Computational Physics and Astrophysics

Transcription

1 Cosmological Inflation Kostas Kokkotas University of Tübingen, Germany and Pablo Laguna Georgia Institute of Technology, USA Spring 2012

2 Our Universe

3 Cosmic Expansion Co-moving coordinates expand at exactly the same rate as the universe. To a good approximation, galaxies following the cosmological flow do not change their co-moving coordinate location. Therefore, the co-moving distance between galaxies is constant. Physical distances on the other hand change because of the cosmological expansion.

4 Cosmic Expansion Co-moving distance between A and B: ( x A x B ) t = ( x A x B ) t+ t Physical distance between A and B: ( r A r B ) t ( r A r B )t+ t From homogeneity and isotropy in the universe, we have that: r A (t) r B (t) = a(t) ( x A x B ) Thus, the scale or expansion factor a(t) encapsulates the dynamics of the universe.

5 Evolution of the Scale Factor Einstein Equations: G µν = 8 π T µν that is Geometry = Matter-Energy. Conservation of Matter-Energy: ν T µν = 0 From Einstein s equations (ȧ ) 2 + k a a 2 = 8 π ρ Friedmann equation 3 with ρ the energy density and k the spatial curvature of the universe, respectively. From conservation of matter-energy ρ = 3ȧ (ρ + p) a with p the pressure in the universe. For matter dominated ρ a 3 and for radiation dominated ρ a 4

6 Curvature The second term in the Friedmann equation is a curvature term H 2 + k a 2 = 8 π 3 ρ where H ȧ/a. k = 0 (flat): Flat Euclidean space. k = +1 (closed): Geometry of a three-sphere. k = 1 (open): Geometry of a three-hyperboloid

7 Friedmann Equation Recall If k = 0, then H 2 + k a 2 = 8 π 3 ρ H 2 = 8 π 3 ρ c with ρ c called the critical density Then H 2 + k a 2 = 8 π 3 ρ 1 + k a 2 H 2 = 8 π 3 H 2 ρ 1 + k a 2 H 2 = ρ ρ c = Ω density parameter

8 The Cosmic Microwave Background Radiation

9 Cosmic Microwave Background Anisotropies T T = 10 5 Ω = 1

10 Inflation The universe seems to have emerged from a very special set of initial conditions A set of initial conditions fined tuned to be patially flat Ω = 1 and highly homogeneous/isotropic T /T = 10 5 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.

11 Flatness Problem The density of the universe ρ seems to be finely tuned to be equal to the critical density ρ c, that is Ω = 1. In other words, the universe seems to be fine-tuned to be flat, that is k = 0. How natural are these values? Recall Friedmann equation 1 + k 1 a 2 H 2 = Ω 1 + k 3 a 2 8 πρ c = Ω 1 + k 3 ρ a 2 8 πρ ρ c = Ω k 1 8 π a 2 ρ Ω = Ω (Ω 1 1)ρ a 2 = 3 k 8 π = const

12 Flatness Problem Given (Ω 1 1)ρ a 2 = 3 k 8 π = const Ω ρ 0 a 2 0 = Ω 1 1 ρ a 2 = const Ω = Ω 1 1 ρ ρ 0 a 2 a 2 0 But thus ( a0 ) 3 ρ = ρ 0 a Ω = Ω 1 1 a 0 a

13 Flatness Problem Consider near the Big Bang ( a 0 /a BB = at the Planck epoch) a small deviation of Ω BB from unity; that is, Ω 1 BB 1 = ɛ with ɛ 1. Thus Ω = Ω 1 BB 1 a 0 a BB Ω = ɛ a 0 a BB Ω = ɛ In order to have today a small deviation δ = Ω , we require ɛ = δ

14 Particle Horizon Horizons exist because there is finite amount of time since the Big Bang. The particle horizon is the maximum, finite distance from which particles (or photons) could have traveled to the observer in this time. It represents the boundary between the observable and the unobservable regions of the universe.

15 The particle horizon is calculated from r h (t) = t 0 dt a Therefore, it depends on the scale factor and thus the matter content of the universe.

16 The particle horizon now is much larger than the particle horizon when the CMB photons where emitted. That is, two widely separated parts of the CMB will have non-overlapping horizons. Horizon Problem: How come then we see them at almost the same temperature. We need a much larger particle horizon when CMB photons are emitted to bring the entire visible universe in causal contact.

17 Flatness and Horizon Problems Solution In the very early universe, We need to drive Ω 1 Ω = 1 + k H 2 a 2 We need to increase the way particle horizon grows. r h (t) = t 0 dt a Therefore, we change how H and thus a evolve. Recall that from Friedmann equation with n = 3, 4 H 2 = 8 π 3 ρ a n As the universe expands, H 2 a 2 decays, and thus 1/H 2 a 2 grows (Flatness problem).

18 A possible solution (Guth) is that H = ȧ/a = constant > 0. That is a e H t. Then k Ω = 1 + H 2 e 2 H t 1 and tend dt r h (inflation) = a eh t r h (today) r init physical coordinates (left), co-moving coordinates (right)

19 Scalar Fields and Inflation Consider a Universe filled with a scalar field φ. Ignoring the curvature term, Friedmann equation reads H 2 = 8 π 3 ρ with ρ = 1 2 φ 2 + V (φ) and V (φ) a potential to be determined. The equation for the dynamics of the scalar field is (assuming a homogenous field) φ + 3 H φ + dv dφ = 0 That is, the equations for a and φ are: φ + 3 H φ + dv dφ = 0 H 2 = 8 π ( ) φ 2 + V

20 The onset of Inflation 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, dv dφ which is equivalent to requiring that the potential energy dominates over the kinetic energy (slow-roll approximation;).

21 The end of Inflation At the bottom of the potential V 0 The field relaxes, converting the energy in the inflation potential into a thermalized gas of matter and radiation (reheating)

22 Phase Transitions in the Early Universe First-order phase transition via bubble nucleation (i.e. boiling water)

23 Phase Transitions in the Early Universe Second-order phase transition, the old phase transforms itself into the new phase in a continuous manner.

24 Domain Walls Recall (one-dimensional case) φ + 3 H φ x 2 φ + dv dφ = 0 H 2 = 8 π ( ) φ 2 + V with a Mexican Sombrero potential V (φ) = λ 8 ( φ 2 η 2) 2

25 Domain Walls Consider the static case solution which has the following solution d 2 dx 2 φ = λ 2 φ ( φ 2 η 2) φ(x) = η tanh [ ] λη 2 (x x 0)

26 Domain Walls Domain Wall: the boundary between regions with φ = ±η. The energy at the wall is V (0) = λ 8 η4

27 Project Solve the equation with φ + 3 H φ 2 x φ + dv dφ = 0 H 2 = 8 π ( ) φ 2 + V V (φ) = λ ( φ 2 η 2) 2 8 Impose periodic boundary conditions in a computation domain of length L = 1024 and grid-spacing dx = 1. Set the values of the parameters λ and η to unity. In calculating H 2, use the average values of φ and φ over the entire computational domain. Initial conditions: φ = 0.01 ξ and φ = 0.01 ξ with ξ [ 1, 1] a random number.