James Rich

Solutions ofFundamentalsof Cosmology

123

Solutions

Chapter 1

1.1 If ΩM = ΩT = 1, then ΩM(a) = ΩT(a) = 1 for all a(t). Structure formationnever ceases as larger and larger regions of negative Newtonian energy detach fromthe expansion.

The values of the Ω’s for the ΩM = ΩT = 0.3 and the ΩM = 0.3,ΩΛ = 0.7models 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 showsΩM(a) for (ΩM = ΩT = 0.3). The universe remains matter dominated for a longer period in thefirst 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−10

∼∫

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−10

∼ a−3/20

∫da

a√

ΩMa−3∼ (2/3)

1

Ω1/2Λ

= 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−10

∼ a−20

∫da

a√

ΩRa−4∼ (1/2)

Ω3/2R

Ω2M

= 5.4 × 10−6 (3)

for ΩR = 1.68Ωγ ∼ 8.5 × 10−5 (three massless neutrino species). Numericalintegration including both radiation and matter gives 4.2 × 10−6. The time wouldnot change by much if you had taken ainf = 0.

The time when the first nuclei formed:

tnuc − tinf

H−10

∼ a−20

∫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 knowledgeof 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 × 1010 L�/(2eV/photon)

4π (zdH )2∼ 100 m−2s−1/z2 . (5)

Solutions 3

The ratio of the flux of nearby galaxies to that of nearby stars is

2 × 1010 L�/(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 ∼ J0 H−10 /(2 eV/photon) ∼ 108 L�Mpc−3 H−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 ∼ 2000 m−3 2 eV

6 MeV∼ 0.6 × 10−3m−3 (8)

or about 0.3 × 10−2 of the available hydrogen. Only a small amount of hydrogenhas 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. Thevolume of the corresponding space is V = (4π/3)z3d3

H ∝ h−370 so the measured

number density is ∝ h370.

Luminosities are determined by multiplying a measured flux by (zdH)2 and are,therefore, proportional to h−2

70 . The luminosity density ∼ ngal Lgal is, therefore, pro-portional to h70.

Galactic masses are determined from the rotation curve, M ∼ v2r/G. The radialdistance r is proportional to the measured angular size and the redshift-determineddistance so the mass is proportional to h−1

70 . Multiplying by ngal gives a mass densityassociated with galaxies proportional to h2

70. Dividing by the critical density givesan 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 21.8 Mpc to NGC1365. For arecession 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−10 × 9885 km sec−1 ∼ 141h−1

70 Mpc . (13)

The full radius of A496 is about 3000 arcsec or about

rc ∼ 2.0 h−170 Mpc . (14)

Galaxy clusters have velocity dispersions that are roughly independent of distancefrom the cluster center implying a density profile of the form

ρ(r ) ∼ M

4πrcr2, (15)

where M and rc are the total mass and radius of the cluster. (Of course this formcannot be accurate near r = 0.) The gravitational energy of the cluster is

Egrav = −∫ rc

0

G M(r )

rρ(r )4πr2dr ∼ −G M2

rc. (16)

By the virial theorem, this must be, in magnitude, twice the total kinetic energyof the cluster. The square of the line-of-sight velocity dispersion, σv , is just twicethe mean-squared velocity (obvious for circular orbits since 〈sin2 ωt〉 = 1/2). We,therefore, have

(1/2)M(2σ 2v ) = (1/2)

G M2

rc(17)

giving

M ∼ 2σ 2v rc

G∼ 4.8 h−1

70 × 1014 M� (18)

for σv = 715 km sec−1. This determination of the total mass clearly supposes thatthe velocity dispersion of the galaxies is the same as that of the dark matter.

For a luminosity of 2 h−270 1012 L�, mass-to-light ratio using the virial mass is

M/L ∼ 230M�/L�. Assuming that this is representative of the universe as a wholewe can estimate the mass density from the luminosity density:

ρM ∼ J0(M/L) ∼ (1.2 h70 × 108L�Mpc−3) × 230

∼ 2.8 h270 × 1010 M�Mpc−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 repre-sentative of the universe as a whole, we find

ΩM ∼ 0.04 h−270

2.4h−150 × 1014 M�

3.45h−5/250 × 1013 M�

∼ 0.47h−1/270 . (21)

2.7 The mean recession velocity of NGC 5033 is ∼ 875 km sec−1 corresponding toa distance of

R = 4300 h−170 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

18012.5 h−1

70 Mpc ∼ 11 h−170 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 masswithin 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 × 1010 m 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 theHipparcos calibration (brighter Cepheids) would then give galactic distances about10% 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= −G M(r )ρ(r )

r2, (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

0P(r )4πr2dr = −

∫ R

0(4π/3)r3 dP

drρ(r )dr , (31)

where in the second form we have integrated by parts and used the fact thatP(R) = 0. Using the hydrostatic equilibrium value for the pressure gradient wefind

PV = 1/3∫ R

0

G M(r )

rρ(r )4πr2dr = −Eg

3, (32)

where Eg is the total gravitational potential energy of the sphere. Using the ideal gaslaw, 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 virialtheorem stating that the kinetic energy particle is 1/2 the magnitude of the potentialenergy 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

p NP

Rβ = 1 . (35)

In a real star, the mass is concentrated near the center so the effective radius isless than R. Hence, a real star would be characterized by β > 1. For example, theslightly 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 numberof photons in the star is approximately

Nγ = (4π/3) (2.4/π2)(β3/1000)N 3p

(m p

mpl

)6

(36)

or

Nγ

Np= (4π/3) (2.4/π2)(β3/1000)N 2

p

(m p

mpl

)6

= 0.30 × 10−3β3

(M

M�

)2

, (37)

where M = Npm p 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 pro-tons, 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 comparableto the electron and proton pressure 2n pkT . At higher masses, stars are unstablebecause both the thermal energy Eth = 3NpkT + ργ V and the gravitational energyEg = −3PV = −6NpkT − ργ V are much greater in magnitude than the totalenergy Eth + Eg = −3NpkT . Under such circumstances, small fluctuations in thephoton density can disrupt the hydrostatic equilibrium and the star will eject massuntil the temperature reaches an acceptable value.

The total thermal energy of the photons is

ργ V ∼ (4π/3)R3 × 2π2

30(kT )4(�c)−3 . (38)

Photons randomly walk through the Sun until escaping at the surface. For a seriesof N steps of size λi , the mean square distance from the point of origin is

⟨(N∑

i=1

λi

)·⎛⎝ N∑

j=1

λ j

⎞⎠⟩

=⟨

N∑i=1

|λi |2⟩

= N 〈|λ|2〉 , (39)

where we have assumed that the scatters completely randomize the direction:〈λi · λ j 〉 = 0 for i �= j . Setting this equal to the square of the solar radius wecan 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 freepath is

λ ∼ (4π/3)R3

Npσ∼ 0.018 m

(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 ion-ized, σ = σT but we write σ = κσT with κ > 1 to remind us of the atoms in theouter 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 anestimate of the luminosity:

L = 32π4

27 × 30∼ 10−4β4κ−1 N 3

p

(m p

mpl

)8�c2

σ

∼ 1.1 × 1026 W

(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−1 M� anddiameter D ∼ 1h−1Mpc is

λ−1 ∼ neσT ∼ 1014h−1 M�m p

(1h−1Mpc)−3σT , (45)

λ

D∼ m p

1014h−1 M�

(1h−1Mpc)2

σT∼ 10−71 × 1073 ∼ 100 . (46)

Clusters are therefore rather transparent to their own photons. This is not surprisingsince 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 ∼ D3n2p

∫ T

0dEγ

⟨v

dσ

dEγ

⟩T

Eγ ∼ n2pαcσT

√meT D3 . (48)

But D = θ R where θ is the angular size of the cluster at a distance R. Furthermore,Np ∼ n p D3 so

L ∼ N 2pαcσT

√meT

R3θ3. (49)

The flux is then given by

φx ∼ N 2pαcσT

√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= −G M(r )ρ(r ) f

r2, (51)

where f is the fraction of the total cluster mass in the form of baryons. Followingthe same reasoning as in Exercise 2.10, we find

6kT

m p= |Egrav|

Mtot∼ G Mtot

R(52)

which determines the total cluster mass.

Chapter 3

3.1 The photon time-of-flight, t f , from ra to rb is given by the integral

t f =∫ rb

ra

dr

rr = 1 − 2G M/r = c(1 − 2G M/rc2) .

The time measured by clock A between the emission of the two photons is

ΔτA = t1√

1 − 2G M/ra . (53)

The time measured by clock B between its reception of the two photons is

ΔτB = t1√

1 − 2G M/rb . (54)

10 Solutions

giving

ΔτB

ΔτA=

√1 − 2G M/rb

1 − 2G M/ra→ ∞ for ra → 2G M . (55)

This is approximately

ΔτB/ΔτA = 1 + G M/ra − G M/rb . (56)

For ra = 6.4 × 106 m (the radius of the Earth), rb = 2.02 × 107 m (the radius ofGPS satellite orbits), and M = MEarth = 6.0 × 1024 kg, this gives

ΔτB/ΔτA = 1 + G M/rearth − G M/rgps ∼ 1 + G M/rearthc2 (57)

∼ 1 + 7 × 10−10 .

Note that the first-order Doppler effect due to satellite motion is of order√

G M/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 ofclock C:

dτ = dt

[(1 − 2G M/r ) − v2

1 − 2G M/r

]1/2

∼ dt[(1 − 2G M/rb)

]1/2.

Integrating from t = t f to t = t f + t1(1 + v) gives

ΔτC ∼ ∼ t1(1 + v)(1 − G M/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 = ημνdxμ xν = ημν

∂ xμ

∂xαdxα ∂ xν

∂xβdxβ (59)

= ηαβdxαdxβ . (60)

Solutions 11

Equation (3.124) is equivalent to

1 =(

∂ x0

∂x0

)2

−3∑

i=1

(∂ x i

∂x0

)2

, (61)

−1 =(

∂ x0

∂x j

)2

−3∑

i=1

(∂ x i

∂x j

)2

j = 1, 3 . (62)

These equations are clearly satisfied for boosts and rotations, e.g.

− 1 = −(cos2 θ + sin2 θ ) (63)

and

1 = γ 2 − β2γ 2 . (64)

3.3 The geodesic equation is

d

dτ

(gμν

dxμ

dτ

)− (1/2)

∂gμλ

∂xν

dxμ

dτ

dxλ

dτ= 0 ν = 0, 3 . (65)

For gμν = ημν all the derivatives of the metric vanish so the geodesic equation is

ημν

d

dτ

(dxμ

dτ

)= 0 (66)

which implies that the four-velocity dxμ/dτ is constant.

3.4 For a slow rocket moving in the vertical (z) direction, the proper time is

dτ = dt[1 + 2gz − z2

]1/2 = dt[1 + gz − (1/2)z2

], (67)

where we neglect higher order terms in gz and z. The Newtonian solution is

z(t) = g(t21 − t2)/2 ⇒ z = −gt (68)

giving

dτ = dt[1 + g2(t2

1 − t2)/2 − g2t2/2]

. (69)

Integrating this over the rocket’s flight from t = −t1 to t = t1 we find Δτ =2t1(1 + g2t2

1 /6) so the rocket-bound clock counts more time than the Earth-boundclock, as expected.

12 Solutions

For the general trajectory, the acceleration γ replaces g in (68) but not in theformula for proper time (67) so

dτ = dt[1 + gγ (t2

1 − t2)/2 − γ 2t2/2]

. (70)

Integration gives (3.17).For an airplane flying at altitude h we have

dτ = dt[1 + gh − v2/2

], (71)

so the comparison between an airplane’s clock and an Earth-bound clock dependson the values of h and v.

3.5 For a slow particle we have dx0/dτ ∼ 1 and dxi/dτ ∼ 0, so the geodesicequation is

d

dτ

(gμν

dxμ

dτ

)= (1/2)

∂g00

∂xνν = 0, 1, 2, 3 . (72)

With g00 = 1 + 2Φ and gii = −1 + 2Φ, the ν = i equation is (neglecting termsproportional to the velocity)

d2xi

dτ 2= −∂Φ

∂xii = 1, 2, 3 , (73)

which is the Newtonian result. The ν = 0 equation gives the relative rates of sta-tionary and moving clocks:

d

dτ

[(1 + 2Φ)

dx0

dτ

]= 0 ⇒ (1 + 2Φ)

dx0

dτ= constant , (74)

which is the relativistic analog of energy conservation.

3.6 Since the lines of constant θ are perpendicular to lines of constant φ, the relation

dS2 = a2(dχ2 + sin2 χdθ2) (75)

follows from the Pythagorean theorem.For r = sin χ , we have dr = √

1 − r2dχ from which it follows that

dS2 = a2

[dr2

1 − r2+ r2dθ2

]. (76)

Consider an object of size dS extending from (χ1, θ ) to (χ1, θ + dθ ) with dS =ar1dθ . Photons emitted toward the pole χ = 0 from the ends of the object will

Solutions 13

follow paths of constant θ so the object will appear to have an angular size

dθ = dS

ar1. (77)

If the sphere is expanding, the photons will still follow lines of constant θ becausethe spherical symmetry (viewed from the center of the sphere) is maintained so thephotons have no preferred direction toward which they can be deviated. In this case,the angular size is

dθ = dS

a(t1)r1, (78)

where t1 is time of emission. This is the same distance–angular size relation as(3.70).

At t0, the photons are spread uniformly over a circle of radius a0χ1 and thereforeof circumference 2πa0r1. The energy flux is then

F = L

2πa0r1(1 + z)2, (79)

where the factors of 1 + z take into account the redshift and “time dilation” due tothe expansion of the sphere. This relation can be compared with (3.63).

3.7 With the constraint x2 + y2 + z2 + w2 = a2, the spatial metric is

dS2 = dx2 + dy2 + dz2 + (xdx + ydy + zdz)2

a2 − x2 − y2 − z2. (80)

Using the transformations x = ra sin θ cos φ, y = ra sin θ sin φ, and z =ra cos θ , we find the Robertson–Walker metric with k = 1.

3.8 All geodesics passing through a point pass through the point’s antipode and viceversa.

3.9 To order z2, we find d1 = dA < d0 < dL . The distances differ in the coefficientof the z2 term so they differ by ∼ 10% at z ∼ 0.1 or R ∼ 430h−1

70 Mpc.

3.10 Taking the original galaxy to be at χ = 0, the initial condition is

[da(t)χ

dt

]t0

= a0χ(t0) = v . (81)

The derivative is with respect to t since this is the time measured by the co-movingclock. Using the metric dτ = dt(1 − a2χ2)1/2, we then find

dχ

dτ(t0) = v

a0

√1 − v2

, (82)

14 Solutions

dt

dτ(t0) = 1√

1 − v2. (83)

Equation (3.87) implies

a2χ ′ = constant = a0v(1 − v2/c2)−1/2 . (84)

Using dτ = dt(1 − a2χ2)1/2 we get

χ = B

a√

a2 + B2B = a0v√

1 − v2. (85)

For B → ∞, we get χ = 1/a as expected for photons whereas for v � 1 we have

χ = a0v

a2. (86)

For a short voyage, (86) integrates to

t = a0χ (t)

v= zdH

v, (87)

where z is the redshift (viewed from χ = 0) of the galaxy at χ (t). For v = 10−1cand z = 10−2 this gives t = tH/10. The voyagers’ proper time is less than this by afactor (1 − v2/c2)1/2.

The most distant galaxy that can be visited has a coordinate

χ = a0v

∫ ∞

t0

dt

a2. (88)

For an empty universe (a ∝ t), this gives

χ = vt0a0

⇒ a0χ = dHv/c (89)

corresponding to a redshift z = v/c (not too surprising). For a critical universe(a ∝ t2/3), the most distant galaxy that can be reached has a redshift of z = 2v/c.In a Newtonian treatment of this problem, the extra distance traveled in a criticaluniverse would be explained by the fact that distant galaxies experience more decel-eration than the explorer because the explorer starts at χ = 0 where the decelerationvanishes because M(χ = 0) = 0.

3.13 The photons follow lines of unit slope so we have

vt2 = t2 − t1 vt3 = t4 − t3 (90)

Solutions 15

which implies

t2 = t11 − β

t3 = t41 + β

. (91)

The time dilation postulate implies that the moving clock measures a time that issmaller by a factor γ :

t ′2 = t1

γ (1 − β)t ′3 = t4

γ (1 + β). (92)

The time t ′ is then given by

t ′ = t ′3 + t ′

2

2= 1

2γ

[t4

1 + β+ t1

1 − β

]= γ

[t4 + t1

2− β(t4 − t1)

2

]

= γ t − βγ x (93)

as expected for a Lorentz transformation. Likewise we have

x ′ = t ′3 − t ′

2

2= 1

2γ

[t4

1 + β− t1

1 − β

]= γ

[t4 − t1

2− β(t4 + t1)

2

]

= γ x − βγ t . (94)

To demonstrate the Lorentz contraction formula, we suppose that B measures thelength of the rod by radar, emitting two photons at t = −L/2 and receiving theechos at t = L/2. B sees a Lorentz contraction because B’s clock measures t ′ =±0.5L

√1 − β2 and therefore concludes that the bar has a length L ′ = L

√1 − β2.

3.14 For an isothermal sphere, the Einstein ring has an angular size

θE = 4πσ 2

v

c2

DLS

DOSisothermal sphere . (95)

A galaxy with a rotation velocity of 200 km s−1 has (σv/c)2 = 2 × 10−7 and θE ∼3 × 10−6 (for DLS ∼ DOS). The volume behind the Einstein ring with redshiftsbetween z and z + Δz is

ΔV = a30r (z)2Δχ ΔΩ , (96)

where r (z) and χ (z) are the co-moving radial coordinates and ΔΩ ∼ πθ2E. Approx-

imating a0r (z) ∼ a0χ (z) ∼ (c/H0)z, this gives

ΔV = (c/H0)3z2Δz ΔΩ ∼ 3 × 10−2Mpc3 (97)

for z = 0.45 and Δz = 0.1. Multiplying by the galaxy number density(∼ 0.015 Mpc−3) gives ∼ 4 × 10−4 for the probability that a galaxy is within the

16 Solutions

Einstein ring. Integrating up to, say, z ∼ 1 suggests that about 0.1% of galaxies havea strongly lensed background galaxy behind it.

Chapter 4

4.1 For a = 0 and for dr = 0 (for a co-moving clock), we have

dT = dt + (1/2)(a/a)2 (ar )2dt = dt + (1/2)v2dt , (98)

where v = H R is the Hubble velocity of the co-moving clock. This gives

dt = dT

1 + v2/2∼ dT (1 − v2/2) . (99)

This is as predicted by special relativity where we expect the time dt measured bythe moving clock to be smaller than the time dT measured by the stationary clockby a factor

√1 − v2 ∼ 1 − v2/2.

To second order, the trajectory of the photon in Fig. 4.1 between ta and t is gov-erned by

a(t)χ ∼∫ t

ta

dt ′ (1 − H (t)(t ′ − t)) = (t − ta) + H (t)

2(t − ta)2 . (100)

For the trajectory between t and tb we have

a(t)χ ∼ (tb − t) − H (t)

2(tb − t)2 . (101)

Taking the sum we find

a(t)χ ∼ (tb − t1)

2+ H (t)

4

[(t − ta)2 − (tb − t)2

]. (102)

Since, to first order, both (t − ta) and (tb − t) are equal to a(t)χ , the second termvanishes to order χ2 so we have

a(t)r ∼ (tb − t1)

2+ O(r3) , (103)

where we use the fact that χ = r + O(r3).Taking the difference between the two trajectories we have

0 = tb + ta − 2t − H (t)

2

[(tb − t)2 + (t − ta)2

]. (104)

Solutions 17

Using (t − ta) ∼ (tb − t) ∼ a(t)r , we find

tb + ta2

= t + (1/2) aar2 + O(r3) (105)

which is equivalent to (4.13).

4.2 The Ricci tensor at the origin of a freely falling coordinate system is given by

Rβγ = ηγ δ Rαβγ δ

= 1/2

[ηγ δ ∂2gβδ

∂xα∂xγ+ ∂2gγ

γ

∂xβ∂xδ− ∂2gα

β

∂xα∂xδ− ∂2gγ

δ

∂xβ∂xγ

]. (106)

The Ricci scalar is

R = ηγ δ Rγ δ = ηαγ ∂2gδδ

∂xα∂xγ− ∂2gγ δ

∂xγ ∂xδ.

The derivatives are

∂ Rμδ

∂xμ= (1/2)

[ημβ ∂3gγ

γ

∂xμ∂xβ∂xδ− ∂3gαμ

∂xμ∂xα∂xδ

](107)

and

ημδ

∂ R

∂xμ= 2

∂ Rμδ

∂xμ(108)

which proves the desired result.

4.3 For a nonrelativistic ideal gas, E ∼ m and 〈p2/2m〉 = (3/2)T from which itfollows that p = nT � nm = ρ. For a relativistic ideal gas, p = E from which itfollows that p = ρ/3.

The 4-divergence of T μν for a collection of particles is

∂T μν

∂xμ=

∫d3 p pμ

[∂ F

∂t+ v · ∇R F

]. (109)

The quantity in brackets vanishes by Liouville’s theorem which proves that the 4-divergence vanishes.

4.4 Consider a freely falling coordinate system x and another coordinate system xdefined by

xμ = ∂xμ

∂ xαxα , xα = ∂ xα

∂xμxμ , (110)

18 Solutions

where the transformation coefficients ∂ xα/∂xμ are constants satisfying

ηαβ

∂ xα

∂xμ

∂ xβ

∂xν= ημν ⇒ ηαβ ∂ xα

∂xμ

∂ xβ

∂xν= ημν . (111)

The proper time is given by

dτ 2 = gμνdxμdxν = gμν

∂ xμ

∂xα

∂ xν

∂xβdxαdxβ (112)

which shows that the metric is a tensor

gαβ = gμν

∂ xμ

∂xα

∂ xν

∂xβ. (113)

Since the x system is freely falling, we have

gαβ = [ηκλ + (1/2)gκλ,μν xμ xν + .....

] ∂ xκ

∂xα

∂ xλ

∂xβ. (114)

Using (111), we see that the x system is also freely falling:

gαβ = ηαβ + (1/2)gκλ,μν xμ xν ∂ xκ

∂xα

∂ xλ

∂xβ+ .....

= ηαβ + (1/2)

[gκλ,μν

∂ xκ

∂xα

∂ xλ

∂xβ

∂ xμ

∂xγ

∂ xν

∂xδ

]xγ xδ + ..... (115)

The quantity in brackets is gαβγ δ so apparently this quantity is a tensor. Since Rαβγ δ

is the sum of terms of this sort, it follows that it is also a tensor.It is easy to show that the “Ricci tensor” is also a tensor:

Rβδ = ηαγ Rαβγ δ = ηαγ ∂ xκ

∂xα

∂ xμ

∂xγRκλμν

∂ xλ

∂xβ

∂ xν

∂xδ

= ηκμ Rκλμν

∂ xλ

∂xβ

∂ xν

∂xδ= Rλν

∂ xλ

∂xβ

∂ xν

∂xδ. (116)

Similarly, one finds the ημνηαβ Rαβ is a tensor, from which it follows that Gμν is a

tensor. The Einstein equation therefore equates two tensors (G and T ) so if it holdin one freely falling frame it must hold in all other frames related to it by a Lorentztransformation.

4.5 The Einstein equation at the origin of a freely falling coordinate system is

Rμν − (1/2)ημν R = −8πGTμν , (117)

Solutions 19

where the Ricci scalar is

R = ημν Rμν . (118)

Contracting the Einstein equation with ημν we find

− R = −8πGημνTμν . (119)

Substituting this into the Einstein equation we find

Rμν = −8πGTμν + 4πGημνηαβ Tαβ , (120)

which is the required result.

4.6 Consider a particle with only gravitational interactions inside a spherical objectof density ρ. The acceleration of the particle is

d2xi

dt2= −4πGρ

3xi . (121)

Two particles initially at rest separated by δxi will have a relative acceleration of

d2δxi

dt2= −4πGρ

3δ . (122)

The invariant separation δ2 = −ημνδxμδxν = (δxi )2 obeys the equation

d2δ2

dt2= 2δ

d2δ

dt2= −2δ

4πGρ

3δ − −2R1010δ

2 , (123)

where we have used dδ/dt ∼ 0 and dt ∼ dτ for particles at rest. We conclude that

Ri0i0 = 4πGρ

3. (124)

This implies

R00 = ημν Rμ0ν0 = −4πGρ . (125)

This is consistent with the second form of the Einstein equation:

R00 = −8πG[T00 − (1/2)g00gμνTμν

] = −4πGρ (126)

if the pressure is negligible.For particles outside the sphere of mass M , the Newtonian acceleration is

d2xi

dt2= −G M

r3xi . (127)

20 Solutions

For two particles separated by δ in the radial direction, the relative accelerationis ∝ Δ(r−2) = −2δ/r3. For two particles separated in a nonradial direction, therelative acceleration is ∝ δ(r−3). This implies

R00 = (−2 + 1 + 1)G M/r3 = 0 . (128)

4.7 For an empty universe we have

R = r t , T = t√

1 + r2 − t0 , (129)

so for a particle at fixed r we have

dR = rdt , dT = dt√

1 + r2 , (130)

implying

dR

dt= r√

1 + r2< 1 (131)

4.8 Using equation (4.107) for the Einstein tensor for the orthogonal coordinatesystem with metric (3.89) we find

G00 = − ∂2g11

∂x2∂x2− ∂2g22

∂x3∂x3− ∂2g33

∂x1∂x1= −2∇2Φ (132)

so the 00 component of the Einstein equation is

∇2Φ = 4πρ . (133)

The other components of Gμν vanish as expected for vanishing pressure.

Chapter 5

5.1 For ΩT = ΩM = 1, we find

a(t) ∝ t2/3 , q0 = 1/2 , t0 = (2/3)H−10 , (134)

a0χ1(z) = a0r1(z) = 2H−10 (1 − 1/

√1 + z) . (135)

For ΩT = ΩΛ = 1, we find

a(t) ∝ exp(H0t) , q0 = −1 , t0 = ∞ (136)

a0χ1(z) = a0r1(z) = zH−10 . (137)

Solutions 21

For ΩT = ΩM = 0, we find

a(t) ∝ t , q0 = 0 , t0 = H−10 (138)

a0χ1(z) = H−10 ln(1 + z) . (139)

Using a0 = H−10 for an empty universe, we find

r1(z) = 1 + z − (1 + z)−1

2. (140)

The age for the ΩT = ΩΛ = 1 universe would obviously be modified by includingradiation energy. The luminosity and angular distances are found by multiplying ordividing a0r1 by 1 + z. For ΩT = ΩM = 1, the minimum angular size occurs forz = 1.25.

5.2 The equation should be a good approximation as long as there are no other rela-tivistic species, i.e., for T � me. After the neutrinos start to become nonrelativistic,the equation is still a good approximation because the ΩM term dominates in anycase.

For trec we can make the approximation that the universe is matter dominated inwhich case the age is just (2/3) the Hubble time at that epoch:

trec = (2/3)H−10 Ω

−1/2M a3/2

rec ∼ 2.6 h−170 × 105 yr√

ΩM. (141)

For teq, we can neglect neither the radiation nor the matter (at teq) so it is necessaryto do a nontrivial integral:

teq = H−10

(1.68Ωγ)3/2

Ω2M

∫ 1

0

xdx√1 + x

∼ 5 × 104yr

(0.3

ΩMh270

)2

. (142)

Neglecting the matter would have given the correct order of magnitude and thecorrect dependence on ΩMh2

70.

5.3 For ΩT = ΩM = 1 + ε we find

t0 = H−10

∫ 1

0

da

a[(1 + ε)a−3 − εa−2

]1/2 , (143)

= H−10

∫ 1

0

a1/2da

[1 + ε(1 − a)]1/2 , (144)

∼ H−10

∫ 1

0daa1/2 [1 − ε(1 − a)/2] ∼ (2/3)H−1

0 (1 − ε/5) . (145)

22 Solutions

5.4 t0 ∼ (2/3)H−10 [1 + ΩΛ/3 + ......]

5.5 The relation

a0χ1(z) ∼ a0r1(z) = H−10 z [1 − (1 + q0)z/2 + .....] (146)

implies

a0dχ1 = H−10 dz [1 − (1 + q0)z + .....] (147)

which implies

a30r2

1 dχ1 ∼ H−30 z2dz [1 − 2(1 + q0)z] . (148)

5.6 The probability for Compton scattering the CMB photons in a galaxy cluster ofdiameter D is

P ∼ σT D〈ne〉 = σT dA(z)Δθ〈ne〉 (149)

where Δθ is the observed angular size of the cluster. The X-ray luminosity of thecluster is

L ∼ ασTc√

mec2kT D3〈n2e〉 (150)

giving an observed energy flux of

φ ∼ ασTc√

mec2kT (dAΔθ )3〈n2e〉d−2

L , (151)

where we ignore the numerical factors. Dividing P2 by φ and rearranging, we find

a0χ1(z) ∼ P2αc√

mec2kT Δθ

(1 + z)3φσT

〈n2e〉

〈ne〉2. (152)

5.7 The luminosity of a blackbody of radius D is L = π D2σ T 4. The measuredenergy flux is

φ = L

4πd2L

= D2σ T 4

4(a0χ1(1 + z))2. (153)

The diameter of a supernova photosphere can be estimated from its photosphereexpansion velocity, v:

D = 2vt

1 + z, (154)

Solutions 23

where t is the observed time since the explosion and the factor (1 + z) takes intoaccount the cosmological time dilation. Combining these two equations we find

a0χ1(z) = vt

(1 + z)2

(σ T 4

φ

)1/2

. (155)

5.8 The horizon for ΩM = ΩT > 1 is given by

a0χhor = H−10

∫ 1

0

da

a2[ΩTa−3 + (1 − ΩT)a−2

]1/2 , (156)

∫ (ΩT−1)/ΩT

0

dx√x(1 − x)

= π/2 − sin−1 [(2 − ΩT)/ΩT] . (157)

The fraction of universe within the horizon is

Vhor

Vtot=

∫ χhor

0 sin2χdχ∫ π

0 sin2χdχ= [χhor − (1/2) sin 2χhor] /π . (158)

For ΩT ∼ 1 this can be approximated by

Vhor

Vtot= 16

3π(ΩT − 1)3/2 [1 + O(ΩT − 1)] . (159)

The horizon at amax can be found by integrating from a = 0 to a = amax or byintegrating from a = 0 to a = 1 and taking ΩT → ∞.

The age at maximal expansion for ΩM = ΩT = 1 is

tamax =∫ amax

0

da

a= H−1

0 Ω−1/2T

∫ ΩT/(ΩT−1)

0

a1/2da[1 − ΩT−1

ΩTa] , (160)

= π

2H−1

0

ΩT

(ΩT − 1)3/2. (161)

5.9 The horizon at time t is given by

a0χhor(t) = H−10

∫ a

0

da

a2[ΩMa−3 + (1 − ΩT)a−2 + ΩΛ

]1/2 . (162)

The integral clearly converges for a → ∞ if and only if ΩΛ > 0.

24 Solutions

5.10 For a universe with ΩT = ΩM < 1 we have

limz→∞ χ1(z) = H−1

0

a0

∫ 1

0

da

a2[ΩTa−3 + (1 − ΩT)a−2

]1/2 . (163)

Using a0 = H−10 /(1 − ΩT)1/2 we find

limz→∞ χ1(z) =

∫ (2−ΩT)/ΩT

1

dx√x2 − 1

= ln

[2 − ΩT + 2

√1 − ΩT

ΩT

](164)

which implies

limz→∞ r1(z) = sinh χ = 2

√1 − ΩT

ΩT(165)

implying

limz→∞ dA(z) = a0r

1 + z= 2H−1

0

ΩTz. (166)

5.11 For ΩT = 1 with ΩΛ � ΩM, the Friedmann equation is

a

a= H0

[ΩMa−3 + ΩΛ

]1/2 ∼ H0

√ΩMa−3/2

[1 + ΩΛ

2ΩMa3

]. (167)

This gives

limz→∞ a0r1(z) = 2H−1

0√ΩM

[1 − ΩΛ

14ΩM

](168)

with dA(z) = a0r1/(1 + z). Supposing that the universe is matter dominated atrecombination with H 2 = H 2

0 ΩMa−3rec , the Hubble angle is

θH ∼ 1

2√

1 + z

[1 − ΩΛ

14ΩM

]. (169)

The factor (1/14) makes the answer relatively insensitive the relative values of ΩM

and ΩΛ for ΩM �= 0.

5.12 Absorption by interstellar matter would not resolve Olbers’ paradox becausethe matter would heat up until it reached a temperature at which it would radiateblackbody photons at the same rate as it absorbed starlight.

Solutions 25

The modern calculation gives a flux per unit solid angle of

dφ

dΩ= n0a3

0 L∫

r21 dχ1

4πd2L

= n0L

4π

∫ t0

0dt1

a(t1)

a0

= n0L

4π

∫ a0

0

da

a0(a/a), (170)

where we have used dL = a0r1(1+z) and dχ = dt/a. Using the Friedmann equationto evaluate a/a we find

dφ

dΩ= (3/5)

n0Lt04π

, t0 = (2/3)H−10 , ΩM = ΩT = 1 . (171)

dφ

dΩ= (1/2)

n0Lt04π

, t0 = H−10 , ΩM = ΩT = 0 . (172)

In both cases the correct calculation adds only a numerical factor to the naive answercalculated assuming that the stars have been burning for a time t0. The fact that thefactors are less than unity is due to the redshift.

In the inflationary model, we find

ρ(a0) = ρV

(ainf

a0

)4

. (173)

We see that the energy density falls as a−4 as expected.

5.13 Equation (86) is equivalent to

dχ = a0vda

a2a, (174)

so the most distant galaxy that can be visited has a coordinate

a0χ = H−10 v

∫ ∞

1

da

a3[ΩMa−3 + (1 − ΩT)a−2 + ΩΛ

]1/2 . (175)

For ΩM = ΩT = 1, this gives a0χ = 2H−10 v, for ΩT = ΩM = 0, a0χ = H−1

0 v,and for ΩT = ΩΛ = 1 it is a0χ = H−1

0 v/2.

5.14 The distance that a photon travels starting at t0 is given by

a0χ = H−10

∫ a

1

da

a2[ΩMa−3 + (1 − ΩT)a−2 + ΩΛ

]1/2 . (176)

For the cases ΩM = ΩT = 1 and ΩM = ΩT = 1, the integral diverges for a → ∞,so the photon can reach the whole universe. For the case ΩΛ = ΩT = 1, the integral

26 Solutions

converges so the photon can reach only galaxies within a finite co-moving volume.This is because of the exponential expansion in a vacuum-dominated universe.

5.15 The coordinates χ , t0, t0 + Δt0, t1, and t1 + Δt1 are related by

χ =∫ t0

t1

dt

a(t)=

∫ t0+Δt0

t1+Δt1

dt

a(t)(177)

⇒ Δt0Δt1

= a(t0)

a(t1)= 1 + z (Δt0 � H−1

0 ) .

This gives

a(t0 + Δt0) ∼ a(t0) + a(t0)Δt0 = a(t0)(1 + H0Δt0)

and

a(t1 + Δt1) ∼ a(t1) + a(t1)Δt1 = a(t1)(1 + H0

√ΩΛ + (1 + z)3ΩMΔt1)

= a(t1)(1 + H0Δt0√

ΩΛ + (1 + z)3ΩM/(1 + z)) .

The new redshift znew measured at t0 + Δt0 (to first order in H0Δt0 � 1) is

1 + znew = a(t0 + Δt0)

a(t1 + Δt1)= (1 + z)

1 + H0Δt0

1 + H0Δt0√

ΩΛ + (1 + z)3ΩM/(1 + z)

∼ (1 + z)[1 + H0Δt0(1 −√

ΩΛ + (1 + z)3ΩM/(1 + z)]

giving

znew − z = (1 + z)H0Δt0[1 −√

ΩΛ + (1 + z)3ΩM/(1 + z] .

For Δt0 = 14yr and z = 1 this gives

znew − z = +1 × 10−9 (ΩΛ = 1)

= −0.414 × 10−9 (ΩM = 1)

= +0.15 × 10−9 (ΩM = 0.27,ΩΛ = 0.73) .

The sign depends on the values of ΩM and ΩΛ. For (ΩM = 0.27,ΩΛ = 0.73), theredshift is rather stable.

Solutions 27

Chapter 6

6.1 It is important to factor out the physical parameters, e.g.,

ρ(T, μ = 0) = g

(2π )3

∫4πp3dp

exp(p/T ) ± 1= gT 4

2π2

∫ ∞

0

x3dx

exp(x) ± 1. (178)

The integrals give only the numerical factors in Table 6.1.For a neutral relativistic gas of electrons and positrons, the ratio between the

potential energy per particle and the kinetic energy per particle is

(e2/4πε0)n1/3

kT∼ e2kT/4πε0�c

kT∼ α . (179)

The gas should be ideal to good approximation. For a nonrelativistic nondegenerategas, the ratio is much smaller so the gas is even more ideal

The particle–antiparticle asymmetry for a relativistic gas in thermal equilibriumin the limit μ/T → 0 is

n − n = g

(2π )3

∫4πp2dp

[1

ep/T (1 − μ/T ) ± 1− 1

ep/T (1 + μ/T ) ± 1

]

gμT 2

π2

∫ ∞

0

x2ex dx

(ex ± 1)2∼ gμT 2

β, (180)

where β = 6 for fermions and β = 3 for bosons. The relative asymmetry is then oforder

n − n

n∼ μ

T. (181)

After the loss of chemical equilibrium between electrons and positrons, The elec-tron to photon ratio is

ne

nγ

= π2

1.2(2π )3/2

(me

T

)3/2exp((μ − me)/T ) , (182)

so the electron chemical potential is given by

me − μ

T= ln(nγ /ne) + (3/2) ln(me/T ) + constant . (183)

28 Solutions

6.2 The energy, entropy, and number of particles in a volume V are given by

E = Vρ(T, μ) ⇒ dE = ρdV + V∂ρ

∂TdT + V

∂ρ

∂μdμ , (184)

S = V s(T, μ) ⇒ dS = sdV + V∂s

∂TdT + V

∂s

∂μdμ , (185)

N = V n(T, μ) ⇒ dN = ndV + V∂n

∂TdT + V

∂n

∂μdμ . (186)

On the other hand, the change in entropy is given by

dS = dE/T + pdV/T − μdN/T . (187)

Substituting in (184) and (186), we get

dS = ρ + p − μn

TdV + ( ) dT + ( ) dμ . (188)

Comparing with (185) we find

s(T, μ) = ρ + p − μn

T. (189)

6.3 Massive particles have momentum-dependent velocities so in Fig. 1.10 thesquare on the left becomes a parallelogram. Since the parallelogram has the samearea as the square, the phase space density is still conserved.

For Fig. 1.11 we must consider the Lorentz-transformed trajectory of the particleof trajectory x(t) = vpt + dx :

(γ βγ

βγ γ

)(t

dx + vpt

)=

(γ t + βγ (dx + vpt)βγ t + γ (dx + vpt)

)=

(t ′

dx ′

). (190)

Setting t ′ = 0 we have the corresponding time in the unprimed frame:

t = −βdx

1 + βvp, (191)

from which we can calculate dx ′:

dx ′ = dx

γ (1 + βvp). (192)

Transforming the momentum four-vector we find

dp′ = γ (1 + βvp)dp . (193)

It follows that dxdp = dx ′dp′.

Solutions 29

6.4 The number of remaining interactions is

∫ ∞

t1

Γ (t)dt =∫ ∞

a1

Γda

a(a/a). (194)

To quickly get the answer, it is a good idea to evaluate a/a with a Friedmann equa-tion normalized at a1, e.g.,

a

a= H1

[ΩR(a1)

(a

a1

)−4

+ · · ·]1/2

. (195)

Substituting this into the integral, we find that the number of interactions is justΓ (t1)H−1

1 times a numerical factor of order unity. Since Γ (t1)H−11 � 1, this proves

the conjecture.

6.5 Numerically, the photon-scattering rate (always dominated by Compton scat-tering) is equal to the expansion rate at T ∼ 0.236 eV. The recombination rate isequal to the expansion rate at T ∼ 0.215 eV. A fraction ∼ 3 × 10−5 of the electronsremains free.

More realistic calculations including all the states of the hydrogen atom give arecombination time of T ∼ 0.26 eV.

6.6 The last annihilations take place at T ∼ 10 keV. A photon of E = 510 keVneeds about 10 collisions to reach a reasonably thermal energy of E ∼ 30 keV. Thetime to do this, ∼ 10(neσTc)−1, is much less than the Hubble time at this epoch.

6.7 The annihilation rate for positrons remains greater than the expansion rate wellinto the matter epoch:

ne〈σv〉 = ne(t0)a−3 πα2

m2ec4

(�c)2c ∼ 5 × 10−14 yr a−3 (196)

which is greater than the expansion rate for a < 10−2. As long as the annihilationrate is greater than the expansion rate we have for the solution to the Boltzmannequation

ne+ ∼ ne+(T )ne−(T )

ne−∼ η−1m3

e exp(−2me/T ) . (197)

For a ∼ 10−2 this gives

ne+ ∼ nγ 10−107. (198)

Since there are only ∼ 1087 photons within our horizon, this means it is extremelyunlikely that there are any remaining positrons.

30 Solutions

6.8 The rate for νν → e+e− takes its maximum value at T ∼ MW with Γ ∼α2m2

ν/mW . Requiring that this be greater than the expansion rate H (mW ) ∼m2

W /m pl sets a lower limit in the mass of the neutrino mν > (mW /α)√

mW /m pl ∼30 keV.

6.9 The baryon number generated per X decay is (2/3)r − (1/3)r = r − 1/3 whilethat per X decay is −r + 1/3. The baryon excess after the decays is

nb − nb = nX (r − r ) . (199)

The relative asymmetry just after the decays is

[nb − nb

nγ

]t1

= nX

nγ

(r − r ) . (200)

Assuming an adiabatic expansion after the decays, the present asymmetry is

[nb − nb

nγ

]t0

=[

nb − nb

nγ

]t1

g(t0)

g(t1). (201)

A reaction is said to violate “C” invariance if the “anti-reaction”, where all parti-cles are replaced by their antiparticles does not, in reality, proceed at the same rateas the original reaction. A reaction is said to violate “P” invariance if the reactionviewed in a mirror does not, in reality, proceed at the same rate as the originalreaction. (Strictly speaking, the P operation reverses all coordinates, while mirrorreflection reverses only the coordinate perpendicular to the mirror. However, mirrorreflection followed by an appropriate spatial rotation can reverse all coordinates.Mirror reflection is, therefore, equivalent to P if rotation symmetry is respected.)Finally a reaction violates “CP” if the process viewed in a mirror with particlesreplaced by antiparticles does not, in reality, proceed at the same rate as the originalreaction.

Consider the reaction

X → de+ branching ratio = 1 − r . (202)

The C-transformed reaction is

X → de− branching ratio = 1 − r . (203)

If this reaction proceeds at the same rate as (202), i.e., if C is not violated, thenclearly we must have r = r .

Reaction (202) viewed in a mirror with particles changed to antiparticles isshown in Fig. 2. The transformed reaction has the relative orientations of spinsand momenta inverted. If the transformed reaction proceeds at the same rate as theoriginal reaction for all possible spin combinations of the original reaction, i.e., if

Solutions 31

X

e

d

X

e

−

+ −

−d

Fig. 2 The decay X → de+ and the antireaction X → de− viewed in a mirror. The antireactionhas the relative orientations of spins and momenta reversed. If, for all possible spin orientations,the first reaction proceeds at the same rate as the transformed reaction (CP invariance), then thespin-averaged branching ratio for X → de+ must be the same as that for X → de−

CP is not violated, then r = r after averaging over all possible spins. For example,in neutron decay, n → pe−νe, the helicity ( p · s) of the ν is opposite the helicity ofthe ν produced in antineutron decay. Since CP is conserved to good approximation,the rates are the same.

A reaction violates “CPT” if a film of the process viewed in a mirror with parti-cles replaced by antiparticles and run in reverse does not, in reality, proceed at thesame rate as the original reaction. Figure 3 shows the reaction

de+ → de+ , (204)

and its CPT-transformed reaction

de− → de− . (205)

The reactions will have a resonant peak at a center-of-mass energy equal to m X forthe first and m X for the second. If CPT invariance is respected, the cross-section forthe first reaction must be equal to that for the second. This would imply that themasses and the total widths (lifetimes) of X and X are equal.

32 Solutions

x x

e

d

d

e

e

de+

+ −

−−

−

−

d

Fig. 3 The reaction de+ → de+ and the antireaction viewed in a mirror with the film run inreverse. The antireaction appears as de− → de−. The two reactions proceed by resonant formationof X or X . If the cross-sections for the two processes are equal (CPT invariance), the masses andtotal widths (lifetimes) of X and X must be equal

6.10 The rate per neutron of np →2 Hγ is

n p〈σv〉 ∼ ηnγ × 7 × 10−26m3 sec−1 ∼ 2 × 1012 sec−1

(T

1 MeV

)3

, (206)

while the expansion rate at the relevant epoch is

H ∼ 0.6 sec−1

(T

1 MeV

)2

. (207)

At T = 60 keV we need η > 4 × 10−12 in order to have Γ > H .The Saha equation follows from the expressions for the chemical potentials in

Table 6.1. Using n p ∼ ηnγ we find

n2

nn∼

(T

m p

)3/2

η exp(B/T ) (208)

which remains small until T ∼ 60 keV.

6.11 The duration of the neutron decay period can be found by integrating the Fried-mann equation. In the approximation of an adiabatic expansion, gs T 3a3 = constant,

Solutions 33

the Friedmann equation is just

− 1

g1/3s T

dg1/3s T

dt=

(8πGgE (T )π2T 4

3 × 30

)1/2

. (209)

This can be integrated numerically using the g(T ) from Fig. 6.1.In the approximation of g = constant, we have

∫ 60 keV

T =800 keVdt ∼ 0.5H (60 keV)−1 ∼ 3 min . (210)

The entropy generated by neutron decay is governed by the equation

dsa3 =∑

i

μi

Td(ni a

3) , (211)

where the sum is over all species. For nonrelativistic particles, the chemical poten-tial is

μi

T= mi

T+ ln

ni

2(mi T/2π )3/2

= mi

T+ ln ni/nγ + (3/2) ln T/mi + constant . (212)

For neutron decays, we take into account the changes in the numbers of neutrons,protons, and electrons (supposing the neutrino chemical potential vanishes) to find

dsa3

dnna3= −mn − m p − me

T− ln

nγ

ne− ln

nn

n p− 3

2ln

me

T+ 0.67 .

For neutron decay occurring near T ∼ 100 keV all the important terms on the rightare of order ∼ 10 so the total generated entropy is of order

Δs

s∼ 10

Δnn

s∼ η . (213)

In the contraction phase of a closed universe, no neutrons are initially presentand the inverse decay pe−νe → n is very unlikely. Neutrons only appear whenthe temperature reaches ∼ 800 keV when they are produced by pe− → nνe andpνe → e+n.

6.12 The freeze-out temperature is proportional to g1/6 so the sensitivity of theneutron–proton ratio at freeze-out to the number of neutrino species is

Δ(n/p)

n/p∼ (1/6)

Δ2Nν

g. (214)

34 Solutions

The addition of one neutrino species would add about 10% to n/p so at this preci-sion it should be possible to set a limit on the number of extra neutrinos of about 2.

Chapter 7

7.1 The mass in a sphere of co-moving radius r and centered at r ′ is

Mr (r ′, t) = a(t)3∫ |r−r ′|<r

ρ(r)d3r

= ρVr + ρV −1/2∑

k

δk exp(i k · r ′)∫ |R|<r

exp(i k · R) d3 R ,

where Vr = 4π (a(t)r )3/3 and where we changed integration variables to R = r−r ′.The integral is elementary

I (k) =∫ |R|<r

exp(i k · R) d3 R = 2π

∫ r

0

∫ 1

−1d cos θ exp(i k R cos θ )

= Vr3

(kr )2

(sin kr

kr− cos kr

)= Vr

3 j1(kr )

(k R)2

∼ Vr kr � 1 ∼ 0 kr � 1 .

We see that the function j1(kr ) effectively truncates the sum so that waves withλ � 2r do not contribute.

To calculate the fluctuations of Mr , we need M2r :

[Mr (r ′)

]2 = ρ2V 2r [1 + 2V −1/2

∑k

δk (I (k)/Vr ) exp(i k · r ′)

+V −1∑

k

∑k′

δk δ∗k′ (I (k)/Vr ) (I (k ′)/Vr ) exp(i(k − k′) · r ′) ] ,

where we use δ−k = δ∗k. Taking the average over r ′, the periodic boundary condi-

tions make each term in the first sum integrate to 0 and leave only the k = k′ termsin the second:

〈M2r 〉 = ρ2V 2

r

[1 + V −1

∑k

(3 j1(kr )

(k R)2

)2

|δk|2]

. (215)

Solutions 35

This gives the variance of Mr :

σ 2r ≡ 〈M2

r 〉 − 〈Mr 〉2

〈Mr 〉2= V −1

∑k

(3 j1(kr )

(k R)2

)2

|δk|2 . (216)

Changing the sum to an integral with the replacement

V −1∑

k

→ 1

(2π )3

∫d3k → 1

2π2

∫ ∞

0k2 dk (217)

and replacing |δk|2 with its average P(k), we recover (7.32).

7.2 The difference between the temperature and the mean temperature is the sumwithout the (nx , ny) = (0, 0) term:

T (θ) − 〈T 〉 = 1

Δθ

∑nx ,ny

ak exp

[2π i

Δθ(nxθx + nyθy)

](218)

with the k = 2πn/Δθ with n = (nx , ny), integer nx , ny . The square is given by asum over (nx , ny, n′

x , n′y):

(T − 〈T 〉)2 = 1

ΔΩ

∑aka∗

k′ exp

[2π i

Δθ((nx − n′

x )θx + (ny − n′y)θy)

].

The average of this is found by integrating over (θx , θy) and dividing by ΔΩ =(Δθ )2. The integral vanishes for n �= n′ and equals ΔΩ for n = n′:

⟨(T − 〈T 〉)2

⟩ = 1

ΔΩ

∑nx ,ny

aka∗k

The number of modes in dkx dky is dnx dny = ΔΩdkx dky/(2π )2 so

⟨(T − 〈T 〉)2

⟩ = 1

(2π )2

∫dkx dky 〈|ak|2〉 = 1

2π

∫k dk 〈|ak|2〉 ,

where the average in the integrand is over modes within dk.The calculation of the correlation function is only a bit more complicated:

[T (θ ′) − 〈T 〉][T (θ ′ − θ ) − 〈T 〉]= 1

ΔΩ

∑k,k′

aka∗k′ exp

[2π i

Δθn′ · θ

]exp

[2π i

Δθ(n − n′) · θ ′

],

36 Solutions

where the sum does not include the k = 0 and k′ = 0 terms. Averaging over θ ′

leaves only the k = k′ terms

1

ΔΩ

∑k

|ak|2eikθ cos φ = 1

(2π )2

∫k dk

∫ 2π

0dφ eikθ cos φ|ak|2 ,

where φ is the angle on the sky between k and θ . Replacing |ak|2 with its averagewe get

ξ (θ ) = 1

2π

∫k dk〈|ak|2〉 1

2π

∫ 2π

0dφ eikθ cos φ

= 1

2π

∫k dk〈|ak|2〉 J0(kθ ) . (219)

J0(0.0025 l )

−0.5

0

0.5

1

l200 6000

J0(0.005 l )J0(0.01 l )J0(0.05 l )(l+1)Cl

104

0.0500

0.1θ

ξ(θ)

(μΚ

2 )

400 800 1000

Fig. 4 The top figure shows J0(�θ) for four values of θ (solid lines) and (� + 1)C� for the standardcosmological parameters. The bottom figure shows the corresponding ξ (θ) for a patch on the skyof ΔΩ = 0.1 × 0.1 rad2

Solutions 37

Figure 4 shows the C� plotted as a function of � along with J0(kθ ) plotted as afunction of � = k for four values of θ . For θ � Δθs ∼ 0.01 the Bessel functionoscillates with a period much shorter than the scale of variations of the C�. We can,therefore, expect that the integral (219) will be small giving small correlations atlarge angular scales, as seen in the bottom panel of Fig. 4.

7.3 The proof follows that for Exercise 7.2. The standard cosmology gives a strongtemperature–polarization correlation at θ ∼ 2.5Δθs because CTE

� is in phase withJ0(�θ ) for this value of θ (Fig. 5). Reversing the sign of the CTE

� for � < 200 wouldeliminate most of this correlation.

J 0 l )(0.025

(l+1)C lTE

−0.5

0

0.5

1

1000l

200 6000 400 800

Fig. 5 The top figure shows J0(�θ) for θ = 0.025 ∼ 2.5Δθs (solid line) and (� + 1)CTE� for the

standard cosmological parameters

7.4 Inverting the Fourier series, one finds

ak = 1

ΔΩ

∫dθx dθy T (θ ) e−ik·θ = 1

ΔΩ

∫θ dθ T (θ )

∫ 2π

0e−ikθ cos φ ,

where φ is the angle between θ and k. The integral over φ is just 2π J0(kθ ):

ak = 2π

ΔΩ

∫θ dθ T (θ )J0(kθ ) .

For a single point-like initial perturbation on our last scattering surface, T (θ ) willhave a minimum at θ = 0 and a maximum at θ = Δθs. The ak will be large for ksuch that J0(kθ ) is in phase with T (θ ). The first minimum of J0(x) is at x = 3.8,suggesting that the ak will be large at kΔθs ∼ 3.8 giving k = � = 3.8/Δθs ∼ 365.This rather overestimates the position of the first acoustic peak (� ∼ 220) which isshifted to lower � by the early time integrated Sachs–Wolfe effect.

38 Solutions

7.5 Multiplying the expression for T (θ, φ) by Y�′m ′(θ, φ) and integrating over dΩ

project out a�m because of the orthogonality of the spherical harmonics. Using theexpansion for exp(ik · r) in terms of the spherical harmonics then yields

a�m(r ′) = 4π

3

∑k

eik·r ′φk j�(kr ) Y�m(Ωk)i� . (220)

The square has a double sum on k and k′ but when averaged over observer positionsr ′ only the k = k′ terms remain:

C� = 〈|a�m |2〉 =(

4π

3

)2 ∑k

|φk j�(kr ) Y�m(Ωk)|2 .

Changing the sum to an integral and replacing k3|φk|2 with its k-independent aver-age 2π2Δ2

k , we find the relation (7.77).

7.6 From

Δ2R(k) = V

12π2 M6pl

(V

V ′

)2

, (221)

we get

aexit

Δ2R

dΔ2R

daexit= −M2

pl

[3

(V ′

V

)2

− 2

(V ′′

V

)]

by using the slow-roll condition and the Friedmann equation. From Fig. 7.9 it is sim-ple to show that for modes entering during the radiation epoch that aexit ∼ a2

1/aenter

where a1 corresponds to the end of inflation. Furthermore, aenter ∝ k−1 for suchmodes. We thus find

ns − 1 ≡ k

Δ2R

dΔ2R

dk= M2

pl

[3

(V ′

V

)2

− 2

(V ′′

V

)].

7.7 The volume of (1/3) of the sky out to z = 0.3 is

V = a30

∫dΩ

∫ χ(z=0.3)

0dχ r (χ )2 ∼ (1/3) (4π/3)(0.3c/H0)3 (222)

∼ 109(h−1Mpc)3 ,

Solutions 39

where we approximate a0χ ∼ (c/H0)z. At k ∼ 0.01(hMpc−1), the number ofmodes in the k interval Δk/k = 0.5 is

NΔk = V k3

4π2

Δk

k∼ 13 (223)

which means that P(k) can be measured to a precision of 1/√

13 ∼ 25%. This isroughly the relative size of the error part of the first point in Fig. 7.6.

For the CMB, there are (2�+1)/2 independent modes available for measuring C�.(The factor 1/2 comes from the reality condition a� −m = a∗

�m .) For the quadrupole,� = 2, this gives a cosmic variance of 1/

√(2.5) = 63% which suggests that the low

value measured by WMAP (Figure 7.11) is not so unreasonable.

7.8 The calculated positions of the first and third peaks are shown in Table 1. Theyagree reasonably well with Fig. 7.11.

Table 1(ΩM,ΩΛ) �s �1 �3

(0.27, 0.73) 293 223 809(1, 0) 255 201 711(0.27, 0) 570 433 1570

7.9 Following the same reasoning as for that in Exercise 7.1, one finds

⟨(Mr − (1/8)M2r )2

⟩ =∫

dk

k|U (k R)|2Δ2(k) , (224)

where

U (kr ) = Vr3 j1(kr )

(k R)2− V2r

3 j1(2kr )

(2k R)2. (225)

The function U (k R) has a peak near k =∼ 2.2/R.

7.10 The longest mode that was never outside the Hubble radius has

λ(a0)a1

a0< dH(a1) , (226)

where a1 refers to the end of inflation. Taking a1/a0 ∼ 1015GeV/10−4eV and usingthe Friedmann equation this gives

λ(a0) ∼ 10−26 H−10 ∼ 1 m . (227)

40 Solutions

Modes with wavelengths shorter than this would not be expected to have a scaleinvariant spectrum.

7.11 The present distance to our LastSS is ∼ 14Gpc for (ΩM,ΩΛ) = (0.27, 0.73).At areion ∼ 0.1, the distance between any point and its LastSS can be calculatedassuming matter domination:

2H−1(treion) = 2c/H0√ΩMa−3/2

reion

∼ 500 Mpc .

The (present) distance to our reioniztion surface is given by the difference between14 Gpc and 500 Mpc scaled to t0: 14 Gpc − 5 Gpc ∼ 900 Mpc. CMB photonsscattering on our reionization surface, therefore, come from angles of ∼ 5000/9000corresponding to multipoles π/θ ∼ 6.

7.12 A photon with a trajectory r(t) = vt +dr has a transformed trajectory given by

⎛⎜⎜⎝

γ βγ 0 0βγ γ 0 00 0 1 00 0 0 1

⎞⎟⎟⎠

⎛⎜⎜⎝

tdz + vz tdy + vytdx + vx t

⎞⎟⎟⎠ =

⎛⎜⎜⎝

γ t(1 + βvz) + βγ dzγ t(β + vz) + γ dz

dy + vytdx + vx t

⎞⎟⎟⎠ . (228)

Setting t ′ = 0 we have the corresponding time in the unprimed frame:

t = −βdz

1 + βvz(229)

from which we can calculate dr ′

dz′ = dz

γ (1 + βvz), (230)

dy′ = dy − vyβdz , (231)

dx ′ = dx − vxβdz , (232)

corresponding to the cube in Fig. 7.22 of volume dxdydz/[γ (1 +βvz)]. Transform-ing the momentum four-vector we find

dp′z = γ (1 + βvz)dpz (233)

and dp′y = dpy and dp′

x = dpx . It follows that d3rd3 p = d3r ′d3 p′.

http://www.springer.com/978-3-642-02799-4