Exam (elaborations) TEST BANK FOR Fundamentals of Cosmology, 2nd Edition By James Rich (Solution Manual)

Exam (elaborations) TEST BANK FOR Fundamentals of Cosmology, 2nd ed By J. (James) Rich (Solution Manual) James Rich Solutions of Fundamentals of Cosmology 123 Solutions Chapter 1 1.1 If ΩM = ΩT = 1, then ΩM(a) = ΩT(a) = 1 for all a(t). Structure formation never ceases as larger and larger regions of negative Newtonian energy detach from the expansion. The values of the Ω’s for the ΩM = ΩT = 0.3 and the ΩM = 0.3,ΩΛ = 0.7 models are shown in Fig. 1. Fig. 1 The solid lines show ΩM(a) and ΩΛ(a) for (ΩM = 0.3,ΩΛ = 0.7). The dashed line shows ΩM(a) for (ΩM = ΩT = 0.3). The universe remains matter dominated for a longer period in the first case 1 2 Solutions 1.2 The time that has passed since the universe became vacuum dominated is (including only the vacuum energy density) t0 − tm=v H−1 0 ∼  da a √ ΩΛ = 1 3 √ ΩΛ ln(ΩΛ/ΩM) = 0.39 . (1) Numerical integration including both matter and vacuum gives 0.32. The duration of the matter-dominated epoch is (including only the matter density) tm=v − tr=m H−1 0 ∼ a−3/2 0  da a  ΩMa−3 ∼ (2/3) 1 Ω 1/2 Λ = 0.78 . (2) Numerical integration including matter and vacuum gives 0.69. The duration of the radiation-dominated epoch is (including only the radiation density) tr=m − tinf H−1 0 ∼ a−2 0  da a  ΩRa−4 ∼ (1/2) Ω 3/2 R Ω2M = 5.4 × 10−6 (3) for ΩR = 1.68Ωγ ∼ 8.5 × 10−5 (three massless neutrino species). Numerical integration including both radiation and matter gives 4.2 × 10−6. The time would not change by much if you had taken ainf = 0. The time when the first nuclei formed: tnuc − tinf H−1 0 ∼ a−2 0  da a  ΩRa−4 ∼ (3 × 10−9)2 2 √ ΩΛ ∼ 4.9 × 10−16, (4) i.e. 3.4 min. 1.3 The universe is expanding today because it was expanding yesterday (see (1.58)). It was expanding yesterday because..... It will be difficult to get an ultimate explanation since it will require knowledge of the physics that was in charge of things before the expansion began. Chapter 2 2.1 The flux from a typical galaxy of redshift z  1 is φ ∼ 2 × 1010L/(2eV/photon) 4π(zdH)2 ∼ 100m−2s−1/z2 . (5) Solutions 3 The ratio of the flux of nearby galaxies to that of nearby stars is 2 × 1010L/(1 Mpc)2 L/(1 pc)2 ∼ 2 × 10−2 . (6) 2.2 The total number of stellar photons can be roughly estimated as follows: nstarlight ∼ J0H−1 0 /(2 eV/photon) ∼ 108LMpc−3H−1 0 /2 ∼ 2 × 103m−3 , (7) which is much less than the number of CMB photons. The number of hydrogen nuclei transformed in order to produce these photons is n p→4He ∼ 2000m−3 2 eV 6MeV ∼ 0.6 × 10−3m−3 (8) or about 0.3 × 10−2 of the available hydrogen. Only a small amount of hydrogen has been transformed since most of it is still in intergalactic space. 2.3 Compton scattering dominates with a mean free path of order (ne(t0)σT)−1 ∼ 600dH , (9) where we have assumed that all matter is ionized (as suggested by the Gunn– Peterson effect). 2.4 It is possible to count the number of galaxies with a redshift less than z. The volume of the corresponding space is V = (4π/3)z3d3H ∝ h−3 70 so the measured number density is ∝ h3 70. Luminosities are determined by multiplying a measured flux by (zdH)2 and are, therefore, proportional to h−2 70 . The luminosity density ∼ ngalLgal is, therefore, proportional to h70. Galactic masses are determined from the rotation curve, M ∼ v2r/G. The radial distance r is proportional to the measured angular size and the redshift-determined distance so the mass is proportional to h−1 70 . Multiplying by ngal gives a mass density associated with galaxies proportional to h2 70. Dividing by the critical density gives an Ω independent of h70. 2.5 For NGC1365 Cepheids (Fig. 2.28) we have V(10 days) ∼ 27.5 , (10) while for LMC Cepheids (Fig. 2.5) we have V(10 days) ∼ 14.3 , (11) 4 Solutions so the ratio of distances is R(NGC1365) R(LMC) ∼ 10(27.5−14.3)/5 ∼ 436 . (12) For LMC distance of 50 kpc, this gives a distance of 21.8Mpc to NGC1365. For a recession velocity of 1441 km sec−1 this gives H0 = 65 km sec−1 Mpc−1. 2.6 The distance to A496 according to Hubble’s law is R ∼ H−1 0 × 9885 km sec−1 ∼ 141h−1 70 Mpc . (13) The full radius of A496 is about 3000 arcsec or about rc ∼ 2.0 h−1 70 Mpc . (14) Galaxy clusters have velocity dispersions that are roughly independent of distance from the cluster center implying a density profile of the form ρ(r ) ∼ M 4πrcr 2 , (15) where M and rc are the total mass and radius of the cluster. (Of course this form cannot be accurate near r = 0.) The gravitational energy of the cluster is Egrav = −  rc 0 GM(r ) r ρ(r)4πr 2dr ∼ −GM2 rc . (16) By the virial theorem, this must be, in magnitude, twice the total kinetic energy of the cluster. The square of the line-of-sight velocity dispersion, σv, is just twice the mean-squared velocity (obvious for circular orbits since sin2 ωt = 1/2). We, therefore, have (1/2)M(2σ2 v ) = (1/2) GM2 rc (17) giving M ∼ 2σ2 v rc G ∼ 4.8 h−1 70 × 1014M (18) for σv = 715 km sec−1. This determination of the total mass clearly supposes that the velocity dispersion of the galaxies is the same as that of the dark matter. For a luminosity of 2 h−2 70 1012L, mass-to-light ratio using the virial mass is M/L ∼ 230M/L. Assuming that this is representative of the universe as a whole we can estimate the mass density from the luminosity density: ρM ∼ J0(M/L) ∼ (1.2 h70 × 108LMpc−3) × 230 ∼ 2.8 h2 70 × 1010MMpc−3 (19) Solutions 5 giving ΩM ∼ 0.2 . (20) If we suppose that the ratio of the baryonic mass to total mass of A496 is representative of the universe as a whole, we find ΩM ∼ 0.04 h−2 70 2.4h−1 50 × 1014M 3.45h−5/2 50 × 1013M ∼ 0.47h−1/2 70 . (21) 2.7 The mean recession velocity of NGC 5033 is ∼ 875 km sec−1 corresponding to a distance of R = 4300 h−1 70 Mpc 875 300 × 103 = 12.5 h−1 70 Mpc . (22) At this distance, the visual radius of 3 arcmin corresponds to a radius of r = 3 60 π 180 12.5 h−1 70 Mpc ∼ 11 h−1 70 kpc . (23) The rotation velocity far from the galactic center is v = 1070 − 690 2 1 sin(65◦) ∼ 210 km sec−1 , (24) where we have used the inclination angle given in the publication. This gives a mass within 6 arcmin of the center of M = v2r G ∼ 2.2 × 1011h−1 70 M (25) The absolute magnitude is MV = 10.1 − 5 log  12.5h−1 70 × 106 pc 10 pc  = −20.4 + 5 log h70 (26) corresponding to a luminosity of L L = 100.4(4.64+20.4−5 log h70) = 1.0 h−2 70 × 1010 . (27) The mass-to-light ratio within 6 arcsec is then M L = 22h70 M L (28) 6 Solutions 2.8 The rotation velocity of the two components is v ∼ 475 − 135 2 ∼ 170 km sec−1 (29) which, for a period of 5.72 days corresponds to an inter-object distance of ∼ 2.7 × 1010m or ∼ 38R. This gives a reduced mass of ∼ 5.8M or about 10M per object. The eclipses last about 0.l5 of a period which gives a stellar diameter of ∼ 1.3 × 1010 m or a stellar radius of ∼ 9R. Using D/R ∼ 9.5 × 10−12, we find a distance to the system of about 43 kpc. Using an LMC distance of 45.7 kpc, the apparent luminosity relation of V = −2.765 log P + 17.044 gives an absolute luminosity of MV = 2.765 log P − 1.256. This is about 0.2 magnitudes fainter than the Hipparcos calibration. Using the Hipparcos calibration (brighter Cepheids) would then give galactic distances about 10% greater or an H0 10% smaller. 2.9 Consider a sphere of ionized hydrogen containing Np protons and Np electrons. The equation for hydrostatic equilibrium for a gravitating sphere is dP dr = −GM(r )ρ(r ) r 2 , (30) where P(r ) is the pressure, M(r ) is the mass contained within the radius r , and ρ(r ) is the mass density. The mean pressure P times total volume V = (4π/3)R3 is PV =  R 0 P(r)4πr 2dr = −  R 0 (4π/3)r 3 dP dr ρ(r)dr , (31) where in the second form we have integrated by parts and used the fact that P(R) = 0. Using the hydrostatic equilibrium value for the pressure gradient we find PV = 1/3  R 0 GM(r ) r ρ(r)4πr 2dr = −Eg 3 , (32) where Eg is the total gravitational potential energy of the sphere. Using the ideal gas law, we get an expression for the mean temperature T : 2NpkT = −Eg/3 . (33) Since the mean kinetic energy per particle is (3/2)kT , this is a form of the virial theorem stating that the kinetic energy particle is 1/2 the magnitude of the potential energy per particle: 2Np(3/2)kT = (1/2)|Eg| . (34) Solutions 7 For a uniform density we can evaluate (32) to find kT = (β/10) Gm2 pNP R β = 1 . (35) In a real star, the mass is concentrated near the center so the effective radius is less than R. Hence, a real star would be characterized by β 1. For example, the slightly more realistic distribution ρ ∝ r−2 gives β = 5/3. The number density of photons in thermal equilibrium is 2.4T 3/π2 so the number of photons in the star is approximately Nγ = (4π/3) (2.4/π2)(β3/1000)N3 p  mp mpl 6 (36) or Nγ Np = (4π/3) (2.4/π2)(β3/1000)N2 p  mp mpl 6 = 0.30 × 10−3β3  M M 2 , (37) where M = Npmp is the mass of the star. For the Sun, β ∼ 2 so Nγ ∼ 0.003Np. Since the number of photons is proportional to the third power of the number of protons, stars with M ∼ 30M will have comparable numbers of photons and protons. It follows that for such stars the radiation pressure ργ /3 ∼ nγ kT is comparable to the electron and proton pressure 2n pkT . At higher masses, stars are unstable because both the thermal energy Eth = 3NpkT + ργ V and the gravitational energy Eg = −3PV = −6NpkT − ργ V are much greater in magnitude than the total energy Eth + Eg = −3NpkT . Under such circumstances, small fluctuations in the photon density can disrupt the hydrostatic equilibrium and the star will eject mass until the temperature reaches an acceptable value. The total thermal energy of the photons is ργ V ∼ (4π/3)R3 × 2 π2 30 (kT )4(c)−3 . (38) Photons randomly walk through the Sun until escaping at the surface. For a series of N steps of size λi , the mean square distance from the point of origin is  N i=1 λi  · ⎛ ⎝ N j=1 λj ⎞ ⎠  =  N i=1 |λi |2  = N|λ|2 , (39) where we have assumed that the scatters completely randomize the direction: λi · λj = 0 for i = j . Setting this equal to the square of the solar radius we can estimate the mean time for a photon to travel to the surface: τ = Nλ/c ∼ (R/c) R λ , (40) 8 Solutions where λ = 1/(nσ) is the photon mean free path. Taking n = Np/V, the mean free path is λ ∼ (4π/3)R3 Npσ ∼ 0.018m  R R 3 M M σT σ (41) giving a mean escape time of τ ∼ (9/4π) Npσ Rc , (42) where σ is the photon-scattering cross section. If the electrons are completely ionized, σ = σT but we write σ = κσT with κ 1 to remind us of the atoms in the outer layers of the Sun. We thus have τ ∼ (9/4π) NpκσT Rc ∼ 0.8 × 104yr κ M M R R , (43) Dividing the total energy of the photons by the mean escape time we get an estimate of the luminosity: L = 32π4 27 × 30 ∼ 10−4β4κ −1N3 p  mp mpl 8 c2 σ ∼ 1.1 × 1026W  M M 3 β4κ −1 (44) which compares favorably with L = 3.8 × 1026 W. 2.10 The mean free path of a photon in a large cluster of mass ∼ 1014h−1M and diameter D ∼ 1h−1Mpc is λ −1 ∼ neσT ∼ 1014h−1M mp (1h−1Mpc)−3σT , (45) λ D ∼ mp 1014h−1M (1h−1Mpc)2 σT ∼ 10−71 × 1073 ∼ 100 . (46) Clusters are therefore rather transparent to their own photons. This is not surprising since we clearly see the galaxies in the clusters. The thermal averaged cross-section times velocity for bremsstrahlung is  v dσ dEγ  T = ασTc Eγ  c v  T ∼ ασTc Eγ  me T 1/2 . (47) Solutions 9 The luminosity is L ∼ D3n2 p  T 0 dEγ  v dσ dEγ  T Eγ ∼ n2 pαcσT  meT D3 . (48) But D = θ R where θ is the angular size of the cluster at a distance R. Furthermore, Np ∼ n pD3 so L ∼ N2 pαcσT √ meT R3θ3 . (49) The flux is then given by φx ∼ N2 pαcσT √ meT R5θ3 (50) which is the desired result. To find the total cluster mass in terms of the X-ray temperature, we modify (30) so that the pressure gradient supports only the baryons: dP dr = −GM(r )ρ(r ) f r 2 , (51) where f is the fraction of the total cluster mass in the form of baryons. Following the same reasoning as in Exercise 2.10, we find 6kT mp = |Egrav| Mtot ∼ GMtot R (52) which determines the total cluster mass. Chapter 3 3.1 The photon time-of-flight, t f , fromra to rb is given by the integral t f =  rb ra dr ˙r ˙r = 1 − 2GM/r = c(1 − 2GM/rc2) . The time measured by clock A between the emission of the two photons is ΔτA = t1  1 − 2GM/ra . (53) The time measured by clock B between its reception of the two photons is ΔτB = t1  1 − 2GM/rb . (54) 10 Solutions giving ΔτB ΔτA =  1 − 2GM/rb 1 − 2GM/ra →∞ for ra → 2GM . (55) This is approximately ΔτB/ΔτA = 1 + GM/ra − GM/rb . (56) For ra = 6.4 × 106 m (the radius of the Earth), rb = 2.02 × 107 m (the radius of GPS satellite orbits), and M = MEarth = 6.0 × 1024 kg, this gives ΔτB/ΔτA = 1 + GM/rearth − GM/rgps ∼ 1 + GM/rearthc2 (57) ∼ 1 + 7 × 10−10 . Note that the first-order Doppler effect due to satellite motion is of order  GM/rgpsc2 which is much greater than the gravitational effect. If clock C recedes slowly, v  1 (v  c), then the second photon is received by clock C at t ∼ t f + t1(1 + v), r ∼ rb + vt1. The metric gives along the path of clock C: dτ = dt  (1 − 2GM/r ) − v2 1 − 2GM/r 1/2 ∼ dt  (1 − 2GM/rb) 1/2 . Integrating from t = t f to t = t f + t1(1 + v) gives ΔτC ∼ ∼ t1(1 + v)(1 − GM/rB) which gives ΔτB/ΔτC = 1 − v . (58) This is the well-known first-order Doppler effect. 3.2 The result follows from the chain rule: dτ 2 = ημνd˜xμ ˜xν = ημν ∂ ˜xμ ∂xα dxα ∂ ˜xν ∂xβ dxβ (59) = ηαβdxαdxβ . (