Gravitation: Cosmology

Transcription

1 An Introduction to General Relativity Center for Relativistic Astrophysics School of Physics Georgia Institute of Technology Notes based on textbook: Spacetime and Geometry by S.M. Carroll Spring 2013

2 Copernican Principle Copernican principle: The universe is pretty much the same everywhere. There should be no special observers. The Earth is not in a central, specially position. The Copernican principle applies on the very largest scales, where local variations in density are averaged over. The Copernican principle has been validated by a number of different observations, such as Number counts of galaxies Oservations of diffuse X-ray and γ-ray backgrounds. Cosmic microwave background radiation. The cosmic microwave background is not perfectly smooth, the deviations from regularity are on the order of 10 5

3 Formally, the Copernican principle is related to isotropy and homogeneity of the universe. Isotropy: a manifold is iotropic at a given point if it looks the same in every direction. More formally, a manifold M is isotropic around a point p if, for any two vectors V and W in T pm, there is an isometry of M such that the pushforward of W under the isometry is parallel with V (not pushed forward). Homogeneity: the metric is the same throughout the space. That is, given any two points p and q in M, there is an isometry which takes p into q. Notice a manifold can be homogeneous but nowhere isotropic (such as R S 2 in the usual metric), or it can be isotropic around a point without being homogeneous (such as a cone). If a space is isotropic everywhere then it is homogeneous. Likewise, if it is isotropic around one point and also homogeneous, it will be isotropic around every point. Observational evidence for isotropy + the Copernican principle implies homogeneity and isotropy everywhere in space in the universe. Since isotropy can be viewed as rotational invariance and homogeneity as translational invariance, thus the space is maximally symmetric; that is, it has the maximum number of Killing vectors. Notice that the expansion of the universe tells us that the space-time is not homogeneous and isotropic.

4 Cosmological Models Observations of distant galaxies show to be receding from us; the universe is apparently not static, but changing with time. We need then a cosmological models homogeneous and isotropic in space, but not in time. That is, we need to foliate the spacetime manifold representing the universe into spacelike slices such that each slice is homogeneous and isotropic. That is, we consider our spacetime to be R Σ, where R represents the time direction and Σ is a homogeneous and isotropic three-manifold. Σ must be a maximally symmetric space. That is space has its maximum possible number of Killing vectors. Therefore the metric takes the form ds 2 = dt 2 + a 2 (t)γ ij (u)du i du j. where t is the timelike coordinate, and (u 1, u 2, u 3 ) are the spatial coordinates on Σ. γ ij is the maximally symmetric metric on Σ. The function a(t) is known as the scale factor. The coordinates used here are known as comoving coordinates. That is, the metric is free of cross terms dt du i and the spacelike components are proportional to a single function of t, An observer who stays at constant u i is called comoving. Only a comoving observer will think that the universe looks isotropic.

5 Recall, for maximally symmetric Euclidean three-metrics γ ij (3) Rijkl = k(γ ik γ jl γ il γ jk ), with k a constant. Thus the Ricci tensor is then (3) R jl = 2kγ jl and (3) R = 6 k the metric can be put in the form dσ 2 = γ ij du i du j = e 2β(r) dr 2 + r 2 (dθ 2 + sin 2 θ dφ 2 ) The components of the Ricci tensor are (3) R11 = 2 r 1β (3) R22 = e 2β (r 1 β 1) + 1 (3) R33 = [e 2β (r 1 β 1) + 1] sin 2 θ Using (3) R jl = 2kγ jl and (3) R = 6 k, one gets β = 1 2 ln(1 kr 2 ). This gives the Friedmann-Robertson-Walker metric. [ ] ds 2 = dt 2 + a 2 dr 2 (t) 1 kr 2 + r 2 (dθ 2 + sin 2 θ dφ 2 )

6 The Friedmann-Robertson-Walker metric can then be rewritten as ds 2 = dt 2 + a 2 (t) dσ 2 where dσ 2 = dr 2 1 kr 2 + r 2 (dθ 2 + sin 2 θ dφ 2 ) is the metric on Σ. One still needs to find from the Einstein s equations an equation that determines the behavior of the scale factor a(t). Notice that the substitutions k k k r k r a a k leave the metric invariant. Therefore the only relevant parameter is k/ k, and there are three cases of interest: k = 1, 0, 1.

7 Case k = 0 (flat): This case corresponds to vanishing curvature on Σ dσ 2 = dr 2 + r 2 dω 2 which is simply flat Euclidean space. Globally, it could describe R 3 or a more complicated manifold, such as the three-torus S 1 S 1 S 1. Case k = +1 (closed): This case corresponds to constant positive curvature on Σ. Defining r = sin χ one gets dσ 2 = dχ 2 + sin 2 χ dω 2 which is the metric of a three-sphere. The only possible global structure is actually the three-sphere. Case k = 1 (open): This case corresponds to constant negative curvature on Σ Define r = sinh ψ to obtain dσ 2 = dψ 2 + sinh 2 ψ dω 2

8 Setting ȧ da/dt, the Christoffel symbols are given by Γ 0 11 = aȧ 1 kr 2 Γ 0 22 = aȧr 2 Γ 0 33 = aȧr 2 sin 2 θ Γ 1 01 = Γ 1 10 = Γ2 02 = Γ2 20 = Γ3 03 = Γ3 30 = ȧ a Γ 1 22 = r(1 kr 2 ) Γ 1 33 = r(1 kr 2 ) sin 2 θ Γ 2 12 = Γ 2 21 = Γ3 13 = Γ3 31 = 1 r Γ 2 33 = sin θ cos θ Γ 3 23 = Γ3 32 = cot θ. The nonzero components of the Ricci tensor are R 00 = 3 ä a aä + 2ȧ 2 + 2k R 11 = 1 kr 2 R 22 = r 2 (aä + 2ȧ 2 + 2k) R 33 = r 2 (aä + 2ȧ 2 + 2k) sin 2 θ, and the Ricci scalar is then R = 6 ä ( ) 2 ȧ a + + k a a 2

9 Energy-momentum Tensor The universe is not empty. We model its matter and energy as a perfect fluid. The energy-momentum tensor for a perfect fluid is given by T µν = (p + ρ)u µu ν + pg µν, where ρ and p are the energy density and pressure, respectively, as measured in the rest frame of an observer with 4-velocity U µ. In comoving coordiantes the fluid is at rest; thus, U µ = (1, 0, 0, 0) and the energy-momentum tensor is ρ T µν = 0 0 g ij p 0 or T µ ν = diag( ρ, p, p, p) Note that T = T µ µ = ρ + 3p

10 Conservation of Energy and Equation of State Consider the zero component of the conservation of energy equation: 0 = µt µ 0 = µt µ 0 + Γ µ µ0 T 0 0 Γ λ µ0 T µ λ = ρ 3 ȧ (ρ + p). a Introduce the equation of state p = wρ where w is a constant independent of time. The conservation of energy equation becomes ρ ρ = 3(1 + w) ȧ a, which can be integrated to obtain ρ a 3(1+w).

11 Cosmological Models DUST or MATTER-DOMINATED: Dust is collisionless, nonrelativistic matter, which obeys w = 0. Examples include ordinary stars and galaxies, for which the pressure is negligible in comparison with the energy density. The energy density in matter falls off as ρ a 3 Notice that this is just the decrease in the number density of particles as the universe expands since for dust the energy density is dominated by the rest energy, which is proportional to the number density. RADIATION-DOMINATED: Radiation includes actual electromagnetic radiation, or massive particles moving at relative velocities sufficiently close to the speed of light that they become indistinguishable from photons. Recall that with the trace given by T µν = 1 4π (F µλ F ν λ 1 4 gµν F λσ F λσ ). T µ µ = 1 [F µλ F µλ 14 ] 4π (4)F λσ F λσ = 0. Thus, T = T µ µ = ρ + 3p implies that p = ρ/3 or w = 1/3 Thus, the energy density in radiation falls off as ρ a 4 Today the energy density of the universe is dominated by matter, with ρ mat /ρ rad 10 6.

12 VACUUM-DOMINATED: Introducing energy into the vacuum is equivalent to introducing a cosmological constant. Einstein s equations with a cosmological constant are G µν = 8πT µν Λg µν This is equivalent to the equations with no cosmological constant but an energy-momentum tensor for the vacuum, T (vac) µν = Λ 8π gµν Recall that Thus or equivalently w = 1. From T µν = (p + ρ)u µu ν + p g µν ρ = p = Λ 8π ρ a 3(1+w). one has that ρ constant Since the energy density in matter and radiation decreases as the universe expands, if there is a nonzero vacuum energy it tends to win out over the long term.

13 Friedmann Equation and Hubble Parameter Starting from R µν = 8π (T µν 12 ) gµν T. The µν = 00 equation is and the µν = ij equations give 3 ä = 4π(ρ + 3p) a ä ( ȧ a + 2 a ) k = 4π(ρ p) a2 It is not difficult to show that these equations yield Friedmann Equations ä a = 4π (ρ + 3p) 3 ( ) 2 ȧ = 8π a 3 ρ k a 2 The rate of expansion is characterized by the Hubble parameter, H = ȧ/a The value of the Hubble parameter at the present epoch is the Hubble constant, H 0. Measurements indicate that H 0 = 70 ± 10 km/sec/mpc. One parametrizes H 0 as H 0 = 100 h km/sec/mpc so that h 0.7

14 More cosmological parameters and scales Hubble length d H = H 1 c = 3, 000 h 1 Mpc 0 Hubble time Deceleration parameter, Density parameter, t H = H 1 0 Ω = = h 1 yr q = aä ȧ 2 8π 3H 2 ρ = ρ ρ crit where the critical density is defined by ρ crit = 3H2 8π

15 Density Parameter Ω Thus ( ) 2 ȧ = a H 2 = 1 = 1 = 8π 3 ρ k a 2 8π 3 ρ k a 2 8 π H 2 ρ k a 2 H 2 ρ k ρ crit a 2 H 2 Thus Ω 1 = k H 2 a 2. The sign of k is therefore determined by whether Ω is greater than, equal to, or less than one. We have ρ < ρ crit Ω < 1 k = 1 open ρ = ρ crit Ω = 1 k = 0 flat ρ > ρ crit Ω > 1 k = +1 closed. The density parameter, then, tells us which of the three Robertson-Walker geometries describes our universe. From the cosmic microwave background observations, we believe that Ω = 1

16 Evolution of the Scale Factor Recall that from ρ = 3(ρ + p)ȧ/a and p = w ρ, we have that ρ a 3(1+w) which can be rewritten as ρ = ρ 0 a n if w = n/3 1 Define the curvature density and its density parameter as ρ c = 3 k 8 π a 2 and Ω c = k H 2 a 2 Therefore, from Friedmann equation H = ȧ ( ) 8 π 1/2 ( ) 8 π 1/2 a = 3 ρ 3 ρ 0a n we get that ȧ a 1 n/2 which yields a t 2/n w n a matter 0 3 t 2/3 radiation t 1/2 curvature t vacuum 1 0 e Ht Notice that all the non-vacuum solutions have a = 0, the Big Bang singularity The matter dominated solution is also called the Einstein-de Sitter universe and the vacuum dominated solution the de Sitter universe

17 From the Friedmann equation H 2 = 8 π 3 ρ i i Dividing by H 2 one gets 1 = i Ω i or 1 = Ω c + Ω where Ω is the sum of all the non-curvature densities. Taking the time derivative of the Hubble parameter Ḣ = ä ( ) 2 ȧ a = 4 π (1 + w i )ρ i a i Since w i 1 then Ḣ 0. That is, the rate of expansion of the universe decreases. The acceleration of the universe refers to ä.

18 From we see that H 2 = 8 π 3 (ρ M0 a 3 + ρ R0 a 4 + ρ Λ0 + ρ c0 a 2) As a 0 the universe becomes first dominated by matter and then by radiation As a the universe is dominated by Λ If Λ < 0 the universe will eventually re-collapse. If Λ > 0 the universe expands for ever or re-collapse depending on the value of Ω M Let s denote by a the value of the scale factor at the turnaround point between perpetual expansion and eventual re-collapse. Then 0 = H 2 = 8 π ( ρ M0 a 3 3 Λ0 + ρ c0 a 2 ) 0 = H 2 H 0 2 = Ω M0 a 3 + Ω Λ0 + Ω c0 a 2 0 = Ω M0 a 3 + Ω Λ0 + (1 Ω M0 + Ω Λ0 ) a 2 0 = Ω M0 + Ω Λ0 a 3 + (1 Ω M0 + Ω Λ0) a Solving this equation for a and finding the value of Ω Λ0 given Ω M0 for which real solutions exists, one gets the condition for the turnaround point.

19 Current observations favor Ω Λ0 0.7 given Ω M0 0.3

20 Redshift and Distances Recall H 0 is related to the age of the universe We would like to also know Ω since among other things determines the curvature k of the universe. In a FRW universe, there are no timelike Killing vectors, but there is a Killing tensor K µν = a 2 (g µν + U µu ν ) where K µν satisfies (σ K µν) = 0. Consider a particle with 4-velocity V µ = dx µ /dλ, thus K 2 = K µν V µ V ν = a 2 [V µv µ + (U µv µ ) 2 ] will be a constant along geodesics. For massive particles, V µv µ = 1, or (V 0 ) 2 = 1 + V 2 where V 2 = g ij V i V j. So V = K a. The particle therefore slows down with respect to the comoving coordinates as the universe expands. Thats is, a gas of particles will cool down as the universe expands.

21 For null geodesics V µv µ = 0, and U µv µ = K a. Recall that the frequency of the photon as measured by a comoving observer is ω = U µv µ. The frequency of the photon emitted with frequency ω em will therefore be observed with a lower frequency ω obs as the universe expands: ω obs = aem ω em a obs The redshift z between two events is defined to be z = λ obs λem λ em = a obs a em 1 Notice that this redshift is not the same as the conventional Doppler effect; it is the expansion of space. To measure z, we know the rest-frame wavelengths and thus λ em and we observe λ obs.

22 Instantaneous Physical Distance and Hubble Law Recall the FRW metric ds 2 = dt 2 + a 2 [ dχ 2 + S 2 k (χ) dω2] where S k = {sin χ, χ, sinh χ} for k = {+1, 0, 1}, respectively. The instantaneous distance is The observed velocity is then thus d P (t) = a(t) χ ȧ v = ḋp (t) = ȧ(t) χ = a d P Hubble law v = H 0 d P

23 Luminosity Distance We define the luminosity distance as d 2 L = L 4πF where L is the absolute luminosity of the source and F is the flux measured by the observer. This is motivated by the fact that for a source at distance d, the flux over the luminosity is F L = 1 A(d) with A(d) the area of a sphere center at the source. In an FRW universe, A = 4 πa 2 0 S2 k, where a 0 is the scale factor when the photons are observed. The flux is diluted by two additional effects: (1) the individual photons redshift by a factor (1 + z), and (2) the photons hit the sphere less frequently, since two photons emitted a time δt apart will be measured at a time (1 + z)δt apart. Therefore or F L = 1 4πa 0 2, S2 (1 + z)2 k d L = a 0 S k (χ)(1 + z) d L is in principle measurable, since there are some astrophysical sources whose absolute luminosities are known standard candles The redshift z is an observable, but χ and a 0 are not.

24 Consider a null radial geodesic thus χ = 0 = ds 2 = dt 2 + a 2 dχ 2 dt a(t) = da z a 2 H = 0 dz H(z ) On the other hand, from Friedmann equation, H 2 = 8 π 3 ρ i i where to simplify our existence we assume ρ i = ρ i0 a n i = ρ i0 (1 + z) n i Thus, we rewrite H(z) = H 0 E(z) where E(z) = i 1/2 Ω i0 (1 + z) n i So the luminosity distance is where d L (z) = a 0 S k (χ) (1 + z) χ = 1 z a 0 H 0 0 dz E(z )

25 Recall, Ω c0 = k/(a 0 H 0 ) 2 therefore a 0 = H 1 k/ωc0 = H 1 / Ω 0 0 c0 Thus, H 1 0 d L (z) = (1 + z) Ωc0 S k (χ) where z χ = Ω c0 0 dz E(z ) and it is understood that Ω c0 = 1 Ω 0 Thus to calculate d L we need the observables Ω 0 and H 0. There seems to be an issue with the above expression for a flat Universe since in that case k = 0 and Ω c0 = 0. However, in this case S k (χ) = χ and z d L (z) = (1 + z) H dz E(z )

26 Lookback time If the age of the Universe today is t 0 and the age when a photon was emitted is t. the lookback time is t 0 t = = t0 dt t a0 da a0 ( ) ( H0 a0 a a H = H 1 0 a H a 0 dz = H 1 0 z (1 + z) E(z) ) ( ) a d a 0 where we have used a/a 0 = 1 + z and H = H 0 E(z) Consider a flat (k = 0) matter-dominated (ρ = ρ M = ρ M0 a 3 ) universe. Then with E(z) = (1 + z) 3/2 one gets 0 t 0 t = H 1 dz 0 z (1 + z) 5/2 2 = [1 3 H 1 (1 + z 3/2] 0 ) Let t 0 or z Age of the Universe in a Matter Dominated Universe t 0 = 2 3 H 1 0 where t H = H 1 0 the Hubble time

27 The Cosmic Microwave Background Radiation

28 Black Body Radiation

29

30 Cosmic Microwave Background Anisotropies Temperature Anisotropies Θ(ˆn) T T (ˆn) = a lm Y lm (ˆn) lm Power Spectrum Θ(ˆn), Θ(ˆn ) = = C l Y lm (ˆn)Y lm (ˆn ) lm (2l + 1) C l 4 π P l (ˆn ˆn ) l where denotes average over all directions and positions with a lm a l m = δ ll δ mm C l We cannot average over all positions, so the observed C l is an average over m C obs l 1 2 l + 1 a lm a lm = 1 d 2ˆn d 2ˆn P l (ˆn ˆn )Θ(ˆn), Θ(ˆn ) m 4 π Cosmic Variance ( C l Cl obs ) 2 = C l 2 2 l + 1

31 Cosmic Microwave Background Anisotropies

32 Cosmic Microwave Background Anisotropies The CMBR spectrum support the view that density perturbations are imprinted on all scales at extremely early times. The observations support support also the view that the Universe is spatially flat; thus, Ω = 1 and then Ω Λ0 = 0.7 ± 01 or equivalently ρ vac 10 8 erg/cm 3. Which is consistent with the Type Ia SN results.

33 Polarization of the Cosmic Microwave Background Radiation

34 The Matter Content in the Universe Matter is equivalent to non-relativistic particles and is approximately Ω M0 = 0.3 ± 0.1 Baryonic matter: stars, planes, gas, dust (i.e. protons and neutrons) is accounts for Ω b = 0.04 ± 0.02 Dark or non-baryonic matter: accounts for Ω M0 Ω b We do not know yet the the origin of dark matter. Coincidence problem: we have that Ω Λ Ω M a 3 Since the vacuum and matter energy are comparable today, the vacuum energy is undetectably small in the past!

35 Inflation The universe seems to have merged from one set of initial conditions A set of initial conditions fined tuned to be spatially flat and highly homogeneous and isotropic Is there a mechanism that could take a wide spectrum of initial conditions and evolve them toward flatness and homogeneity/isotropy? The answers is YES. The inflationary universe scenario provides such mechanism.

36 Flatness Problem The densiy of the universe seems to be finely tuned to be equal to the critical density ρ c. Thus, the universe is fine-tuned to be flat Recall Friedmann equation H 2 8 π = 3 ρ κ a 2 3 H 2 a 2 = ρ a 2 3 κ 8 π 8 π ρ c a 2 = ρ a 2 3 κ 8 π (ρ c ρ) a 2 = 3 κ 8 π (Ω 1 1)ρ a 2 = 3 κ 8 π = const At a 0, the term ρ a ρ 0 a 2 0 Therefore, at a 0, the factor (Ω 1 1) (Ω 1 1) 0 Thus, any small deviation from Ω = 1 will yield a huge value of Ω 0

37 Particle Horizon Horizons exist because there is finite amount of time since the Big Bang. The photons travel only a finite distance during this time. Consider a photon along a radial null geodesic in a matter dominated, flat universe. Thus, from 0 = ds 2 = dt 2 + a 2 dr 2 with H 2 = H 2 0 a 3, the comoving distance traveled by this photon in an interval of time [t 1, t 2 ] is t2 dt a2 r = t 1 a = da a2 a 1 H a 2 = da a 1 H 0 a 1/2 = 2 H 1 ( a 0 2 a 1 ) The comoving horizon distance traveled since the Big Bang to a time in which a = a is r hor (a ) = 2 H 1 a 0 The physical horizon distance is then the proper distance traveled by the photon, namely d hor (a ) = a r hor (a ) = 2 H 1 One can show that in general d hor (a ) H 1 = d H (a ) where d H is the Hubble distance.

38 Horizon Problem Given r = 2 H 1 ( a 0 2 a 1 ) Consider a photon originated at the CMB. The comoving distance between a point in the CMB (a 1 = a CMB 1/1200) and us (a 2 = a 0 = 1) is r = 2 H 1 (1 a 0 CMB ) 2 H 1 0 The comoving horizon distance between a point in the CMB and the Big Bang r hor (A CMB ) = 2 H 1 acmb H Thus, two widely separated parts of the CMB will have non-overlapping horizons. How come then we see them at almost the same temperature.

39 Inflation Recall, Friedmann equation H 2 = 8 π 3 ρ κ a 2 For matter dominated ρ a 3 and for radiation dominated ρ a 4 Thus, as the Universe expands (i.e. a grows), the curvature term becomes more dominant. We need a type of matter for which 8 π 3 If ρ = constant with 8 π 3 ρ κ a2, for instance ρ = constant ρ κ t a2, one has that H = ȧ/a = constant > 0 and thus a eh Notice that in this case one would have a period of accelerated expansion (ä > 0). Since Ω c = 1 Ω = κ H 2 a 2, with a period of inflation (i.e. a eh t ), one can potentially drive Ω 1 and solve the flatness problem

40 The same acceleration could also grow our horizon in such a way that all of the observable universe today was in causal contact at the time when the CMB was generated. That is, One can then select H inf such that r(inf dec) = tdec dt adec t inf a = da a inf H a 2 = H 1 (a 1 a 1 inf inf dec ) H 1 a 1 inf inf r(dec today) = t0 dt t dec a 2 H 1 0 r(inf dec) H 1 a 1 r(dec today) 2 H 1 inf inf 0

41 How to start and stop Inflation Consider a Universe filled with a scalar field φ. Then, Friedmann equation reads H 2 = 8 π 3 ρ + κ a 2 with ρ = 1 φ 2 + V (φ) 2 and V a potential to be determined. The equation for the dynamics of the scalar field is (assuming a homogenous field) φ + 3 H φ + V (φ) = 0 Ignoring for the moment the curvature term, the equations for a and φ are then: φ + 3 H φ + V (φ) = 0 H 2 = 8 π 3 ( 1 2 ) φ 2 + V (φ) Recall that we need a type of matter such that ρ constant, so H constant and then a e Ht Therefore, we require that φ 2 V φ 3 H φ, V which is equivalent to requiring that the potential energy dominates over the kinetic energy (slow-roll approximation;).

42 Therefore Define 3 H φ V H 2 8 π 3 V ɛ = 1 ( ) V 2 η = 1 ( ) V 16 π V 8 π V where ɛ measures the slope of the potential and η its curvature We will see that the necessary conditions for inflation (slow-roll parameters) requires ɛ < 1, η < 1 Inflation also requires acceleration; that is, ä > 0. Notice that But Thus ä Ḣ a = Ḣ + H2 > 0 H 2 < 1 Ḣ = (H2 ) 2 H = 4 π 3 ( ) V φ H = 4 π 9 ( ) V 2 = 1 (V ) 2 ( ) Ḣ H 2 < 1 1 V 2 = ɛ < 1 16 π V H 6 V

43 e-foldings The amount of inflation is given by the number of e-foldings N ln a(te) a(t i ) = te t i φe V H dt 8 π φ i V dφ The minimum amount of inflation required to solve the various cosmological problems is about 70 e-foldings. Inflation ends when V 0

44 Example Consider the case of power-law inflation, V φ n Then φe N = 8 π φ i V dφ = 8 π V φe φ i φ n dφ = 8 π ( φ 2 ) e n φ2 i = 8 π n φ2 i where we set φ e = 0, so V (φ e) 0 In order to get N = 70, one needs φ i 70 n 8 π 1.7 n From the slow-roll parameters one also needs, ɛ = 1 ( ) 2 n < 1 η = 1 16 π φ i 8 π [ ] n (n 1) < 1 φ i or φ i > n n (n 1) φ i > 16 π 8 π

45 Revisit Horizon Problem Recall, particle horizon size tdec dt d H (inf dec) = a dec H 1 inf t ( inf adec 1 a inf adec a = a dec a ) inf H 1 inf da H a 2 ( e H inf t ) 1 t0 dt d H (dec today) = a 0 t dec a 2 H 1 0 Then d H (inf dec) H 1 e H inf t r(dec today) 2 H 1 inf 0

46 Relic Fluctuations from Inflation Let s consider the following perturbation of the scalar field φ(t) φ 0 (t) + δφ(t, x) The perturbation δφ will induce a perturbation in the stress-energy tensor, and therefore, through Einstein s equation, there will also be a perturbation in the metric. In a flat FRW model, the perturbed metric will read ds 2 = (1 + 2 Φ) dt 2 + (1 2 Φ) a 2 (t) (dx 2 + dy 2 + dz 2 ) with Φ(t, x) the metric perturbation associated with the scalar perturbation δφ. The perturbed part of the Einstein equations reads δg 0 0 = 2( 2 Φ 3 H Φ 3 H 2 Φ) δg 0 i = 2 i ( Φ + H Φ) δg i i = 2[ Φ + 4 Φ H + (2 Ḣ + 3 H2 ) Φ] Notice that from 3 H φ 0 = V and H 2 = 8 π V /3 one has that Ḣ = 4 π φ 2 0

47 The perturbed part of the stress-energy tensor reads δt 0 0 = φ 2 0 Φ + φ 0 δ φ + V δφ δt 0 i = φ 0 i (δφ) δt i i = φ 2 0 Φ φ 0 δ φ + V δφ From the δg ν µ = 8 π δt ν µ (in units 8 π = 1) 2 Φ 3 H Φ 3 H 2 Φ = 4 π( φ 2 0 Φ + φ 0 δ φ + V δφ) Φ + H Φ = 4 π φ 0 δφ Φ + 4 Φ H + (2 Ḣ + 3 H2 ) Φ = 4 π( φ 2 0 Φ φ 0 δ φ + V δφ) Using the equations for φ 0 and Ḣ, one can show that the third equation can be obtained from the time derivative of the first equation.

48 The equations for Φ and δφ are then Φ + H Φ = 4 π φ 0 δφ ( ) ( 4 π φ ) Φ = 4 π φ 2 d δφ 0 dt φ 0 In Fourier space, these equations read Φ k + H Φ k = 4 π φ 0 δφ k ( 1 k 2 ) π φ 2 a 0 2 Φ k = d ( ) δφk dt φ 0 with k the wave number. Important to point out that Φ k and δφ k are not gauge invariant.

49 Introduce the following gauge invariant perturbation ξ k Φ k + H δφ k φ 0 Which leads to ξ k + ( φ 2 H + 2 φ 0 φ H ) ξ k + k2 a 2 ξ k = 0 This equation and the equations H 2 = 8 π 3 ( ) 1 φ V φ H φ 0 + V = 0 are sufficient to obtain the evolution of a, φ 0 and ξ k.

50 If we convert from cosmic time t to conformal time η such that dt = a dη, the metric becomes ds 2 = a 2 ( dη 2 + dx 2 + dy 2 + dz 2 ) If we introduce the definitions α = a φ 0 /H and χ k = α ξ k, the equation for ξ k becomes ( ) χ k + k 2 α χ k = 0 α where primes denote derivatives with respect to conformal time. If k 2 α /α, one can integrate the equation to get χ k α or equivalently ξ k = constant. In the slow-roll approximation, α /α 2 a 2 H 2 and χ k + (k 2 2 a 2 H 2) χ k = 0 If k a H (sub-horizon scale perturbations), the perturbations χ k oscillate. The equation becomes that of a harmonic oscillator. If k a H (super-horizon scale perturbations), the perturbations χ k exponentially grow.

51