Sound waves before recombination 25 Feb 2010

Transcription

1 Sound waves before recombinaion 25 Feb 2010 ü Physical condiions a recombinaion ü A recombinaion, which has he greaer mass densiy, pressureless maer or radiaion? W m0 = 0.26 pressureless maer, mosly dark maer, maer ha does no inerac wih ligh W b0 = baryons, ordinary maer W r0 = 1.2μ10-5 radiaion r b =r b0 a -3 r r =r r0 a -4 r m ê r r =r m0 ê r r0 a A a = , he mass-energy densiy of pressureless maer and radiaion are equal. Q: A recombinaion, which has greaer mass densiy, pressureless maer or radiaion? We will discuss sound waves, which has o do wih radiaion and maer. Q: Does dark maer paricipae in he sound waves? A a eq = (z = 3600), he mass-energy densiy of baryonic maer and radiaion are equal. Q: A recombinaion, are elecrons pressureless? The energy of a CBR phoon is 2.3μ10-4 ev. ü A recombinaion, which has greaer number densiy, baryonic maer or radiaion? A he presen ime, he mass of baryonic maer is 938MeV. The mass of a phoon is 2.73 K êh K êev = 2.3μ10-4 ev. The number densiy n r ên b = μ938 MeVê2.3μ10-4 ev = 1.1μ10 9. More precisely, becuase phoons have differen energies, I need o inegrae he Planck number specrum. n r = 0.41μ10 9 phoon m -3 HT ê2.725 K 3 n b = 0.25 nucleon m -3 HW b0 ê.043hh 0 ê72 kmêsêmpc 2 n r ên b = 1.64μ10 9. The number of phoons and baryons do no change. As he universe expands, he number of baryons in a coexpanding box does no change. The number of baryons enering mus equal he number exiing, because of homogeneiy. Same argumen is rue for phoons. Q: The number of phoons and baryons in his room do change when I urn on he ligh. In wha sense is he previous saemen correc?

2 2 W04RWMeric.nb Sound speed before recombinaion The emperaure a recombinaion is 3000K There are many phoons for every baryon or elecron. n r ên b = 1.64μ10 9 A a eq = (z = 3600), he mass-energy densiy of baryonic maer and radiaion are equal. The speed of sound v s = J dp d r N1ê2 where he derivaive is for adiabaic changes. Proof: Newon's 2nd law F = ma deermines he movemen of a sound wave, The force is due o an excess pressure. The mass is due o an excess densiy. Consider a slab of gas beween x and x + dx. Because of he presence of he disurbance, x moves o x +ch, x. The ma erm becomes Hr 0 dx 2 c. 2 The force comes from he difference in pressure. The force erm is - P dx. x I need o relae pressure o mass densiy: P =- P r c r 0 x Collec all; cancel r 0 and d x: 2 c 2 = P r 2 c x 2 The speed of sound is v s = J dp d r N1ê2 The derivaive is aken wih no hea flow, if he wavelengh is large compared o he mean-free pah. Q: Jus before recombinaion, does dark maer paricipae in he sound waves? ü Values ρcriical = *^-29 Gram ë HCeni Meer Gram Ceni 3 Meer 3 ConverAH `*^-33 Kilogram ë IKelvin 4 Meer 3 MH2.73 Kelvin 4, Gram ë HCeni Meer 3 E Gram Ceni 3 Meer 3 % ê %%

3 W04RWMeric.nb 3 1.2*^-5 ê ê % *^-5 ê ê %

4 4 W04RWMeric.nb Calculaion of he sound speed è The composiion of he gas is ordinary maer and phoons. è Q: Does maer or radiaion provide more pressure? The answer is deailed, bu i is based on a principle. Wha principle is he basis for calculaing he answer? è Pressure P = np x v x, where n is number densiy, p and v are momenum and speed. For maer, P = nmv 2 x = 1 3 nmv2. For radiaion, P = 1 ne. 3 è Equipariion: In hermal equilibrium, he energy of each paricle is he same. (More precisely, he energy of each degree of freedom is 1 kt. In QM, degrees of freedom may be frozen wih 0 energy.) 2 è Maer: P = 2 n 3 kt= nkt. 3 2 P = u, where u is he energy densiy. è Radiaion: P º 1 n 1 kt. More accuraely, P = 0.90 nkt. 3 2 P = 1 3 u. è There are 10 9 phoons for every baryon. è Radiaion also dominaes he energy densiy. (No he case if he res mass densiy is added in.) Consider a box of gas wih a fixed number of paricles. The box expands or shrinks because of he sound wave. 1. du= dq-pdv=-pdv Because here is no hea flow, dq= 0. du=-pdv. Recall u = a B T 4 and P = 1 a 3 B T 4 d HuV = Vdu+ udv =-PdV du=-hu + P dvêv 4 T 3 dt=-jt T4 N dvêv 3 dtêt =-dvêv 2. dp= 4 a 3 B T 3 dt. 3. d r=d r b + d r r d r b =-r b dvêv, since mass of he baryons in he box (r b V) is unchanged. d r b = 3 r b dtêt d r r = 4 a B T 3 dt 4. Gaher all: v 2 s = P = J 4 a r 3 B T 3 dtni3 r b dtêt + 4 a B T 3 dtm -1 = J r b N -1 r r v s H1 + RD -1ê2, where R = 3 r b 4 r r Q: If R ` 1, how fas do sound waves ravel? Q: Why do baryons slow he speed of sound? Recall v s = J dp d r N1ê2.

5 W04RWMeric.nb 5 Number densiy of phoons: InegraeAx 2 ë H x 1, 8x, 0, <E 2 Zea@3D Energy densiy: InegraeAx 3 ë H x 1, 8x, 0, <E π 4 15 Average energy: % ê %% êê N XE\ is % ê

6 6 W04RWMeric.nb Calculaion of he horizon, sep 1 How far does a sound wave ravels from he big bang (=0) o he ime of recombinaion? e a be he expansion parameer a las scaering (recombinaion) a E be he expansion parameer a epoch when r r = r b. The disance of he horizon is d = Ÿ v s d. v s d is how far he sound wave moves. As he wave is moving, he saring poin is moving oo. Calculaion is wrong. Beer posed: Wha is he comoving coordinae r of a sound wave ha ravels from he big bang (=0) o he epoch of recombinaion? v s d= adr r = Ÿ 0 v s a -1 d. Q: Given r, how do you ge he disance of he horizon?

7 W04RWMeric.nb 7 Calculaion of he horizon, sep 2 How far does a sound wave ravels from he big bang o he epoch of recombinaion? e a be he expansion parameer a las scaering (recombinaion) a E be he expansion parameer a epoch when r r = r b. The disance of he horizon is d = Ÿ v s d. v s d is how far he sound wave moves. As he wave is moving, he saring poin is moving oo. Calculaion is wrong. Beer posed: Wha is he comoving coordinae r of a sound wave ha ravels from he big bang o he epoch of Çrecombinaion? v s d= adr r = Ÿ 0 v s a -1 d. Q: Given r, how do you ge he disance of he horizon? d = a r d = a Ÿ 0 v s a -1 d. The sound speed v s H1 + RD -1ê2 depends on R = 3 4 r b r r = 3 4 J a a E N. ü To gain undersanding, consider his simplified case: R ` 1. Then d = a v s Ÿ 0 a -1 d Use Friedman's equaion J da H 0 d N2 = IW k0 +W m0 a -1 +W r0 a -2 +W v0 a 2 M ØW r0 a -2 d = a v s Ÿ 0 a -1 d = H -1-1ê2 a 0 a v s W r0 Ÿ0 da = H -1 0 a 2-1ê2 v s W r0 Apply F's eqn a a HHa = H 0 W 1ê2-2 r0 a o ge he ransparen resul d = H -1 Ha v s Q: Inerpre he formula for d. A dense region produces a sound wave ha goes in all direcions o cover a lengh 2 d. The angle subended is q= 2 d ra = 2 H-1 Ha v s a rha Q: Inerpre he formula for q.

8 8 W04RWMeric.nb ü Resuls for bes cosmological values d = a Ÿ 0 a -1 v s Ha d Change d= H -1 a -1 da, and inegrae o ge (Weinberg 2008, Cosmology, p. 145) d = 2 H -1 0 a -3ê2 H3 R W m0-1ê2 ln 9AH1 + R 1ê2 + HR E + R 1ê2 EëI1 +, R E M= For W m0 = 0.26, W v0 = 0.74, W b0 = 0.043, R = 0.62 R E = 0.21 d = 1.16 H -1 3ê2 0 a q= 2 d rha a = 1ê48 = 1.2