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Inflation and the cosmological constant problem Larissa Lorenz Sebastian Sapeta Krzyzowa 18.−28. September 2002 Contents Standard model of cosmology and its problems The inflationary paradigm Review of the most important inflationary scenarios Cosmological constant and inflation Experimental determination of cosmological constant The cosmological constant problem Big Bang N S φ0 φ(T) = φ0 φ(T) = 0 φ(T) = φ0φ(T) = φ0 About 15 billion years ago, time and space both began in a cosmological singularity. The standard model of cosmology t=0 t = 10−43s Big ρ= t = 10−10s t = 1 billion years MP g 94 = 10 l P3 cm3 á! il Vo t = 10−35s Bang −infinitely high temperature T −infinitely large density ρ −average particle energy exceeds MPc2 = 1019GeV quantum gravitation era of unity of all interactions except for gravity Experimental evidence: the expansion of the universe the abundance of elements the cosmic microwave background −baryon creation era of unity of electromagnetic and weak interaction symmetry breaking between electromagnetic and weak interaction −galaxy and star formation Problems: ¥ the flatness problem ¥ the horizon problem ¥ the monopole problem ¥ the uniqueness problem Why was the early universe so extremely flat? The flatness problem Let ρc be the critical density of the universe and ρ0 its present day value. For the parameter Ω0 = ρ0/ρc, we assume (1) SvOutPlaceObject The development of the deviation from the critical value |Ω − 1| in the course of time is given by for the radiation dominated universe SvOutPlaceObject | Ω − 1 |~ t 3 2 for the matter dominated universe In a radiation dominated universe of the age of t0 = 1017s, (1) leads to SvOutPlaceObject | Ω − 1 | < 10 − 16 | Ω − 1 |< 10 − 28 | Ω−1|< 10−50 Extremely flat! today nucleosynthesis electro−weak symmetry breaking 10−34s Why is the 3K radiation so highly isotropic? The horizon problem Any observers event horizon is given by t d H (t) = R (t)∫ 0 dt ′ R ( t ′) (2) Therefore, two observer separated by s ≥ 2dH(t) are completely independent at any time. dH dH s Owen Lets assume that 10−35s after Big Bang the temperature T was approx. 1015GeV and that R ~ 1/T; using the present day value of R and (2) , we find that today s = 2.4m! Oscar We should be observing different causally independent regions! The microwave background radiation is isotropic at a scale of 1028cm (thermal equilibrium)! Where have all the monopoles1) gone? The monopole problem All Grand Unified Theories predict superheavy stable particles carying magnetic charge. mm = 1016 mproton S These magnetic monopoles should be as abundant as protons. N This means that the density in the universe at present should be 1015 times higher that observed! predicted 1) and other relics g ρ =10 cm3 −14 g ρ =10 cm3 −29 observed What I am really interested in is whether God could have created the world differently. A. Einstein The uniqueness problem Elementary particle theories turn out to present many different, yet equivalent solutions. So why looks low energy physics around as the way it does and not completely different? Why is space−time four−dimensional? 134 136 135 138 137 Why has the fine−structure constant this value? SvOutPlaceObject Slightly different values of physical constants would have led to a totally different universe. The inflationary paradigm provides answers to the standard models problems. Problems and inflationary answers In classical big bang theory, |Ω−1| moves away from 0 in the course of time, The flatness problem but during the inflationary period we find |Ω−1| approaching 0. If it gets close enough, it stays close during all of the post−inflationary period. No need for the early universe to be extremely flat! The inflationary paradigm provides answers to the standard models problems. Problems and inflationary answers A small area in the very early universe which reached thermal equilibrium could today be larger than the oberserved universe due to the dramatical increase during the inflationary period. The horizon problem Thermal equilibrium is the simplest and correct explanation for the isotropy of 3K radiation! today observed universe time INFLATION small area The inflationary paradigm provides answers to the standard models problems. Problems and inflationary answers The monopole problem While the universe is expanding exponentially, the monopole densitiy decreases much faster than the energy density. Monopole density in the universe can be neglected! The uniqueness problem Maybe the reason why α equals 1/137 is that if it was otherwise, our kind of life could not exist in the universe. The metastabile, supercooled state of the universe causes inflation. Guths inflationary scenario Let φ be a scalar field with its minima for T > Tc (superhigh temperature) T > Tc T < Tc at φ(T) = 0 φ(T) = 0, φ(T) = φ0 T < Tc φ(T) = φ0 φ(T) = 0 phase transition from φ(T) = 0 into ground state φ(T) = φ0 via new ground state restored symmetry, ground state The universe remains in this metastabile (supercooled) state. The universe expands. This leads to exponential expansion. Problems: ¥collision of bubble walls destructs homogeneity and isotropy ¥supercooled state has to be stable to create enough inflation bubble formation φ0 φ(T) = φ0 φ(T) = 0 φ(T) = φ0φ(T) = φ0 The universe gets hot and can be described by standard big bang cosmology. Inflation can not only take place in a supercooled state, but also during the process of φ approaching the value φ0. The new inflationary universe scenario barrier height ~ T4 minimum of potential V(φ) T = 1015 GeV φ=T Coleman−Weinberg potential φ+3Hφ+V(φ)=0 T < 109 GeV: tunnelling probality reaches 100% At about T < 109 GeV, the barrier becomcs unstable. Oscillations around the minimum are damped by expansion and the production of relativistic particles. This part has to be flat enough to ensure a sufficiently long time of rolling downhill = time of inflation (much longer than H− 1). Problems: elementary particle theory required whose effective potential satifies many unnatural constraints start of inflation depends on drop of temperature which takes 6 orders of magnitude longer than Planck time Thermalzation leads to temparatures like at the beginning of inflation (T < 109 GeV); future development is described by the standard big bang model. Instead of assuming a certain value, the scalar fields φ initial distribution is regarded as chaotic. The evolution of a scalar field The chaotic inflation .. . φ with mass m << MPl is described by the Klein−Gordon equation:φ + 3H φ = −m 2φ scenario (3) In the chaotic inflation scenario, no specific initial value is assumed but a chaotic distribution of values of φ0. If the initial value of the field φ0 exceeds 1/5 MPl, the friction term is big enough to make the solution rapidly approach ϕ ( t) = ϕ 0 − the regime: for t< mM Pl 2 3π and R(t ) = R0 exp( MP4−Planck density φ An island of classical space−time rises out of the space−time foam; typical initial value of φ is: 2 M Pl m (4) ϕ0 4π mφ (t) ⇒ R(t) = R0 exp( t ) = R0 exp( H (ϕ )t) mM Pl 3 M Pl V(φ) ϕ0 = t Inflation takes places while the field is rolling downhill with friction; typical expansion is 1010^9! Inflation stops when φ reaches its minimum, friction becomes negligible and φ performs oscillations round the minimum; energy is used for particle creation. 2π [ϕ02 − ϕ 2 (t )]) 2 M Pl (6) (5) quasi−exponential expansion ! This scenario offers some amazing features: Processes separated by at least H 1 are completely independent. ⇓ Any inflationary domain of initial size exceeding 2H 1 can be considered as a separate mini− universe! That means, a region of size H 1 which expands exponentially during the period of inflation creates a lot of new mini−universes. Of those, exponentially many have smaller than the mother universe, and φ0 also Only this model of inflation can explain why the observable part of the universe is so homogeneous, but from it also follows exponentially many have bigger φ0. that on a much larger scale the universe is extremely inhomogeneous. Moreover, realistic models of elementary particles consider many kinds of scalar fields whose potential energy may have several different minima. So ⇒ The Universe is divided into domains with various laws of particles physics or even dimensionality. Cosmological constant may be identified with a vacuum energy density. Why do we bother with Λ ? Originally Introduced by Einstein as a free parameter to the field equations to balance an attractive gravitational force and to allow a static universe. 1 Rµν − Rg µν + Λg µν = 8πGTµν 2 (7) The idea of Λ came back in the context of modern quantum field theories in which the vacuum is not necessarily a state of zero energy but it is defined as a state of the lowest energy. Due to the Lorentz invariance of the ground state the vacuum energy momentum tensor has to be proportional to . (8) SvOutPlaceObject SvOutPlaceObject The effect of an energy−momentum tensor of the form (8) is equivalent to that of a cosmological constant from (7) and this is the origin of the identification of the cosmological constant with the energy of the vacuum. Λ ρ vac = ρV = (9) 8πG The vacuum can therefore be thought of as a perfect fluid with the equation of state: (10) There exist three different contributions to Λ: SvOutPlaceObject the static cosmological constant Λgeo quantum fluctuations Λfluc additional contributions due to currently unknown particles and interactions Λinv (11) SvOutPlaceObject Non−vanishing Λ may be responsible for inflationary expansion. Cosmological constant and inflation. Under the assumption of an homogeneous isotropic universe and non−vanishing cosmological constant Einstein equations can be reduced into Friedmann equations: 2 R. 8πG k Λ H 2 ≡ = ρ− 2 + 3 R 3 R time SvOutPlaceObject .. 4πG R =− (ρ + 3 p )+ Λ 3 3 R SvOutPlaceObject First two terms in (12) decrease quickly in the expanding universe while the third remains constant. At last cosmological constant begins to dominate and we can write: 2 R. Λ H 2 ≡ = 3 R SvOutPlaceObject This leads to the exponential expansion: Λ R(t ) ∝ exp t 3 SvOutPlaceObject How does the history of the universe depend on Λ value ? Model universes and their fates The Friedmann equation: A positive cosmological constant accelerates the expansion, .. R 4πG (ρ + 3 p )+ Λ =− R 3 3 while a negative Λ and ordinary matter decelerate it. Λ c = 4(8πGM ) −2 − the value of Einsteins static universe R(t) R(t) 0<Λ<Λc,k=1 Λ<0,k=−1,0,+1 Λ=0,k=1 R(t) Λ=0,k=−1 t t Λ>0,k= 0,− R(t) Λ=Λc(1+ε),k=1 1 Λ>Λc,k=1 ε <<1 Λ=0,k= 0 t t Critical density and deceleration parameter. Alternative notation for Friedmann equations. Friedmann equation: 2 R. 8πG k Λ H2 ≡ = ρ− 2 + 3 3 R R SvOutPlaceObject (16) may be write in a form: SvOutPlaceObject where and ΩM = ρ ρc Ωk = − k a2 H 2 ΩΛ = Λ 3H 2 ρc− critical density of matter in the universe The universe is flat provided that ΩM +ΩΛ = 1 Let us introduce a deceleration parameter: & &R 1 R q ≡ − 2 = Ω M − ΩΛ R& 2 (17) Positive q cause the universe to decelerate while negative q leads to acceleration Observational tests implies (ΩM=0.3 , ΩΛ=0.7) Determination of Λ using type Ia supernovae The luminosity distance is defined as: d l2 = L 4πF (18) where L is the luminosity and F the measured flux of the galaxy One can show that: H 0d l = z + 1 (1 − q0 )z 2 + ... 2 (19) We can obtain the value of q0 by measuring the luminosity distance and red−shift. The best method to determine dl is to use the standard candle properties of type Ia supernovae. Observational test implies non vanishing Λ Traditional model of flat universe without Λ is not favoured. There seems to be strong evidence for a positive Λ and accelerating universe. Independent measurements suggest the universe with non−vanishing Λ. Determining Λ by measuring angular diameter of objects. The angular diameter distance is defined as: (20) SvOutPlaceObject where D is the proper diameter of an object and Θ its apparent angular size It can be shown that: H 0d A = z − 1 (3 − q0 )z 2 + ... 2 (21) Measuring the angular diameter of objects and the red−shift allows a determination of q0 Data from 82 compact radio sources results in evidence for q0 of approximately 0.5. There are many ways in which cosmological constant can manifest itself. Alternatives to determine Λ. Observations of numbers of galaxies as a function of red−shift Counting of galaxies are sensitive test of ΩΛ Examination of 1000 infrared galaxies results in: Ω0 = 0.9 +−00..75 → q0 = 0.45+−00..35 25 An accelerating universe seems to be trustworthy. Alternatives to determine Λ. The lens probability rises dramatically as ΩΛ is increased to Gravitational lensing unity as we keep Ω fixed. The existing data allow us to place an upper limit on ΩΛ < 0.7 in a flat universe Determining of ΩM by weighing clusters of galaxies. Alternatives to determine Λ. Many cosmological tests constrain some combinations of ΩΛ Matter density and ΩM. It is useful to determine ΩM independently by adding masses of clusters of galaxies. Measurements imply: SvOutPlaceObject Estimated values of Λ are in total contradiction to reality The Λ problem. A relaivistic field can be considered as a sum of harmonic oscillators of all possible frequencies ω. In the case of a scalar field with mass m, the vacuum energy is the sum of all contributions: 1 E0 = ∑ hω j j 2 One can show that: 4 E0 k max ρV = lim 3 = L →∞ L 16π 2 (22) (23) Assuming the validity of the general theory of relativity up to the Planck scale kmax= ρV ≈ 10 92 gcm−3 lPl we get: while the experimental value is of order SvOutPlaceObject 121 orders difference between experimental and theoretical value of ρV !!! Quantum fluctuations exist due to virtual particle creation. Let us assume that these particles take up for a short time their Compton volume Lc3. h m c 3m 4 → ρV = 3 = 3 Lc = mc Lc h (24) This should produce effects noticeable on scales of meters to kilometers while there is no evidence for any effect of Λ at the distances of 1028 cm. Effects of Λ should be observed in today known universe but they are not. Several suggested solutions for the Λ problem. Supersymmetry In this theory for every boson there is fermion which is its supersymmetric partner.This special symmetry is associated with a supercharges Qa. Hamiltonian has a form: } (25) Qα 0 = Qα+ 0 = 0 (26) α In a supersymetric state: Which implies { H = ∑ Qα , Qα+ 0H 0 =0 (27) Contributions from bosons are canceled by contributions from fermions and the energy of the vacuum state vanishes. The problem is that supersymmetry seems to be broken in the observed world. String theory The search is on for a four−dimensional string theory with broken supersymmetry and vanishing or very small cosmological constant. Several suggested solutions for the Λ problem. Feynmans path integrals and the principle of least action Using Feynmans path integral formalism and the principle of least action one can get that the wavefunction of the universe Ψ has a form: Ψ ≈ e3π / h GΛ (28) If we consider Λ as a free parameter, it turns out that this expression has a prominent maximum for Λ = 0, which would solve the cosmological constant problem. Holographic theory (speculative) The number of degrees of freedom in a region grows as the area of its boundary, rather than its volume. Therefore the conventional computations of Λ involves a vast overcounting of degrees of freedom.