1 Problem set 6 for Cosmology (ns-tp430m) Problems are due at Thu Apr 4. 14. Units and the age of the Universe. (8 points) 1 parsec (1pc) is defined as the distance from which 1 astronomical unit (1 AU) subtends an angle of 1 arc second (1′′). 1 AU is defined as the mean distance between the Earth and the Sun, 1 AU 1.5 108 km (more precisely 149 597 870.691 kilometers). ≃ × (a) (1 point) Express 1pc in terms of kilometers and light years. (b) (2 points) The Hubble parameter today is H0 = 72 3 km/s/Mpc. Express H0 in inverse ± −42 seconds, and in giga-electron volts (GeV). Show that ~H0 = 2.1332h 10 GeV, where h is × defined as, H0 = 100h km/s/Mpc, such that h = 0.72 0.03. (Often we shall work in units ± in which the reduced Planck constant ~ = 1.) (c) (5 points) Estimate the age of the Universe as t0 1/H0 (express it in giga-years, Gy). ≃ In order to find out a more precise value of the age of the Universe, solve the Friedmann equation, a˙ 2 8πG Λ H2 = = N ρ + (1) a2 3c2 3 and show by integrating equation (1) that the scale factor reads 3 8πGN 2 √3Λ t a (t)= 2 ρ0 sinh , (2) c Λ 2 3 where a(0) = 0, a(t0) = 1, ρ(t)= ρ0/a(t) (nonrelativistic matter). By inverting this expres- sion, show that the age of the Universe can be written as, 1 1 1+ √ΩΛ t0 = ln , (3) 3H0 √ΩΛ 1 √ΩΛ − 2 where ΩΛ = Λ/(3H ). Do you get the expected result in the limit when ΩΛ 0? What is 0 → t0 for ΩΛ = 0.73 0.04 (WMAP value)? How does your result compare with the age of the ± Universe quoted by the WMAP team, t0 = 13.7 0.2 Gy? ± 15. Particle Horizon. (4 points + 3 bonus points) Particle horizon is defined as the distance the light travels from some initial time to today. (a) (2 points) 2 Show that in the FLRW space-time with the metric metric, g = diag(1, a2, a2, a2), (4) µν − − − particle horizon, when expressed in physical units, equals t cdt′ ℓphys(t)= a(t) ′ . (5) Z0 a(t ) In comoving coordinates, particle horizon (when expressed in comoving coordinates) is simply, t ′ ′ ℓ = ℓphys/a(t)= 0 cdt /a(t ). R Show also that, when the metric is expressed in terms of conformal time, the (comoving) particle horizon is simply, ℓ(t)= c(η(t) η(0)) . (6) − (b) (2 points) If the particle horizon at the photon-electron decoupling is ℓ = rs = 147 Mpc (WMAP result) (the redshift at decoupling equals z = 1089), what is particle horizon (a) at the time of the +194 matter-radiation equality, when z = 3233 −210 , (b) today (at z = 0). For simplicity assume that the Universe is matter dominated, a t2/3. ∝ (c∗) (3 bonus points) Repeat the calculation from 15.(b) by calculating in a Universe dominated by a cosmological term Λ and nonrelativistic matter. Evaluate particle horizon at zeq = 3233 and at z = 0. For your numerical estimates, take ΩΛ = 0.74, and Ωm = 0.26 and radiation density, Ωγ = Ωm/(1+ zeq). You may perform approximate calculations, so your answers need not be exact, yet they should represent a good approximation to the correct answers..