.26 QUIZ 3 SOLUTIONS, FALL 21 p. 3 (5 points) Which of following describes Sachs-Wolfe effect? Photons from fluid which had a velocity toward us along

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1 INSTITUTE OF TECHNOLOGY MASSACHUSETTS Department Physics.26 The Early Universe December 19, 21 Physics Alan Guth Prof. QUIZ 3 SOLUTIONS Quiz Date December 5, 21 to Remove Blank Pages Reformatted answer all questions in this stapled booklet. Please PROBLEM 1 DID YOU DO THE READING? (2 points) (5 points) Which one of following statements about CMB is NOT correct? The dipole distortion is a simple Doppler shift, caused by net motion of relative to a frame of reference in which CMB is isotropic. observer After dipole distortion of CMB is subtracted away, mean temperature over sky is <T> 2.725K. averaging After dipole distortion of CMB is subtracted away, temperature of (iii) CMB varies by.3 microkelvin across sky. The photons of CMB have mostly been traveling on straight lines since y last scattered at t 37; yr, at a location called surface of last were scattering. The actual variation is about 3 microkelvin, or maybe a few times that [Comment Ryden quotes COBE root mean square fractional variation of CMB much. temperature as ffit * T Eq. (.) (2nd Edition), which gives a value of about 3 microkelvin, given that as 3 K. In Lecture Notes 2 we quoted a value of computed from Planck T The root mean square fluctuations increase with better angular resolution, data. fluctuations with small angular wavelengths are not seen unless resolution is high..26 QUIZ 3 SOLUTIONS, FALL 21 p. 2 (5 points) The nonuniformities in cosmic microwave background allow us to (b) ripples in mass density of universe at time when plasma measure to form neutral atoms, about 3, - 4, years after big bang. combined ripples are crucial for understanding what happened later, since y are These which led to complicated tapestry of galaies, clusters of galaies, and seeds Which of following sentences describes how se ripples are created in voids. contet of inflationary models Magnetic monopoles can form randomly during grand unified ory phase resulting in nonuniformities in mass density. transition, Cosmic strings, which are linelike topological defects, can form randomly during grand unified ory phase transition, resulting in nonuniformities in mass density. (iii) They are generated by quantum fluctuations during inflation. Since early universe was very hot, re were large rmal fluctuations ultimately evolved into ripples in mass density. which (5 points) In Chapter of The First Three Minutes, Steven Weinberg describes future of universe (assuming, as was thought n to be case, that constant is zero). One possibility that he discusses is that cosmic cosmological density could be greater than critical density. Assuming that we live in matter such a universe, which of following statements is NOT true? The universe is finite and its epansion will eventually cease, giving way to an contraction. accelerating Three minutes after temperature reaches a thousand million degrees (1 9 K), laws of physics guarantee that universe will crunch, and time will stop. During at least early part of contracting phase, we will be able to observe (iii) redshifts and blueshifts. both When universe has recontracted to one-hundredth its present size, radiation background will begin to dominate sky, with a temperature of about 3 K. Weinberg is very clear no speculations about end of universe are [Comment to be true Does time really have to stop some three minutes after guaranteed reaches a thousand million degrees? Obviously, we cannot be sure. All temperature uncertainties that we met in preceding chapter, in trying to eplore first of a second, will return to perple us as we look into last hundredth of hundredth asecond."]

2 .26 QUIZ 3 SOLUTIONS, FALL 21 p. 3 (5 points) Which of following describes Sachs-Wolfe effect? Photons from fluid which had a velocity toward us along line of sight appear of Doppler effect. redder Photons from fluid which had a velocity toward us along line of sight appear of Doppler effect. bluer Photons traveling toward us from surface of last scattering appear redder (iii) of absorption in intergalactic medium. Photons traveling toward us from surface of last scattering appear bluer of absorption in intergalactic medium. Photons from overdense regions at surface of last scattering appear redder (v) y must climb out of gravitational potential well. Photons from overdense regions at surface of last scattering appear bluer (vi) y must climb out of gravitational potential well. [Comment Ryden discusses Sachs-Wolfe effect on pp (2nd Edition). End of Problem QUIZ 3 SOLUTIONS, FALL 21 p. 4 2 TIME EVOLUTION OF A UNIVERSE INCLUDING A HY- PROBLEM KIND OF MATTER (3 points) POTHETICAL that a flat universe includes nonrelativistic matter, radiation, and also mysticium, Suppose where mass density of mysticium behaves as / 1 ρmyst 5 (t) a as universe epands. In this problem we will define (t) ; t is present time. For following questions, you need not evaluate any where integrals that might arise, but y must be integrals of eplicit functions with of limits of integration; remember that is not given. You may epress your eplicit in terms of present value of Hubble epansion rate, H, and various answers contributions to present value of Ω Ωm;, Ωrad;, and Ωmyst;. (7 points) Write an epression for Hubble epansion rate H(). (b) (7 points) Write an epression for current age of universe. (3 points) Write an epression for time t() in terms of value of. (3 points) Write an epression for total mass density ρ() as a function of. (1 points) Write an epression for present value of physical horizon distance, (e) `p;hor(t). Since universe is flat, first Friedmann equation becomes but n we can write ρ as Now use 2» a(t ) (ρm; G H 3 H 2 3 Gρ ; 3» a(t ) + ρrad; 3H2 ρc; and Ω ρ ; ρc G 4» a(t ) + ρmyst; 5 )

3 .26 QUIZ 3 SOLUTIONS, FALL 21 p. 5 so Finally, 2 H2 H ρc; ( ρm;» ) a(t 3» a(t ) + ρrad; H ρ 2 Ω m; 3 + Ω rad; 4 + Ω myst; 5 H H() 2 r (b) To find current age t, we start with H _a a _ ff 4» a(t ) + ρmyst; +Ωrad; + Ω myst; Ωm; H ) dt ) H dt So t can be found by integrating over range of, from to 1 t H() 1 H q +Ωrad; + Ωmyst; Ωm; To find time t corresponding to some value of or than 1, one simply integrates from to dt t() Z 1 H H( ) Z From first Friedmann equation, H 2 3 q +Ωrad; + Ωmyst; Ωm; 3 ρ Gρ H2 () ) G 5 ).26 QUIZ 3 SOLUTIONS, FALL 21 p. 6 Given answer in part, this becomes 3 ρ() G 2 H 4» + Ω myst; +Ωrad; Ωm; (e) The general formula for physical horizon distance is given on formula sheet `p;hor(t) Z t c dt ) a(t we are not given function, but we can change variable of integration Here integrate over to So dt dt da da 1 _a da 1 a `p;hor(t) Z c 2 H() c H c da a a 2 H da da ah() _a q +Ωrad; + Ωmyst; Ωm;

4 .26 QUIZ 3 SOLUTIONS, FALL 21 p. 7 PROBLEM 3 PROPERTIES OF BLACK-BODY RADIATION (25 points) The following problem was Problem 6 of Review Problems for Quiz 3. answering following questions, remember that you can refer to formulas on In formula sheets, circulated separately. Since you were not asked to bring calculators, you may leave your answers in form of algebraic epressions, such as 32 p 5 (3). (5 points) For black-body radiation (also called rmal radiation) of photons at T, what is average energy per photon? temperature (b) (5 points) For same radiation, what is average entropy per photon? (5 points) Now consider black-body radiation of a massless boson which has spin so re is only one spin state. Would average energy per particle and zero, per particle be different from answers you gave in parts and (b)? If entropy how would y change? so, (5 points) Now consider black-body radiation of electron neutrinos at temperature T. These particles are fermions with spin 1/2, and we will assume that y massless and have only one possible spin state. What is average energy per are for this case? particle (5 points) What is average entropy per particle for black-body radiation of (e) as described in part? neutrinos, Solution The average energy per photon is found by dividing energy density by number The photon is a boson with two spin states, so g g Λ 2. Using density. formulas on front of eam, E 2 g 3 4 (kt) 3 (μhc) 4 kt 3 (3) You were not epected to evaluate this numerically, but it is interesting to know that E 271 kt that average energy per photon is significantly more than kt, which is often Note as a rough estimate. used.26 QUIZ 3 SOLUTIONS, FALL 21 p. The method is same as above, ecept this time we use formula for entropy (b) density S 22 g k 4 T k (3) Numerically, this gives 362 k, where k is Boltzmann constant. In this case we would have g g Λ 1. The average energy per particle and average entropy particle depends only on ratio gg Λ, so re would be no difference from answers given in parts and (b). For a fermion, g is 7/ times number of spin states, and g Λ is 3/4 times of spin states. So average energy per particle is number Numerically, E kt. E 2 g 3 4 (kt) 3 (μhc) (kt) 3 (μhc) 3 (3) 2 (kt) kt 1 (3) Mamatician General has determined Warning memorization of this number may adversely that affect your ability to remember value of. one takes into account both neutrinos and antineutrinos, average energy per If is unaffected energy density and total number density are both particle doubled, but ir ratio is unchanged.

5 .26 QUIZ 3 SOLUTIONS, FALL 21 p. 9 that energy per particle is higher for fermions than it is for bosons. This Note can be understood as a natural consequence of fact that fermions must result eclusion principle, while bosons do not. Large numbers of bosons can obey collect in lowest energy levels. In fermion systems, on or hand, refore low-lying levels can accommodate at most one particle, and n additional are forced to higher energy levels. particles (e) The values of g and g Λ are again 7/ and 3/4 respectively, so S Numerically, this gives S 422k. 22 g k 4 T k 4 T 3 3 (3) 2 (kt) k 135 (3).26 QUIZ 3 SOLUTIONS, FALL 21 p. 1 PROBLEM 4 THE CONSEQUENCES OF AN ALT-PHOTON (25 points) that, in addition to particles that are known to eist, re also eisted Suppose alt-photon, which has eactly properties of a photon it is massless, has two spin an (or polarization states), and has same interactions with or particles that states do. Like photons, it is its own antiparticle. photons (5 points) In rmal equilibrium at temperature T, what is total energy density alt-photons? of (5 points) In rmal equilibrium at temperature T, what is number density of (b) alt-photons? (1 points) In this situation, what would be temperature ratios TνTfl and today? TνTaltfl (5 points) Would eistence of this particle increase or decrease abundance of or would it have no effect? helium, Solution The energy density will be same as for photons, since re is no difference. The formula is general 2 g u 3 4 (kt) 3 ; (μhc) given on formula sheets, and g 2 for alt-photons (or photons), since re as two polarization states, and particles are bosons. So are ualtfl 2 (b) For number density, general formula is 15 g Λ (3) n 2 (kt) 3 4 (kt) 3 (4.1) (μhc) ; where g Λ 2 since again alt-photons are bosons with two polarization states. So 2 (3) naltfl 2 (kt) 3 (4.2)

6 .26 QUIZ 3 SOLUTIONS, FALL 21 p. 11 As in actual scenario, event that causes a temperature difference is of electron-positron pairs from rmal equilibrium mi, which disappearance as kt changes from values large compared to mec MeV to values occurs are small compared to it. The key point is that this disappearance occurs after that neutrinos have decoupled from or particles, so all of entropy from pairs is given to photons, and none is given to neutrinos. electron-positron In this case entropy isgiven to both photons and alt-photons. general formula for entropy density is on formula sheet, and it can be The as rewritten where s AgT 3 ; (4.3) 22 A The value of A will in fact not be needed for this problem. k 4 (4.4) neutrinos have decoupled bytimee + e pairs disappear, entropy Since neutrinos and entropy of everything else will be separately conserved. Entropy of means that entropy per comoving volume does not change. During conservation period before e + e freeze-out, g is constant, so constancy of entropy per comoving volume implies that S svphys gt 3 AVphys ga 3 T 3 A ; (4.5) S const implies that a 3 T 3 is constant, and so at is constant. Here T so common temperature of photons, alt-photons, electrons and positrons, and is all of which were in rmal equilibrium during this period. Since at is neutrinos, during this period, we can give constant a name, constant For neutrinos, formula sheet tells us that while 7 gν z } Fermion factor e + e 7 g z } Fermion factor 3 species 3 νe;νμ;νfi at [at ]before (4.6) 1 Species 2 Particle antiparticle 2 Particle antiparticle 1 Spin states 2 Spin states 21 ; (4.7) (4.).26 QUIZ 3 SOLUTIONS, FALL 21 p. 12 Thus gelse gfl + galtfl + g e + e (4.9) Thus before e + e freezeout, two conserved quantities were Sν Agν[aT ] 3 before ; Selse Agelse[aT ] 3 before (4.1) e + e freezeout, temperature of neutrinos Tν will no longer be After as temperature Tfl of photons and alt-photons, and of course e + e same will no longer be present. But Tfl and Taltfl will be equal to each or, since pairs have same interactions; we know that interactions of photons keep y in rmal equilibrium until tdecoupling ο 3; years, so both photons m alt-photons will remain in rmal equilibrium until long after era of and + e freezeout, which is of order 1 1 seconds. Thus two conserved quantities e be will Sν 3 Selse after ; Agν[aTν] A(gfl + galtfl)[atfl] 3 after (4.11) By equating values of Sν before and after, we see that [atν]after [at ]before ; (4.12) and n by equating values of Selse before and after, we see that [atfl]after gelse gfl + galtfl 13 [at ]before gelse gfl + galtfl where we used Eq. (4.12) in last step. It follows that» ν T Tfl after g fl + galtfl gelse [atν]after ; (4.13) 13 (4.14) It would increase abundance of helium. The main effect of alt-photon would to increase epansion rate of universe, which in turn would cause be to decouple earlier from rmal equilibrium mi, which in turn would neutrinos that ratio nnnp, ratio of neutrons to protons, would become frozen mean a larger value. The increased epansion rate would also mean less time available at free neutron decay, which furr increases number of neutrons that remain for temperature falls low enough for helium formation to complete. Essentially when neutrons become bound into helium, so more neutrons implies more helium. all

7 .26 QUIZ 3 SOLUTIONS, FALL 21 p. 13 Problem Maimum Score Initials TOTAL 1