Inflation Cosmology. Ch 06 - Initial Conditions. Korea University Eunil Won. Korea U/Dept. Physics, Prof. Eunil Won (All rights are reserved)

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1 Infltion Cosmology Ch 6 - Initil Conditions Kore University Eunil Won 1

2 The Einstein-Boltzmnn Equtions t Erly Times For nine first-order differentil equtions for the nine perturbtion vribles, we need initil conditions to solve. - Let us consider the Boltzmnn equtions t very erly times: for ny k-modes of interest, kη 1 Θ + ikµθ = Φ ikµψ τ Θ Θ + µv b 1 P (µ)π (see ch4) LHS: the first term is order Θ/η the second term is order kθ Θ/η kθ All terms in the Boltzmnn equtions multiplied by k cn be neglected t erly times. Physiclly, this mens tht, t erly times, ll perturbtions of interest hve wvelength (~k -1 ) much lrger thn the distnce over which cusl physics opertes. - Therefore, the perturbtions to the photon nd neutrino temperture evolve ccording to (Θ: monopole) Θ + Φ = N + Φ = - The sme principles cn be pplied to the mtter distributions. The over-density equtions reduce to δ = 3 Φ δ b = 3 Φ The velocities re comprble to the first moments of the rdition distributions, so they re smller thn the over-densities by fctor of order kη. τ v b + ȧ v b = ikψ + τ R [v b +3iΘ 1 ] v b = 3iΘ 1 The lrgeness of in ensures tht At erly times, higher moments re ll negligibly smll.

3 The Einstein-Boltzmnn Equtions t Erly Times Now for the Einstein equtions t erly times. From k Φ +3ȧ cn be neglected (k ) so, 3ȧ Φ Ψȧ Φ Ψȧ But since rdition domintes, =4πG [ρ dm δ + ρ b δ b +4ρ γ Θ +4ρ ν N ] two mtter terms cn be neglected t erly times (rdition domintes) =16πG [ρ γ Θ + ρ ν N ] η nd ȧ/ =1/η Φ η Ψ η = 16πGρ ρ γ 3 ρ Θ + ρ ν ρ N = ρ γ η ρ Θ + ρ ν ρ N ȧ = 8πG 8πG ρ 3 3 ρ = 1 η (for erly times) ρ ν To simplify further, we define f ν ρ γ + ρ ν ρ =1 ρ ν ρ =1 f ν Φη Ψ = [1 f ν ]Θ + f ν N ρ γ Then, By differentiting both sides, we get Φη + Φ Ψ = [1 f ν ] Θ + f ν N = [1 f ν ]( Φ )+f ν ( Φ ) Φη + Φ Ψ = Φ 3

4 The Einstein-Boltzmnn Equtions t Erly Times The second Einstein eqution ws: - eliminting Ψ, we get (difficult to show) Φη +4 Φ = k (Φ + Ψ) = 3πG [ρ γ Θ + ρ ν N ] (see ch5). - now setting Φ = η p leds to p(p 1) + 4p = so p= or -3. p = -3: the decying mode. If it is excited very erly on, it will quickly die out nd hve no impct on the universe. p = : it is the mode tht we re interested in. - so we consider p = only from now on. In tht cse, Φη Ψ = [1 f ν ]Θ + f ν N becomes Φ = [1 f ν ]Θ + f ν N Both Θ nd N re lso constnt in time. In most models of structure formtion, they re equl since whtever cuses perturbtions tends not to distinguish between photons nd neutrinos. Therefore, we will set Θ (k, η i )=N (k, η i ) Φ(k, η i )=Θ (k, η i ) or : ηi indictes n erly time (so the eqution is initil conditions). The initil conditions for mtter, both δ nd δ b depend upon the nture of the primordil perturbtions. From Θ + Φ = δ = 3 Φ Θ 1 3 δ = for the drk mtter over-density. (lso sme for the bryon over-density) Primordil perturbtions re often divided into the constnt bove: constnt = : dibtic perturbtion constnt : iso-curvture perturbtion 4 δ =3Θ + constnt Adibtic perturbtions hve constnt mtter-to-rdition rtio since nd we focus on dibtic initil conditions (iso-curvture perturbtions hve not been very successful). n dm n γ = n() dm 1+δ n () 1+3Θ γ

5 The Einstein-Boltzmnn Equtions t Erly Times For the most prt, velocities nd dipole moments re negligibly smll in the very erly universe. However, we will encounter situtions where we need to know the initil conditions for these s well. They re (exercise 3) Θ 1 = N 1 = iv b 3 = iv 3 = kφ 6H 5

6 The Horizon Let s revisit the condition kη<<1: The wve number k 1 λ of the mode in question. kη is the rtio of the (comoving horizon)/(the comoving wvelength) of the perturbtion. If kη << 1, the mode hs wvelength so lrge tht no cusl physics could possibly hve ffected it. T (GeV) Comoving Distnce (Mpc) Qudrupole Scle Tody Scle of Non-linerity Tody Horizon Photons Decoupled The horizon grows s the scle fctor increses. Comoving wvelengths remin constnt. The CMB comes from the lst scttering surfce 1-3. At tht time, the lrgest scles (e.g., the one lbeled qudrupole ) were still outside the horizon. The horizon problem sks how regions seprted by distnces lrger thn the horizon t the lst scttering surfce cn hve the sme temperture

7 Infltion An erly epoch of rpid expnsion solves the horizon problem. Negtive pressure is required to produce this rpid expnsion. Negtive pressure is esy to ccommoe in sclr field theory. A solution to the horizon problem. Let s write the comoving horizon s η = d 1 H( ) The comoving horizon then is the logrithmic integrl of the comoving Hubble rdius, 1/H. - comoving horizon : η - comoving Hubble rdius: 1/H If prticles re seprted by distnces greter thn η, they never could hve communicted with one nother. If prticles re seprted by distnces greter thn 1/H, they cnnot tlk to ech other now. A solution to the horizon problem: perhps erly on, the universe ws not dominted by either mtter or rdition. How must the sclr fctor evolve in order to solve the horizon problem? If the comoving Hubble rdius is to decrese, then H must increse: d d/ = d > So, to solve the horizon problem, the universe must go through period which it is ccelerting, expnding even more rpidly. This is the origin of the term infltion. 7

8 Negtive Pressure Infltion Using the time-time & spce-spce components of the zero-order Einstein equtions (ch): d/ = 8πG 3 ρ d / + 1 d/ = 4πGP d / = 4πG (ρ +3P) 3 Accelertion mens d / >, so ρ +3P <, P < ρ 3 Since the energy density ρ is lwys positive, the pressure must be negtive. Implementtion with sclr field We wnt to know if sclr field - which we will cll - cn hve negtive. So our first tsk is write down the energy momentum tensor for φ(x, t) T α β αν φ = g x ν φ x β gα β φ(x, t) ρ +3P 1 φ gµν x µ φ x ν + V (φ) where V(φ) is the potentil for the field. (exmple, for free field w/ mss m hs V(φ)=m / φ ). We will ssume tht φ is mostly homogeneous, consisting of zero-order prt φ () (t) nd first-order perturbtion δφ(x,t). For the homogeneous prt of the field, dφ () =, φ () x i = 8

9 Infltion The energy-momentum tensor then reduces to T ()α β α dφ() = g = g α g β φ x β gα β dφ () + g α β Since the time-time component of T = - ρ, ρ = 1 dφ () 1 g dφ () 1 + V (φ () ) dφ () V (φ () ) V (φ () ) or ρ = 1 = g α σ g dφ () σ dφ() kinetic energy density gβ dφ () + V (φ () ) + gβ α 1 dφ () V (φ () ) potentil energy density The pressure for the homogeneous field is P = T ()i g i i i = g µi g µi = g ji g ik = 1 since =1 P = 1 dφ () V (φ () ) A field configurtion with negtive pressure is therefore one with more potentil energy thn kinetic. An exmple is: flse vcuum 9

10 Infltion Since φ () = constnt, its energy density remins constnt with time. The densities of both mtter nd rdition fll off very rpidly s the universe expnds. Thus the universe will quickly dominted by the vcuum energy. Einstein s eqution for the evolution of is d/ = 8πGρ 3 = constnt H ρ 1/ nd tht field trpped in flse vcuum produces exponentil expnsion, with constnt. The primordil comoving horizon, tht generted before the end of infltion, is obtined by integrting the inverse of (t) = e e H(t t e) η prim = te t b 1 (t) = t<t e te t b over time, e H(t te) e = e H(t te) e H t e = 1 H e e t b e H(t e t b ) 1 where tb is the beginning of the infltion. So if the field is trpped for t lest 6 e-foldings problem is solved. (H(t e t b ) > 6), the horizon To void the problem of the universe never reching its true vcuum stte, models of infltion mde use of sclr field slowly rolling towrd its true ground stte. To determine the evolution of φ (), we strt from d/ = 8πG 3 ρ If the dominnt component in the universe is φ, the energy density becomes ρ = 1 dφ () + V (φ () ) Now, differentiting the bove eqution, d/ d / d/ = 8πG 3 dφ () d φ () + V dφ() 1

11 From d / = 4πG (ρ +3P) 3 Infltion nd d/ 11 = 8πG 3 ρ LHS becomes d/ 4πG 8πG (ρ +3P) 3 3 ρ = From P = 1 dφ () ρ = 1 dφ () + V (φ () ) nd 8πGH(ρ + P) = 8πGH dφ () dφ () 8πGH = 8πG dφ () d φ () 3 d φ () Using conforml time, +3H dφ() + V = dη = 1, dφ () = dη dφ () = 1 dφ () dη dη d φ () = d 1 dφ () = d 1 dφ () dη dη + 1 = 1 d φ () dη H dφ () dη d/ 8πG ρ3 P 3 ρ = 8πGH(ρ + P) V (φ () ) dη + V dφ() d φ () dη

12 Infltion Therefore, d φ () +3H dφ() + V = becomes 1 d φ () dη H dφ () dη + 3H dφ () dη + V = φ () +H φ () + V = Most models of infltion re slow roll models, in which the zero-order field nd hence the Hubble rte, vry slowly. During infltion, η e d H = 1 H e d = 1 H ( e ) e :ttheendofinfltion To quntify slow roll, we define two vribles which vnish in the limit tht φ remins constnt. First, let s define d 1 H = Ḣ H Since H is lwys decresing, ε is lwys positive. During infltion, ε << 1 During the rdition er ε= Note tht for the rdition er, t 1/,H = 1/t 1/ t 1/ = 1 t 1, = d 1 H = d (t) = One definition of n infltionry epoch is one in which ε<1. A complementry vrible which lso quntities how slowly the field is rolling is: δ 1 d φ () / H dφ () / = 1 H φ () H φ () φ () = In most models, δ is smll (δ hs nothing to do with the overdensities). 1 H φ 3H φ () + V () 1

13 Quntizing Hrmonic Oscilltor Grvity Wve Production In order to compute the quntum fluctutions in the metric, we need to quntize the field. A simple hrmonic oscilltor with unit mss nd frequency ω is governed by the eqution d x + ω x = Upon quntiztion, x becomes quntum opertor : ˆx = v(ω,t)â + v (ω,t)â where â : quntum opertor cting on the stte of the system v(ω,t) : solution to the simple hrmonic eqution, v(ω,t) e iωt â nnihiltes the vcuum stte >, in which there re no prticles. It lso stisfies the commuttion reltion [â, â ] ââ â â =1 nd is equivlent to [ˆx, ˆp] =i s long s v is normlized vi v(ω,t)= 1 e iωt ω These fcts enble us to compute the quntum fluctutions of the opertor ˆx in the ground stte >: ˆx = ˆx ˆx = (v â + vâ)(vâ + v â ) = v(ω,t) ââ = v(ω,t) [â, â ]+â â = v(ω,t) â = â = ˆx =1/ω 13

14 Tensor perturbtions Grvity Wve Production Recll tht tensor perturbtions to the metric re described by two functions h+ nd hx nd ech of which obeys ḧ +ȧ ḣ + k h = To quntize h, we define h h 16πG Derivtive of h, with respect to conforml time cn be written s nd ḧ 16πG = h ȧ ä h h + (ȧ) h 3 ḣ 16πG = d dη h = h ȧ h Hving ll into the differentil eqution bove, we get h ȧ h ä = 1 h + h + (ȧ) h 3 +ȧ k ä h = h ȧ h + k h It hs no dmping term so we write down n expression for the quntum opertor ˆ h = v(k, η)â k + v (k, η)â k where the coefficients of the cretion nd nnihiltion opertors stisfy v + 14 k ä v =

15 Grvity Wve Production Let s see how the eventul solution determines the power spectrum of the fluctutions of the tensor perturbtions. We write the vrince of perturbtions in the field s h ˆ h ( k, η)ˆ h( k, η) = v( k, η) (π) 3 δ 3 ( k k ) : there is one difference between h nd quntum simple hrmonic oscilltor. A quntum field is defined in ll spce, so it cn be considered s collection of n infinite collection of oscilltors, ech t different sptil position. Reclling tht h h 16πG, we see tht ĥ ( k, η)ĥ( k, η) = 16πG v( k, η) (π) 3 δ 3 ( k k ) (π) 3 P h (k)δ 3 ( k k ) where the second line defines the power spectrum of the primordil perturbtions to the metric. v(k, η) P h (k) =16πG Now, the problem of determining the spectrum of tensor perturbtions produced during the infltion becomes solving nd order differentil eqution. To solve this, we first need to evlute during the infltion. ȧ = H = /η ä = 1 d = dη η η So, the eqution for v becomes: v + k η ä/ ( η = 1/H) v = 15

16 Grvity Wve Production The initil conditions to solve this eqution come from considering v t very erly times before infltion hs done most of its work. At tht time, -η is lrge, ~ηprim so the k term domintes, nd the eqution reduces to tht of the simple hrmonic oscilltor. In tht cse, v = e ikη / k This knowledge enbles us to choose the proper solution the originl eqution: v + v = e ikη 1 i k kη After infltion hs worked for mny e-folds k η becomes very smll. i lim v(k, η) =e ikη kη k kη k η v = The primordil power spectrum for tensor modes, which scles s v /, is therefore constnt in time, fter infltion hs stretched the mode to be lrger thn the horizon. P h (k) = 16πG 1 k 3 η = 8πGH 1 k 3 where η = H We ssume tht H is constnt: H is to be evluted t the time when the mode of interest leves the horizon. 16

17 Sclr Perturbtions Sclr perturbtions The gol of this prt is to find the perturbtion spectrum of Ψ (or Φ; we ssume throughout tht they re equl in mgnitude) emerging from infltion. Sclr field perturbtions round smooth bckground Let s decompose the sclr field into zero-order homogeneous prt nd perturbtion φ(x, t )=φ () (t)+δφ(x, t ) nd find eqution governing δφ in the presence of smoothly expnding universe: g = -1, gij = δij Consider the conservtion of energy-momentum tensor: T µ ν;µ = Tµ ν x µ +Γµ αµt α ν Γ α νµt µ α = ν = component: using Γ ij = δ ij H Γ = Γ i j = Γ i j = δ ij H Γ i = Γ i = T µ x µ +Γµ αµt α Γ α µt µ α = T T + Ti x i +Γ αt α +Γ i αit α Γ α T α Γ α it i α = + Ti x i ++ Γi i T +Γ i jit j 3H Γ it i Γ j i Tj i = δ ij H 17

18 Sclr Perturbtions So, expressing the perturbtion in terms of δt μ ν, δt Let s compute δt i From T α β = g + ik i δt i +3HT HT i i = αν φ x ν φ x β gα β T i = g iν φ,ν φ, g i = = g iν φ,ν φ, 1 φ gµν x µ 1 φ gµν x µ φ x ν + V (φ) φ x ν + V (φ) Since g iν = δ iν nd φ (),i = we hve φ,i δφ,i, φ (), = φ () / T i = ik i δφ φ () / = ik i 3 φ () δφ The time-time component of the energy-momentum tensor: T = g (φ, ) g 1 φ φ gµν x µ x ν + V (φ) = g (φ, ) 1 φ gµν x µ φ x ν V (φ) Setting φ = φ () + δφ leds to (g = -1) T = 1 (φ(), + δφ,) 1 δφ,iδφ,i V (φ () + δφ) Neglecting the higher order term, δt = φ (), δφ, V δφ = φ () δ φ V δφ 18

19 Sclr Perturbtions Similrly, one cn show tht the spce-spce component is δt i j = δ ij φ() δ φ Now the conservtion eqution Using 1 1 η +3H V η = η φ () δ φ Multiplying 3, we get V δφ δt φ () δ φ V = V φ φ φ η φ () δ φ + ik i δt i +3HT HTi i = V δφ + φ () δ φ ȧ 3 = V φ() k 3 φ () δφ 3H φ () δ φ +3HV δφ = φ () δ φ + δ φ becomes + ik i ik i 3 φ () δ φ 3H bove eqution becomes φ() δ φ V φ() δφ V δ φ 3H φ () δ φ V δφ 3HV δφ φ () 4H φ () V + δφ V φ() k φ() = V is typiclly smll, proportionl to the slow-roll vrible ε nd δ, so it cn be neglected. φ () +H φ () + V = Using nd dividing by we get δ φ +Hδ φ + k δφ = φ () = P δφ = H k 3 = dη η = 1 η This eqution for perturbtions to δφ is identicl to the tensor perturbtions to the metric. So is immeditely obtined. 19

20 Super-Horizon Perturbtions Until now, we neglected the metic perturbtions. When the wvelengths of the perturbtions is of order the horizon or smller, this pproximtion is vlid. Let s begin by rewriting the eqution for conservtion of energy. This time in the presence of the metric perturbtion: δt Sclr Perturbtions + ik i δt i +3HδT HδT i i = 3(P + ρ) Ψ To check the vlidity of neglecting Ψ in the previous cse, Ψ δt P + ρ δg = 6HΦ, +6ΨH k Φ =8πGδT k Ψ +3H( Ψ + HΨ) =4πG δt needs to be hold. To see this, from (ch5) (Φ = - Ψ is used) LHS is of order k Ψ ~ H Ψ. Therefore, Ψ GδT H δt ρ P + ρ ρ = P + ρ ρ is very smll since δt P + ρ P ρ during the infltion. So, t lest in slow-roll models of infltion, we re justified in neglecting perturbtions when computing the spectrum of δφ. The bove rgument holds only for modes tht hve not yet pssed outside the horizon. Super-horizon modes, on the other hnd, require more creful tretment. At some point before infltion ends, perturbtions to Ψ must be grow in importnce reltive to those in the energy-momentum tensor.

21 Let s define ζ ik iδt i H k (ρ + P) Ψ to del with the coupling between the metric perturbtion nd energy density perturbtion. - Ψ is responsible for sub-horizon modes nd those which just left the horizon. dφ () + V (φ () ) P = 1 φ() = - From ρ = 1 nd P + ρ = dφ () dφ () V (φ () ) - We know δt i = ik i nd note tht φ () 3 δφ δti = g ν φ,ν φ, gi So, Sclr Perturbtions ζ = ik i( ik i φ() δφ)h/ k ( φ () /) = Hδφ φ () 1 gµν φ,µ φ,ν + V = ik i φ () δφ round the time of horizon crossing. After infltion ends, rdition. Since P = 1, 3 ρ r ζ = 4kρ rθ 1 H k ( φ () /) Ψ = 3HΘ 1 k = Φ Ψ Ψ ik i δt i =4kρ r Θ 1 Θ 1 = kφ 6H, proportionl to the dipole of the Φ = Ψ = 3 Ψ (post infltion) 1

22 Sclr Perturbtions The vrible ζ is importnt becuse it is conserved when the perturbtion moves outside the horizon. Now, Ψ = δφ H 3 post infltion φ () Equivlently, P Ψ = 4 9 H φ () horizon crossing P δφ = 4 9 H=k H H φ () k 3 = H 9k 3 φ () H=k Another wy to express the power spectrum of sclr perturbtions is to eliminte φ () in fvor of the slow-roll prmeter ε. It cn be shown s P Ψ = P Φ = 8πG H 9k 3 H=k Since P h (k) = 8πGH k 3, the rtio of sclr to tensor is or order 1/ε. One cn lso show tht P Φ (k) = 18π G 9k 3 H V V H=k It remins to prove tht ζ is conserved on super-horizon scles: from conservtion eqution δt + ik i δt i +3HδT HδTi i = 3(P + ρ) Ψ k i δt i k - on lrge scles nd so cn be neglected. This leves T +3HδT HδT i i = 3(P + ρ) Ψ

23 ik i δti H k = δt 3 On lrge scles, (exercise 13) δt ρ + P +3HδT HδTi i = 3(P + ρ) ζ on lrge scles ζ = Ψ 1 3 δt δt 3H + 1 ρ + P +(ρ + P) dρ + dp HδTi i = 3(P + ρ) ζ dρ = 3H(ρ + P) (ch ) nd ζ = δt ρ + P 1 3(ρ + P) H(ρ + P)δTi i δt dp Now, two terms in the brckets cncel for the clss of perturbtions we re considering. Since H(ρ + P) = 1 3 dρ H(ρ + P)δT i i δt dp δt i i 3 + dp dρ δt = δp dp dρ δρ = Thus, ζ is indeed conserved on lrge scles. Sclr Perturbtions 3

24 ik i δti H k = δt 3 On lrge scles, (exercise 13) δt ρ + P +3HδT HδTi i = 3(P + ρ) ζ on lrge scles ζ = Ψ 1 3 δt δt 3H + 1 ρ + P +(ρ + P) dρ + dp HδTi i = 3(P + ρ) ζ dρ = 3H(ρ + P) (ch ) nd ζ = δt ρ + P 1 3(ρ + P) H(ρ + P)δTi i δt dp Now, two terms in the brckets cncel for the clss of perturbtions we re considering. Since H(ρ + P) = 1 3 dρ H(ρ + P)δT i i δt dp δt i i 3 + dp dρ δt = δp dp dρ δρ = Thus, ζ is indeed conserved on lrge scles. Sclr Perturbtions 4

25 Sptilly Flt Slicing Sclr Perturbtions Consider guge with sptilly flt slicing, with the sptil pert of the metric gij = δij. In this guge, the line element is ds = 1(1 + A) B,i dx i + δ ij dx i dx j δ φ +Hδ φ + k δφ = In this cse, those in the grvittionl metric. We hve v = ikb ik φ () δφ (ρ + P) (sptilly flt slicing) when δt i = ik i 3 φ () δφ is given exctly (exercise 16): the perturbtion to the sclr field do not couple to is used. So, ζ Φ H ih is guge invrint nd in sptilly flt slicing is equl to (sptilly flt slicing) k v ζ = H δφ φ () H P ζ = P δφ φ () 5