MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.286: The Early Universe December 21, 2013 Prof. Alan Guth QUIZ 3 SOLUTIONS
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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physis Department Physis 8.286: The Early Universe Deember 2, 203 Prof. Alan Guth QUIZ 3 SOLUTIONS Quiz Date: Deember 5, 203 PROBLEM : DID YOU DO THE READING? (35 points) (a) (5 points) Ryden summarizes the results of the COBE satellite experiment for the measurements of the osmi mirowave bakground (CMB) in the form of three important results. The first was that, in any partiular diretion of the sky, the spetrum of the CMB is very lose to that of an ideal blakbody. The FIRAS instrument on the COBE satellite ould have deteted deviations from the blakbody spetrum as small as ɛ/ɛ 0 n,wheren is an integer. To within ±, what is n? Answer: n =4 (b) (5 points) The seond result was the measurement of a dipole distortion of the CMB spetrum; that is, the radiation is slightly blueshifted to higher temperatures in one diretion, and slightly redshifted to lower temperatures in the opposite diretion. To what physial effet was this dipole distortion attributed? Answer: The large dipole in the CMB is attributed to the motion of the satellite relative to the frame in whih the CMB is very nearly isotropi. (The entire Loal Group is moving relative to this frame at a speed of about ) () (5 points) The third result onerned the measurement of temperature flutuations after the dipole feature mentioned above was subtrated out. Defining δt T (θ, φ) T (θ, φ), T T where T =2.725 K, the average value of T, they found a root mean square flutuation, ( ) 2 /2 δt, T equal to some number. To within an order of magnitude, what was that number? Answer: (δt ) 2 /2 = T
2 8.286 QUIZ 3 SOLUTIONS, FALL 203 p. 2 (d) (5 points) Whih of the following desribes the Sahs-Wolfe effet? (i) Photons from fluid whih had a veloity toward us along the line of sight appear redder beause of the Doppler effet. (ii) Photons from fluid whih had a veloity toward us along the line of sight appear bluer beause of the Doppler effet. (iii) Photons from overdense regions at the surfae of last sattering appear redder beause they must limb out of the gravitational potential well. (iv) Photons from overdense regions at the surfae of last sattering appear bluer beause they must limb out of the gravitational potential well. (v) Photons traveling toward us from the surfae of last sattering appear redder beause of absorption in the intergalati medium. (vi) Photons traveling toward us from the surfae of last sattering appear bluer beause of absorption in the intergalati medium. (e) (5 points) The flatness problem refers to the extreme fine-tuning that is needed in Ω at early times, in order for it to be as lose to today as we observe. Starting with the assumption that Ω today is equal to within about %, one onludes that at one seond after the big bang, Ω t= se< 0 m, where m is an integer. To within ± 3, what is m? Answer: m = 8. (See the derivation in Leture Notes 8.) (f) (5 points) The total energy density of the present universe onsists mainly of baryoni matter, dark matter, and dark energy. Give the perentages of eah, aording to the best fit obtained from the Plank 203 data. You will get full redit if the first (baryoni matter) is aurate to ±2%, and the other two are aurate to within ±5%. Answer: Baryoni matter: 5%. Dark matter: 26.5%. Dark energy: 68.5%. The Plank 203 numbers were given in Leture Notes 7. To the requested auray, however, numbers suh as Ryden s Benhmark Model would also be satisfatory. (g) (5 points) Within the onventional hot big bang osmology (without inflation), it is diffiult to understand how the temperature of the CMB an be orrelated at angular separations that are so large that the points on the surfae of last sattering was separated from eah other by more than a horizon distane. Approximately what angle, in degrees, orresponds to a separation on the surfae last sattering of one horizon length? You will get full redit if your answer is right to within a fator of 2. Answer: Ryden gives as the angle subtended by the Hubble length on the surfae of last sattering. For a matter-dominated universe, whih would be a good model for our universe, the horizon length is twie the Hubble length. Any number from to 5 was onsidered aeptable.
3 8.286 QUIZ 3 SOLUTIONS, FALL 203 p. 3 PROBLEM 2: FREEZE-OUT OF MUONS (25 points) See the solutions to Problem Set 7, Problem 2 (203). PROBLEM 3: THE EVENT HORIZON FOR OUR UNIVERSE (25 points) (a) In a spherial pulse eah light ray is moving radially outward, so dθ =dφ =0. A light ray travels along a null trajetory, meaning that ds 2 =0,sowehave from whih it follows that ds 2 = 2 dt 2 + a 2 (t)dr 2 =0. (3.) dr dt = ±. (3.2) We are interested in a radial pulse that starts at r =0attimet = t 0,sothe limiting value of r is given by r max = t 0 dt. (3.3) (b) Changing variables of integration to x =, (3.4) a(t 0 ) the integral beomes dt dt r max = =, (3.5) a(t 0 ) x whereweusedthefatthatt = t 0 orresponds to x = a(t 0 )/a(t 0 )=. As given to us on the formula sheet, the first-order Friedmann equation an be written as x = H 4 0 Ω m,0 x +Ω 2 rad,0 +Ω va,0 x +Ω k,0 x. (3.6) dt Using this substitution, r max = a(t0 )H 0, Ω +Ω 4 m,0x +Ω rad,0 va,0 x (3.7)
4 8.286 QUIZ 3 SOLUTIONS, FALL 203 p. 4 wherewehaveusedω k,0 = 0, sine the universe is taken to be flat. () To find the value of the redshift for the light that we are presently reeiving from oordinate distane r max, we an begin by notiing that the time of emission t e an be determined by the equation whih implies that the oordinate distane traveled by a light pulse between times t e and t 0 must equal r max. Using Eq. (3.2) for the oordinate veloity of light, this equation reads t0 t e dt = rmax. (3.8) The half-redit answer to the quiz problem would inlude the above equation, followed by the statement that the redshift z eh an be determined from a(t0) z = a(t e ). (3.9) The full-redit answer is obtained by hanging the variable of integration as in part (b), so Eq. (3.8) beomes dt r max = x e (3.0) dt =, a(t 0 ) x where x e is the value of x orresponding to t = t e. Then using Eq. (3.6) with Ω k,0 = 0, we find x e r max =. (3.) a(t0 )H 0 x e Ω m,0 x +Ω rad,0 +Ω 4 va,0x To omplete the answer in this language, we use z = xe. (3.2) Eqs. (3.) and (3.2) onstitute a full answer to the question, but one ould go further and replae r max using Eq. (3.7), finding Ωm,0 x +Ω rad,0 +Ω va,0 x 4 (3.3) =. x e Ω m,0 x +Ω rad,0 +Ω va, 0 x 4
5 8.286 QUIZ 3 SOLUTIONS, FALL 203 p. 5 In this form the answer depends only on the values of Ω X,0. You were of ourse not asked to evaluate this formula numerially, but you might be interested in knowing that the Plank 203 values Ω m,0 =0.35, Ω va,0 =0.685, and Ω rad,0 = lead to z eh =.87. Thus, no event that is happening now (i.e., at the same value of the osmi time) in a galaxy at redshift larger than.87 will ever be visible to us or our desendants, even in priniple. PROBLEM 4: BEHAVIOR OF Ω IN A UNIVERSE DOMINATED BY MYSTERIOUS STUFF (5 points) (a) From the Friedmann equation, 2 8π k 2 H = Gρ, 3 a 2 and the definition of ρ, H 2 8π = Gρ, 3 we have 8π k 2 8π Gρ = Gρ 3 a 2 3 whih an be rearranged to give Dividing both sides by ρ we find 3k 2 ρ ρ =. 8πGa 2, ρ ρ Ω k = 3 2 =. ρ Ω 8πGa 2 ρ (The middle expression above is obtained from the expression on the left by multiplying both the numerator and the denominator by /ρ.) Now multiply both the numerator and denominator of the expression on the right by T 2, reognizing that at is approximately onstant. We then have Ω T 2 = A, Ω ρ where 3k 2 A =. 8πG(aT ) 2
6 8.286 QUIZ 3 SOLUTIONS, FALL 203 p. 6 (b) Combining the above equation with our knowledge that T /a and ρ /a 5, we have immediately that Ω /a 2 Ω a3. /a 5 To determine how a depends on time, we an solve the Friedmann equation for a flat universe dominated by mysterious stuff: Thus and then ( ) 2 ȧ onst da onst = = a a 5 =. dt a 3/2 a 3/2 da = onst dt 2 a 5/2 = onst t, 5 where the onstant of integration is set to zero by using the onvention that a =0whent =0. Thus a t 2/5, and then Ω a 3 t 6/5. Ω
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