Inflation Cosmology. Ch 02 - The smooth, expanding universe. Korea University Eunil Won

Transcription

1 Infltion Cosmology Ch 02 - The smooth, expnding universe Kore University Eunil Won

2 From now on, we use = c = k B = The metric Generl Reltivity - in 2-dimensionl plne, the invrint distnce squred is (dx) 2 +(dy) 2 - if we use polr coordintes, the invrint distnce squred becomes (dr) 2 + r 2 (dθ) 2 =(dr) 2 +(dθ) 2 So, mthemticlly, d 2 = i,j=,2 In cse of 2D polr coordintes g ij dx i dx j where gij is clled metric x = r, x 2 = θ g =, g 22 = r 2 This is when coordintes re curved Another wy of thinking bout the metric: let s think of contours of mountin This is when spce itself is curved () (b) Even though () nd (b) pss sme number of contours, their distnce is different but their verticl distnce is sme The gret dvntge of the metric is tht it incorportes grvity (not s n externl force but through the geometry) 2

3 Generl Reltivity Specil reltivity is described by Minkowski spce-time with metric gμν=ημν: η µν = Now, wht is the metric which describes the expnding universe? (0,0) t (,0) (0,0) t2 (,0) t3 (0,0) (,0) - from our rguments with the scle fctor nd comoving distnce (in chpter 0) physicl distnce: (t2) physicl distnce: (t3) comoving distnce = comoving distnce = time comoving distnce = - it seems resonble to ssume tht the metric of n expnding, flt universe is g µν = (t) (t) (t) nd is clled the Friedmnn-Robertson-Wlker (FRW) metric. 3

4 Generl Reltivity The geodesic geodesic: the pth followed by prticle in the bsence of ny force d 2 x dt 2 =0 d 2 x i - in Eucliden 2D plne, the geodesic equtions become: dt 2 =0 - but in polr coordintes x i =(r, θ) the eq. for free prticle look significntly different Strting with x i x j Now using dx i dt = xi dx j x j dt nd : trnsformtion mtrix becomes x = x cos x 2, x 2 = x sin x 2 x i cos x x j = 2 x sin x 2 sin x 2 x cos x 2 d x i dt x j = dx i x j = x i dx k dt x j x k = 2 x i dx k dt x j x k dt d 2 x i dt 2 =0? The 2D geodesic in polr coordintes become d 2 x i dt 2 = d x i dx j dt x j = xi d 2 x j dt x j dt x i dx k x j x k dt dx j dt =0 If I multiply the inverse of the trnsformtion mtrix to the bove: x x xi d 2 x j x x j dt 2 + x i 4 i 2 x i x j x k dx k dt dx j dt =0

5 Generl Reltivity The geodesic - we get the following form: d 2 x x 2 x i dx k dx j dt 2 + x x j x k =0 dt dt i So, it is convenient to define the Christoffel symbol Γ l jk s Γ x jk x Note tht Γ l jk =0 for Crtesin coordintes system so we recover d 2 x i dt 2 =0 i 2 x i x j x k For the generl reltivity, we need two smll chnges to mke from here: ) Allow indices to rnge 0,,2,3 (spce-time) 2) t λ where λ is evolution prmeter which monotoniclly increses long the prticle s pth. x µ (λ i ) x µ (λ f ) λ Then the geodesic eq. becomes in generl reltivity s d 2 x µ dλ 2 = Γµ αβ dx α dλ dx β dλ Yes, for now let us ssume this: rigorous proof will be dded lter (not importnt for future discussions) 5

6 Generl Reltivity The geodesic One cn get the Christoffel symbol directly from the metric (to prove, it requires bit of elbortion so we skip here for now) Γ µ αβ = gµν g αν 2 x β + g βν x α g αβ Note: g μν is the inverse of gμν so g μν in the flt FRW metric is x ν identicl to gμν except tht its sptil elements re / 2 insted of 2. - Let s compute Γ μ αβ for the flt FRW metric Γ 0 αβ : metric is digonl, so g 00 =- nd g 0i =0 Γ 0 αβ = g α0 2 x β + g β0 x α g αβ x 0 = = 0 0 ( g 00 = - ) therefore, So, Γ 0 00 = 2 Γ 0 αβ = 2 g 00 t g αβ x 0 =0 nd re nonzero only if α nd β re both sptil indices. Γ 0 0i =Γ 0 i0 =0 Γ 0 ij = 2 g ij t = 2 2ȧδ ij =ȧδ ij 6

7 The geodesic Generl Reltivity Γ i αβ : is non-zero only when one of its lower indices is zero nd one is sptil Γ i αβ = giν g αν 2 x β + g βν x α g αβ x ν For α=j, β=k, we get Γ i jk = giν g jν 2 x k Γ i 00 = giν 2 g 0ν x 0 + g kν x j g jk x ν + g 0ν x 0 g 00 x ν : re ll zero becuse the metric hs no spcil dependence (first two terms) nd ν 0 due to g iν (3rd term) =0 (obvious) The only non-zero terms re then Γ i 0j = Γ i j0 = giν g 0ν 2 x j + g jν x 0 If ν=k, Γ i 0j = gik 2 g 0k x j + g jk x 0 g 0j x ν g 0j x k nd re zero if ν=0 (obvious) So, Γ i 0j = gik 2 = g jk x 0 = 0 0 ( g 00 = - ) = 2 δ ȧ ijȧ = δ ij 7

8 Generl Reltivity Let s see how prticle s energy chnges s the universe expnds With the four-dimensionl energy-momentum vector: P α =(E, P ) let us define the prmeter λ implicitly s P α = dxα dλ The 0th component of the geodesic eq. becomes then LHS: d 2 x 0 dλ 2 d 2 x 0 dλ 2 = d dλ = Γ0 αβ E de dt = Γ0 ijp i P j E dx0 dt = dx α dλ = E de dt dx β dλ RHS: so tht Γ 0 ijp i P j d dλ = dx0 dλ since d dx 0 = E d dt Γ 0 0i =Γ 0 i0 =0, P i = dxi dλ For mssless prticle E de dt = Γ0 ijp i P j P µ P µ = g µν P µ P ν = E 2 + δ ij 2 P i P j =0 nd from Γ 0 ij = δ ij ȧ becomes E de dt = δ E 2 ijȧ 2 δ ij de dt + ȧ E =0 nd the solution to tht is E The energy of mssless prticle should decrese s the universe expnds since it is inversely proportionl to. (Wvelength is stretched when the universe expnds) 8

9 Generl Reltivity Einstein Equtions Einstein equtions relte the geometry to the energy-momentum: G µν R µν 2 g µνr =8πGT µν geometry energy-momentum G µν : Einstein tensor R µν : Ricci tensor R : Ricci sclr, defined s R = g µν R µν G : Newton s constnt : Energy-momentum tensor (will be defined soon) T µν Ricci tensor is defined s R µν =Γ α µν,α Γ α µα,ν +Γ α βαγ β µν Γ α βνγ β µα where Γ α µν,α Γα µν x α (note tht Γ α μν contins the st derivtive of the metric so Ricci tensor hs both 2nd nd st derivtive of the metric) I know it looks quite depressing (complicted) but we lredy did lot of work for FRW universe. Most of them ctully vnish (see the next slide). 9

10 Generl Reltivity Einstein Equtions It turns out tht there re only two sets of non-vnishing components of the Ricci tensor: one with μ=ν=0 nd μ=ν=i. i) μ=ν=0 R 00 =Γ α 00,α Γ α 0α,0 +Γ α βαγ β 00 Γα β0γ β 0α R 00 = Γ i 0i,0 Γ i j0γ j 0i = = 0 0 ( Γ α 00=0 ) ( Γ 0 i0=γ 0 0i=0 ) Now using Γ i 0j =Γ i ȧ j0 = δ ij we get R 00 = Γ i 0i,0 Γ i j0γ j 0i = ȧ ȧ δ ii δ ij t 2 äȧ ȧ 2 ȧ = = 3ä ii) μ=ν 0 One cn lso show tht R ij = δ ij 2ȧ 2 + ä Now we hve expressions for Rμν nd R for flt FRW universe! (LHS for Einstein equtions) Now the Ricci sclr becomes R g µν R µν = g 00 R 00 + g ij R ij = R R ii =3ä + 2 3[2ȧ2 + ä] =6 ä ȧ 2 + 0

11 Generl Reltivity Einstein Equtions To understnd the evolution of the scle fctor in homogeneous universe, we need to consider only time-time component of the Einstein equtions: R 00 2 g 00R =8πGT 00 where LHS: R 00 2 g 00R = 3ä 2 ( )6 (ρ: energy density of ll species: mtter, rdition nd the T 00 = ρ drk energy) ä ȧ 2 ȧ 2 + =3 RHS: 8πGT 00 =8πGρ ȧ 2 8πG = 3 ρ So, the evolution of the scle fctor (or universe) of flt FRW is determined by Einstein equtions nd is determined by the energy density of the universe. Since we defined ρ cr = 3H2 0 8πG in ch0, dividing bove by H 2 0 gives H 2 (t) H 2 0 = ρ ρ cr is finlly obtined. (note tht ssumed the universe is flt: so there is no term corresponding to the curvture of the universe)

12 Distnces Two possible wys to mesure distnce in n expnding universe ) The comoving distnce which remins fixed s the universe expnds 2) The physicl distnce which grows simply becuse of the expnsion - Neither of two mesures ccurtely describes the process of interest: ex) Light leving distnt qusi-stellr object (QSO) t z=3 strts towrd us when (z)=/4 rrives us tody. (So the light rrives us when the universe is expnded by 4 times: +z=/, so =/(+3) = /4) In this cse, which distnce do we use to relte the luminosity of the QSO to the flux we see? At the time of (t)=/4 us Tody (t)= us - The fundmentl distnce mesure, from which ll others cn be clculted, is the distnce on the comoving grid: in time dt, light trvels comoving distnce dx = dt (c =) The totl comoving distnce of light could hve trveled is η t 0 dt (t ) (definition of conforml time) 2

13 Conforml time η t 0 dt (t ) Distnces - No informtion could hve propgted further thn η since the beginning of the time. (Regions seprted by distnce greter thn η re not cuslly connected.) - We cn think of η s the comoving horizon. - η is clled the conforml time, nd cn be used to discuss the evolution of the universe. Another importnt distnce: distnce between distnt emitter nd us: χ() = t0 t() dt (t ) = d 2 H( ) H = ȧ d, Hdt = d, dt = H - Typiclly, we cn see object out to z 6; t these lte times rdition cn be ignored. If the universe is purely mtter dominted t such times, then H 3/2 (see chpter 0) H = C 3/2 d, χ() = 2 C = (/2) d = 3/2 C C 2(/2) = C 2( /2 ) = 2 H 0 /2 χ Flt,MD () = Note tht 2 H 0 /2 = 2 H 0 H 0 = C 3/2 0 = C ( 0 =) +z : this comoving distnce goes s z/h0 for smll z nd symptotes to 2/H0 s z gets very lrge. 3

14 Distnces Angulr dimeter distnce - A clssicl wy to determine distnces in stronomy is to mesure the ngle θ subtended by n object of known physicl size l. θ d A : ngulr dimeter distnce - To compute the ngulr dimeter distnce in n expnding universe, comoving size of the object is / (physicl size = x comoving size) comoving distnce to the object is χ() So the ngle subtended is θ = (/) χ() d flt A = θ = χ = χ +z In n open or closed universe, the curvture density is defined s Ωk = - Ω0 If Ωk 0, d A = H 0 Ω k Note tht the ngulr dimeter distnce is equl to the comoving distnce t low z, but decreses t very lrge redshift. sinh [ ΩH0 χ] Ω k > 0 sin [ Ω k H 0 χ] Ω k < 0 4 where Ω 0 = ρ totl ρ crit ρ totl = rdition, mtter, cosmologicl constnt

15 Distnces Angulr dimeter distnce If Ωk 0, d A = H 0 Ω k sinh [ ΩH0 χ] Ω k > 0 sin [ Ω k H 0 χ] Ω k < 0 - We re not redy yet to derive this (will try lter). But wht does it men nywy? d A Ω k > 0 : open universe Ω k < 0 : closed universe χ So the closed universe will eventully collpse while the open universe will expnd forever. 5

16 Luminosity distnce Distnces - Another wy of inferring distnces in stronomy is to mesure the flux from n object of known luminosity. For non-expnding universe, the observed flux F t distnce d from source of known luminosity L is For n expnding universe (ssuming the photons re monochromtic): working on the comoving grid, F = F = L 4πd 2 L(χ) 4πχ 2 () where L(χ) is the luminosity through comoving sphericl shell with rdius χ() If we let L to be the luminosity t source: L(χ) =L 2 i) # of photons crossing the shell is smller by ii) the energy of photon is smller tody by L L F = L2 4πχ 2 () = 4π χ() 2 4πd 2 L where we define the luminosity distnce: d L χ Distnce (/H0) z d L Ω Λ =0.7 Ω Λ =0 0 0 χ d A Three distnce mesures in flt expnding universe. From top to bottom, the luminosity distnce, the comoving distnce, nd the ngulr dimeter distnce. Mtter only (light curves) nd 70% cosmologicl constnt (hevy curves) re shown. 6

17 Evolution of Energy Energy-momentum tensor - We introduced the energy-momentum tensor on the RHS of the Einstein equtions. First consider the cse of perfect isotropic fluid. Then, T µ ν = ρ P P P P : pressure of the fluid - How do the components of the energy-momentum tensor evolve with time? Consider first the cse where there is no grvity nd velocities re negligible. ρ t =0 P : continuity eqution x i =0 : Euler eqution 7 T µ ν x µ =0 In n expnding universe, the conservtion criterion must be modified: conservtion implies the covrint derivtive is zero. T µ ν;µ =0 T ν;µ µ Tµ ν x µ +Γµ αµtν α Γ α νµt α µ =0 (w/o proof) is four seprte equtions. T µ Let s consider the ν=0 component: 0 x µ +Γµ αµt0 α Γ α 0µT α µ =0 ρ t Γµ 0µ ρ Γα 0µT α µ =0 Γ i 0j =Γ i ȧ From j0 = δ ij we get ρ t 3ȧ ρ 3ȧ P =0 or ρ t + ȧ [3ρ +3P] =0 Rerrnging this 3 [ρ3 ] t = 3ȧ P 3 [ρ3 ] t = ρ t +3ȧ ρ

18 Evolution of Energy Energy-momentum tensor - Mtter hs effectively zero pressure, so: 3 [ρ m 3 ] t =0, ρ m 3 - Rdition hs P = ρ, so 3 ρ r t = ȧ 4ρ r = 4 [ρ r 4 ] t =0 ρ r 4 Throughout most of the erly universe, rections proceeded rpidly enough to keep prticles in equilibrium, different species shring common temperture: it is convenient to introduce the occuption number, or distribution function of species. - By Heisenberg s principle, no prticle cn be loclized into region of phse spce smller thn (2π) 3 so the number of phse spce elements in d 3 xd 3 p is d 3 xd 3 p/(2π) 3 nd the energy density is ρ i = g i i g i f i d 3 p (2π) 3 f i(x, p )E(p) : lbels different species : degenercy of the species (e.g., equl to 2 for the photon for its spin sttes) : distribution function ( = ) 8

19 Evolution of Energy (dvnced: skip if you cnnot follow) In equilibrium t T, bosons hve Bose-Einstein distributions nd fermions hve Fermi-Dirc distributions f BE = f FD = e (E µ)/t e (E µ)/t + with μ the chemicl potentil. (These distributions do not depend on position vector or the direction of momentum, simply on the mgnitude p.) The pressure cn be similrly expressed s in integrl over the distribution function: P i = g i d 3 p (2π) 3 f i(x, p ) 3E(p) p 2 (w/o proof: if you red Weinberg, Cosmology, detiled cn be found) P To good pproximtion μ<<t nd one cn show tht i T = ρ i + P i T Now, 3 [ρ3 ] = 3ȧ t P cn be written s 3 [(ρ + P)3 ] t Using P t = dt P we get dt T 3 [(ρ + P)3 ] dt ρ + P t dt T P t =0 = 3 T t (ρ + P) 3 T =0 So the entropy density s ρ + P T must scle 3 9

20 Cosmic Inventory Now we discuss the question of how much energy is contributed by the different components of the universe. - Photons The temperture of the CMB photons hs been mesured precisely: T = 2.725±0.002K. The energy density if this rdition is: d 3 p ρ γ =2 (2π) 3 e p/t p, which is from ρ i = g i d 3 p (2π) 3 f i(x, p )E(p) (We neglected the chemicl potentil μ, nd present experimentl limit is μ/t<9x0-5 so we ignore μ) nd cn be integrted out nlyticlly: x p/t ρ γ = 8πT 4 (2π) 3 0 dx x 3 e x ρ γ = π2 5 T 4 Note tht since the energy density of rdition scles s -4, the temperture of the CMB must scle s -. It is useful to hve ll energy densities in the sme units, by dividing ll densities by the criticl density tody: 4 ρ γ = π K ρ cr h 2 ev 4 where T = 2.725K ρ cr = h 2 ev 4 = h K = ev re used. There s no sptil dependence in the Bose-Einstein function for the photons (zero-oder). A smll perturbtions do hve sptil dependence nd correspond to the nisotropies in the CMB.

21 Cosmic Inventory - Bryons Unlike CMB, bryons cnnot be described s gs with temperture nd zero chemicl potentil. The bryon density must be mesured directly. All mesurements indicte tht: ρ b ρ cr =Ω b 3 Ωb: rtio of the bryon density to the criticl density tody Ω b h The bryon density is roughly 2-5% of the criticl density. - Mtter The mount of mtter cn be inferred from the grvittionl effects in glctic systems: Ω m h 0.2 ρ m ρ cr =Ω m 3 The mtter density is roughly 30% of the criticl density (drk mtter). - Neutrinos Unlike the CMB nd bryons, cosmic neutrinos hve not been observed. So rguments re theoreticl (but strong): For the mssless boson ρ γ = π2 per ech spin stte. Using, we get s = ρ + P T 30 T 4 P = ρ γ 3 = π 2 T = 2π2 45 T 3 per spin stte. 2

22 - Neutrinos (cont.) Cosmic Inventory Mssless fermions contribute 7/8 from wht mssless bosons do. (w/o proof) mssive prticles contribute negligibly. 7/8 hs to do with the fctor: (cn you prove?) Now, before nnihiltions of e - nd e + - fermions - electrons (2 spins) nd positrons (2 spins), neutrinos (3 genertions, spin), nti-neutrinos (3) - bosons - photons (2 pins) So t before nnihiltion, f FD = e (E µ)/t + s( )= 2π2 45 T 3 [2 + (7/8)( )] = 43π2 90 T 3 where T is the common temperture t. After nnihiltion e - nd e + re gone wy nd the photon nd neutrino tempertures re no longer identicl: s( 2 )= 2π2 45 [2T 3 γ T 3 ν] Using s( ) 3 = s( 2 ) 3 2 Now, using T = 2 T ν ( 2 ) 43 Tγ 3 2 ( T ) 3 2 =4 + (T ν ( 2 ) 2 ) 3 T ν 8 T ν T γ = 4 /3 Neutrino temperture is lower by fctor of (4/) /3 since photons re heted by e + e - nnihiltion. 22

23 - Neutrinos (cont.) Cosmic Inventory Fermi-Dirc integrl is smller by fctor of 7/8 from e p/t + term. Note tht the energy density of mssless prticle scles s T 4. So the neutrino energy density is smller thn the photon density by (4/) 4/3. ρ ν ρ γ = /3 If there were three species of mssless neutrinos tody, their contribution to the energy density is Ω ν ρ 4/3 ν =3 7 4 ργ 4/3 = ρ cr 8 ρ cr 8 h 2 = h 2 (where mν=0 is ssumed) tody = - Drk energy Evidence pointing the existence of something else beyond the rdition nd mtter - Simple budgetry shortfll: the totl energy density of the universe ~ criticl (by both theoreticlly nd from observtion) but totl mtter density ~ /3 of the criticl density : the remining 2/3 of the density must be in some smooth, unclustered form: drk energy. 23