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1 Special Topics in Nuclear and Particle Physics Astroparticle Physics Lecture 12 Cosmology II Nov. 24, 2015 Sun Kee Kim Seoul National University

2 ( ) ( ) ( 3 ) : (38 )

3 ( Big Bang Nucleosynthesis) 1948 Ralph Alpher, Hans Bethe, George Gamow - A=5, 8 Li

4 ( Big Bang Nucleosynthesis) t :1s ~ 3min kt < MeV : D, He-3, He-4, Li-7(Be-7) ( ) : Y = m He n He m N (n n + n p )

5 ( Big Bang Nucleosynthesis) p + n d + γ Li-6 d + 4 He 6 Li + γ 3 He-3 d + p He + γ d + d 3 He + n H-3 n + d 3 H + γ d + d 3 H + p n + 3 He 3 H + p He-4 n + 3 He 4 He + γ d + 3 He 4 He + p p + 3 H 4 He + γ d + 3 H 4 He + n Be-7 p + 6 Li 7 Be + γ 7 Be + e 7 Li +ν : 53.2d Li-7 n + 7 Be 7 Li + p Li p + 6 Li 4 He + 3 He p + 7 Li 4 He + 4 He p + 7 Li 8 Be( 4 He + 4 He) + γ

6 / ( Big Bang Nucleosynthesis) ν e + n p + e e + + n p +ν e n n n p e (m n m p )/T m n m p = MeV : σ G F 2 T 2 ( ): n T 3 H = 1.66 g * T 2 m pl : Γ = n συ G F 2 T 5 g * = N ν

7 ( Big Bang Nucleosynthesis) / T 2 Γ = n συ G 2 F T 5 H = 1.66 g * m pl Γ < H. Tf~0.8 MeV Tf~0.8 MeV / n n n p e 1.3/

8 ( Big Bang Nucleosynthesis) (Deutron bottleneck) p + n d + γ d 2.2 MeV (photo disintegration). γ + d p + n 2.2 MeV. n ( E > 2.2MeV) ~ 9 γ b n γ n ~ 10 d 0.1 MeV n n n p e (m n m p )/T e 180/ / 7

9 ( Big Bang Nucleosynthesis). p + n d + γ 3 d + p He + γ 3 4 d + He He + / n n n p e (m n m p )/T e 180/τ n 1/ 7 n He4 = n n / 2 p Y = m He n He m N (n n + n p ) = 4m n / 2 N n m N (n n + n p ) = 2n n / n p (1+ n n / n p ) 0.25 ~0.06

10 ( Big Ban Nucleosynthesis) / BBN / 2014 PDG 2014 : ±1.1 MeV/c 2

11 ( Big Ban Nucleosynthesis)

12 (CMB) BBN baryon pdg2014

13 Damped Lyaman Alpha(DLA) system Quasar (QSO) : QSO Lyα QSO A " /

14 Damped Lyaman Alpha(DLA) system

15 Damped Lyaman Alpha(DLA) system 5 precision sample

16 ( Big Ban Nucleosynthesis) 7 7 g * = gi + g j = 2 + (4 + 2Nv ) 8 8 bosons fermions H 2 T = 1.66 g * à freeze-out temperature m pl n n ( m m ) / T n p n p f = e à affect the 4 He abundance N ν = 3.27 ± 0.24 R.H.Cyburt, et al., Astropart. Phys. 23, 313 (2005)

17 ( Big Ban Nucleosynthesis) Lithium - problem? 5 sigma /???

18 Baryon asymmetry baryon anti-baryon n b n b n γ T 3 anti-baryon. A b = n n b b n + n b b = n b n 2n b ~ n 2 b γ n γ 0 ~ 10 9 η b = n b n γ ~ 10 9 Sakharov condition (1967) -Baryon number violating processes -C and CP violation -Out of thermal equilibrium

19 Antiproton/proton ratio (cosmic ray) AMS collab

20 20

21 (Recombination) (CMB) à à p + e H + γ binding energy = 13.6 ev n ( T > 13.6eV) ~ n ~ 10 At T ~ ev ( 3000K ) 9 γ b n γ R 0 R = λ 0 λ = 1+ z = T T 0 = = R rec t 0 t rec R 0 3/2 t rec = t 0 R rec R 0 3/2 = /2 380,000years

22 (CMB) Thomson ( ) (horizon) à 1+z ~ 1100 and t dec ~ 380,000 years 380,000.

23 (CMB) comoving radial coordinate re te te+δte. t0 t0+δt0. ds=0 geodesic. RW metric ds 2 = 0 = c 2 dt 2 R(t) 2 t 0 t e dt R(t) t 0 +Δt 0 t e +Δt e = 1 c dt R(t) 0 dr 1 kr 2 r e = 1 c 0 dr 1 kr 2 r e dr 1 kr 2 re comoving. t 0 +Δt 0 t e +Δt e dt R(t) t 0 dt = 0 t e R(t)

24 (CMB) t 0 +Δt 0 t e +Δt e dt R(t) t 0 dt = 0 t e R(t) t 0 +Δt 0 t 0 = te t e +Δt e t e +Δt e t 0 +Δt 0 + te t0 t e +Δt e dt = t e R(t) t 0 +Δt 0 t 0 dt R(t) ( ) R(t) Δt e R(t e ) = Δt 0 R(t 0 ) Δt = 1 ν = λ c Δt 0 Δt e = λ 0 λ e = R(t 0) R(t e ) = 1+ z z : cosmological red shift

25 (CMB) 1965 : Penzias & Wilson : 3K microwave background 1992 : COBE full spectrum : T=2.725 K Planck, angular resolution ~ 7 o : MAXIMA, BOOMERANG(Balloon), DASI(ground) : Anisotropy 2003 : WMAP CMB T(θ,φ) direction dependence, angular resolution ~0.2 o 2013 : Planck CMB,, angular resolution ~5

26 (CMB) (Planck HFI ) Focal plane Bolometer 1micron silicon nitride + NTD Ge sensor

27 (CMB) n γ (ν)dv = 8π 2 c ν 2 dν e hν /kt 1 T 0 = ± K COBE (2009)

28 (CMB) T 0 = ± K ρ r0 = π 2 30 g* T 4 0 = ev/m 3 Ω r = ρ r0 / ρ c = h 2 N γ = g 2π T = / m 3 = 411/ cm 3

29 (CMB)

30 (CMB) : 10 1 δt T ~ 10 5, : 2? (horizon problem)

31 (particle horizon) (Particle Horizon) Particle horizon :, (causal contact) t=0 t=t t=t ls d H (t ls ) cdt t=0 LS(t=t ls ) t=t cdt cdt R(t cd t ls ) R( t ) d H (t ls ) = R(t ls ) t ls 0 cd t R( t )

32 (Particle Horizon) d H (t ls ) = R(t ls ) t ls 0 cd t R( t ) R t 1/2 d H = 2ct R t 2/3 d H = 3ct LSS d H (t ls ) ~ 3ct ls LSS proper distance t 0 cd t d p (t ls ) = R(t 0 ) d p (t ls ) ~ 3ct 0 t ls R( t ) LSS LSS d p (t ls ) R(t ls ) R(t 0 ) θ = d H (t ls ) d p (t ls ) R(t 0 ) R(t ls ) = d (t ) H ls d p (t ls ) (1+ z ) = t ls ls (1+ z ls ) ~ 1.7! t 0

33 (CMB) 10 1 δt T ~ 10 5 T ( θ, φ) = a Y lm l l= 0 m= l lm ( θ, φ) higher l à smaller structure l =0 : monopole, l=1 : dipole,... * a lm = Y lm (θ,φ)t (θ,φ)dω C l 1 represents the level of structure 2 l = a lm found at 2l + 1 m= l Δθ ~ o 180 C 0 : average temperature over all direction < T >= ± K l

34 (CMB) (multipole) l l. θ λ = 2π l = 360! l l, θ res = π l = 180! l. ) 7 (COBE) : l =26, 0.23 (WMAP) : l =783

35 (CMB) l=1 dipole anisotropy ΔT T ~ cosθ Milky way galaxy is moving with v=550 km/s relative to the local cluster. Local cluster is moving with v=630 km/s relative to the super cluster vector sum of the velocity : 369 km/s

36 (CMB) (Acoustic peaks) ( ) + + : CMB - P = 1 3 ρc2 c s = dp dρ = c 3

37 (CMB) (Acoustic peak) λ n = λ 1 / n Fourier mode ~ t ls c λ 1 = R(t ls ) s d t = R(t ls ) 0 R( t ) 3 R t 2/3 t ls 0 cd t R( t ) λ 1 d H (t ls ) 3, α = λ 1 d p (t ls ) (1+ z ls ) ~ 1! l ~ 200

38 (CMB) (Acoustic peak) Ω>1, θ l Ω<1, θ l Ω0 = 1.03 ± 0.03

39 (CMB) Sacks-Wolfe ( ). CMB. Sacks-Wolfe.. Intgrated Sacks-Wolfe (ISW).. ISW CMB hot spot. Damping tale damping

40 (CMB)

41 (CMB) CMB only CMB + BAO PDG 2014 from S. Serjeant (2010)

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