Solutions to exercises Solutions to exercises Exercise 1.1 When we calculated that the skyisasbrightasthe Sun, we assumedthatthe lineofsight stoppedonthe star,i.e.starsare opaque.Whenwe calculated the brightness of thesky for a S−5/2 powerlaw,weintegrated down to zeroflux, which (for anyparticular typeofstar)meansintegratingtor=∞. So thelines of sight don’tstoponstarsinthe latter case; starsare treated as transparent. 2 Exercise 1.2 We use E = γm0c ,where E is theenergy, m0 is the rest mass, γ =(1−v2/c2)−1/2,and c is thespeed of light.Wehavethat 1020 eV = γc2 × 938.28 MeV/c2 1 γ × 109 eV .The quoted accuracy of the energy doesnot justifycarrying more than just thefirstsignificantfigure on the proton’srestmass. The γ factor is then just γ =1020/109 =1011.The cosmic rayismovingatveryclose to thespeed of light,soitwouldtakeabout 100 000 yearsfor theproton to crossthe Galaxyinthe Galaxy’srestframe.But moving clocksrun slow,soitwouldtake 100 000/γ yearsinthe proton’srestframe,i.e. 105/1011 years, or 10−6 years, or about 30 seconds! Exercise 1.3 First we differentiate Equation1.7 with respect to time t to get 8πG ! F 2Λc2RR˙ 2R˙ R¨ = ρR˙ 2 +2ρRR˙ + , 3 3 (S1.1) wherewewrite ρ = ρm + ρr for brevity andthe ‘dot’ notation is used to indicate differentiation with respect to time, i.e. R˙ =dR/dt and R¨ =d2R/dt2.The conservationofmatter energy gives d 1 ( ρc2R3 = ρc˙ 2R3 +3ρc2R2R˙ dt d(R3) = −p dt = −3pR2R,˙ so ρc˙ 2R3 = −3pR2R˙ − 3ρc2R2R.˙ EquationS1.1 hasaterm ρR˙ 2,sowerearrangethe above to find −3pRR˙ ρR˙ 2 = − 3ρRR˙ 2 c C D 3p = −RR˙ +3ρ . c2 SubstitutingthisintoEquationS1.1 gives 5 C D+ 8πG 3p 2Λc2RR˙ 2R˙ R¨ = 2ρRR˙ − RR˙ +3ρ + 3 c2 3 C D 8πGRR˙ 3p 2Λc2 = 2ρ − − 3ρ + RR˙ 3 c2 3 C D −8πGRR˙ 3p 2Λc2 = ρ + + RR˙ 3 c2 3 C D 3p RR˙ 2Λc2 = −8πG ρ + + RR.˙ c2 3 3 291 Solutions to exercises Dividing this by 2R˙ gives C D 3p R Λc2R R¨ = −4πG ρ + + c2 3 3 C D 3p R Λc2R = −4πG ρ + ρ + + , m r c2 3 3 as required. Exercise 1.4 Λ=0 Ω If ,then Λ is always zero(Equation1.17). From 2 3 Equation1.33,wethereforehavethat (H/H0) =(1+z) when Ωm =1and Λ=0.Now,Equation1.28 tells us that −1 dz H = , 1+z dt so C D 1 1 dz 2 =(1+z)3, 2 2 H0 (1 + z) dt which we maywrite more simply as dz/dt ∝ (1 + z)5/2,ordt/dz ∝ (1 + z)−5/2. −3/2 Integratingthiswith respect to z,weget t ∝ (1 + z) .But 1+z=R0/R,so t∝R3/2,or R=αt2/3, (S1.2) where α is some constant. In particular,atthe current time t = t0 we have 2/3 R0 = αt0 , (S1.3) 2/3 anddividingEquationS1.2 by EquationS1.3 gives R/R0 =(t/t0) . Exercise 1.5 We can rearrangeEquation1.35 to read C D t 2/3 R = R0 . (S1.4) t0 From Equation1.12,wehavethat H =(1/R)dR/dt.Differentiating EquationS1.4, we get dR 2 R = 0 t−1/3. dt 3 2/3 t0 At atime t = t0,thisisjust $ dR$ 2 R $ = 0 . dt $ 3 t t=t0 0 Therefore theHubbleparameter at atime t = t0 in this modelUniverseis $ 1 dR$ 1 2 R 2 H = $ = 0 = , 0 R dt$ R 3 t 3t 0 t=t0 0 0 0 −1 −1 or t0 =2/(3H0) as required.PuttinginH0=72±3km s Mpc ,wefind t0 =9.1±0.4Gyr. Exercise 1.6 Theangular diameter in degreeswill be inversely proportional to dA (Equation 1.47),sothe angular area (e.g. in square degrees) will vary as 292 Solutions to exercises θ2 ∝ d−2 d2 A .The fluxwill be inversely proportionalto L(Equation 1.49),i.e. S ∝ d−2 S/θ2 ∝ d2 /d2 .The surface brightness will therefore vary as A L.But L 2 −4 dL =(1+z) dA(Equation 1.50),sosurface brightness must vary as (1 + z) . −1 −1 Exercise 1.7 In Section 1.5 we aregiven that H0 =72±3km s Mpc 16 and ΩΛ,0 =0.742 ± 0.030.One parsec is 3.09 × 10 m, so in SI −18 −1 units, H0 =(2.3±0.1) × 10 s .Equation1.17 relates thesetwo 2 2 2 2 quantities to Λ: ΩΛ,0 =Λc/(3H0 ),soΛ=3ΩΛ,0 H0 /c .Puttinginthe −52 −2 -numbers,weget Λ=(1.3 ± 0.2) × 10 m .The horizon size will be 3/Λ=(1.5 ± 0.1) × 1026 m, or 4900 ± 300 Mpc. This cosmological event horizon will be exceedingly distant; for comparison, thecurrentradius of the observableUniverseinSection 1.9 is about 3.53c/H0 =14900 Mpc. Exercise 2.1 The 13.6 eV photon does ionize anotheratom. However, the process of recombinationneedn’tresultinthe emission of just one photon. Sometimes theelectron will bind first in ahighenergystate (releasingone photon with an energy < 13.6 eV), then releasethe remainingenergyinstages as the electron drops down theenergylevels of thehydrogen atom.Each of thesestages will involvethe releaseofaphoton, butnone of thesephotons will have enough energy on its owntoionize hydrogen atoms. Exercise 2.2 We aregiven that T =2.725 ± 0.001 K, so theenergydensity 2 4 −8 4 8 must be ρr,0 c =4σT /c =4×5.67 × 10 × 2.725 /(3.00 × 10 ) joules per 2 −14 −3 cubicmetre, i.e. ρr,0 c =4.17 × 10 Jm ,ormass-equivalentdensity of −31 −3 ρr,0 =4.64 × 10 kg m .Applying Equation1.16,and remembering that −1 −1 −18 −1 H0 =100h km s Mpc =3.24 × 10 h s ,wefind that 8πGρr,0 −2 −5 Ωr,0 = 2 =2.47h × 10 . 3H0 So 2 −5 Ωr,0 h 1 2.5 × 10 , as required. Exercise 2.3 Thematter energy density scales as R−3,while the photon/neutrino energy density scales as R−4.Thereforefrom Equations 1.15 and1.16, Ωr/Ωm =(1+z)Ωr,0/Ωm,0.From Exercise2.2 andthe text following 2 −5 4 it, we have that Ωr,0 h 1 4.2 × 10 (TCMB,0/2.725 K) .The epochof matter–radiation equality must by definitionsatisfy Ωr/Ωm =1,so Ωm,0 1+zeq = Ωr,0 h2 = Ω (T /2.725 )−4 4.2 × 10−5 m,0 CMB,0 K 2 −4 1 23 800 Ωm,0 h (TCMB,0/2.725 K) , as required. Exercise 2.4 Theanalysisisthe same up to Equation1.30,where ρ this time 3 3 is ρr.However,instead of ρ = ρ0 × R0/R ,wemustalsotakeintoaccount the 4 4 fact that photons lose energy from redshifting, so ρr = ρ0 × R0/R .With Λ setto zero, theequivalentofEquation1.32 comesout as C D 2 1 ( H 2 2 =(1+z) 1−Ωr,0+Ωr,0(1 + z) , H0 293 Solutions to exercises 2 −2 2 andinserting Ωr,0 =1andusing H =(1+z) (dz/dt) ,wefind that C D dz 2 = H2(1 + z)6 dt 0 so dz d(1 + z) = = H (1 + z)3. dt dt 0 Nowthe dimensionlessscale factor a is related to redshift via a =1/(1 + z),so we couldwrite this as d(a−1) = H a−3 dt 0 thus da −a−2 = H a−3 dt 0 hence a da ∝ dt. Integratingthisgives a2 ∝ t,ora∝t1/2as required. Exercise 2.5 ! is measured in Js.AJoulehas dimensions of energy (like 1 2 2 −2 2 mv )soithas dimensions ML T ,where we write Mfor thedimension of mass,Lfor length, andTfor time. (Notethatnumerical constants areignored in dimensionalanalysis.)Thereforewecan write thedimensions of ! as [!]=ML2T−1.Similarly,the dimensions of c are [c]=LT−1.Tofind the dimensions of G,wecan startwith thefamiliar equation F = GMm/r2,and note that forceismasstimes acceleration, so ma = GMm/r2 or G = ar2/M ,sothe dimensions of G are [G]=LT−2L2/M = M−1L3T−2.Now let’ssupposethatthe Planck timeisgiven by aformulaofthe form !xcyGz,where theconstants x, y and z aretobedetermined. Theresultmusthavethe dimensions of time, so 1 (x 1 (y 1 (z T = ML2T−1 LT−1 M−1L3T−2 . Multiplyingthisout andrearranginggives T = Mx−zL2x+y+3zT−x−y−2z. Theleft-hand side hasnomassM,sox−zmust equalzero, i.e. x = z.The left-hand side alsohas no lengthL,so2x+y+3z=0.The left-hand side has exactly one powerofT,so−x−y−2z=1.Wehavethree simultaneous equations for threeunknowns. Substitutinginx=zinto theother twoequations gives 5x + y =0and −3x − y =1.Therefore y = −1 − 3x = −5x,or x=1/2.Since x = z,wehave z =1/2.Finally,any of theequations involving y implythat y = −5/2.Therefor-ethe characteristic timemustbeofthe form !xcyGz = !1/2c−5/2G1/2 = !G/c5,asrequired. Exercise 2.6 We have alreadythat (1/R)d2R/dt2 = α(α − 1)t−2.Since t is positive and α>1,the right-hand side must be positive.Thereforethe left-hand side must alsobepositive.Since R is alsopositive, d2R/dt2 > 0. Exercise 2.7 We startwith 3Hφ˙ = −V % (Eqn 2.23) 294 Solutions to exercises andthenuse 8π H2 = V. 3m2 (Eqn 2.24) Pl Nowthe H dt terminthe integral in thequestioncan alsobeexpressedas dt dφ Hdt=H dφ=H . dφ φ˙ Next we useEquation2.23 to get dφ dφ H dt = H = −3H2 . (−V %/3H) V % Finally,using Equation2.24 this comesout as C D −8π V H dt = dφ, m2 V % Pl so we reach the required integral: * −8π φ1 V N = dφ. m2 V % Pl φ2 % Forthe next part, we set V 1 V/φ and φ1 =0(asadvisedinthe question) to write this as * −8π 0 V N = φ dφ. m2 V Pl φ2 Evaluatingthisintegralgives C √ D 4π 2 πφ 2 N = φ2 = 2 .