PHY 375. Homework 5 solutions. (Due by noon on Thursday, May 31, 2012) (a) Why is there a dipole anisotropy in the Cosmic Microwave Background (CMB)?

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1 PHY 375 (Due by noon on Thursday, May 31, 2012) 1. Answer the following questions. (a) Why is there a dipole anisotropy in the Cosmic Microwave Background (CMB)? Answer is in lecture notes, so won t be repeated here. (b) Differentiate between the epoch of recombination, the epoch of photon decoupling, and the epoch of last scattering. Answer is in lecture notes, so won t be repeated here. (c) Smarty P. Antz tells you that recombination took place when the mean energy per CMB photon fell below the ionization energy of hydrogen (13.6 ev). A quick calculation shows you that this would correspond to a temperature of 60,000 K. Yet we know that recombination took place long afterward, when the Universe had cooled down to 3800 K. Why was this the case (i.e., what is wrong with Smarty P. Antz s reasoning)? Even though the mean energy per CMB photon is below the ionization energy, there are still a large number of photons in the high energy tail of the distribution, more so because there are almost a billion photons for every baryon. This high energy tail of photons continues to break up any hydrogen atoms that form, until the Universe has cooled to a much lower temperature. (d) Read Section 11.3 of Ryden, and state (in a very concise form a few sentences, at most) what is meant by the magnetic monopole problem. Then, explain how inflation takes care of this problem (Section 11.4). Answer is in lecture notes, so won t be repeated here.

2 2. When the Universe was fully ionized, photons interacted primarily with electrons via Thomson scattering, for which the cross-section is σ e m 2. (a) Find the average distance traveled by a photon between collisions, also known as the mean free path: λ mfp 1/n e σ e, at the time of radiation-matter equality when a rm When fully ionized, n e n bary. But n bary 1/a 3, so we get n bary n bary /a 3, assuming a 0 1. Therefore, with a a rm at the time of radiation-matter equality, we get λ mfp 1 ( ) a3 rm n bary /a 3 rm σ n bary σ e e ( ) 3 (0.22 m 3 )( m 2 ) m (b) The Friedmann equation for a radiation-dominated flat universe is H 2 H 2 0 Ω r,0 a 4 Use this to find the Hubble parameter at the time of radiation-matter equality when a rm Take the value of Ω r,0 from the Benchmark model. H rm H 0 Ωr,0 a 2 rm (70 km s 1 Mpc 1 ) ( ) km s 1 Mpc 1 (c) The photons remain coupled to the electrons as long as their mean free path λ mfp is shorter than the Hubble distance c/h. Verify that they are coupled at the time of radiation-matter equality. Converting H rm to SI units, we find c/h rm m. Therefore, clearly ( ) λ rm m < c ( ) m H rm and so the photons are coupled to the electrons at the time of radiation-matter equality. Page 2 of 8

3 3. Given that the Universe is described by the Benchmark model, and the redshift of the last scattering surface is z ls 1100, (a) Find the angular diameter distance to the surface of last scattering. From equation (6.42) in Lecture 11, we know that the horizon distance in the Benchmark model is given by d hor (t 0 ) 3.24 c H 0 14,000 Mpc And, from equation (7.41), we know that as z, the angular distance d A d hor(t 0 ) z Therefore, the angular distance to the surface of last scattering will be d A d hor(t 0 ) z ls 14,000 Mpc Mpc (b) Find the luminosity distance to the surface of last scattering. From equation (7.37) in Lecture 12, we get ) 2 ( 2 d L d A (1+z 13 Mpc ) Mpc (c) Find the proper distance d p (t 0 ) to the surface of last scattering. From equation (3.28) in Lecture 5, we get d p (t 0 ) r Meanwhile from equation (7.36) in Lecture 12, we get d A S κ(r) 1+z r 1+z because S κ (r) r in a flat universe, which is the case here since we re assuming the Benchmark model. Combining the two equations above, we get ) d p (t 0 ) d A (1+z 13 Mpc ( ) ,000 Mpc Page 3 of 8

4 4. Big Bang Nucleosynthesis of 4 He was a race against time to bind neutrons before they decayed to protons. (a) First, prove that the imum possible value of the primordial 4 He fraction is 2f 1+f where f n n /n p 1 is the neutron-proton ratio at the time of nucleosynthesis. First, since f n n /n p, we get n n fn p. Now, we get the imum 4 He fraction if all available n n neutrons bind to protons. Since there are 2 neutrons (and 2 protons) in one 4 He nucleus, this implies that we would get n n /2 nuclei of 4 He. Furthermore, since one 4 He nucleus has a mass of 4m p, where m p is the mass of a proton, n n /2 such nuclei will have a mass of (n n /2)4m p. We are assuming m p m n. And, since we have a total of (n n +n p ) neutrons and protons, their mass will be (n n +n p )m p. From the above considerations, we get (n n/2)4m p (n n +n p )m p 2n n n n +n p 2(fn p) fn p +n p Therefore, canceling common terms and rearranging, we get 2f 1+f (b) Assuming that the neutron-proton ratio remained constant at f n n /n p 1/5 after freezeout, and that all available neutrons were incorporated into 4 He, find the value of. For f 1/5, we get 2f 1+f 2(1/5) 1+1/5 2/5 6/ Therefore, we get Page 4 of

5 (c) In reality, was less than this value because some of the free neutrons decayed into protons, thereby decreasing the number of neutrons available to combine with protons to form 4 He. Show that if nucleosynthesis starts after a time delay of t nuc, neutron decay makes the neutron-to-proton ratio decrease from its freeze-out value of n n /n p 1/5 to n nf where τ n is the decay time of the neutron. exp ( t nuc /τ n ) [ 5+ 1 exp ( t nuc /τ n ) At freeze-out, we have n ni /n pi 1/5, so n pi 5n ni. Given thatifyoustartoutwithapopulationoffreeneutronsn ni, thenumberoffreeneutrons remaining after time t nuc will be ( n nf n ni exp t ) nuc τ n The remaining neutrons become protons, so the number of protons after time t nuc is n pi + [n ni n nf Putting n pi 5n ni and the expression for n nf into this expression, we get [ 5n ni + n ni n ni exp( t nuc /τ n ) and this may be written as n ni {5+ [ } 1 exp( t nuc /τ n ) From the expressions above, we obtain n nf n ni exp( t nuc /τ n ) [ } n ni {5+ 1 exp( t nuc /τ n ) Therefore, we get the desired expression: n nf exp ( t nuc /τ n ) [ 5+ 1 exp ( t nuc /τ n ) Page 5 of 8

6 (d) As a result of the process described in part (c), find the neutron-proton ratio f new n nf /, and the corresponding value of, if nucleosynthesis starts after a delay of t nuc 200 s, and the decay time of the neutron is τ n 890 s. Assume that all available neutrons (that haven t decayed) are incorporated into 4 He nuclei. Given t nuc 200 s, and τ n 890s, we get f new n nf exp ( 200/890) [ exp ( 200/890) and New 2f new 1+f new 2(0.15) (e) Suppose, instead, that the neutron decay time were τ n 89 s, with all other physical parameters unchanged (including that we still have t nuc 200 s). Calculate, again assuming that all available neutrons are incorporated into 4 He nuclei. With t nuc 200 s, and a smaller τ n 89s, we get f new n nf exp ( 200/89) [ exp ( 200/89) and New 2f new 1+f new 2(0.018) This makes sense the neutrons decay faster, so there are fewer neutrons left to form 4 He, so the imum 4 He fraction is smaller. Continued on the next page... Page 6 of 8

7 5. Consider the inflaton field V(φ) Aφ 4 Equation (11.53) can be used to find when the slow roll condition breaks down by setting ( EP V ) 2 dv 1 where E P ( c 5 /G) 1/2 is the Planck energy. (a) Use the limiting condition given above to find the value of φ at which the slow roll conditions break down; this marks the end of the period of inflation. Leave your answer as an expression, don t substitute numerical values. Setting dv 4Aφ3 in the expression we get E 2 P V 2 ( ) 2 dv 1 c 5 /G ( ) 2 4Aφ 3 1 (Aφ 4 ) 2 which works out to c 5 (16)φ 6 Gφ 8 1 or Therefore, we get 16 c 5 G φ8 φ 6 ( 16 c 5 φ G ) 1/2 We can either leave it in this form, or simplify by inserting E P ( c 5 /G) 1/2 to get ( 16 c 5) 1/2 φ 4E P G Page 7 of 8

8 (b) Find the number of e-foldings N. Equation (11.48): 3H φ c 3dV with dv/ 4Aφ 3 gives 3H c 3 (4Aφ 3 ) dt Then, we get N ln [ a(tf ) tf H(t)dt a(t i ) t i φf H [ 3H c 3 (4Aφ 3 ) where we have put H (8πGV/3c 2 ) 1/2. φf 3H 2 4A c 3 φ 3 φf Putting V Aφ 4, and canceling terms we get N 2πG c 5 φf 3 4A c 3 φ 3 φ 2πG c 5 [ 8πGV 3c 2 [ φ 2 2 φf Therefore N πg [ φ 2 c 5 f φ2 i We can leave it in this form, or use the fact that inflation ends of φ f 4E P, together with E 2 P c5 /G and write this as N π E 2 P [ [ φ φ 2 i φ 2 2 f π i EP 2 16 Page 8 of 8

Lecture notes 20: Inflation

Lecture notes 20: Inflation The observed galaxies, quasars and supernovae, as well as observations of intergalactic absorption lines, tell us about the state of the universe during the period where z