( ) Note 4 19 11 22 6 6.1 1 Ω m = 1 Ω m.3 6.1.1 : ( ) r-process α 1: 2 32T h(t 1/2 = 1.45 1 1 y) 2 38U(t 1/2 = 4.468 1 9 y) 2 35U(t 1/2 = 7.38 1 8 y) 2 44Pu(t 1/2 = 8.26 1 7 y) β / (J.A.Johnson and M.Bolte: APJ 554 (21) 888) 11.8 Gy t G 19.8 Gy (Gy = 1 9 y) 14 Gyr HR 13 ± 3 ( 1 ) 1
1: ( ) ( ) ( ) 4 13±3 (R.Jimenez and P.Padoan: APJ 498 (1998) 74, E. Carretta et al., :APJ 533 (2) 215) 2: (M15) - ( ( )- ) ( ) ( ) ( Gyr) M.Salaris et al.,astrophys.j.479 (1997):665E672 6.1.2 (ȧ ) 2 H 2 = = 8π kc2 Gρ a 3c2 a 2 + Λc2 3 ρ = ρ m + ρ r + ρ Λ, ρ Λ = Λc4 8πG (1) t = (z = ) H 2 Ω = ρ/ρ c, ρ c = 3H2 c2 8πG Ω k k a 2 H2 t = Z t dt = Z a da da dt = = 1 Ω m Ω r Ω Λ (2a) Z a Z da ah = dz z (1 + z)h 1 + z = a /a ρ m a 3, ρ m = ρ m (a /a) 3 (1) H 2 = 8πG [ ( a ) ] 3 3c 2 ρ m + ρλ k [ { 8πG ( a a 2 = a ) } 3 H2 ρ m + ρλ a H = H [ Ωm (1 + z) 3 + Ω Λ + Ω k (1 + z) 2] 1/2 3H 2 c2 k a 2 H2 ( a ) ] 2 a (3) (4a) (4b) (3) t = 1 H Z z dz (1 + z)[ω m (1 + z) 3 + Ω Λ + Ω k (1 + z) 2 ] 1/2 (5) 2
z 3 Ω Λ = Ω k = Ω Λ = 3: ( ) Ω m = Ω m = 1, Ω Λ = Ω m > 1, Ω Λ = Ω m.26,ω Λ.74, Ω k ( ) Ω m w { (13)} w=-1 WMAP === 2 1 1 3 H (1 + z) 3/2, t Ω m === 1 1 H 1 + z t Ω m=1 (6) z= t = 1 H 132 Ω k = 1, Ω Λ = Ω m = (7a) t = 2 1 88 Ω k = Ω Λ =, Ω m = 1 (7b) 3 H Ω k, Ω Λ t = 2 [ ] 1 1 + ΩΛ Ω log Λ =.74 === 1.4H 3 H ΩΛ 1 1 = 137 (8) ΩΛ 6.1. Ω k = (5) [ t = 2 { }] 1 ΩΛ log (1 + z) 3/2 + (1 + z) 3 H ΩΛ Ω 3 + Ω m m Ω Λ (9) 3
( ) 12 15 Ω m = 1 t 2/3H 9 1 6.2 Ia Ia 199 (z 1 Ia (Si) ( 4 ) ( 4 ) ( 5 ) II I Ia 4: Ia ) SNIa Si Ia CCD 3 2 ( 5) 199 The Supernova Cosmology Project The Hi-z Supernova Search Team 2 z 1 Ia * 1) * 2) (HST= Hubble Space Telescope) 5 HST ( 6) * 1) The Hi-z Supernova Search Team; A.G.Riess et al., ; Astronom. J. 116 (1998) 19-138, The Supernova Cosmology Project; S.Perlmutter et al.,; Astrophys. J. 517 (1999) 565-586 * 2) 4
図 5: (a) Ia 型超新星は白色矮星が発達した伴星を持ち 伴星からの物質降着が炭素の爆発的燃焼を引き起こす 後には何も残らない (b) 超新星探索法: CCD カメラで日を置いて撮影し 最初の映像 (3 週間前) から現在の映像を引き算する コンピューター処理による大量処理 が可能である なお 地上観測より衛星観測の方が遙かに鮮明な写真が得られる 右上に同じ映像をハッブル望遠鏡で撮影した図を示す (http://www-supernova.lbl.gov/public/figures/bigcomposite.jpg) 図 6: 加速膨張の証拠: 見かけの明るさを距離指数 (µ = m M 式 (36) 図 12 参照) で表し z の関数としてプロット 挿入図は 等速一様 膨張 (ΩM = ) と仮定したときの距離指数との差 点線より上が見かけ上暗い加速膨張を 点線から下が見かけ上明るい減速膨張を示す 超 新星は 赤方遷移ｚの関数として 近傍 (z. 1) で暗く遠くで明るい すなわち 宇宙初期は減速膨張 途中から加速膨張に転じた 挿入図 一番上の線は 当初疑われた星間塵の影響曲線 A.G.Riess et al. The Astrophys. J. 659 (27) 98-121 図 6 は 超新星光度の見かけの明るさを距離指数 [µ = (m M) 式 (36) 参照] で表したデータを さらに等速膨張 (ハッブル定数が定数) の時の値を差し引いて減速から加速への変化を見やすくした図を挿入図として入れてある 5
6.3 ä a = 4πG Λc2 (ρ + 3P) + 3c2 3 ä = G M ( a 2 1 + 3 P ) + Λc2 ρ 3 a, M = 4π ( ρ ) 3 a3 c 2 1 * 3) (11) * 4) 6.2. (P vac = ρ vac ) (1) (11) ( 4 ) (1) P = wρ, w < 1 3 ( w = 1) (13) 7: SNIa( ) CMBR( ) Ω Λ Ω m (Ω m + Ω Λ + Ω k = 1) D.N.Spergeletal.,; Astrophys.J.S17(27)377 48 * 3) * 4) CMBR (Ω = 1) Ω M.3 WMAP 22 Ω m + Ω Λ + Ω k = 1 7 H 2 = 8π kc2 Gρ 3c2 a 2 + Λc2 (12) 3 (1) Λ ρ Λ = Λc 2 /8πG P Λ = ρ Λ 6
6.4 12 (fine-tuning problem) (E Pl = G 1/2 1 19 GeV ) E 4 ρ V (predicted) M 4 Pl = 176 GeV 4 (14) ρ V (observed).75 2 1 29 h 2 g c 2 cm 3 = (3meV ) 4 1 47 GeV 4 (15) 12 14 12 (SUSY) SUSY 1GeV 5-6 Ω Λ = Why now (coincidence problem) * 5) a 3 a 4 const. ρ m ρ Λ 12 why now ( ) (Quintessence) * 6) w = p Q = Q 2 /2 V (Q) 1 Q 2 << V (Q) ρ Q Q 2 (16) /2 +V (Q) +1 Q 2 >> V (Q) w w < ( Q 2 >> V (Q)) ( Q 2 << V (Q)) ( 8 ) (V = V e λκq ) * 7) (attractor ) ( 8 ) attractor Q Q m Q = V (Q) H = 1 33 ev * 8) * 5) z Ω m.3(1+z) 3, Ω Λ =.7 = const. Ω Λ = Ω m z.3 z >> 3 Ω Λ /Ω m <<.1 T E Planck 1 19 GeV z 1 32 (T.1MeV ) z 1 9 * 6) 4 5 5 * 7) 1 Q * 8) 7
8: ( ) Q ( ) (attractor) ( ) attractor ( ) T.Barreiro, E.J.Copeland and N.J.Nunes; Phys. ReV. D61 (2) 12731 2 Q (3 mev) SUSY (sneutrino) * 9) Quintessence K-essence Tachyon field Dilaton field Chaplygin Gas Phantom (w < 1) w a w(a) = w + w 1 z, or w(a) = w + w a (1 a/a ) = w + w a z 1 + z (17) 6.5 P = wρ, (w = ) ( ) ρ X dρ ρ + P + 3da a = (18) ρ X a 3(1+w) (19) w = 1/3,, 1 * 1) (23) d L = a r(1 + z) Ω k= === 1 + z H Z z cdz [Ω m (1 + z) 3 + Ω X (1 + z) 3(1+w) ] 1/2 (2) * 9) E.J.Copeland, M.Sami and S. Tsujikwa Int. J. Mod. Phys. D15: 1753-936 (26) [hep-th/6357]. * 1) 6.6 8
w = 1, Ω X Ω Λ Ia GRB (Gamma Ray Burst) (LSS=Large Scale Structure) w 9 WMAP3 9 1 CMBR LSS z 2 1 GRB z 6 z w = 1 9: ( ) ( )WMAP,3 ( ) w =.926 +.51.75 ApJS 17 27 377-48 1: ( ) CMB LSS z 1σ 1,2,3 1 (P = ρ w ) w (z.5) (Y.Wang and M.Tegmark; Phys. Rev. Lett. 92 (24) 24132) ( ) GRB (Gamma Ray Burst) z 7 z Ω m =.369 Λ CDM Ω m =.416, Ω tot = 1.115 Ω m = 1 (F.Wright; astro-ph/71584) 9
6.6 z=4 1/5 5 r (luminosity disatance) I d L L L = 4πd 2 LI (21) d L r t t + t (r=) t t + δt r = t = t 4π(a r) 2 n γ T 3 a 3 E γ /hν ν/ν = δt /δt = 1 + z (21) Lδt hν = 4πa2 r 2 Iδt hν (22) d L = a r(1 + z) (23) r (1+z) (angular diameter distance) D δθ d A = D/δθ d A = d L r s = rδθ t D = a(t)s d A = D δθ = a(t)rδθ δθ r = a(t)r = a r 1 + z = d L (1 + z) 2 (24) 6.6.1 z r z r r l ( = ) dl = dr 1 kr 2 ds 2 = dl dt [dl = cdt/a(t)] a z t= r t r= l = Z r Z dr t = cdt Z a 1 kr 2 a(t ) = (4b) === 1 a H Z z (25) cda Z z a(t) a 2 H(a ) = cdz H(z (26) ) cdz [Ω m (1 + z) 3 + Ω Λ + Ω k (1 + z) 2 ] 1/2 Ω k = 1 Ω m Ω Λ (27) 1
l z r z (k= r = l) z l = cz [ ( 1 1 a H 2 + Ω m 4 Ω ) ] Λ z + O(z 2 ) = cz 2 a H d L = a r(1 + z) = cz [ ( ) ] 1 q 1 + z + O(z 2 ) a H 2 r d A = a 1 + z = cz [ ( ) ] 3 + q 1 z + O(z 2 ) a H 2 [ 1 ( 1 + q 2 ) ] z + O(z 2 ) (28a) (28b) (28c) z d L d A c H z + O(z 2 ) (29) q ä/ȧ ȧ/a = ä a 2 H (3) ä a = 4πG Λc2 (ρ + 3P) + 3c2 3 ( q = 1 + 3P ) Ωm ρ m 2 Ω Λ (31) q = Ω m 2 Ω Λ (32) z 2 r = l r l Z r l = dr 1 kr 2 c sinh[(h H a Ωk /c) Ω k a l] Ω k > = r = l Ω k = c sin[(h H a Ωk /c) Ω k a l] Ω k < r Ω k = kc 2 /(a H ) 2 k (33) r = l[1+o(l 2 )] O(z 3 ) z (28b)(28c) 11 z z 1 z (33) I L m M L d L m = 2.5log 1 I I = 2.5log 1 L + 5log 1 d L +C I = 2.53 1 8 Wm 2 * 11) (34) 11
11: d A d L z Ω Ω m λ Ω Λ (http://www-utap.phys.s.utokyo.ac.jp/ suto/) d L = 1pc M = 2.5log 1 L L L = 78.7L, M = 4.74 (35) L 1pc I m M = 5log 1 d L 1pc (distance modulus) (36) m M (m-m) d L 12 (m-m) z 12: (m-m) z * 11) α ( ) 12
6.7 6.1: k= t = t(z) y = ΩΛ Ω m (1 + z) 3/2 t = 1 H Z z dz (1 + z)[ω Λ + Ω m (1 + z) 3 ] 1/2 (37) Z Ω Λ Ωm (1+z) 3/2 t = 2 1 dy 3 H = 2 1 [ log(y + ] Ω Λ y ΩΛ 1 + y 2 3 2 Ωm + 1) (1+z) 3/2 H ΩΛ [ = 2 { }] 1 ΩΛ log (1 + z) 3/2 + (1 + z) 3 H ΩΛ Ω 3 + Ω m m Ω Λ (38a) (38b) 6.2: S V E = ρ V V F x E = ρ V xs ( ) P P = F S = 1 E S x = ρ V (39) 13: 2 1 du = PdV, U = ρv dρ dv P = ρ = (ρ + P) (4) 13