + 2 r2 dω 2 FRW METRIC where a(t) is the scale factor and k is the curvature parameter. Comments: 1) The line element has a rescaling symmetry: a λa, r r/λ, k λ 2 k, which we can use to set a(t 0 ) 1 today. 2) r is a comoving coordinate. Physical results depend only on the physical coordinate and the physical curvature The physical velocity of an object is r phys = a(t)r k phys = k/a 2 (t). v phys dr phys dt = a(t) dr dt + da dt r v pec + Hr phys, H 1 da a dt, where H is the Hubble parameter. 3) comoving time is called conformal time τ = dt/a(t): ] ds 2 = a 2 (τ) [dτ 2 dr2 1 kr 2 r2 dω 2 = a 2 (τ) static metric 4) sometimes it is convenient to use χ = dr/ 1 kr 2 : sinh 2 χ ds 2 = a 2 (τ) dτ 2 dχ 2 χ 2 dω 2 k = sin 2 χ 1 0 +1 3

1.2. KINEMATICS How do particles evolve in the FRW universe? In the absence of non-gravitational forces, particles move along geodesics = curves of least action, or extremal proper time s/c. Consider a particle of mass m. The four-velocity of the particle is U µ dxµ ds. Geodesic motion is defined by the geodesic equation where du µ ds + Γµ αβ U α U β = 0, Γ µ αβ 1 2 gµλ ( α g βλ + β g αλ λ g αβ ) g µλ g λν = δ µ ν λ / X λ Christoffel symbol Using d ds U µ (X α (s)) = dxα ds U µ X = U α Xµ α X α ) U α ( U µ X α + Γµ αβ U β, we get U α α U µ = 0. covariant derivative In terms of the four-momentum, P µ mu µ, we have P α P µ X α = Γµ αβ P α P β. This form of the geodesic equation holds also for massless particles! 4

Geodesic Motion in FRW Ex: Show that ds 2 = dt 2 a 2 (t)γ ij dx i dx j Γ µ 00 = Γ0 0β = 0 Γ 0 ij = aȧγ ij Γ i 0j = ȧ a δi j Γ i jk = 1 2 γil ( j γ kl + ) (= 0 for E 3 ) In FRW, i P µ = 0 (because of homogeneity), so the geodesic equation becomes We discover two facts: P 0dP µ dt = Γ µ αβ P α P β = ( 2Γ µ 0j P 0 + Γ µ ij P i) P j ( ) 1) particles at rest stay at rest : P j = 0 dp i 2) momentum decays : Consider µ = 0 of ( ): E energy E P 0 dt = 0 de dt = Γ0 ijp i P j = ȧ a p2 ( ) physical 3-momentum p 2 a 2 γ ij P i P j Recall that m 2 = g µν P µ P ν = E 2 p 2. This implies E de = pdp, so that ( ) becomes ṗ p = ȧ a p(t) 1 a(t) physical 3-momentum decays! For massless particles: p = E 1 a energy decays For massive particles: p = mv 1 v 2 1 a peculiar velocity decays 5

REDSHIFT wavelength In QM, λ = h / p momentum Since p(t) 1, we find λ(t) a(t), a(t) i.e. light emitted at t 1 with λ 1 is observed at t 0 with λ 0 = a(t 0) a(t 1 ) λ 1 > λ 1. This redshift is quantified by z λ 0 λ 1 λ 1 If a(t 0 ) 1, then 1 + z = 1 a(t 1 ). For nearby sources, we can expand a(t 1 ) as [ a(t 1 ) = a(t 0 ) 1 + (t 1 t 0 ) H 0 ] + HUBBLE CONSTANT H 0 ȧ(t 0) a(t 0 ) The distance to the source is d c(t 1 t 0 ), so we find z = H 0 d + HUBBLE S LAW H 0 = 100 h km s 1 Mpc 1 h = 0.67 ± 0.01 For distant sources, we have to be more careful about what we mean by distance (see lecture notes). 6

1.3. DYNAMICS The dynamics of a(t) is determined by the Einstein equation: G µν [a(t)] = 8πG T µν CURVATURE MATTER ENERGY-MOMENTUM TENSOR number current four-vector energy-momentum tensor N µ = (N 0, N i ) = T µν = ( ) T00 T 0i T i0 T ij = ( number ) density, flux energy density momentum density energy flux stress tensor A comoving observer sees a homogeneous and isotropic universe iff: Any 3-scalar is only a function of time N 0 n(t) T 00 ρ(t) Any 3-vector vanishes N i 0 T 0i 0 Any 3-tensor is proportional to g ij T ij P (t)g ij ρ Hence, N µ = (n, 0) T µ ν g µλ P T λν = perfect P fluid P A general observer sees where N µ = nu µ n : no. density ρ : energy density P : pressure U µ : relative 4-velocity T µ ν = (ρ + P )U µ U ν P δ µ ν in the rest frame of the fluid For U µ = (1, 0, 0, 0) this reduces to the previous results. For U µ = γ(1, v i ) we get boosted quantities: e.g. N 0 = γn, etc. 7

CONSERVATION LAWS 1) Particle number In Minkowski, the conservation of particle number implies ṅ = i N i µ N µ = 0 In curved spacetimes, this becomes µ N µ = 0 = µ N µ + Γ µ µλ N λ Using N i = 0 and Γ µ µ0 = Γi i0 = ȧ ṅ a δi i = 3ȧ, we get a 2) Energy and momentum n = 3 ȧ a n a 3. In Minkowski, the conservation of energy and momentum implies continuity } ρ = i π i Euler π i = i P µ T µ ν = 0 In curved spacetimes, this becomes µ T µ ν = 0 = µ T µ ν + Γ µ µλ T λ ν Γ λ µνt µ λ Consider the ν = 0 component in FRW: Since T i 0 = 0, this reduces to µ T µ 0 + Γ µ µλ T λ 0 Γ λ µ0t µ λ = 0 dρ dt + Γi i0 ρ Γ i j0t j i = 0 Using T j i = P δ j i and Γ i i0 = 3ȧ, we get a ρ + 3ȧ a (ρ + P ) = 0 continuity equation or d(ρa 3 ) dt = P d(a3 ) dt du = P dv 8

COSMIC INVENTORY Classify sources by their equation of state w P/ρ For w = const. we can integrate ρ ρ = 3(1 + w)ȧ a, to get ρ a 3(1+w). Name w ρ Examples m MATTER 0 a 3 non-relativistic particles Cold Dark Matter (CDM) Baryons (nuclei + electrons!) c b r RADIATION 1 3 a 4 relativistic particles Photons Neutrinos Gravitons γ ν g Λ DARK ENERGY 1 a 0 What the hell!? Vacuum Energy Modified Gravity Λ 9

SPACETIME CURVATURE EINSTEIN TENSOR G µν = R µν RICCI TENSOR 1 2 g µν R RICCI SCALAR Ex: Show that where R µν λ Γ λ µν ν Γ λ µλ + Γλ λρ Γρ µν Γ ρ µλ Γλ νρ R g µν R µν. R 00 = 3ä a, R 0i = 0, R ij = R = 6 [ ä a + (ȧ a ) ] 2 + k a 2 [ ä a + 2 (ȧ a ) ] 2 + 2 k g a 2 ij Ex: Compute G µ ν = g µλ G λν. You should find [ (ȧ ) ] 2 G 0 0 = 3 + k a a 2 = 8πGT 0 0 = 8πGρ [ (ȧ ) ] 2 G i j = 2ä a + + k δ i a a 2 j = 8πGT i j = 8πGP δj i 10

We get the FRIEDMANN EQUATIONS (ȧ a ) 2 = 8πG 3 ρ k a 2 ä a = 4πG ȧ (ρ + 3P ) ρ = 3 3 a (ρ + P ) where ρ ρ γ + ρ ν + ρ c + ρ b + ρ }{{}}{{} Λ. ρ r ρ m In terms of the Hubble parameter, the first Friedmann equation becomes H 2 = 8πG 3 ρ k a 2 ( ) We will use a subscript 0 to denote quantities today, at t = t 0. A flat universe (k = 0) corresponds to a critical density ρ crit,0 = 3H2 0 8πG = 1.9 10 29 h 2 grams cm 3 = 2.8 10 11 h 2 M Mpc 3 For each component I = r, m, Λ,..., define the fractional density today as Ω I ρ I,0 ρ crit,0 Eq. ( ) then becomes H 2 (a) = H 2 0 [Ω r a 4 + Ω m a 3 + Ω k a 2 + Ω Λ ], where a 0 1 and Ω k Observations show that k (a 0 H 0 ) 2. Ω k 0.01, Ω r = 9.4 10 5, Ω m = 0.32, Ω Λ = 0.68, We will from now on set Ω k 0. Ω b = 0.05, Ω c = 0.27, 0.001 < Ω ν < 0.02. 11

Often the energy density is dominated by a single component: first radiation, then matter, then dark energy radiation matter cosmological constant The Friedmann equation then reduces to which has the following solution a(t) t 2/3(1+w I) ( ) 2 1 da = H0 2 Ω I a 3(1+wI), a dt w I 1 t 2/3 t 1/2 MD RD e Ht w I = 1 ΛD or, in conformal time, a(τ) τ 2/(1+3w I) w I 1 τ 2 τ MD RD ( τ) 1 w I = 1 ΛD 12

Chapter 2. INFLATION Why is the universe homogeneous and isotropic? Why is the CMB so uniform? 2.1. THE HORIZON PROBLEM Consider the propagation of light in the FRW spacetime ] ds 2 = a 2 (τ) [dτ 2 dχ 2 Sk(χ)dΩ 2 2. Because of isotropy, we can focus on purely radial geodesics (dθ = dφ = 0): ] ds 2 = a 2 (τ) [dτ 2 dχ 2. Photons travel on null geodesics, ds 2 = 0 χ = ± τ (straight lines) Consider the FRW universe in these coordinates: comoving particle outside the particle horizon at p event horizon at p p particle horizon at p 13

It is the particle horizon that is relevant for the horizon problem: χ ph (τ) = τ τ i = t t i dt a a(t) = da ln a a i aȧ = (ah) 1 d ln a ( ) ln a i comoving Hubble radius e.g. for a fluid with w = P/ρ: (ah) 1 = H 1 0 a 1 2 (1+3w). Ordinary matter satisfies the Strong Energy Condition (SEC): 1 + 3w > 0. (ah) 1 increases and ( ) is dominated by late times. For the fluid, we have [ χ ph (a) = 2H 1 0 a 1 2 (1+3w) a 1 2 (1+3w) i 1 + 3w ] a i 0, w> 1 3 = 2H 1 0 1 + 3w a 1 2 (1+3w) 2 1 + 3w (ah) 1 [NB: In the standard cosmology: particle horizon Hubble radius horizon ] The singularity is at τ i 2H 1 0 1 + 3w a 1 i 2 (1+3w) = 0 and the particle horizon is finite and grows as χ ph = τ. 14

This has an important consequence: most parts of the cosmic microwave background (CMB) have never been in causal contact. D TODAY CMB PHOTON p q RECOMBINATION SINGULARITY p D q SURFACE OF LAST-SCATTERING The CMB is made of 10 4 causally disconnected regions, yet it is observed to be almost perfectly uniform!? = horizon problem 15

2.2. A SHRINKING HUBBLE SPHERE A simple solution to the horizon problem is to postulate a shrinking Hubble sphere d dt (ah) 1 < 0 SEC-violating fluid 1 + 3w < 0 This implies that τ i 2H 1 0 1 + 3w a 1 i 2 (1+3w) w< 1 3 There was more (conformal) time between the singularity and recombination than we had thought! D BIG BANG p q RECOMBINATION END OF INFLATION INFLATION CAUSAL CONTACT SINGULARITY 16

HUBBLE RADIUS vs. PARTICLE HORIZON (ah) 1 distance over which particles can travel in one expansion time χ ph distance over which particles can travel in the history of the universe Consider two particles separated by a distance λ: λ > (ah) 1 particles cannot talk to each other now λ > χ ph particles could never have communicated Conservative solution to the horizon problem: The observable universe was inside the Hubble radius at the beginning of inflation (a 0 H 0 ) 1 < (a I H I ) 1 scales inflation standard Big Bang inflation Big Bang reheating time How much inflation do we need? (a I H I ) 1 > (a 0 H 0 ) 1 e 60 (a E H E ) 1 During inflation H const., so we find a E > e 60 a I 60 e-folds 17

CONDITIONS FOR INFLATION The shrinking Hubble sphere is equivalent to other ways of describing inflation: Proof Accelerated expansion ä > 0 d(ah) 1 dt = d dt (ȧ) 1 = ä (ȧ) 2 < 0 ä > 0 Slowly-varying Hubble ε Ḣ H 2 < 1 d(ah) 1 dt = ȧh + aḣ (ah) 2 = (ε 1) a < 0 ε < 1 Exponential expansion ds 2 dt 2 e 2Ht dx 2 ε 1 H = ȧ a const. a(t) = eht Negative pressure w P ρ < 1 3 ä a = ρ + 3P 6M 2 pl > 0 ρ + 3P < 0 Constant density d ln ρ d ln a = 2ε < 1 H 2 ρ 2HḢ ρ ε = 1 ρ 2 Hρ 18

2.3 THE PHYSICS OF INFLATION Inflation occurs: ε = Ḣ ln H = d H2 dn < 1, where dn d ln a = Hdt. Inflation lasts: η = d ln ε dn = ε Hε < 1 What microphyscis leads to {ε, η } < 1? Scalar Field Dynamics Inflation is often modelled by the evolution of a scalar field φ (the inflaton) with energy density V (φ) (the inflaton potential): The stress-energy tensor associated with the inflaton is ( ) 1 T µν = µ φ ν φ g µν 2 gαβ α φ β φ V (φ) Let us evaluate this for a homogeneous field φ = φ(t): ρ φ T 0 0 = 1 2 φ 2 + V (φ) (= KE + PE) P φ 1 3 T i i = 1 2 φ 2 V (φ) (= KE PE) 19

We then feed this into the Friedmann equations H 2 = ρ φ 3Mpl 2 = 1 [ ] 1 3Mpl 2 2 φ c 2 + V (F 1 ) M pl 8πG Ḣ = ρ φ + P φ 2M 2 pl = 1 2 φ 2 M 2 pl (F 2 ) Take a time derivative of (F 1 ) 2HḢ = 1 [ φ φ + V φ] 3Mpl 2, where V dv dφ and use (F 2 ) to get the Klein-Gordon equation: φ + 3H φ = V (KG) ACCELERATION FRICTION FORCE The ratio of (F 2 ) and (F 1 ) gives ε = 1 2 φ 2 M 2 pl H2 < 1. Inflation occurs if the KE is small = slow-roll inflation For inflation to last the acceleration should be small: It is easy to show that η = ε Hε = 2(ε δ). δ φ H φ < 1 The conditions {ε, δ } 1 imply {ε, η } 1. 20

So far, this was exact. Now, we make the slow-roll approximation: 1) ε = 1 2 φ 2 M 2 pl H2 1 H2 V 3M 2 pl (F) SR 2) δ = φ H φ 1 3H φ V (KG) SR (KG) SR 1 2 Hence, we find ε = φ 2 Mpl 2 H2 (F) SR M 2 pl 2 ( ) V 2 ɛ v V Next, we consider d dt (KG) SR 3Ḣ φ + 3H φ = V φ, which leads to ε + δ Mpl 2 V V η v Successful SR inflation occurs when {ɛ v, η v } 1 : 21

The total amount of inflation is N tot = ae a I d ln a = te t I Hdt Using Hdt = Ḣ φ dφ = ± 1 2ε dφ M pl 1 2ɛv dφ M pl, we find N tot = φe φ I 1 dφ 60 2ɛv M pl to solve the horizon problem. Inflation ends when the inflationary energy becomes dominated by kinetic energy: ε(t E ) 1 The field starts oscillating and behaves like matter: P φ 0 ρ φ a 3 This energy needs to be converted to Standard Model degrees of freedom = reheating This initiates the Hot Big Bang. 22

Chapter 3. THERMAL HISTORY 3.1. THE HOT BIG BANG Two key concepts: Particles are in local thermal equilibrium as long as Γ H interaction rate expansion rate relativistic particles dominate non-relativistic particles are Boltzmann-suppressed : n e m/t. Particles decouple from the thermal bath when Γ H Non-equilibrium phenomena make the world interesting: dark matter freeze-out Big Bang nucleosynthesis recombination 23

3.2. EQUILIBRIUM 3.2.1. Equilibrium Thermodynamics Consider a particle in a volume V = L 3 : In QM, the momentum eigenstates have a discrete spectrum The density of states in momentum space is L3 h 3 = V h 3. in phase space is 1 h 3. Including g internal degrees of freedom (e.g. spin), we get density of states = g = h 3 g (2π) 3. h/2π 1 Now, consider a gas of particles : The probability of a state being occupied is given by the (phase space) distribution function f(x, p, t) = f(p). homogeneity + isotropy 24

Integrating over momentum, we find: number density n = g (2π) 3 energy density ρ = g (2π) 3 pressure P = g (2π) 3 d 3 p f(p) d 3 p f(p)e(p) p 2 d 3 p f(p) 3E(p) where E(p) = m 2 + p 2 for a weakly interacting gas of particles. For particles in kinetic equilibrium (maximum entropy), the distribution functions are f(p) = 1 e (E(p) µ)/t ± 1 + fermions bosons T (t) : temperature (k B 1) µ(t ) : chemical potential ds = du + P dv µdn T Each particle species i (with m i, µ i, T i ) has its own f i n i, ρ i, P i. For species in chemical equilibrium, the chemical potentials are related: 1 + 2 3 + 4 implies µ 1 + µ 2 = µ 3 + µ 4. photons : µ γ = 0, because photon number is not conserved: e + p e + p + γ. antiparticles : µ X = µ X, because X + X γ + γ. 25

Thermal equilibrium = kinetic equilibrium + chemical equilibrium holds when Γ i > H. interaction rate expansion rate Species then share a common temperature T i = T. Species decouple from the thermal bath when Γ i (T dec,i ) H(T dec,i ). After decoupling, T < T dec,i, the interactions are too rare to maintain equilibrium (see 3.3). 3.2.2. Densities and Pressure Our goal is to find n(t ), ρ(t ) and P (T ). At early times, µ/t 1 for all particles, so we set µ 0 (for now): n = ρ = g p 2 dp 2π 2 0 exp[ p 2 + m 2 /T ] ± 1 g p 2 p dp 2 + m 2 2π 2 0 exp[ p 2 + m 2 /T ] ± 1 Define x m/t and ξ p/t, to write this as n = ρ = g 2π 2T 3 I ± (x) g 2π 2T 4 J ± (x) where I ± (x) J ± (x) 0 0 ξ 2 dξ exp[ ξ 2 + x 2 ] ± 1 ξ 2 ξ dξ 2 + x 2 exp[ ξ 2 + x 2 ] ± 1 The functions I ± (x) and J ± (x) have analytic expressions in certain limits. 26

1) Relativistic Limit (T m or x 0) Consider I ± (0) = 0 dξ ξ 2 e ξ ± 1. For bosons, this is a standard integral: I (0) = 2ζ(3) = 2.4... For fermions, we get I + (0) = 3 4 I (0). [ Trick : 1 e ξ + 1 = 1 e ξ 1 2 ] e 2ξ 1 Hence, n = ζ(3) π 2 gt 3 { 1 bosons 3 4 fermions. Ex: Show that ρ = π2 30 gt 4 { 1 bosons 7 8 fermions. and P = 1 3 ρ ( radiation ). Ex: Using that the temperature of the cosmic microwave background is T 0 = 2.73 K, show that n γ,0 = 2ζ(3) π 2 T 3 0 410 photons cm 3, ρ γ,0 = π2 15 T 4 0 4.6 10 34 g cm 3 Ω γ h 2 2.5 10 5. 27

2) Non-Relativistic Limit (T m or x 1) Consider I ± (x 1) = 0 dξ e ξ 2 ξ2 +x 2 (same for bosons and fermions). Most of the contribution to the integral comes from ξ x, so we can use ξ 2 + x 2 x + ξ2 2x to write I ± (x) e x dξ ξ 2 e ξ2 /(2x) 0 = (2x) 3/2 e x dξ ξ 2 e ξ2 = 0 π 2 x3/2 e x. Hence, n = g ( mt 2π ) 3/2 e m/t Since E(p) m, the energy density is ρ mn. Ex: Show that P = nt ρ = mn ( matter ). 28

Relativistic particles ( radiation ) dominate the early universe: ρ r = i ρ i = π2 30 g (T ) T 4, where g (T ) is the effective number of relativistic degrees of freedom. We can have two types of contributions: Relativistic species that are in equilibrium with photons, T i = T m i, g th (T ) = i=b g i + 7 g i. 8 Relativistic species that are not in equilibrium with photons, T i T m i, g dec (T ) = i=b In the Standard Model we have: i=f ( ) 4 Ti g i + 7 T 8 i=f ( ) 4 Ti g i. T 29

Above 10 5 MeV, all particles are relativistic: g b = 28 photons (2), W ±, Z 0 (3 3), gluons (8 2), Higgs (1) g f = 90 quarks (6 2 3 2), charged leptons (3 2 2), neutrinos (3 1 2) and hence g = g b + 7 8 g f = 106.75. When T m i, a species becomes non-relativistic and isn t counted in g. 3.2.3. Entropy Ex: Show that in equilibrium (and for µ = 0) P T = ρ + P T ( ) Consider the 2nd law of thermodynamics: ds = du + P dv T U=ρV 1 ( d [ (ρ + P )V ] ) V dp T = 1 T d[ (ρ + P )V ] V T 2(ρ + P ) dt [using ( )] [ ] ρ + P = d V. T The time derivative of the entropy is ds dt = d [ ] ρ + P V dt T = V [ dρ T dt + 1 ] dv V dt (ρ + P ) + V T }{{} = 0 by continuity eqn. = 0. [ dp dt ρ + P ] dt T dt }{{} = 0 by ( ) Entropy is conserved in equilibrium. 30

It will be convenient to work with the entropy density s = ρ + P T. For a collection of species, we have s = i ρ i + P i T i = 2π2 45 g S(T ) T 3, where g S (T ) are the effective no. of degrees of freedom in entropy : g S th (T ) = gth (T ) g dec S (T ) = i=b g i ( Ti T ) 3 + 7 8 i=f ( ) 3 Ti g i g dec (T ) T The conservation of entropy implies s a 3 N i n i s (= const., when particle no. is conserved) T g 1/3 S a 1 ( a 1, when g S = const.) ρ r T 4 a 4 (as expected for radiation) Ex: Using the Friedmann equation, show that T 1 MeV 1.5g 1/4 ( ) 1/2 1sec. t 31

3.2.4. Neutrino Decoupling Neutrinos are coupled to the thermal bath by weak interactions: e.g. ν + e + ν + e + with interaction rate Γ = 2 dimensional analysis G 2 F T 5 Decoupling occurs when Fermi s constant G F 10 5 GeV 2 Γ H G2 F T 5 T 2 /M pl ( ) 3 T 1 T dec 1 MeV. 1 MeV Shortly after neutrino decoupling, electrons and positrons annihilate. This transfers entropy to the photons, but not to the decoupled neutrinos. Photons are heated (relative to neutrinos). neutrino decoupling photon heating electron-positron annihilation 32

To compute the photon heating, we consider conserved quantities before and after e + e annihilation: Hence, we find g th S = (T ν a) before = (T ν a) after [free streaming] [ ] [ g (th) S (T γa) 3 = g (th) S (T γa) 3 [entropy] before ]after { 2 + 7 8 (2 2) = 11 ( ) 2 T m 1/3 e 4 T ν = T γ 2 T < m e 11 and g (T m e ) = 2 + 7 8 (3 2) ( 4 11) 4/3 = 3.36. 3.2.5. Cosmic Neutrino Background Ex: Show that n ν,0 = 3 4 3 4 11 n γ,0 335 neutrinos cm 3 and ρ ν,0 = ( ) 4/3 7 4 8 3 ρ γ,0 Ω ν h 2 1.7 10 5 (m ν = 0) 11 ν m νn ν,0 Ω ν h 2 mν 94 ev (m ν 0) Observational constraints on neutrino masses, 0.05 ev m ν 1 ev, imply 0.001 < Ω ν < 0.02. 33

3.3. BEYOND EQUILIBRIUM 3.3.1. Boltzmann Equation In the absence of interactions: With interactions: dn i dt + 3ȧ a n i = 0 1 d(a 3 n i ) a 3 dt = C i [{n j }] COLLISION TERM BOLTZMANN EQUATION Let us guess the form of the RHS for We focus on species 1: 1 + 2 3 + 4 1 d(a 3 n 1 ) a 3 dt = α n 1 n 2 + β n 3 n 4 ( ) DESTROY CREATE where α σv is the thermally averaged cross section. The interaction rate of species 1 is Γ 1 n 2 σv cosmology particle physics We relate β to α by noting that the collision term has to vanish in equilibrium: ( ) n1 n 2 β = α. n 3 n 4 ( ) can therefore be written as [ 1 d(a 3 ( ) n 1 ) n1 n 2 = σv n a 3 1 n 2 dt n 3 n 4 eq eq n 3 n 4 ] 34

or, in terms of N i n i /s, d ln N 1 d ln a = Γ 1 H interaction efficiency [ ( ) N1 N 2 1 N 3 N 4 eq N 3 N 4 N 1 N 2 deviation from equilibrium ] Γ 1 H : System is quickly driven towards chemical equilibrium Γ 1 H : Species 1 freezes out : N 1 = n 1 s n 1a 3 const. relativistic non-relativistic freeze-out relic density equilibrium 1 10 100 35

3.3.2. Dark Matter Relics Consider Weakly Interacting Massive Particles: X + X l + l strongly interacting particles (e.g. charged leptons) n l n eq l Assume: no initial asymmetry, i.e. n X = n X. The Boltzmann equation for N X n X /s can then be written as dn X dt [ ] = s σv NX 2 (N eq X )2 ( ) Defining x M X /T and using H = H(M X )/x 2 (RD), we can write ( ) as dn X dx = λ x 2 [ N 2 X (N eq X )2 ] RICCATI EQUATION where λ 2π2 45 g MX 3 σv S H(M X ) particle physics cosmology. Find the solution for λ const. 36

First, numerical: 1 10 100 Then, analytical: For x x f, we have N eq X N X and hence dn X dx λn 2 X x 2. Integrating from x f to x = +, we get 1 N X 1 N f X = λ x f. For N X N f X, this becomes N X x f λ, where we could estimate x f from Γ(x f ) H(x f ). 37

THE WIMP MIRACLE We wish to estimate the dark matter density today: Ω X ρ X,0 ρ crit,0 = M Xn X,0 3M 2 pl H2 0 = M XN X,0 s 0 3M 2 pl H2 0 = M X N X s 0 3M 2 pl H2 0, where we have used N X,0 = N X. Substituting N X = x f/λ and s 0 s(t 0 ), we find ( Ω X h 2 xf ) ( ) 1/2 10 10 8 GeV 2 0.1 10 g (M X ) σv. This reproduces the observed DM density if σv 10 4 GeV 1 0.1 G F WIMP miracle. 38

3.3.3. Recombination Above 1 ev, photons are in equilibrium with electrons and protons: e + p + H + γ, where ( ) 3/2 ( ) n eq mi T µi m i i = g i exp 2π T with µ p + µ e = µ H., i = {e, p, H} To remove the dependence on µ i, we consider ( ) nh = g ( ) 3/2 H mh 2π e (m p+m e m H )/T. n e n p eq g e g p m e m p T 4 1 n p = n e B H m p + m e m H = 13.6 ev 2 2 m e Hence, we get ( nh ) n 2 e eq = ( ) 3/2 2π e B H/T m e T ( ) We define the free electron density as X e n e n b. Ignoring heavy nuclei, we have n b n p + n H = n e + n H and hence 1 X e X 2 e = n H n 2 e n b = n H n 2 e η n γ where η = 5.5 10 10 (Ω b h 2 /0.020) is the baryon-to-photon ratio. Substituting ( ) and n γ (T ), we get ( ) 1 Xe X 2 e eq = 2ζ(3) π 2 ( 2πT η m e ) 3/2 e B H/T SAHA EQUATION Smallness of η 10 9 suppresses recombination until T B H. 39

recombination decoupling CMB Saha Boltzmann plasma neutral hydrogen Recombination Let us define recombination as X e (T rec ) = 10 1. Using the Saha equation, we find Photon decoupling T rec 0.3 ev B H z rec 1320 z eq t 0 t rec 290 000 yrs (1 + z rec ) 3/2 Thomson scattering, e + γ e + γ, occurs with rate Γ γ = n e σ T, where σ T = 8π 3 αe 2 m 2 e = 2 10 3 MeV 2. Photon decoupling happens at Γ γ (T dec ) H(T dec ), or ( ) 3/2 Tdec n b (T dec )X e (T dec )σ T H 0 Ωm. T 0 40

Using the Saha equation for X e (T dec ), we find T dec 0.27 ev z dec 1100 t dec 380 000 yrs Electron freeze-out Solving the Boltzmann equation gives Xe 10 3. The residual electron density creates the polarization of the CMB. 3.3.4. Big Bang Nucleosynthesis Light elements (H, He, Li) were synthesised in the Big Bang. Goal: explain one number: n He n H 1 16 Summary: Step 0: Equilibrium Step 1: Neutron Freeze-Out Step 2: Neutron Decay Step 3: Helium Fusion Fractional Abundance equilibrium Temperature [MeV] 41

Step 0: EQUILIBRIUM 1) Neutron-to-proton ratio Consider n + ν e p + + e If {µ e, µ ν } T, then µ n = µ p and ( nn ) n p eq = g n g p ( mn m p ) 3/2 e (m n m p )/T = 1 1 Q m n m p = 1.30 MeV ( nn ) n p eq = e Q/T 2) Deuterium Consider n + p + D + γ Since µ γ = 0, we have µ n + µ p = µ D and ( nd n n n p ) eq = g ( ) 3/2 D md 2π e (m D m n m p )/T g n g p m n m p T 3 2 2 2 m p B D m n +m p m D = 2.22 MeV ( nd ) n p eq = 3 4 neq n ( ) 3/2 4π e B D/T m p T 42

To get an order of magnitude estimate, we use and hence n n n b = η n γ = η 2ζ(3) π 2 T 3, ( nd n p ) eq ( T η m p ) 3/2 e B D/T. Smallness of η 10 9 suppresses deuterium until T B D. Initial condition: only neutrons and protons. Step 1: NEUTRON FREEZE-OUT It is convenient to define the neutron fraction as X n n n n n + n p X eq n (T ) = e Q/T 1 + e Q/T. Neutrons freeze-out when the weak interactions become inefficient (cf. neutrino decoupling). We can estimate the relic abundance of neutrons by their equilibrium abundance at neutrino decoupling X n X eq n (0.8 MeV) = 0.17 1 6. Step 2: NEUTRON DECAY Neutrons are unstable: X n (t) = X n e t/τ n = 1 6 e t/τ n, τ n 900 sec. 43

Step 3: HELIUM FUSION Helium can only form after deuterium is produced = deuterium bottleneck n + p D + γ D + p 3 He + γ D + 3 He 4 He + p We estimate the time of nucleosynthesis as ( ) nd 1 T nuc 0.06 MeV n p eq Hence, we find X n (t nuc ) 1 6 e 300/900 1 8. ( ) 2 1 MeV t nuc 1 sec 300 sec. T nuc Since virtually all neutrons go into 4 He, we get n He (t nuc ) = 1 2 n n(t nuc ), or n He n H = n He n p 1 2 X n(t nuc ) 1 X n (t nuc ) 1 2 X n(t nuc ) 1 16, as we wished to show. Sometimes, the result is expressed as the mass fraction of helium, 4n He n H 1 4. 44

Chapter 4. COSMOLOGICAL PERTURBATION THEORY 4.1. NEWTONIAN PERTURBATION THEORY Newtonian gravity is an adequate description on small scales (< H 1 ) and for non-relativistic matter (CDM + baryons after decoupling). The equations for a fluid with mass density ρ, pressure P and velocity u are: t ρ = r (ρu) (C) Continuity equation [conservation of mass] ( t + u r ) u = rp ρ r Φ (E) Euler equation [conservation of momentum] F=ma 2 r Φ = 4πGρ (P ) Poisson equation We wish to consider small fluctuations around a homogeneous background: and linearise the fluid equations. ρ(t, r) = ρ(t) + δρ(t, r),... 1) Static space without gravity The background solution is ρ = const., P = const. and ū = 0. The linearised evolution equations are t δρ = r ( ρu) ρ t u = r δp (C) (E) Combining t (C) and r (E), we get 2 t δρ 2 r δp = 0. 45

speed of sound For adiabatic fluctuations (see later), we have δp = c 2 s δρ and hence ( ) 2 t c 2 s r 2 δρ = 0 wave equation Solution: δρ = Ae i(ωt k r), where ω = c s k, with k k. fluctuations oscillate 2) Static space with gravity For Φ 0, we get ( ) 2 t c 2 s r 2 δρ = 4πG ρ δρ from ρ 2 r δφ Solution: δρ = Ae i(ωt k r), where ( ) ω 2 = c 2 sk 2 4πG ρ c 2 s k 2 kj 2 pressure gravity Jeans scale: k J 4πG ρ/c s On small scales, k > k J, fluctuations oscillate On large scales, k < k J, fluctuations grow exponentially 46

3) Expanding space In an expanding space, we have r = a(t)x and u ṙ = Hr + v Rather than labelling events by t and r, it is convenient to use t and x: For spatial gradients, this means using r = a 1 x ( ) For time derivatives, this means using ( ) ( ) ( ) ( ) x = + x = t r t x t r t ( ) = t x x ( (a 1 ) (t)r) + t Hx x Notation alert: we will drop the subscripts x from now on! x ( ) x With this in mind, we look at the fluid equations in an expanding universe: Continuity equation: Substituting ( ) and ( ) into (C), we get [ ] [ ρ(1 t Hx ] 1 + δ) + a [ ρ(1 + δ)(hax + v) ] = 0, where δ δρ/ ρ is the density contrast. At zeroth-order, we find ρ t + 3H ρ = 0, where we used x x = 3. As expected, we have ρ a 3. At first-order, we get [ ρ + 3H ρ t ] } {{ } = 0 δ + ρ δ t + ρ a v = 0. 47

Hence, we find δ + 1 a v = 0 (C) where δ ( δ t ) x. Euler equation: Similar manipulations of the Euler equation lead to v + Hv = 1 a ρ δp 1 a δφ (E) For δp = δφ = 0, this implies v a 1 (cf. Ch. 1). Poisson equation: It takes no work to show that 2 δφ = 4πGa 2 ρ δ (P ) Combining t (C) with (E) and (P ), we get δ + 2H δ c2 s a 2 2 δ = 4πG ρδ. Hubble friction The Jeans scale k J (t) 4πG ρ(t)/c s (t) is time-dependent. k > k J : fluctuations oscillate with decreasing amplitude k < k J : fluctuations grow as a power law Ex: Show that long-wavelength fluctuations in a matter-dominated universe, 4πG ρ = 3 2 H2 Ω m, have the following power-law solutions δ = a, a 3/2. 48

4.2. RELATIVISTIC PERTURBATION THEORY GR is required on large scales and for relativistic fluids. Perturb metric: g µν = ḡ µν + δg µν Perturb matter: ρ = ρ + δρ,... Linearise evolution equations. 4.2.1. PERTURBED METRIC The most general perturbations around FRW are ] ds 2 = a 2 (τ) [(1 + 2A)dτ 2 2B i dx i dτ (δ ij + h ij )dx i dx j, where A(τ, x), etc. Convention: Latin indices are raised and lowered with δ ij : e.g. h i i = δ ij h ij. Write B i i B + ˆB i scalar vector : i ˆBi = 0 h ij 2Cδ ij + 2 i j E + 2 (i Ê j) + 2Êij scalar scalar vector tensor where i j E ( i j 1 3 δ ij 2) E (i Ê j) 1 2 ( iêj + j Ê i ) i Ê i = 0 i Ê ij = Êi i = 0 10 = 4 scalar modes : A, B, C, E + 4 vector modes : ˆBi, Êi + 2 tensor modes : Ê ij Theorem: At first order, scalars, vectors, and tensors don t mix! We can treat them separately. 49

The Gauge Problem Perturbations depend on the choice of coordinates ( gauge choice ). Consider an unperturbed FRW universe, ds 2 = a 2 (τ) [ dτ 2 δ ij dx i dx j], ρ = ρ(t),. A change of spatial coordinates, x i x i x i + ξ i (τ, x), implies dx i = d x i τ ξ i dτ k ξ i dx k, which results in fictitious metric perturbations ds 2 = a 2 (τ) [ dτ 2 2ξ id x i dτ ( ) δ ij + 2 (i ξ j) d x i d x j] + O(ξ 2 ) fake perturbations (gauge modes) Similarly, a change of time slicing, τ τ + ξ 0 (τ, x), induces fictitious density perturbations ρ(τ) ρ(τ + ξ 0 ) = ρ(τ) + ρ ξ 0. We need a way to identify true perturbations. Gauge Transformations Consider X µ X µ X µ + ξ µ (τ, x), where ξ 0 T, ξ i i L. To see how the metric transforms, note that ds 2 = g µν (X)dX µ dx ν = g αβ ( X)d X α d X β. Writing d X α = ( X α / X µ )dx µ (and similarly for d X β ), we find g µν (X) = X α X µ X β X ν g αβ( X). 50

As an example, let us take µ = ν = 0: g 00 (X) = X α τ X β τ g αβ( X) ( ) 2 to first order τ g 00( τ X). We can write this as a 2 (τ) ( 1 + 2A ) = ( 1 + T ) 2 a 2 (τ + T ) ( 1 + 2Ã) = ( 1 + 2T + )( a(τ) + a T + )2( 1 + 2 Ã ) = a 2 (τ) ( 1 + 2HT + 2T + 2Ã + ), where H a /a. This implies that A Ã = A T HT. Ex: Show that B B = B + T L C C = C HT 1 3 2 L, E Ẽ = E L. Gauge-Invariant Variables Ex: Show that are gauge-invariant. Ψ A + H(B E ) + (B E ), Φ C H(B E ) + 1 3 2 E, The Bardeen potentials Ψ and Φ represent true perturbations that cannot be removed by a coordinate transformation. 51

Gauge Fixing An alternative solution of the gauge problem is to fix the gauge, follow all perturbations (metric and matter), compute observables. Two popular gauges are: Newtonian gauge: B = E 0. Spatially flat gauge: C = E 0. 4.2.2. PERTURBED MATTER In a homogeneous universe, T µ ν = ( ρ + P )Ū µ Ū ν P δ µ ν, where Ū µ = a 1 δ µ 0 for a comoving observer. In a perturbed universe, T µ ν = T µ ν + δt µ ν = T µ ν + (δρ + δp )Ū µ Ū ν + ( ρ + P )(δu µ Ū ν + Ū µ δu ν ) δp δ µ ν Π µ ν anisotropic stress (mostly negligible) To derive δu µ, we consider g µν U µ U ν = 1: At first order, this implies δg µν Ū µ Ū ν + 2ŪµδU µ = 0. Using Ū µ = a 1 δ µ 0 and δg 00 = 2a 2 A, we find δu 0 = Aa 1. Writing δu i = v i /a, where v i dx i /dτ, we get U µ = a 1 [1 A, v i ]. Ex: Show that U µ = a[1 + A, (v i + B i )]. 52

Hence, we find T 0 0 = ρ + δρ T i 0 = ( ρ + P )v i q i 3-momentum density T 0 j = ( ρ + P )(v j + B j ) T i j = ( P + δp )δj i In a multi-component universe, we have T µν = I T µν I and hence δρ = I δρ I δp = I δp I q i = I q i I (note: velocities don t add!) Gauge Transformations X µ X µ = X µ + (T, i L) implies T µ ν(x) = Xµ X α X β X ν T α β( X). and δρ δρ T ρ, δp δp T P, v v + L, where v i i v. A gauge-invariant density fluctuation is ρ δρ + ρ (v + B). Popular matter gauges are: comoving-gauge density contrast Uniform density gauge: δρ 0 Comoving gauge: q 0 + B 0. 53

Adiabatic Fluctuations are predicted by inflation. What are they? Consider a mixture of fluids, with ρ I, P I, etc. Adiabatic perturbations are perturbations induced by a common, local shift in time of all background quantities: e.g. δρ I (τ, x) = ρ I (τ + δτ(x)) ρ I (τ) = ρ I δτ(x) same for all I. Hence, δτ = δρ I ρ I = δρ J ρ J, for all I and J. Using ρ I = 3H(1 + w I ) ρ I and δ I δρ I / ρ I, we find δ I 1 + w I = δ J 1 + w J. Adiabatic perturbations satisfy δ r = 4 3 δ m. Whatever dominates the background also dominates fluctuations: δρ tot = ρ tot δ tot = I ρ I δ I similar for all I. 54

4.2.3. LINEARISED EVOLUTION EQUATIONS See: www.damtp.cam.ac.uk/user/db275/cosmology/cpt.nb (thanks to Yi Wang) We will work in Newtonian gauge: g µν = a 2 ( 1 + 2Ψ 0 0 (1 2Φ)δ ij ), and ignore anisotropic stress: Π ij = 0. Inverse metric Christoffel symbols At first order, we find g µν = 1 a 2 ( 1 2Ψ 0 0 (1 + 2Φ)δ ij Ex: Show that Γ 0 0i = i Ψ, ) + O(Ψ 2, Φ 2 ). Γ 0 00 = 1 2 g00 0 g 00 = 1 2a 2(1 2Ψ) 0[a 2 (1 + 2Ψ)] Γ i 00 = δ ij j Ψ, = H + Ψ. Γ 0 ij = Hδ ij [Φ + 2H(Φ + Ψ)] δ ij, Γ i j0 = Hδ i j Φ δ i j, Γ i jk = 2δ i (j k)φ + δ jk δ il l Φ. 55

PERTURBED CONSERVATION EQUATIONS Consider µ T µ ν = 0 = µ T µ ν + Γ µ µαt α ν Γ α µνt µ α. The ν = 0 component leads to the continuity equation: At zeroth-order, we find ρ = 3H( ρ + P ). At first-order, we get δρ = 3H(δρ + δp ) + 3Φ ( ρ + P ) q dilution by expansion perturbed expansion: fluid flow a eff a(1 Φ) For adiabatic perturbations, we can write this as δ + (1 + w) ( v 3Φ ) + 3H(c 2 s w)δ = 0 (C) For non-relativistic matter (w = c 2 s 1) on sub-hubble scales (Φ v Ch. 5), this reproduces the Newtonian continuity equation δ + v 0. The ν = i component leads to the Euler equation: v = Hv redshifting + 3H P ρ v δp Ψ ρ + P relativistic correction For adiabatic perturbations, we get pressure gravity v + H(1 3c 2 s)v = δ Ψ (E) 1 + w For non-relativistic matter (w = c 2 s 1), this reproduces the Newtonian Euler equation. c2 s 56

PERTURBED EINSTEIN EQUATIONS Γ λ µν(φ, Ψ) R µν G µν R µν 1 2 g µνr Ex: Show that (see lecture notes) G 00 = 3H 2 + 2 2 Φ 6HΦ G 0i = 2 i (Φ + HΨ) G ij = (2H + H 2 )δ ij + i j (Φ Ψ) + [ 2 (Ψ Φ) + 2Φ + 2(2H + H 2 )(Φ + Ψ) + 2HΨ + 4HΦ ] δ ij Consider the trace-free part of the ij-einstein equation: Next, look at the 00-equation: i j (Φ Ψ) = 0 Φ = Ψ. G 00 = 8πGT 00 3H 2 + 2 2 Φ 6HΦ = 8πGg 0µ T µ 0 = 8πG ( ) g 00 T 0 0 + g 0i T i 0 = 8πGa 2 (1 + 2Φ)( ρ + δρ) = 8πGa 2 ρ(1 + 2Φ + δ). At zeroth-order, we recover the Friedmann equation: 3H 2 = 8πGa 2 ρ. At first-order, we get 2 Φ = 4πGa 2 ρδ + 3H(Φ + HΦ) ( ) Moving on to the 0i-equation: G 0i = 8πGT 0i Φ + HΦ = 4πGa 2 q ( ) 57

Substituting ( ) into ( ), we find 2 Φ = 4πGa 2 ρ Poisson equation where is the comoving-gauge density contrast. Finally, we look at the trace of the ij-equation: G i i = 8πGT i i. At zeroth-order, we recover the second Friedmann equation: 2H + H 2 = 8πGa 2 P. At first-order, we get an evolution equation for Φ: Φ + 3HΦ + (2H + H 2 )Φ = 4πGa 2 δp. 58

4.3. CONSERVED CURVATURE PERTURBATION A very important quantity is the comoving curvature perturbation R. It is conserved on superhorizon scales (k H = ah) and therefore provides the link between fluctuations created by inflation (Ch. 6) and those observed in the late universe (Ch. 5). Consider the perturbed 3-metric γ ij = a 2 [(1 + 2C)δ ij + 2E ij ], and the corresponding 3-curvature a 2 R (3) = 4 (C 2 1 ) 3 2 E. The comoving curvature perturbation is [ R C 1 ] 3 2 E q i = B i = 0 or, in gauge-invariant form R = C 1 3 2 E + H(v + B). Evaluated in Newtonian gauge, this is R = Φ + Hv = Φ H(Φ + HΦ) 4πGa 2 ( ρ + P ). Using the Einstein equations, we can show that (see lecture notes) 4πGa 2 ( ρ + P )R = 4πGa 2 H (δp P ) ρ δρ + H P ρ 2 Φ. The first term on the r.h.s. vanishes for adiabatic perturbations and we get ( ) 2 d ln R k d ln a k H 0 H i.e. R is conserved on superhorizon scales! 59

Chapter 5. STRUCTURE FORMATION Radiation Neutrions Photons Dark Energy Thomson Scattering Metric Electrons Dark Matter Baryons Coulomb Scattering Protons Matter Perturbed Einstein equations: (E1) 2 Φ 3H(Φ + HΦ) = 4πGa 2 ρδ (E2) Φ + HΦ = 4πGa 2 ( ρ + P )v 2 Φ = 4πGa 2 ρ (P) (E3) Φ + 3HΦ + (2H + H 2 )Φ = 4πGa 2 δp Perturbed conservation equations: (C) δ I + 3H(c 2 s,i w I )δ I = (1 + w I )( v I 3Φ ) (E) v I + H(1 3c 2 s,i)v I = c2 s,i 1 + w I δ I Φ where H and Φ are sourced by all components. 60

5.1. INITIAL CONDITIONS inflation R CMB + LSS Φ δ m, δ r When the universe is dominated by a single component, we have Φ + 3(1 + w)hφ + wk 2 Φ = 0. On superhorizon scales, this implies Φ = const. δ = 2Φ = const. δ m = 3 4 δ r 3 2 Φ RD = R(0). To follow the evolution of superhorizon modes, we use the conservation of the curvature perturbation ( ) 2 Φ R = Φ 3(1 + w) H + Φ k H 5 + 3w 3 + 3w Φ. Through the radiation-to-matter transition, we have R = 3 2 Φ RD = 5 3 Φ MD Φ MD = 9 10 Φ RD. We wish to understands what happens to the superhorizon initial conditions when modes enter the horizon. 61

5.2. EVOLUTION OF FLUCTUATIONS 5.2.1. GRAVITATIONAL POTENTIAL In the radiation era, we have Φ + 4 τ Φ + k2 3 Φ = 0, whose solution is ( ) sin x x cos x Φ k (τ) = 2R k (0) Outside the horizon, this reproduces Φ = const. x 3, where x 1 3 kτ. Inside the horizon, we get Φ k (τ) kτ 1 6R k (0) cos ( 1 3 kτ ) (kτ) 2 oscillations a 2 (τ). In the matter era, we have Φ + 6 τ Φ = 0. The growing mode is Φ = const. on all scales. Summary: 62

5.2.2. RADIATION In the radiation era, perturbations in the radiation density dominate. They are then directly determined by the potential (via the Poisson equation) r = 2 3 ( ) 2 k Φ = 2 H 3 (kτ)2 Φ. In the subhorizon limit, we therefore find δ r r 4R(0) cos ( ) 1 3 kτ, which is the solution to δ r 1 3 2 δ r = 0 Fluctuations oscillate around δ r = 0. In the matter era, radiation perturbations are subdominant. Their evolution then has to be determined from the conservation equations. On subhorizon scales, we have (C) (E) δ r = 4 3 v r v r = 1 4 δ r Φ δ r 1 3 2 δ r = 4 3 2 Φ = const. Fluctuations oscillate around δ r = 4Φ MD (k). The acoustic oscillations of δ r (τ) are the origin of the peaks in the spectrum of CMB anisotropies (see Advanced Cosmology). 63

5.2.3 DARK MATTER I. Early Times At early times, the universe was a mixture of radiation (r) and matter (m). Ex: Show that H 2 = H2 0Ω 2 m Ω r ( 1 y + 1 ) y 2, y a a eq. Consider the conservation equations for matter (on subhorizon scales): (C) δ m = v m (E) v m δ m + Hδ m = 2 Φ. = Hv m Φ Since radiation fluctuations rapidly oscillate, the time-averaged gravitational potential is only sourced by the matter fluctuations. Hence, we get δ m + Hδ m 4πGa 2 ρ m δ m 0, which can be written as the Mészáros equation (see Problem Set 3) d 2 δ m dy + 2 + 3y dδ m 2 2y(1 + y) dy 3 2y(1 + y) δ m = 0. The solutions are 2 + 3y δ m (2 + 3y) ln ( 1 + y + 1 1 + y 1 ) 6 1 + y. In the limit y 1 (RD), the growing mode is δ m ln y ln a. In the limit y 1 (MD), the growing mode is δ m y a. 64

II. Intermediate Times The solution in the matter era also follows directly from Φ = const.: m = 2 3 ( ) 2 k Φ = 1 H 6 (kτ)2 Φ a (on all scales!) On subhorizon scales, δ m m a (as before). III. Late Times At late times, the universe is a mixture of matter (m) and dark energy (Λ). Since dark energy doesn t have fluctuations, we still have 2 Φ = 4πGa 2 ρ m m m /a. Pressure fluctuations are negligible, so the Einstein equations give Φ + 3HΦ + (2H + H 2 )Φ = 0, ( m /a) + 3H( m /a) + (2H + H 2 )( m /a) = 0, which rearranges to m + H m + (H H 2 ) m = 0. Ex: Show that H H 2 = 4πGa 2 ( ρ + P ) = 4πGa 2 ρ m. Hence, we find m + H m 4πGa 2 ρ m m = 0. In the matter era, this reproduces m a. In the dark energy era, we have H 2 4πGa 2 ρ m and hence m = const. Matter fluctuations stop growing when Λ comes to dominate. 65

Power Spectrum observed fluctuations m,k (z) transfer function = T (k, z) R k initial conditions A key observable is the power spectrum P (k) m,k 2 = T (k) 2 R k 2. For scale-invariant initial conditions, k 3 R k 2 = const., we predict: large scales small scales Ex: Explain the asymptotic scalings of the matter power spectrum P (k) = { k k < keq. k 3 k > k eq 66

Chapter 6. INITIAL CONDITIONS FROM INFLATION Quantum fluctuations during inflation give rise to primordial perturbations. The mechanism is the following: The inflaton is a CLOCK measuring the time to the END OF INFLATION By the uncertainty principle, precise timing is not possible. Locally, inflation ends at different times...... which induces curvature perturbations after inflation. dark matter (Chapter 5) CMB (Adv.Cosmo.) 67

The Big Picture is: comoving scales classical stochastic field subhorizon superhorizon quantum fluctuations horizon exit reheating horizon re-entry CMB today time compute evolution from now on 6.1. CLASSICAL EVOLUTION Before we quantise the fluctuations, we look at their classical dynamics. It will be useful to derive this from the inflaton action S = dτd 3 x [ ] 1 g 2 gµν µ φ ν φ V (φ), g det(g µν ). For the homogeneous field φ(τ), this implies φ + 2H φ + a 2dV d φ = 0. To determine the evolution of fluctuations, we substitute φ(τ, x) = φ(τ) + f(τ, x) a(τ) and g µν = ḡ µν + δg µν. In spatially flat gauge, the metric fluctuations are small in the slow-roll limit. We will ignore them for now and use g µν = ḡ µν. 68

At quadratic order in f, the action is S (2) = 1 dτd 3 x [ (f ) 2 ( f) 2 2Hff + ( ) H 2 a 2 V ],φφ f 2 2 = 1 [ ( ) ] a dτd 3 x (f ) 2 ( f) 2 + 2 a a2 V,φφ f 2. During slow-roll inflation, we have V,φφ H 2 3M 2 pl V,φφ V = 3η v 1. Since a = a 2 H, with H const., we also have Hence, we can drop the V,φφ term S (2) 1 2 Euler-Lagrange implies or a a 2a H = 2a 2 H 2 a 2 V,φφ. f 2 f a a f = 0 f k + ] dτd 3 x [(f ) 2 ( f) 2 + a a f 2 ) (k 2 a f k = 0, where a In the subhorizon limit, k H, we get f k + k 2 f k 0. time-dependent mass Mukhanov-Sasaki equation a a 2H2 2 τ 2.. de Sitter Inflaton fluctuations are a collection of harmonic oscillators! 69

6.2. QUANTUM OSCILLATORS Consider a one-dimensional harmonic oscillator in quantum mechanics: coordinate q mass m 1 potential V = 1 2 ω2 q 2 q + ω 2 q = 0. The action is S[q] = 1 dt [ q 2 ω 2 q 2] and the conjugate momentum is p L 2 q = q. Canonical Quantisation classical fields q, p quantum operators ˆq, ˆp canonical commutation relation [ˆq, ˆp] = i (I) initial conditions Mode expansion: ˆq(t) = q(t) â + q (t) â (II) mode function satisfies q + ω 2 q = 0 Substituting (II) into (I), we get [ˆq, ˆp] = (q q qq ) [â, â ] = i. }{{} iw [q, q ] W.l.o.g., we can set W [q, q ] 1 (III), so that [â, â ] = 1 (IV). We define the vacuum state via â 0 = 0 (V). But, we haven t fixed q(t) completely, so â and 0 aren t fixed either. 70

Choice of Vacuum The preferred vacuum is the ground state of the Hamiltonian. Consider Ĥ = 1 2 ˆp2 + 1 2 ω2ˆq 2 and act on 0 : = 1 2 [ ( q 2 + ω 2 q 2 )ââ + ( q 2 + ω 2 q 2 ) â â + ( q 2 + ω 2 q 2 )(ââ + â â) ] Ĥ 0 = 1 2 ( q2 + ω 2 q 2 ) }{{} 0 â â 0 + 1 2 ( q 2 + ω 2 q 2 ) 0. q = ±iωq W [q, q ] = 2ω q 2 Since W > 0, we have q = iωq q(t) = 1 2ω e iωt. Zero-Point Fluctuations positive frequency solution This implies Ĥ 0 Ĥ 0 = 1 2 ω, ˆq 0 ˆq 0 = 0 qâ + q â 0 = 0, ˆq 2 0 ˆq ˆq 0 = 0 (q â + qâ)(qâ + q â ) 0 = q(t) 2 0 ââ 0 = q(t) 2 0 [â, â ] 0 = q(t) 2. ˆq 2 = q(t) 2 = 2ω (VI) 71

6.3. QUANTUM FLUCTUATIONS IN DE SITTER Let s return to S[f] = 1 dτd 3 x [ (f ) 2 + ]. 2 The conjugate momentum is π L f = f. Canonical Quantisation f, π ˆf, ˆπ [ ˆf(τ, x), ˆπ(τ, x) ] = i δ(x x ) (I ) In Fourier space, we find [ ˆf d k (τ), ˆπ k (τ)] 3 x = (2π) 3/2 = i d 3 x (2π) 3 e i(k+k ) x locality d 3 x (2π) [ ˆf(τ, x), ˆπ(τ, x )] 3/2 }{{} iδ(x x ) e ik x e ik x = i δ(k + k ) (I ) initial conditions Mode expansion: ˆfk (τ) = f k (τ) â k + fk (τ) â k (II ) mode function satisfies f k + Substituting (II ) into (I ), we get (k 2 a a ) f k = 0 W [f k, f k ] [â k, â k ] = δ(k + k ). Setting W [f k, f k ] 1 (III ), we find [â k, â k ] = δ(k + k ) (IV ). We define the vacuum state via â k 0 = 0 (V ). As before, we need to fix f k (τ) to uniquely determine â k and 0. 72

Choice of Vacuum At early times, the modes are deep inside the horizon and satisfy f k + k 2 f k 0 (SHO in flat space!) We therefore match to the preferred vacuum of the harmonic oscillator lim f k(τ) = 1 e ikτ τ 2k ( ) Bunch-Davies initial condition Consider the MS equation in ds: f k + ( k 2 2 τ 2 ) f k = 0. which has the exact solution ( f k (τ) = α e ikτ 1 i 2k kτ ) ( + β eikτ 1 + i ) 2k kτ. Eq. ( ) selects the positive frequency solution: α = 1, β = 0: ( f k (τ) = e ikτ 1 i ) 2k kτ Bunch-Davies mode function ( ) 73

Zero-Point Fluctuations Finally, we can predict the quantum statistics of the operator d ˆf(τ, 3 k [ ] x) = f (2π) 3/2 k (τ)â k + fk (τ)a k e ik x. We can compute the variance: ˆf 2 0 ˆf (τ, 0) ˆf(τ, 0) 0 d 3 k d 3 k = (2π) 3/2 (2π) 0 ( f 3/2 k â k + f )( ) kâ k fk â k + fk â k 0 d 3 k d 3 k = (2π) 3/2 (2π) f 3/2 k (τ)f k (τ) 0 [â k, â k ] 0 = = d 3 k (2π) 3 f k(τ) 2 d ln k k3 2π 2 f k(τ) 2 d ln k 2 f(k, τ), where we have defined the (dimensionless) power spectrum as 2 f(k, τ) k3 2π 2 f k(τ) 2. (VI ) Using the BD mode function ( ), we get 2 δφ(k, τ) = a 2 2 f(k, τ) = The power spectrum at horizon crossing is ( ) ( 2 ( ) ) 2 H k 1 + 2π ah 2 δφ(k) ( ) 2 H. 2π k=ah k ah ( ) 2 H. 2π 74

6.4. PRIMORDIAL PERTURBATIONS FROM INFLATION 6.4.1. CURVATURE PERTURBATIONS We want to switch from δφ to R at horizon crossing. Recall that R = C 1 3 2 E + H(B + v) Hence, we have spatially flat (C=E=0) inflation (B+v = δφ/ φ ) δφ. H φ 2 R = 1 2ε 2 δφ M 2 pl, where ε = 1 2 φ 2 M 2 pl H2. Evaluating this at horizon crossing, we find 2 R(k) = 1 1 8π 2 ε H 2 Mpl 2 k=ah A s ( ) ns 1 k k amplitude: A s = 2.2 10 9 spectral index: n s = 0.96 ± 0.01 Ex: Show that 2 R = 1 12π 2 V 3 M 6 pl (V ) 2. Since H(t) and ε(t) depend on time, we expect a small scale-dependence: n s 1 d ln 2 R d ln k = d ln 2 R dn dn d ln k. Using d ln 2 R dn d ln k dn ln H = 2d dn = d ln(ah) dn d ln ε dn = 2ε η, = 1 + d ln H dn = 1 ε, 75

we get n s 1 2ε η (to first order in SR). Ex: Show that n s 1 3M 2 pl ( V V ) 2 + 2Mpl 2 V V. Summary: Initial Conditions Late-time Observables: Matter CMB 6000 4000 2000 2 500 2000 76

6.4.2. GRAVITATIONAL WAVES ] Recall ds 2 = a 2 (τ) [dτ 2 (δ ij + 2Êij)dx i dx j. Substituting this into the Einstein-Hilbert action, gives S (2) = M 2 pl 8 dτ d 3 x a 2 [ (Ê ij) 2 ( Êij) 2 ]. It is convenient to define M pl 2 aêij 1 f + f 0 f f + 0, 2 0 0 0 so that S (2) = 1 2 I=+, dτd 3 x [ ] (f I) 2 ( f I ) 2 + a a f I 2 = two copies of the scalar action!, The power spectrum of tensors therefore is ( ) 2 2 2 t 2 2 = 2 2 Ê f. am pl Substituting our previous result, gives 2 t (k) = 2 H 2 π 2 Mpl 2 k=ah A t ( ) nt k. k Ex: Show that ( ) V 2. n t 2ε M 2 pl V 77

An important quantity is the tensor-to-scalar ratio and a crucial plot is: r A t A s 0.25 0.20 0.15 concave convex 0.10 small-field large-field 0.05 0.00 natural 0.94 0.96 0.98 1.00 78