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 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 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 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 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 . m2 2 m Pl Pl √ √ Thus to have N>60 we need φ2 >mPl 60/(2 π),orinother words, φ2 > 2.2mPl. Exercise 2.8 No, notimmediately.Atfirstthe CMB will appear very uniform, as you receive light from onlyyour immediate neighbourhood. As timeprogresses you will receive light from largerand more distantparts of theUniverse. You’ll onlybeabletosee the structureswith wavelength λ oncelight hashad timeto travel the distance λ,i.e.after atime δt = λ/c,where c is thespeed of light.The size of thelargest acoustic peak is setbythe sound horizon after inflation.Once light hashad timetotravelthisdistance, all theacoustics will starttobecome visible. Also,the acoustic peakswill have adifferent angular size on thesky, because thesurface of lastscattering wascloser. Finally,the CMB wouldn’thave peaked at microwavewavelengths then,soperhaps we shouldn’tcall it the CMB then!

Exercise 2.9 We found in Section 2.7 that theparticle horizon radi√ us at recombinationwas 2c/H =0.46 Mpc. Thesound speed is√cs = c/ 3,sothe sound horizon will be 2cs/H =(2c/H) × (cs/c)=0.46/ 3 Mpc =0.27 Mpc. Exercise 2.10 clumpsthrough gravitation, while dark energy appearstobesmoothlydistributed through space. Dark matter is alsoessentially pressureless, with Ωm dominated by therestmassofthe dark matter particles, 295 Solutions to exercises while dark energy hasastrongnegative pressure.Darkmatter makesupabout 20%ofthe total energy density of theUniverse, andatrecombinationmadeup about 70%. Dark energy,meanwhile, wasnegligible at recombinationand yet dominates thepresent-day energy density of theUniverse. (One hopesthatit will soon be possibletoadd that thedarkmatter particle hasbeen directly detected,though that is not yettrue at the timeofwriting; certainly,the proposed particle physics mechanisms for generatingdarkmatter anddarkenergyare very different.) Exercise 2.11 Oneparsecisabout 3.09 × 1016 m, so 3 6 16 −18 −1 H0 =72×10 /(10 × 3.09 × 10 ) 1 2.33 × 10 s .InChapter 1wesaw 2 2 2 2 that ΩΛ,0 =Λc/(3H0 ),soΛ=3ΩΛ,0H0/c .Puttinginthe numbers gives Λ=1.3×10−52 m−2. Exercise 3.1 Theluminosity contributed by ashell of radius r → r +drwill be I(r) times theareaofthe shell, 2πr dr.Summing theseshells,the total ) ∞ luminosity will be L = 0 I(r)2πr dr.Let’sdefine L0 to be theluminosity with I0 = r0 =1,i.e. * ∞ L0 = f(r)2πr dr. 0 Nowlet’scalculate the luminosity in themore general case: * C D ∞ r L = I0 f 2πr dr 0 r0 * ∞ C D C D 2 r r r = I0r0 f 2π d . 0 r0 r0 r0

But this integral has the same form as the integral defining L0,which also 2 integrates from 0 to ∞,soL=I0r0L0. Exercise 3.2 Ashell of thickness dr andradius r will have mass dM =4πr2ρdr.The gravitationalpotential energy of this shell will be −GM(

SubstitutingthisintoEquationS3.1 gives −G 4 πr3ρ dE = 3 dM = −G4πr2ρ × 4πr2ρdr GR r 3 so 1 ( 4 2 2 dEGR = −3G × 3 πr ρ dr.

296 Solutions to exercises

Integratingthisfrom radius 0 to radius R gives * R 1 ( 1 ( 5 4 2 2 4 2 R EGR = −3G 3 πr ρ dr = −3G 3πρ 0 5 −3G 1 ( = 4πR3ρ 2 5R 3 3GM 2 = − , 5R 4 3 where M = 3 πR ρ is thetotal mass of thesphere. 3 Exercise 3.3 Thekinetic energy will be EK = 2 NkT,where N is thenumber of gasparticles.Virial equilibrium is 2EK = −EGR,i.e. 3GM 2 3NkT = . 5 R Therequirementfor gravitationalcollapse is therefore 3GM 2 3NkT< . 5 R To reach Equation3.7, we need to eliminate N and R.Toagood approximation, at recombinationwecan assume that thegas particle masses arethe proton mass mp,sothe numberofparticles must be N = M/mp.Wecan alsouse 4 3 1/3 M = 3 πR to eliminate R,since R =(3M/4πρ) .Inserting these substitutions gives C D M 3 4πρ 1/3 3 kT < GM 2 , mp 5 3M which when rearranged in terms of M givesthe required equation. Thecurrenttemperature of theCMB is about 2.7 K, andthe redshift of recombinationisabout z =1000,sothe photon temperature at recombination must be T =2.7(1+z)13000 K.Matter andradiation will just have been in thermalequilibrium,sothiswill have been the matter temperature too. The baryonic density will be proportionalto(1 + z)3,and usingEquation1.26 and ρb =Ωbρcrit (Equation 1.22),wehavethatthe baryonic density at z =1000 will be 3 ρb = ρb,0(1 + z) 3 = ρcrit × Ωb,0 (1 + z) −26 2 3 −3 =1.8789 × 10 × Ωb,0 h (1 + z) kg m =1.8789 × 10−26 × 2.273 × 10−2 × (1 +1000)3 kg m−3 1 4.3 × 10−19 kg m−3. Puttinginthe numbers gives C D C D 5×(1.381 × 10−23 −1)×3000 3/2 3 1/2 M> JK K × (6.673 × 10−11 Nm2kg−2) × (1.673 × 10−27 kg) 4π × 4.3 × 10−19 kg m−3 1 2 × 1036 kg, 6 or M>10 M),asrequired. 297 Solutions to exercises

Exercise 3.4 Foraflat universe, the comoving distance is the same as the proper motiondistance (Equation 1.56).Thisisn’t true in general (watch out!) but it’strue in aflat universe. Theproper motiondistance is related to theangular diameter distance dA by Equation1.50,which gives dA = dcomoving/(1 + z).The definitionofangular diameter distance in Equation1.47 givesusarelationship between thesize of an object as it wasatthe timeofredshift z andthe angular size as it appearstoday. Theproper size of theBAO wiggles is justthe comoving size dividedby(1 + z),i.e. LBAO/(1 + z).The angular diameter distance to redshift z is therefore dA =(LBAO/(1 + z))/θBAO.The comoving distance to redshift z must therefore be dcomoving = dA × (1 + z)=LBAO/θBAO,asrequired. Exercise 3.5 Here thetrick is to useEquation1.43.Itfollows from that relationthatasmall comoving intervalalong theredshiftaxismustequal δdcomoving = cδz/H(z).Setting this comoving intervaltoLBAO givesus LBAO = cδz/H(z),soH(z)=cδz/LBAO,asrequired. Exercise 3.6 No.The amplitude of thefluctuations coulddependonthe bias, butthe scale length itself is bias-independent. Exercise 4.1 First,weneed to getEquation1.7 into aform wherethe only time-dependent parameter is R.The density ρ is time-dependent andvaries as 3 ρ = ρ0(R0/R) (where subscript 0 indicates present-dayvalues), so we have C D C D dR 2 8πG R 3 8πGρ R3 = ρ 0 R2 − c2 = 0 0 R−1 − c2 dt 3 0 R 3 (where we’veused k =+1). If we set dR/dt =0andsolve,wefind that 3 2 Rmax = R =8πGρ0R0/(3c ).Therefore C D dR 2 R = max c2 − c2. dt R Using thechain rulewehavethat C D C D C D C D C D dR 2 dR 2 dt 2 dR 2 R 2 = = dθ dt dθ dt c andso C D C D C D dR 2 R 2 R = max c2 − c2 = R R − R2, dθ c R max as required. We’reasked to verify that Equation4.2 works rather than proving it, so all we have to do is substitute it in.DifferentiatingEquation4.2 with respect to θ gives dR R = max sin θ dθ 2 so C D dR 2 R2 R2 1 ( = max sin2 θ = max 1 − cos2 θ . dθ 4 4

298 Solutions to exercises

Meanwhile, R2 R2 R R − R2 = max (1 − cos θ) − max (1 − cos θ)2 max 2 4 R2 R2 1 ( = max (2 − 2cos θ) − max 1+cos2 θ − 2cos θ 4 4 R2 1 ( = max 2 − 2cos θ − 1 − cos2 θ +2cos θ 4 R2 1 ( = max 1 − cos2 θ , 4 which equals (dR/dθ)2 as above. Finally,wejustneed to differentiate Equation4.3, which gives dt R R = max (1 − cos θ)= , dθ 2c c as required. Therefore Equations 4.2 and4.3 areasolution. Exercise 4.2 To show this,we’ll first getthings in terms of H.It’saflat 2 2 matter-dominated universe, so Ωm =1=8πGρm/(3H ),thus 4πGρm =3H /2. We alsoknowthat H(t)=a/˙ a.SubstitutingthisintoEquation4.9, we have δ¨ +2H(t)δ˙=3H2(t)δ/2.

Next we use H(t)=2/(3t) to reformulate this in terms of adifferential equation involvingjust δ andtime: C D 4 3 2 2 2 δ¨+ δ˙ = δ = δ. 3t 2 3t 3t2 Next,let’stry powerlaw solutions δ = btc where b and c areconstants.Then δ˙ = bctc−1 and δ¨ = bc(c − 1)tc−2.Substitutingin, we find 4 2 bc(c − 1)tc−2 + bctc−1 = btc. 3t 3t2 Collecting the termstogether,wefind that c−2 4 c−2 2 c−2 bc(c − 1)t + 3 bct = 3 bt , anddividingthrough by btc−2 gives 4 2 c(c − 1) + 3 c = 3 . Thesolutiontothisquadratic equationisc=2/3or c = −1.The −1 solutionis known as the decaying mode,and is not physically relevant in this universe(it decaysmore rapidlythanthe growing mode grows andisquickly negligible).The 2/3 powerlaw time-dependence(which we found ultimately from linearized fluid dynamicequations)isidentical to Equation4.8, which is whythe latter is known as the linear theory. Exercise 4.3 Thereddercolour will be theone with thelargerV-band to B-band fluxratio SV/SB.The fluxesare related to themagnitudesby

299 Solutions to exercises

V = −2.5log10 SV + cV andB=−2.5log10 SB + cB,where cV and cB are constants (notnecessarily identical).Therefore

(B–V)=−2.5log10 SB + cB +2.5log10 SV − cV

= −2.5(log10 SB − log10 SV)+(cB−cV) =−2.5log10(SB/SV)+(cB−cV) =2.5log10(SV/SB)+(cB−cV), which gives

2.5log10(SV/SB)=(B–V) − (cB − cV) so

log10(SV/SB)=(B–V)/2.5 − (cB − cV)/2.5 thus

(B–V)/2.5−(cB−cV)/2.5 (SV/SB)=10 ( )/2.5 −(c −c )/2.5 =10B–V × 10 B V =10(B–V)/2.5 × constant.

Therefore thelargerthe valueof(B–V),the largerthe valueofSV/SB.Therefore (B–V)=1is redder than (B–V)=0. Exercise 4.4 We haven’tspecified the geometryyet, so let’skeep things simple. Let’stakethe dustand starstobeinacylinderfacing us,with cross-sectional area A.Let’sset the length of thecylindertobeh,and measure distances alongthislengthwith thevariable x.Aninfinitesimal layer wouldhave thickness dx andvolume A dx.The biggerthe volume,the more starsitwill contain, so let’sset the luminosity of theshell to be dL = ρA dx,where ρ is a constant(theluminosity density). By the time thelight emergesfrom theend of cylinder, it will have been extinguished by afactor of eτ(x),where τ(x) is theoptical depthatadistance x into thecylinder. This optical depthmustbe proportionaltox,because each increment δx will suppressthe light by thesame factor,which we couldwrite as eδτ ,solet’swrite that as τ = kx.Wecould, for example, write theoptical depthfrom oneend of thecloud to theother as τ total = kh.The light that emergesfrom theshell at x → x +dxwill therefore be −τ(x) −kx dLout = L × e = ρA dx × e .Ifweintegrate that from x =0to x = h, we get * h ! F −kx ρA −kh Lout = ρA e dx = 1 − e . x=0 k Some quick checks: notethat k hasdimensions of one overlength(because τ = kx and τ is dimensionless),soA/k has dimensions of volume,and so ρA/k is luminosity density times volume,which is aluminosity.Notealsothat kh is dimensionless. Now, what wouldhappenifthere were no dust? Theluminosity wouldjustbe Lno dust = ρAh.The dusthas therefore reduced the output luminosity by afactor L ρA/k ! F 1 ! F out = 1 − e−kh = 1 − e−kh . Lno dust ρAh kh 300 Solutions to exercises

This ratio is independent of thegeometrical crosssection A andofthe luminosity density ρ.Ifthe cloudisdeep enough, then theterminbrackets is 1 1,sowejust have Lout/Lno dust =1/(kh).Wecan nowwrite this for Hα light: L ( α) 1 out H = . Lno dust(Hα) kHα h

ForHβ,wehavethat τ Hβ 1 1.45 τ Hα,sokHβ =1.45 kHα,thus L ( β) 1 1 1 L ( α) out H = = = out H . Lno dust(Hβ) kHβ h 1.45 kHα h 1.45 Lno dust(Hα) Therefore L (Hα) L (Hα) out =1.45 no dust . (S4.1) Lout(Hβ) Lno dust(Hβ) This is independent of h,sowe’ve nowremovedall dependenceonthe geometry. So even if kh is enormous and Lout * Lno dust,the luminosity ratio of Hα andHβ is onlyever 1.45 times theratio that you getwith no dust, when enough dustis evenly mixedwith thegas emitting theemissionlines. Nowsupposethatyou wronglyassumedthatit’sasimple dustscreen with an optical depthofτHαfor Hα and τ Hβ =1.45 τ Hα for Hβ.Your luminosities wouldbe

−τ Hα Lout(Hα)=Lno dust(Hα) × e ,

−1.45 τ Hα Lout(Hβ)=Lno dust(Hβ) × e , so theluminosity ratio wouldbe

Lout(Hα) Lno dust(Hα) 0.45 τ = e Hα . (S4.2) Lout(Hβ) Lno dust(Hβ) 0.45 τ Comparing this to EquationS4.1, we have 1.45 =e Hα ,or τHα=ln(1.45)/0.45 1 0.83.Since τ Hα 1 0.7AV,wehave AV 1 1.2.So, if you have an optically-thick cloudinwhich the dustiswell-mixedwith thegas,but you wronglyassumedaforegrounddustscreen,you’dinfer aV-band extinctionofjust 1.2 magnitudes, regardlessofwhatthe realextinction τtotal is from oneend of the cloud to theother. Exercise 4.5 Astronomical absolute magnitudesare definedas m=−2.5log10 L + constant, so d(ln L) −2.5 1 dm = −2.5d(log L)=−2.5 = dL. 10 ln 10 ln 10 L (S4.3) Therefore dN − ln 10 dN = L . dm 2.5 dL (S4.4) The − sign just indicates that the magnitude increment dm is in theopposite sensetothe luminosity increment dL,and is usually neglected. Exercise 4.6 Thevariance of aprobabilitydistribution p(x) is themean of the squares minus thesquare of themean,i.e. * 1 C* 1 D2 Var(x)= x2p(x)dx− xp(x)dx . 0 0 301 Solutions to exercises

Now, our probability distributionisuniform, so p(x)=1for all x from 0 to 1, hencethisisjust * 1 C* 1 D2 Var(x)= x2dx− xdx 0 0 : L &C D @x=1 x3 x=1 x2 2 = − 3 2 x=0 x=0 1 1 1 = 3 − 4 = 12 , as required.The standard deviation is the square root√ of thevariance, so the standard deviation of theuniformdistributionis1/ 12.The central limit theorem states that if you have N measurements,each with an uncertainty σ (i.e.taken from thesamedistributionwith standard deviation√σ), then thestandard deviation of themean averageofthese measurements is σ/ N.Now,ifour nullhypothesis holds,then V/Vmax is uniformlydistributed,soeach√measurementofV/Vmax is takenfrom adistributionwith standard deviation 1/ 12.Therefor√ethe standard deviation of theaverage N measurements of V/Vmax must be 1/ 12N,as required. Exercise 4.7 Yes, providedthatthe selectionfunctionhas been correctly stated. Exercise 4.8 No,not necessarily.Supposethatyou hadavolume-limited sample with Vmax = V (zmax) for all galaxies.Now supposethathalf the galaxies existatexactly z =0,half areatz=zmax,and thereare none in between. Clearly,the numbers of galaxies areevolvingverystronglyand discontinuously, but (V/Vmax/ =1/2. Exercise 4.9 Theamount of light emitted perunitvolumewill be givenbythe numberdensity of galaxies multiplied by their luminosity,i.e. L × φ(L).At luminosities farbelowthe break, φ(L) ∝ L−α,soLφ(L)∝L1−α.Since we’re giventhatthe faint-endslope α satisfies α<1,thismustbeincreasing with luminosity.Atthe bright endwehavethat φ ∝ exp(−L/L∗),which tendstozero faster than 1/L,soLφ(L)(which is proportionaltoLexp(−L/L∗))mustalso tendtozero. We’d expect one turning point—butwhere? We can differentiate Lφ(L),set the result equaltozeroand rearrange.Thisgives d(Lφ) = φ (−e−L/L∗ (L/L )−α+1 +(1−α)e−L/L∗ (L/L )−α)=0. dL ∗ ∗ ∗ −L/L 1−α −α Dividing by φ∗e ∗ gives (L/L∗) =(1−α)(L/L∗) .Furtherdividing −α by (L/L∗) gives L/L∗ =1−α,orL=(1−α)L∗.The galaxies that dominate thecosmicluminosity density arethereforethosewith luminosities of (1 − α)L∗. Exercise 4.10 PDE is vertical translations, while PLEishorizontal translations. Exercise 4.11 Active galaxies can be seen to much higherredshifts than the elliptical galaxies used in the Tolman testinChapter 3, andasthe predicted redshift-dependence of surface brightness is strong, i.e. (1 + z)4,itmight appear that the radiolobesofradiogalaxies have astrong advantage. Theattractionofthe Tolman testisthatthe (1 + z)4 surface brightness prediction is independent of thecosmological parameters. In order to applyit, we need apopulation of objects whoseluminosity perunitarea(in,for example, square )is 302 Solutions to exercises constant. In this case, rearrangingthe relation in the questiongives us L/r2 ∝ Q7/6r−4/3ρ7/12.Wemight hope to findactive galaxies with thesame Q on averageifwematch otherproperties of thecentral engine (e.g. optical emission lines andcontinuum)onaverage.Wemight alsobeabletocalibrate out anyvariationsindensity through otherobservations as indicated in the question, butwe’re still leftwith asurface brightness that depends on thelinear size of the system.Without additionally having astandard rod as acomparison, we can’t applythe Tolman testasitstands. Exercise 4.12 Thereare 60 × 60 =3600 arcseconds in adegree, so thereare 36002 1 1.30 × 107 square arcseconds in asquare degree. Therefore thenumberofrandom 5σ noise spikes in one square degree would be (1.30 × 107)/(3.5 × 106) 1 3.7.Sowe’dexpect one 5σ noise spikein 1/3.7square degrees, or about 0.27 square degrees. In practice, noise spikes can occurmore frequently than this for avariety of reasons (including instrumental effects). Exercise 4.13 Suppose that your cameraordetector covers an area A on the sky. Let’ssay that you invest all your timeinapencil-beam survey,and it reaches aflux S.The numbercountsare Euclidean,soN(>S)=kS−1.5,where k is some constant. Therefore thenumberofgalaxies seen in the pencil-beam survey is −1.5 npencil = A × N(>S)=AkS . Nowsupposethatinstead of doing apencil-beam survey,you spread your integrationtimeover m fieldsofview, each of which hasarea A.The total√area that you coverism×A,but theimageswouldbeshallowerbyafactor of m,so thetotal numberofgalaxies in the wide-field survey wouldbe √ −1.5 −0.75 −1.5 0.25 −1.5 nwide = mA( mS) = mAm S = m AS . 0.25 Comparing thistonpencil,wesee that nwide = m npencil,sothe wide-field survey finds more galaxies by afactor of m0.25. Asimilar calculationshows that if thesource countsare steeper than N(>S)∝S−2,thenthe pencil-beam survey wouldsee more.However,only rarely aresource countsthatsteep (we’llsee an exampleinChapter 5). In thevast majority of cases,wide-field surveysfind more objects in agiven observing time than pencil-beam surveys. In practice, though, there’soften alimit to howwide you can make asurvey,because thetimespent simply moving thetelescope or reading out thedetector becomessignificant(we’veneglected botheffects here).

Exercise 5.1 Iν dν is thebackground intensity in an interval ν → ν +dν. Thebackground perdecadeisthe background in alogarithmic interval, ν → ν +dlog10 ν.Let’swrite this as B dlog10 ν.Ifwecan setthis equaltosomethingtimes dν,thenthatsomethingmustbeIν.Now, dlog10 ν =(dlnν)/ln(10),soBdlog10 ν =(B/ln(10))dlnν.But dlnν =(1/ν)dν,so 1 Bdlog ν = B dν. 10 ν ln(10) Therefore 1 I = B , ν ν ln(10) 303 Solutions to exercises

so B =ln(10) νIν. Therefore thebackground intensity perdecadeoffrequency is proportionaltoνIν.Looking at Figure 5.1, we seethatthe far-infraredbumphas a similar height andareatothe optical/near-infraredbump, each overroughly thesamelogarithmic frequency intervalofabout Δlog10 ν =1.5.Therefore there’sabout thesameenergyoutput in thefar-infraredbumpasinthe optical/near-infraredbump.

Exercise 5.2 This will be theone in which Sν dN/dlnSν is amaximum,and −1 2 since dlnSν =Sν dSν,wecan alsoexpressthisasSνdN/dSν.Thisissimilar (thoughnot quite identical) to Figure 5.2. Exercise 5.3 Theangular resolutioninradians is 1.22λ/D =1.22 × 500 × 10−6/3.5=1.7429 × 10−4 (we’ll carry some extrasignificantfigures until theend of thecalculation). In degreesthis is 1.7429 × 10−4 × 360◦/(2π)=0.009 985 8◦.Inarcseconds this is 0.009 985 8 × 3600 =35.95%%,or36.0%% to theaccuracy of theinitial numbers. Exercise 5.4 (a) Thefractional rangewouldbe0.09/0.15 =0.6or 60%, which we couldalsoquoteasapossiblevariation of afactor of 1/0.6=1.7. (b) Thevariation in β changesthe extrapolation from the 800 µmquoted to the rest frame,which is 850/(1 + z) µm =850/4 µm =212.5 µm. Thewavelength dependenceisλ−β,so C D k (800 µ ) 800 −β d m = =3.765−β , kd(212.5 µm) 212.5 i.e. 0.0705–0.2656 when β =1–2,orafurthervariation of afactor of 3.8.The total variation so faris1.7×3.816.5. (c)Using theblack body spectrum giveninEquation2.2 andputtinginthe numbers for awavelength of 212.5 µm(i.e. ν = c/212.5 µm =2.998 × 108ms−1/(212.5 × 10−6 m)=1.411 × 1012 Hz)and temperatures of T =20Kand 40 K, we findthat B(1.411 , 40 ) exp(hν/kT ) − 1 THz K = 1 B(1.411 THz, 20 K) exp(hν/kT2) − 1 exp(6.626 × 10−34 ×1.411 × 1012 /1.381 × 10−23 −1 ×20 ) − 1 = Js Hz JK K exp(6.626 × 10−34 Js×1.411 × 1012 Hz/1.381 × 10−23 JK−1 ×40 K) − 1 =6.435.

Therange of allowedtemperatures therefore givesanadditionalfractional range of 6.4,sothe total fractional rangeis1.7×3.8×6.4141,i.e.wecannot even quoteadustmasstowithin an order of magnitude! However, if we measure fluxesatmore wavelengths,wemight be able to reduce theseuncertainties by constrainingthe valueofβon theRayleigh–Jeanstail, and determining thetemperature from thewavelength λmax of thelocation of thepeak of thespectral energy distribution. This is quantified with theWien displacement law, which can be expressedinastrophysically-useful quantities as λ 20 max =1.45 K. 100 µm T Thereis, however, still theissuethatgalaxies do not have singletemperatures. 304 Solutions to exercises

Exercise 5.5 Supposethatthere were no background. In some fixed observing time, supposethatwecollect N photons from adistantobject. Using Poissonstatistics,the variance on√this numberwill alsobeN,sothe standard deviation√ (i.e.t√he noise)will be N.The signal-to-noiseratio will therefore be N/ N = N.Now supposethatthere’sastrongbackground, so N + N N 0 N we√ observe √ back photo√ns,with back .The noise on this will be N + Nback 1 Nback 0 N.WhatwewantisNandnot N + Nback,sowe have to observe an additionalblankbit of skytoestimate Nback.Thiscan be done if we have asmall object in our camera, so we can useblankbits of theimage,but if our detectorhas onlyone or asmall numberofpixels,wehavetospend extra N timeobserving blanksky.However,evenneglectingt√he uncertainty on our back estimate, we√still have asignal-to-noiseratio of N/ Nback,which is much less than the N/ N that we’d have in thecaseofnobackground. So once Nback ≥ N we enter the background-limited regime wheregood signal-to-noiseisharderto get. In thecaseofthe SCUBAcamera, the faintestobjects are 1 105–106 times fainter than thesky background. Worse, the background varies on timescales of lessthanasecond, so observing techniquesatsubmmwavelengths areoften geared towards making thebestbackground subtraction. Exercise 5.6 SeeFigure S5.1.

1012 ) 1 − sr 1011 ! L

( 220

/ Arp ty si

no 1010 mi lu M82 109

108 0 1 2 3 redshift, z

FigureS5.1 This is thesameasFigure 5.15, butwith theapproximate location of one possibleflux limit marked as athick black line.

Exercise 6.1 Supposethatwewanted to separateahuman beingintoprotons andelectrons,thenholdthemone metreapart. Fora60 kg mass,the force 2 2 required wouldbeF =(ne) /(4πε0r ),where r =1m, n =60kg/mp and ε0 is the vacuum permittivity of free space. This comesout as agigantic 29 −2 26 F 1 3 × 10 kg ms .The luminosity of theSun is L) =3.83 × 10 W, so the 18 −2 momentum fluxfrom theSun is L)/c =1.28 × 10 kg ms .Ifwecould 305 Solutions to exercises employ all the momentum fluxfrom all the 1 1011 starsinthe Galaxyinkeeping the positive andnegative parts separate, it wouldbejustsufficienttomaintain a 1 mseparation for just 60 kg. Thepotential barrier for separatingthe charged componentsofaplasma accreting around ablack holeisclearly insuperablefor radiation pressure. Exercise 6.2 Puttingthe numbers into Equation6.6 gives 4π × (6.67 × 10−11 2 −2) × (3.00 × 108 −1)×(1.99 × 1030 ) × (1.67 × 10−27 ) L = Nm kg ms kg kg E 6.65 × 10−29 m2 =1.26 × 1031 W. Theluminosity of theSun is 3.83 × 1026 W, which is farbelowthe Eddington limit. Exercise 6.3 Assuming that themassofa100 Wlight bulb is (say)about 50 g, we getanEddington limit of just 0.2 W.Clearly,alight bulb radiates at much more than theEddington limit. Light bulbsdon’tblowthemselves apartbecause they arenot gravitationally bound. Exercise 6.4 To obtain Equation6.26 we startwith Equation1.53,thenuse Equation1.41.Itimmediately follows that 4πcdz c dz dV = d2 (1 + z)3 =4πd2 (1 + z)2 . A (1 + z) H(z) A H(z) (Weignore the − sign, which just refers to thedirections in which the infinitesimal increments aremeasured.) Next,puttinginthe relationship between 2 angular diameter andluminosity distance, dL =(1+z) dA(Equation 1.50),gives 4πd2 c dz 4πd2 c dz dV = L (1 + z)2 = L . (1 + z)4 H(z) (1 + z)2 H(z)

Dividing by dz andmultiplyingbyH0/H0 gives dV 4πcd2 c 4πd2 = L = L , 2 2 dz (1 + z) H(z) H0 (1 + z) H(z)/H0 as required. We can rearrangethisas 4πd2 H(z) L =(1+z)2 . dV/dz c Finally,weuse Equation1.28: |dz/dt| =(1+z)H(z)(again we’llnot worry about thesign). Therefore 4πd2 1 L dt = (1 + z)dz, dV/dz c which is Equation6.24,asrequired.

Exercise 6.5 Theangular size θ will satisfy θ 1 tan θ = rh/D, where D =10Mpc and rh is givenbyEquation6.29: 8 8 −2 rh =(10 /10 ) × (220/200) pc =0.83 pc.Plugging in thenumbers,we have θ 1 r/D =0.83 pc/10 Mpc =8.3×10−8 radians. In arcseconds this is 306 Solutions to exercises

θ =8.3×10−8 × (360◦/2π) × 60 × 60 =0.017%%.Thisisclearly alot smaller than theseeing limit of ground-basedtelescopes. Exercise 6.6 The e-folding timescale for Eddington-limited black holegrowth 8 is theSalpeter timescale tE dividedbythe efficiency η,i.e. te-fold =4×10 /η yr. 9 Therehavebeen 3 × 10 /te-fold e-foldingssince thestartofthe Universe,or 6 6 1 0.75(η/0.1) e-foldings.Toreach 10 M),one needs loge(10 /10 )=11.5 e-foldings.Evenifη=1,you have only 7.5e-foldings, so 3 Gyrisnot long enough.

Exercise 7.1 Comovingdistances add, so rS = rL + rLS.Therefore rLS = rS − rL.Inflat space, angular diameter distance is simply comoving distance dividedby(1 + z) (Chapter 1), butinthiscaseweneed the redshift of thebackground source as seen from thelens.Wecouldwrite this factor as (1 + zLS).Thisisthe factor by which theUniverseexpandedbetween the source redshift andthe lensredshift, i.e. RL/RS,where R is thescale factor.But R R /R R /R L = L 0 = 0 S RS RS/R0 R0/RL

(where thesubscript 0 refers to thepresent day), so (1 + zLS)=(1+zS)/(1 + zL). Therefore our final expression for theangular diameter distance DLS is

(1 + zL) DLS =(rS−rL)× . (1 + zS)

Exercise 7.2 First,matchingdistances along thetop of Figure 7.7 showsthat θDS = βDS + αKDLS.But α = αKDLS/DS,soθDS=βDS+αDS.Dividing out thescalar DS gives θ = β + α,which we can rearrangetoβ=θ−α,as required. Exercise 7.3 We set β =0in Equation7.8. We can rearrangethistoshowthat # 4GM D θ = LS . 2 c DLDS But what wouldthislook like? Thebackground object is exactly behind thelens andit’sdeflected by an angle θ.Isitdeflected to the leftorright or up or down? In fact, thereisnothing to give thedeflection anyparticular direction, so the background source is lensedintoaring.These areveryrare, butanexample is showninFigure S7.1. Exercise 7.4 β 2 +4θ2 J Eis always positive,but thesquare root of it can be β2 +4θ2 >β θ =0 positive or negative. E unless E ,sothe negative rootmust always give anegative θ.Thisisindeed aphysical solutionand representsan anglemeasured in theopposite direction: as showninFigure 7.7, theimage is on theother side of thelens. Note that one imageisatθ>θEandthe otherisat θ<θE,unless θ = θE andthe system is an Einstein ring.

307 Solutions to exercises

FigureS7.1 Thegravitationallens0038+4133 (an Einstein ring)from the COSMOS survey, taken by theHST .The imageis15%% by 15%%.

Exercise 7.5 Fromthe previous exercise, asource can have multiple images,so thereisnot necessarily aunique imageposition θ for agiven source position β. In mathematical terms, we wouldspeak of themapping β → θ as being one-to-many. However, each imageposition θ does mapinaone-to-one wayonto asource position β,i.e.each imagepositioncan correspond to onlyone position in thebackground source.Tosee why, consider Equation7.4. Thefunction α(θ) must be asingle-valued function,i.e.any particular input θ can give onlyone possibleoutput α.Thereforethere can be onlyone valueofβfor agiven input θ. Exercise 7.6 We’reasked to differentiate Equation7.12,which gives dβ/dθ =1+(θ2/θ2) E .Thisgives us one of thefractionsinEquation7.16.The magnification is therefore C D C D C D θ dθ θ θ2 −1 θ2 −1 θ2 −1 = 1+ E =θ θ− E 1+ E β dβ β θ2 θ θ2 C D C D C D θ2 −1 θ2 −1 θ2 θ2 θ4 −1 = 1− E 1+ E = 1+ E − E − E θ2 θ2 θ2 θ2 θ4 C D θ4 −1 = 1− E , θ4 as required. Exercise 7.7 Anegative magnification meansthatthe imageismirror-reversed. Forexample, apositive change dβ wouldhaveacorresponding dθ in theopposite direction, so dθ is negative.Therefore dθ/dβ is negative in Equation7.16. Exercise7.8 We startfrom Equation7.24.The mass enclosed is Σπξ2 andwe set ξ = DLθ: 4GM(ξ) 4G 4G αK = = × Σπξ2 = × Σπ × D θ. c2ξ c2ξ c2 L

308 Solutions to exercises

Now, D α = LS α,K (Eqn 7.6) DS so 4πGΣ D D α = L LS θ, 2 c DS as required.

If we then set Σ=Σcr,wefind that α(θ)=θfor any θ,soβ=0.Thismeans that the gravitationallensisacting as aperfect focusing lens! However, this is a very special case—gravitationallensesingeneral do not focus light.As‘lenses’ in theoptical sense, theyhaveall formsofaberration,exceptofcoursechromatic aberrationsince gravitationallensing is strictly achromatic. Exercise 7.9 Fromlefttoright,theyare asaddlepoint,amaximumand a minimum. Exercise 7.10 Thetimedelay of theimage at the centreincreases. In adiagram likeFigure 7.15, thecentral panelshowing thegravitationaltimedelay wouldbe acquiring asharper andhigherpoint in thecentre. When the lens potential becomesasingular isothermal sphere, thetimedelay becomesinfinite, so the imagedisappears. Photons wouldtakeaninfinite amount of timetoclimbout of theinfinitely-deep potential well, and(by symmetry) spendanotherinfinite amount of timefalling in beforehand. But amore thoughtful answeristhatthis deep potential well wouldform ablack hole. Right from Equation7.1, we’ve been assuming aweak-field limit, so abetter answeristhatthese simple assumptions breakdownasthe potential becomesmore extreme. Exercise 7.11 Thebackground objects have thesameredshift, so we couldthink of theluminosity function as differential source counts, thus dN/dS ∝ S−α.Thereforethe numberofobjects perunitareabrighter than a 1−α flux S0 will be N(>S0)∝S0 ,which we couldwrite as 1−α N(>S0)=kS0 . If thebackground galaxies aregravitationally magnifiedbyafactor of µ,the intrinsicfluxes will be Sintrinsic = S/µ,while thecomovingvolumesampled will be smaller by afactor of 1/µ.Thereforethe numberofgalaxies brighter than an observed flux S0 will be C D k S 1−α N (>S )= 0 =kµ−1S1−αµα−1 = kS1−αµα−2 = N(>S )µα−2. lensed 0 µ µ 0 0 0 Therefore for amagnification of µ (where µ>1), thelensing changesthe number of background galaxies perunitareabyafactor of µα−2.For this factor to be biggerthan 1 we need µα−2 > 1, so log(µα−2) > log(1) =0, thus (α − 2) log(µ) > 0. We alreadyknowthat log(µ) > 0 (because µ>1), so this can happenonlyif α>2.For example, if the source countshaveaEuclidean slope (α =2.5), then 309 Solutions to exercises lensing wouldincrease thenumberofobjects.The effect of samplingless volumesdue to lensing, andsofinding fewerobjects than the fluxmagnification on its ownwouldsuggest,isknown as the Broadhurst effect.(SeeBroadhurst, T.J.,Taylor,A.N. andPeacock,J.A., 1995, AstrophysicalJournal, 438,49.) Exercise 8.1 There’snoguarantee that the re-emitted photon comesout in the same direction—in fact, it probablywon’t. Acorollaryisthatany Lyman α cloudshouldglowfaintly in Lyman α light in all directions from thesere-emitted photons,evenifthe cloudisnot interceptingour lineofsight to aquasar(because therewill always be some lineofsight that does).Thisre-emission is in general toofaint to detect. However, Lyman α emission can sometimes be seen if thereare internalionizing sources(e.g. star formation)within damped Lyman α systems, which you will meet later in the chapter. Exercise 8.2 Thecolumndensity through thecentrewill be thesameasthat seen through acubical cloudwith aside 2 Mpc,facing the observer(because the absorptiondoesn’t depend on thedistributionofmaterial that the light doesn’t pass through). One Mpc is about 3 × 1024 cm,sowecan write thedensity as (3 × 1024)3 cm−3 =2.7×1073 Mpc−3.The total numberofneutral hydrogen atomsinthe cube must be 2.7 × 1073 Mpc−3 × 8 Mpc3 =21.6×1073,which is spread overaprojected area of 2 × 2 Mpc2 =36×1048 cm2.Thereforethe columndensity must be 21.6 × 1073/(36 × 1048) cm−2 1 6 × 1024 cm−2. Exercise 8.3 In orderfor ahydrogen atom to absorb an Hα photon, the photon must have theright energy,and theremustbeanatomwith an electron in the n =2energy levelreadytoabsorb thephoton. This energy levelisat E=−13.6/n2 eV = −13.6/4 eV = −3.4 eV. In order to be in such astate, the atommusthaveabsorbed aphoton of energy (−3.4 eV) − (−13.6 eV)=10.2 eV. Photons of this energy require ablack body temperature of order 10.2 eV × 1.602 × 10−19 JeV−1 T 1 E/k = =120 000 K. 1.381 × 10−23 JK−1 This is hotter than thesurface of an Ostar,and is much hotter than thetypical temperatures in theintergalactic medium.Lyman α clouds aretoo cold to have many atomswith electrons alreadyexcited to the n =2level, so theclouds have almostnoHαabsorption. −3 Exercise 8.4 We can write σ(ν)=σ0(ν/νlimit) ,where −22 −2 σ0 =7.88 × 10 m ,and νlimit is thefrequency of theLyman limit. Writing −α Jν = kν andplugging theterms in,wefind ) ) ∞ ∞ −3 −α−1 ν (σJν/(hν)) dν ν (ν/νlimit) kν dν τ = N ) limit = N σ limit ) H I ∞ H I 0 ∞ −α−1 ν (Jν/(hν)) dν ν kν dν lim)it limit ∞ −α−4 −α−3 N σ0 ν ν dν N σ0 ν α = H I ) limit = H I limit ν−3 ∞ ν−α−1 dν ν−3 α +3 ν−α limit νlimit limit limit N σ α = H I 0 , α +3 wherewefirstcancelled the h terms, then cancelled the k terms. Setting τ>1, N > 1.3((α+3)/α) × 1021 −2 we find H I m ,asrequired.

310 Solutions to exercises

Exercise 8.5 Equation1.28 relates dz/dt to H(z).Taking the modulus andreciprocal of that equationgives (1 + z) |dt/dz| =1/H(z). Apopulation with constantproper sizes hasconstant A in Equation8.2, andaconstantcomovingdensity is constant nco in thesameequation. Therefore d2N∝(1 + z)3 |dt/dz|∝(1 + z)2/H(z).Ifwewrite 2 dX/dz =(1+z) H0/H(z),then $ $ $ 1 $ d2N = n A × (1 + z)2c $ $ dN dz co $H(z)$ H I gives c d2N = n A dX dN , co H I H0 which is constant. Exercise 8.6 Gravitationallensing of thebackground quasarbythe damped Lyman α system couldcause such an effect. Thestrengthofthiseffect, andthe biases that it creates on themeasured cosmicevolutionofneutral gas, arestill the subject of debate. However, it turns outthatthisisprobably onlya10–20%effect Ω z>2 on H I at . Exercise 8.7 Dust in thedampedLyman α systemsshouldreddenthe quasar spectra, so one might comparethe optical spectral indices or B–V colours of quasarswith andwithout damped Lyman α absorbers.However,ifdamped systemsare very dusty, they mayinducesomuchreddening that thequasars drop outofthe parent sample, so bright quasarcatalogueswouldbebiased to detecting low-reddening systems. Statistical analyses suggest that this latter effect doesnot dominate, butdirect results on quasarreddening arecurrently conflicting. Exercise 8.8 Theenergyofthe hydrogen Lymanlimit is E =13.6eV, i.e. E =13.6×1.602 × 10−19 J =2.179 × 10−18 J. This corresponds to afrequency of ν = E/h,where h is Planck’s constant, which comes out as ν =3.289 × 1015 Hz.The wavelengthofthislight is λ = c/ν, where c is the ,which comesout as λ =9.116 × 10−8 m, or 91.2 nm (i.e. 912 A)˚ to threesignificantfigures.For thehelium Lymanlimit, λHe = λ × 13.6/54.4=22.8 nm. Theredshifted hydrogen Lymanlimit in Figure 8.20 is at awavelength of 912 × (1 + z) A˚ =912 × (1 +3.2) A˚ =3830 A.˚

311