Inflation and the cosmological constant problem Larissa Lorenz Sebastian Sapeta Krzyzowa 18. 8. September 00
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 φ(t) = φ 0 φ 0 φ(t) = 0 φ(t) = φ 0 φ(t) = φ 0
About 15 billion years ago, time and space both began in a cosmological singularity. The standard model of cosmology t = 0 Big Bang ρ = M t = 10 4 s l P 94 g = 10 P cm infinitely high temperature T infinitely large density ρ average particle energy exceeds MPc = 10 19 GeV 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 t = 10 5 s Voilá! baryon creation era of unity of electromagnetic and weak interaction t = 10 10 s t = 1 billion years 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 ρ c be the critical density of the universe and ρ 0 its present day value. For the parameter Ω 0 = ρ 0 /ρ c, we assume (1) The development of the deviation from the critical value Ω 1 in the course of time is given by for the radiation dominated universe Ω 1 ~ t for the matter dominated universe In a radiation dominated universe of the age of t0 = 10 17 s, (1) leads to Ω 1 < 10 16 Ω 1 < 10 8 Ω 1 < 10 50 Extremely flat! today nucleosynthesis electro weak symmetry breaking 10 4 s
Why is the K radiation so highly isotropic? The horizon problem Any observer s event horizon is given by d H ( t ) = t d t R ( t ) R ( t 0 ) Therefore, two observer separated by s d H (t) are completely independent at any time. () d H d H Oscar s Owen Let s assume that 10 5 s after Big Bang the temperature T was approx. 10 15 GeV and that R ~ 1/T; using the present day value of R and (), we find that today s =.4m! We should be observing different causally independent regions! The microwave background radiation is isotropic at a scale of 10 8 cm (thermal equilibrium)!
Where have all the monopoles 1) gone? The monopole problem All Grand Unified Theories predict superheavy stable particles carying magnetic charge. S 16 m m = 10 m proton N These magnetic monopoles should be as abundant as protons. This means that the density in the universe at present should be 10 15 times higher that observed! g ρ= 10 cm g ρ= 10 cm 14 9 predicted observed 1) and other relics
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? 14 16 18 Why has the fine structure constant this value? 15 17 Slightly different values of physical constants would have led to a totally different universe.
The inflationary paradigm provides answers to the standard model s problems. Problems and inflationary answers The flatness problem In classical big bang theory, Ω 1 moves away from 0 in the course of time, 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 model s problems. Problems and inflationary answers The horizon problem 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. Thermal equilibrium is the simplest and correct explanation for the isotropy of K radiation! today observed universe time INFLATION small area
The inflationary paradigm provides answers to the standard model s 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 1/17 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. Guth s inflationary scenario Let φ be a scalar field with its minima for T > Tc φ(t) = 0 T < T at c φ(t) = 0, φ(t) = φ0 T > Tc (superhigh temperature) φ(t) = 0 restored symmetry, ground state The universe expands. T < Tc φ(t) = φ 0 new ground state The universe remains in this metastabile (supercooled) state. This leads to exponential expansion. phase transition from φ(t) = 0 into ground state φ(t) = φ 0 via bubble formation φ 0 φ(t) = 0 φ(t) = φ 0 φ(t) = φ 0 φ(t) = φ 0 Problems: collision of bubble walls destructs homogeneity and isotropy supercooled state has to be stable to create enough inflation 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 φ 0. The new inflationary universe scenario barrier height ~ T 4 minimum of potential V(φ) T = 10 15 GeV φ = T Coleman Weinberg potential T < 10 9 GeV: tunnelling probality reaches 100% φ +Hφ +V (φ)=0 Oscillations around the minimum are damped by expansion and the production of relativistic particles. At about T < 10 9 GeV, the barrier becomcs unstable. 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 < 10 9 GeV); future development is described by the standard big bang model.
Instead of assuming a certain value, the scalar field s φ initial distribution is regarded as chaotic. The chaotic inflation scenario The evolution of a scalar field φ with mass m << M Pl is described by the Klein Gordon equation: φ+ H φ = m φ In the chaotic inflation scenario, no specific initial value is assumed but a chaotic distribution of values of φ 0. If the initial value of the field φ 0 exceeds 1/5 M Pl, the friction term is big enough to make the solution rapidly approach the regime: mm Pl ϕ( t) = ϕ π 0 t and R( t) = R0 exp( [ ϕ0 ϕ ( t)]) π M Pl (4) (5)... () ϕ t < mm An island of classical space time rises out of the space time foam; typical initial value of φ is: 4π mφ( t) R( t) = R0 exp( t) = R0 M 0 t Pl Inflation takes places while the field is rolling downhill with friction; typical expansion is Pl exp( H ( ϕ ) ) for (6) quasi exponential expansion! V(φ) φ M P 4 Planck density Inflation stops when φ reaches its minimum, friction becomes negligible and φ performs oscillations round the minimum; energy is used for particle creation. This scenario offers some amazing features: Processes separated by at least H 1 are completely independent. Any inflationary domain of initial size exceeding H 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. M ϕ Pl 10 10^9! Of those, exponentially many have smaller 0 = φ0 m than the mother universe, and 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 φ that on a much larger scale the universe is extremely inhomogeneous. Moreover, realistic models of elementary particles 0. 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 R + Λ = 8 µν Rgµν gµν πgtµν 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) 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 = ρ = (9) vac V 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: the static cosmological constant Λgeo quantum fluctuations Λ fluc additional contributions due to currently unknown particles and interactions Λinv (11) (7)
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: H. R R 8πG = k ρ R Λ + time.. R 4πG = p R ( ρ + ) Λ + First two terms in (1) decrease quickly in the expanding universe while the third remains constant. At last cosmological constant begins to dominate and we can write: H. Λ R R = This leads to the exponential expansion: R( t) exp Λ t
How does the history of the universe depend on Λ value? Model universes and their fates The Friedmann equation:.. R 4πG = p R ( ρ + ) Λ + Λ c = 4 A positive cosmological constant accelerates the expansion, while a negative Λ and ordinary matter decelerate it. ( 8π GM ) the value of Einstein s static universe R(t) Λ<0,k= 1,0,+1 R(t) 0<Λ<Λ c,k=1 Λ=0,k=1 R(t) Λ=0,k= 1 Λ>0,k= 0, 1 Λ>Λ c,k=1 t R(t) Λ=Λ c (1+ε),k=1 t ε <<1 Λ=0,k= 0 t t
Critical density and deceleration parameter. Alternative notation for Friedmann equations. Friedmann equation: may be write in a form: H. R k Λ R 8πG = ρ R + (16) where Ω M = ρ ρ c Ω k = a k H Ω Λ = Λ H and ρ c critical density of matter in the universe The universe is flat provided that Ω M +Ω Λ = 1 Let us introduce a deceleration parameter: q RR & R& = 1 Ω M Ω Λ (17) Positive q cause the universe to decelerate while negative q leads to acceleration
Observational tests implies (Ω M =0., Ω Λ =0.7) Determination of Λ using type Ia supernovae The luminosity distance is defined as: L d (18) l = 4πF where L is the luminosity and F the measured flux of the galaxy 1 0 l 0 + One can show that: H d = z + ( 1 q ) z... We can obtain the value of q 0 by measuring the luminosity distance and red shift. The best method to determine d l is to use the standard candle properties of type Ia supernovae. Observational test implies non vanishing Λ (19) 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: (0) where D is the proper diameter of an object and Θ its apparent angular size 1 0 A 0 + It can be shown that: H d = z ( q ) z... (1) Measuring the angular diameter of objects and the red shift allows a determination of q0 Data from 8 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. Counting of galaxies Observations of numbers of galaxies as a function of red shift are sensitive test of Ω Λ Examination of 1000 infrared galaxies results in: Ω + 0.7 + 0.5 0 = 0.9 0.5 q0 = 0. 45 0.5
An accelerating universe seems to be trustworthy. Alternatives to determine Λ. Gravitational lensing The lens probability rises dramatically as Ω Λ is increased to 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 Ω M by weighing clusters of galaxies. Alternatives to determine Λ. Matter density Many cosmological tests constrain some combinations of Ω Λ and Ω M. It is useful to determine Ω M independently by adding masses of clusters of galaxies. Measurements imply:
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 lim 0 = = V L k 16π 4 max One can show that: ρ Assuming the validity of the general theory of relativity up to the Planck scale k max = 9 l Pl we get: ρ V 10 gcm while the experimental value is of order E L 11 orders difference between experimental and theoretical value of ρ V!!! Quantum fluctuations exist due to virtual particle creation. Let us assume that these particles take up for a short time their Compton volume L c. L h m c = ρv = = mc Lc 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 10 8 cm. c h m 4 () () (4) 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 Q a. Hamiltonian has a form: H + = {, } In a supersymetric state: Q α 0 Q 0 = 0 α = + α Q α Q α (5) (6) Which implies 0 H 0 = 0 (7) 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. Feynman s path integrals and the principle of least action Using Feynman s path integral formalism and the principle of least action one can get that the wavefunction of the universe Ψ has a form: Ψ e π / hgλ (8) 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.