Cosmology for Physicists (Solutions, Instructor Solutions Manual) [1 ed.] 9781498755313, 1498755313

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David H. Lyth

Cosmology for Physicists solutions

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1.1 Both have dimensions 1/E 1.2 This question applies only to the first three lines, since lines four and five are merely definitions. To verify the first line, we should first use line four to convert GeV to Joules, and then use the value of ~ given in Table A.2. This gives 1 m = 5.068 × 1015 × 6.242 ∗ 109 × 1.055 × 10−34 × 2.998 × 108 = 1.0006 m which is in adequate agreement as we work only to four significant figures. The proof for the second and third lines is similar. 1.3 Six quarks, six leptons, four gauge bosons and the Higgs making seventeen in all. 1.4 Charge conservation forbids electron decay because lighter particles have zero charge. 1.5 According to Table 5 and its caption, the nucleon mass is m = 939 × 1.78 × 10−30 = 1.67 × 10−27 kg. Number density of nucleons is n = M/mV where M is the Earth mass and V is its volume. The Earth’s radius is 6400 km (not 6400 m as stated in the question). This gives n = 4.7 × 1030 m−3 . Going to natural units using the first line of Table A.3 gives n = 3.5 × 10−17 GeV3 . As seen around Eq. (8.17), working in natural units, Γ = σn ∼ G2F E 2 n ∼ 10−27 E 2 GeV−1 . In MKS units, time for passage of a neutrino through the earth is t = 6.4×106 m/c s. Using the second line of Table A.3, this is in natural units t ' 1022 GeV−1 . We are asked to compare this with Γ−1 = 1027 E −2 GeV = 1045 ( eV/E)2 GeV−1 . The two would be equal if E ∼ 1012 eV. (The above calculation is an exercise in the use of the expression σ ∼ G2F E 2 given before Eq. (8.17), that being the only −1/2 expression given in this book. The expression is actually valid only for E  GF ∼ 11 10 eV. For bigger E, σ is smaller being σ ∼ GF . It follows that a typical neutrino actually passes straight through the earth no matter how high is its energy.) 2.1 We can choose the origins of x and t so that V = x/t. Then Eq. (1.3) gives V0 =

x0 x + vt V +v = = . 0 t t + vx 1 + vV

The non-relativistic expression V 0 = V + v is recovered when vV  1. Using dimensional analysis to restore c, this criterion becomes vV  c2 . 2.2 The generic Lorentz transformation for time is t − vx/c2 t0 = p . 1 − v 2 /c2 Taking the unprimed frame to be the astronaut’s, we can set the position x of

2 

the astronaut to zero. Taking t0 to be the earth time, the amount by which the astronaut has aged is   p 1 v2 2 2 t = 20 1 − v /c ' 20 × 1 − = 19.9999995 yr. 2 c2 This takes the turnaround of the astronaut to be instantaneous and ignores the effect of the acceleration during the turnaround. 2.3 Since gµν is symmetric and g µν is the inverse of gµν we have gµν g µν = gµν (g −1 )µν = gµν (g −1 )νµ = δµµ = 4. 2.4 Writing ds2 first in generic coordinates and then in locally orthonormal coordinates we have 0 ds2 = gαβ dx0α dx0β

= gµν dxµ dxν ∂xµ ∂xν = gµν 0α 0β dx0α dx0β ∂x ∂x ∂xµ ∂xν = ηµν 0α 0β dx0α dx0β . ∂x ∂x Equating the first and last expressions gives the desired result. 0 2.5 By definition A0µ = gµν A0ν . Using Eq. (2.6) this gives A0µ = = =

∂xβ 0ν ∂xα g A αβ ∂x0µ ∂x0ν ∂xα gαβ Aβ ∂x0µ ∂xα Aα . ∂x0µ

2.6 Using Eq. (2.7), Eq. (2.12) becomes  0β  ∂xµ ∂x µ α Dν A = ∂ν A ∂x0β ∂xα β

β

∂xµ ∂x0 ∂xµ ∂ 2 x0 α ∂ A + Aα ν ∂x0 β ∂xα ∂x0 β ∂xν ∂xα = δαµ ∂ν Aα + Γµνα Aα =

= ∂ν Aµ + Γµνα Aα . 2.7 Only the space-space components of gµν are affected by the transformation. Denote them by gij . Cartesian coordinates are given in terms of polar coordinates by x = r sin θ cos φ, y = r sin θ sin φ and z = r cos θ. In Cartesian coordinates gij is

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the unit matrix. Therefore  2  2  2 ∂x ∂y ∂z grr = + + ∂r ∂r ∂r 2 2 2 = sin θ cos φ + sin θ sin2 φ + cos2 θ = 1 ∂x ∂x ∂y ∂y ∂z ∂z grθ = + + ∂r ∂θ ∂r ∂θ ∂r ∂θ = r sin θ cos φ cos θ cos φ + r sin θ sin φ cos θ sin φ − r cos θ sin θ = 0. Evaluating the other components in the same way, one finds that the other nonzero components are gθθ = r2 and gφφ = r2 sin2 θ. The corresponding spatial line element is
d`2 = dr2 + r2 dθ2 + sin2 θdφ2 which is usually derived geometrically. 2.8 Using Eq. (2.19), the nonzero components are Γθrθ = 1/r, Γφφr = 2/r and Γφφθ = 2 cot θ (and the equal quantities that are obtained by swopping the lower indices). The only non-zero components of the curvature tensor have all space indices. Also, from the Γ’s one can see that all Rrijk vanish, and so do all Rθijk if any lower index is φ. Of the remaining components, some have all terms vanishing, and the rest have a cancellation as for instance in
2
2 Rθθθr = Γθθr − Γθθr = 0 . 2.9 On the right hand side of Eq. (2.15), it should be Aµ not Aµ . To answer the question we need the following relation which is derived in the same way as Eq. (2.17): Dµ (Aα B β ) = ∂µ (Aα B β ) + Γνµα Aν B β + Γβµν Aα B ν . Replacing B β by Aµ , Aα by Dβ and Dµ by Dα , this gives Dα Dβ Aµ = ∂α ∂β Aµ + (∂α Γµνβ )Aν + Γµνβ ∂α Aν

+ Γναβ ∂ν Aµ + Γµγν Aγ + Γµαν ∂β Aν + Γνγβ Aγ . This leads to Eq. (2,41), after noting some cancellations. 2.10 Eq. (2.40) corresponds to Eq. (2.38). In the latter, the symmetric matric 0 gµν has 10 independent components and the index τ takes on 4 values, which means that there are 10 × 4 = 40 independent equations. 2.11 A suspended rod will point to the earth if its suspension is sufficiently free of friction. 3.1 The trace of a matrix Aij is A11 + A22 + A33 . In Eq. (3.1), the trace of Tij is 3P which means that Tij specifies P . Also, Σij = Tij − P δij which means that Tij also specifies Σij . 3.2 To first order in v, the Lorentz boost Eq. (1.3) becomes x0 = x − vt,

t0 = t − vx.

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For a boost in an arbitrary direction we have (with {x1 , x2 , x3 } = {x, y, z}) x0i = xi − v i t,

t0 = t − vi xi ,

these being the only expressions which transform correctly under rotations and reduce to Eq. (1) for a boost along the x axis. (As seen in Section 1.1, the placement of the spatial indices is irrelevant but the pairs to be summed over are conventionally chosen to be one up and one down.) For a tensor formed from two 4-vectors we have (still to first order in v) A00 B 00 = A0 B 0 − vi Ai B 0 − vi A0 B i A00 B 0i = A0 B i − vj Aj B i − v i A0 B 0 A0i B 0j = Ai B j − v i A0 B j − v j Ai B 0 . Since all tensors transform in the same way, this gives T 000 = T 00 − 2vi T i0 T 00i = T 0i − vj T ji − v i T 00 T 0ij = T ij − v i T j0 − v j T i0 . In the rest frame Ti0 = 0 and we are supposing that Tij = P δij . Then these expressions become T 000 = T 00 T 00i = T 0i − v i P + T 00




T 0ij = T ij , which is the required result. 3.3 The first sentence of the question is superfluous because the proof that dr/dt = 0 for a photon at the surface of a black hole is already in the text. Therefore, the position of a photon at the surface never changes and the photon cannot emerge. For a massive object, ds2 is positive which means that dr/dt vanishes at some position r > rs and the object cannot even emerge from that position which means a fortiori that it cannot emerge from the surface of the black hole. 3.4 The value of G given in Table A.2 is incorrect. It should be G = 6.67 × −11 3 10 m kg −1 s−2 which means that in MKS units 2GM = 2.67 × 1020 m3 s−2 . The Schwarzschild radius of a black hole with mass M is 2GM , evaluated as a length in units that make c = 1. The quantity 2GM /c2 has the dimension L, and reduces to 2GM in units that make c = 1. Therefore, with any choice of units, GM /c2 = 2.97 × 103 m is the Schwarzschild radius of a black hole with mass M . A black hole with mass 106 M therefore has Schwarzschild radius 106 GM /c2 = 2.97 × 109 m. 3.5 To answer this question we should use Eq. (3.39). After a coordinate time interval t, the time interval τ measured by a fixed clock is p τ = 1 + 2φt ' (1 + φ)t.

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At a distance r from the centre of the earth, φ = −GM/r. The bottom of the floor can be taken to have r = R where R is the radius of the earth. Then the time intervals measured by the clocks at the bottom and top of the room (during equal intervals of coordinate time) are in the ratio τbottom τtop

=

1− 1−

GM R+h GM R

GM GM − R R GM h = 1+ , R R



' 1+

1−

h R



where h is the height of the room. The fraction by which the lower clock runs slower is therefore equal to the dimensionless quantity (GM/R)(h/R). This use of Eq. (3.39) requires units that make c = 1. The quantity (GM/R)(h/R)/c2 is dimensionless and reduces to (GM/R)(h/R) in units that make c = 1. The fraction with any choice of units is therefore (GM/R)(h/R)/c2 . Using MKS units and taking h = 5 m the fraction is found to be 1.47 × 10−16 . To justify the comparison of the clocks after equal amounts of coordinate time t, suppose that a light signal is passed from one clock to another, so as to compare the readings of the clocks. The trajectory of a photon satisfies ds2 = 0 which means s dr 1 + 2φ = ' 1 + 2φ. dt 1 − 2φ This is independent of coordinate time, which means that the coordinate time taken for the signal to travel becomes negligible after a large amount of coordinate time. 3.6 Eq. (2.11) is incorrect; from Eq, (2,8) it should read 0 Bµν =

∂xα ∂xβ Bαβ . ∂x0µ ∂x0µ

In Eq. (3.19), the argument of ξ µ can be taken as either the unprimed or the primed coordinate because we are working only to first order. Taking it as the primed coordinate we have ∂xα ∂ξ α α = δ − . µ ∂x0µ ∂x0µ Then the corrected version of Eq. (2.11) gives    ∂ξ α ∂ξ β 0 α β hµν = δµ − 0µ δν − 0ν hαβ . ∂x ∂x Going now to the unprimed coordinate, this gives to first order h0µν = hµν −

∂ξ α ∂ξ β h − hβµ . αν ∂xµ ∂xν

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But as we work to first order, hµν in the second and third terms can be replaced by ηµν , giving Eq. (3.21). From that we get h0µν = hµν −

∂ξ β ∂ξ α ν δα − η νγ γ ηµβ , µ ∂x ∂x

and ∂ξ α − ∂xα ∂ξ α = h− α − ∂x ∂ξ α = h − 2 α. ∂x

h0 = h −

∂ξ β γµ η ηµβ ∂xγ ∂ξ β γ δ ∂xγ β

Combining these expressions leads immediately to Eq. (3.23). 3.7 On the right hand side of Eq. (3.26), Tµν should be replaced by its first-order perturbation δTµν . The question should say ‘Calculate to first order the perturbations in the Levi-Civita ...’. Since the unperturbed Γµνβ vanishes (flat spacetime with Minkowski coordinates), the definition (2.19) gives (working always to first order in the perturbations) 1 Γµνβ ' η µγ (∂β hγν + ∂ν hγβ − ∂γ hνβ ) , 2 and

1 ∂α Γµνβ ' η µγ (∂α ∂β hγν + ∂α ∂ν hγβ − ∂α ∂γ hνβ ) . 2 Since the unperturbed Rµνβα vanishes we find using the definition (2.42) Rµνβα = Rνα = R =

1 µγ η (∂α ∂ν hγβ − ∂α ∂γ hνβ − ∂β ∂ν hγα + ∂β ∂γ hνα ) , 2 1 µγ η (∂α ∂ν hγµ − ∂α ∂γ hνµ − ∂µ ∂ν hγα + ∂µ ∂γ hνα ) , 2
1 µγ νβ η η ∂ν ∂β hγµ − η νβ ∂ν ∂γ hβµ − η νβ ∂µ ∂β hγν + ∂µ ∂γ h . 2

The left hand side of the field equation (3.9) is (taking account of the cancellation of two terms). 1 Rµν − ηµν R = 2 −

1 [∂ν ∂µ h − η σγ (∂ν ∂γ hµσ + ∂σ ∂µ hγν − ∂σ ∂γ hµν )] 2
1 ηµν η σγ η τ β ∂σ ∂β hγτ + η σγ ∂σ ∂γ h . 4

(1)

¯ µν . From Eq. (3.22) we learn that h ¯ = −h, We want to evaluate this in terms of h which means that ¯ ¯ µν − 1 ηµν h. hµν = h 2

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¯ µν =0 implies ∂µ h ¯ µ = 0 and therefore ∂µ h ¯ = 0. Using these Also, the condition ∂µ h ν results, we find that 1 ¯ µν . Rµν − ηµν R = h 2 3.8 In Eq. (3.38), the final expression should be multiplied by −1. To answer the question, we need to convert the upper-index quantities given by Eqs. (3.36)–(3.38) to lower-index quantities. h00 = η0µ η0ν hµν = h00 = −2φ hi0 = ηiµ η0ν hµν = −hi0 = 0 hij = ηiµ ηjν hµν = hij = −2δij φ. Substituting these into Eq. (3.18) gives the desired result. 3.9 The statement to be verified is the second sentence of the paragraph following ¯ 0i = 0, the upper-index version of hµν Eq. (3.42). In that sentence it should be h ¯ not being needed. Also, it should be h instead of h. The choice of µ which satisfies these conditions, and Eq. (3.44), is achieved by Eq. (3.42) with 0 = −i (2A00 + A11 + A22 ) / (4k) 1 = −iA0i /k 2 = −iA02 /k 3 = −i (2A00 − −A11 − A22 ) / (4k) . 4.1 Take the waves to emerge from a sphere with radius 1 km. Then at emergence |h+,× | ∼ 10−22 ×

109 light-years 3 × 1017 × 3 × 107 m = 10−22 × ' 100 . 1 km 103 m

The wave, on emergence, therefore corresponds to a metric perturbation of order 1. That is to be expected, because the metric just outside a black hole (Eq. (3.16)) differs from the flat-space metric by an amount of order 1. 4.2 Using Table A.5, the mass of the proton is m = 1.78 × 10−30 × 938 = 1.67 × −27 10 km. The gravitational force between two protons separated by a distance r = 10−10 m is Gm2 = 1.88 × 10−44 N. r2 The electric force between them is 2.3 × 10−8 N, not 2.3 N as stated in question. The electric force between a proton and an electron is equal and opposite to that between two protons. If there are p/e protons per electron, the effective electric force between two protons (taking account of the opposite force generated by the electrons) is p  − 1 × 2.3 × 10−8 N. e This is equal in magnitude to the gravitational force if p | − 1| = 8.2 × 10−37 . e

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We know from observation that the gravitational force dominates, which implies that |(p/e) − 1| is much less that the above value. 4.3 For this question and the next one we need to pretend that the galaxy and the galaxy cluster are spherically symmetric. The kinetic energy of an object with mass m and speed v  c is mv 2 /2. Its potential energy at a distance r from a galaxy with mass M is −GM m/r. When r = ∞ the potential energy vanishes which p means that the escape velocity (required if the object to reach r = ∞) is v = 2GM/R where R is the radius of the galaxy. Bearing in mind Table 4.1, let’s take M = 1010 M and R = 25 kpc. Then the escape velocity is 5.9 × 105 m s−1 . The speed of the CDM particles must be smaller. 4.4 An object in a circular orbit with radius r and speed v has acceleration a = v 2 /r. If the object goes around an object with mass M we also have a = GM/r2 . p Equating these expressions gives v = GM/r. For a galaxy, using the mass and radius estimate of the previous question, this gives v = 4 × 105 m s−1 . For a galaxy cluster let’s take M = 1014.5 M and R = 5 Mpc. This gives v = 8 × 103 m s−1 . When an object moves within a galaxy, 1 GM (r) mv 2 (r) − = E, 2 r where E is its energy, r is its distance from the galaxy centre, v(r) is its speed and M (r) is the galaxy mass enclosed within a sphere of radius r. We need E < 0 so that the object does not escape from the galaxy, and we need −GM/R < E because v 2 is positive. It follows that 1 GM (r) GM (r) GM − < v 2 (r) < . r R 2 r Unless r is close to 0 or close to R, this implies v 2 (r) ∼ GM/R which is about the same as if were moving in a circular orbit. The spread of redshifts is 2v/c. For a galaxy this is roughly 3 × 10−3 . For a galaxy cluster it is roughly 5 × 10−5 . 5.1 Using these numbers the galaxy number density is roughly 1011 1 ' 4 4 3 40 Mpc3 3 π(10 Mpc) and so the typical distance between them is about 401/3 ' 3 Mpc. When they are created at z ' 10 the typical distance is about 0.3 Mpc. Taking the galaxy size as 0.025 Mpc as in solution 4.3, they would be touching when z = 3/.025 ' 102 . But the temperature then was about 102 T0 ' 300 ◦ K where T0 ' 3 ◦ K is the present temperature of the CMB. They could not have existed then because they contain gas at temperatures far below 300 ◦ K. 5.2 This question and the next require material from Chapter 8. Last scattering takes place during matter domination. To simplify things I will ignore the cosmological constant so that matter domination persists to the present epoch. Then a = (t/t0 )2/3 where t0 is the present age of the Universe, and t0 1 H = = 3/2 . H0 t a

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Last scattering is at a−1 = 1100 and so we have cH −1 = (1100)3/2 cH0−1 . Using cH0−1 = 4.46 Gpc (given in Table A.4) this gives cH −1 = 0.12 Mpc at last scattering. The distance to the horizon is then 2cH −1 . 5.3 The present distance to the surface of last scattering is almost the same as the distance to the edge of the presently observable Universe which is 2cH0−1 . At last scattering this distance is 2cH0−1 a = 2cH0−1 /1100. As seen in the solution to the previous question, the Hubble distance at that epoch is cH0−1 /11003/2 . The required angle (ignoring the curvature of the segment of the last scattering surface that we deal with) is therefore cH0−1 /11003/2 = 0.015 radians. 2cH0−1 /1100 That the angle is much less than 2π is important, because the Hubble distance at any epoch is roughly the distance that anything can have travelled since the beginning of the Big Bang. The near-homogeneity of the CMB (indicating the near-homogeneity of the Universe at last scattering) cannot therefore have been established during the Big Bang. As we shall see, it was instead established during the earlier era of inflation. 5.4 The k.e. released is the sum of the initial rest energies minus the sum of the final rest energies. Using a subscript to denote particle species, taking the proton and neutron rest energies given in Table A.5 to be exact and setting the neutrino masses to zero, these are as follows. In the first two lines of Table 5.2, 2me c2 = 1.02 MeV. In the third line, mn − mp − me = 0.79 MeV. In the fourth line, mn + me − mp = 1.81 MeV. 6.1 Let n be the mass density of neutrons and p be the mass density of protons. It is sufficiently accurate to take the proton and neutron masses to be equal. Then the mass density of 4 He is 2n and the total mass density is p + n which means that the helium fraction is 2n 2 2 1 = = = . p+n p/n + 1 8 4 6.2 As seen in this chapter, the redshift at matter-radiation equality is z = 3380. Proceeding as in Question 5.2, the Hubble distance is then cH0−1 /33803/2 = 0.023 Mpc. If a gravitational wave then has that wavelength, its wavelength now is 0.023 × 3380 = 78 Mpc and its frequency is c/78 Mpc = 1.2 × 10−16 s−1 . 6.3 With N neutrino species Eq. (6.10) would give (using Tables A 3 and A 4) aeq =

Ωγ + N3 Ων 5.4 + 1.2N = 10−5 × Ωb + Ω c 0.313

To get aeq = 1/1100 we would need N = 19.2. 6.4 I estimate the slope to be eln(32000) /500 = 64 kms−1 Mpc−1 . This corresponds to 1 Mpc 1 = s = 1.5 × 1010 yr. H0 64 km

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7.1 Equating the forward and backward rates gives fa fb (1 ± fc )(1 ± fd ) = fc fd (1 ± fa )(1 ± fb ). Dividing both sides by fa fb fc fd gives Eq. (7.1). 7.2 It is convenient to work with x ≡ E/T . The first part of the question can be answered directly by plotting ns (x) and ρs (x). Alternatively one can proceed as follows for ns (x) (and similarly for ρs (x)). To find the maximum of ns (x), we need to set dns /dx = 0 and we should exclude the case x = 0 solution to that equation which exists for the positive sign. The position of the maximum is therefore given by ex (x/2 − 1) = ±1. If x  1 this equation can be satisfied only for the above-mentioned x = 0 case. If x  1 it cannot be satisfied at all because the right hand side is much bigger than 1. Therefore the maximum of ns (x) occurs when roughly x = 1 corresponding to E ∼ T . From the form of ns (x), one can see that a change ∆x from the peak position makes hardly any difference if |∆x|  1, but significantly reduces ns (x) if that condition is violated, which means that the width of the peak corresponds to |∆x| ∼ 1. To answer the rest of the question we set the chemical potential to zero because the question focussespon the mass. With zero chemical potential, Eqs. (7.8) and (7.9) are valid with E = p2 + m2 . These expressions are significantly different from zero only if E ∼ T . Therefore, if the mass is much less than T it is also much less than E in the regime of interest, which makes E = p a good approximation. 7.3 In Exercise 3 it should be 108 not 104 . Also, in Eqs. (7.10) and (7.11) the factor 2 after the second equality should not be there. These equations are setting kB = 0, which according to Table A.1 means that T = 2.725 ◦ K corresponds to T = 2.348 × 10−3 GeV. Inserting this into the corrected Eq. (7.10), and converting the result from GeV−1 to m using again Table A.1, gives nγ = 4.104 × 108 m−3 . (The four figures are justified because T0 = 2.725 ± 0.001 ◦ K as seen in Table A.3.) 7.4 In the question, it should say ρ0 = 4.850 GeV m−3 (as given in Table A.4). Using Eq. (7.10) with Tν3 /Tγ3 = 4/11 we find nν /nγ = 9/11. which corresponds to nν = 3.358 × 108 m−3 . The present energy density is given in natural units by Eq. (8.4), in which c = 1. The right hand side of that equation has dimension M L−3 , therefore the present energy density with any choice of units is ρ0 = 3H02 c2 /8πG. Taking H0 from Table A.3 and converting from Joules to GeV, this gives ρ0 = 4.80 GeV m−3 . If one or more of the neutrino species is at present matter as opposed to radiation, the contribution of those species to the present energy density fraction is P P nν c2 mν /3 mν c2 Ων = = . ρ0 42.90 eV (This number updates the one given in Exercise 7.4, which corresponds to an earlier and less accurate value for H0 .) 7.5 The average photon energy is ργ /nγ = 2.702T . In this formula kB is set to 1, and in general the average energy is 2.702kB T . 8.1 With Rµν defined by Eq. (2.43), the left hand side of the Einstein equation

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(3.9) should be multiplied by −1. To arrive at (3.9) itself, one should instead define Rµν as Rλµνλ which has the opposite sign. (The first convention, that I have adopted, is used by Weinberg. The second is used by Schutz.) To answer the question, note first that the metric corresponds to gµν diagonal with g00 = −1 and gij = a2 (t)δij . Inverting the matrix gµν shows that g µν is diagonal with g 00 = −1 and g ij = a−2 (t)δ ij . Using Eq. (2.19) the nonzero components of Γµαβ are seen to be Γ0ij = ∂0 gij /2 = aaδ ˙ ij and Γi0j = Γij0 = δij a/a ˙ ≡ Hδij . The definition (3.9) gives R00 = ∂0 Γµ0µ − ∂µ Γµ00 + Γµσ0 Γσ0µ − Γµσµ Γσ00 . Using the above Γ’s, Eq. R00 = 3∂0 (a/a) ˙ + Γij0 Γj0i a˙ 2 a˙ a˙ a ¨ = 3 − 3 2 + δij δij a a a a a ¨ = 3 . a Evaluating R = g µν Rµν we get R = −R00 + a−2 δij Rij . Keeping only the nonzero Γ’s in the second term, this gives
a ¨ R = −3 + a−2 −3a¨ a − 3a˙ 2 + 3a˙ 2 + 3a˙ 2 − 9a˙ 2 a a˙ 2 a ¨ = −6 − 6 2 . a a

(2)

Finally we get 1 a ¨ a ¨ R00 − g00 R = 3 − 3 − 3H 2 = −3H 2 , 2 a a which when inserted into Eq. (3.9) gives Eq. (8.4) with the left hand side multiplied by −1. 8.2 In Exercise 8.2 it should say Appendix A2, not Appendix B. To calculate H we use H ∝ 1/t. During matter domination H ∝ 1/a3/2 and during radiation domination H ∝ 1/a2 . When T = 1 MeV, a = T0 /1 MeV = 2.36 × 10−4 . Matterradiation equality is at a = 1/3389 and so when T = 1 MeV we have H=

 a 2 H 1 0 eq = . 3/2 a 13.9 yr aeq

8.3 In Table 5.1, the present age of the Universe should be 1.38 × 1010 yr as stated in the text. In the second column of Table 5.1, E is proportional to a which allows one to read off a for each row. For the last five rows, the epoch is after

12 

the epoch of matter-radiation equality. Assuming complete matter domination, we can use Eq. (8.14) to calculate t(a). (For Ω  1 this equation becomes t = 2/3H corresponding to a ∝ t2/3 .) For the last five rows this gives (from the bottom upwards) t = 1.38×1010 yr, 9.3×109 yr, 5.39×109 yr, 2.47×108 yr and 4.76×105 yr. The last three are in only approximate agreement with the table, because the the contribution of the radiation is not completely negligible. For an accurate result one should calculate t(a) by numerically integrating the expression s s da ρ(a) Ω0 /a3 = aH(a) = aH0 = aH0 . dt ρ0 Ω0 /a3 + 1 − Ω0 For the first three rows of the table, t is given by Eq. (8.16) with kT = E/2.7 ◦ K. Concerning the epoch of last scattering, it should be noted that the statements after Eq. (7.30) are not correct. In fact, X = 0.1 corresponds to z ' 1250. Also, the final number in Eq. (7.31) should be z = 1333. The redshift z = 1100 is actually the one at which the mean free path of a photon corresponds to a photon travel time that is equal to the age of the Universe. That is the definition of ‘last scattering’ adopted in this book. 8.4 Eq. (8.6) gives 8πGρa2 a˙ 2 = − K, 3 and differentiating this gives 2a¨ ˙a =


8πG d a˙ ρa2 . 3 da

Dividing by 2a˙ then gives Eq. (8.7). 8.5 To get Eq. (8.7), multiply Eq. (8.6) by a2 , and differentiate the result to get 2a¨ ˙a =


8πG d a˙ ρa2 . 3 da

Dividing this by 2a˙ gives Eq. (8.7). To get Eq. (8.9), multiply that equation by a to get
a˙ 2 4πG d d(a/a) ˙ + = a2 ρ(a) , a dt a 3 da and use d(a/a) ˙ da˙ a˙ a = − a. ˙ dt dt a 8.6 Einstein’s proposal was for a Universe containing just matter plus a cosmological constant. Then ρ = ρ0m /a3 + ρΛ , and Eq. (8.7) reads  ρ  0m a ¨ = 4πG − 2 + 2aρΛ . a This gives a ¨ = 0 at the present epoch if ρΛ = ρ0m /2.

 13

8.7 Using Eq. (8.4), the required equation is r Z 1 3 t(z) = [aρ1/2 (a)]−1 da. 8πG 1/(z+1) with ρ(a) = ρ0m /a3 + ρΛ . Setting ρΛ to zero this gives 2 t(9) = 3

r

   1 3 1− . 8πGρ0m 10

Using Eq. (8.4) this becomes t(9) =

2 1/2 3Ω0 H0

= 1.68 × 1010 yr.

(Note thatp the symbol da is missing from Eq. (8.11). Also, the prefactor in Eq. (8.10) should be 8πG/3.) 8.8 In Eqs. (8.20) and (8.21), the lower limit should be a(tbb ). In Eq. (8.21) we have to change the limits, and insert a factor c which has been suppressed. It then gives present distance of an object with redshift z, Z 1 da x(z) = c 2 (1+z)−1 a H(a) Z 1 r ρ0 da c = . H0 (1+z)−1 ρ(a) a2 Ignoring the cosmological constant (and the radiation which anyway is completely negligible) this gives r Z 1 c 1 da x(z) = H0 Ω0 (1+z)−1 a1/2 " r  1/2 # 2c 1 1 = 1− . H0 Ω0 1+z This gives x(9) = 11.0 Gpc. 8.9 Using Eqs. (8.11) and (8.4) the required age is 1 t0 = H0

1

Z 0



ρ0 aρ(a)

1/2 da.

Keeping only the matter contribution gives ρ(a) = ρ0 /a3 and t0 =

1 1/2 Ω0 H0

Z

1

ada = 0

1 1/2 2Ω0 H0

= 1.30 × 1010 yr.

14 

8.10 We deal with the era when the energy density is given by ρ = ρm + ρΛ , with ρm = ρm0 /a3 the matter contribution and ρλ the cosmological constant contribution. Since√Ω ≡ ρm /ρ we have 1 − Ω = ρΛ /ρ. Since H 2 = 8πGρ/3 (Eq. (8.4)), the product H 1 − Ω is constant, so that Eq. (8.14) gives √   2 1 d 1+ 1−Ω dt √ √ = ln . da 3H 1 − Ω da Ω Also, d ln dΩ



1+

√  1−Ω √ = Ω

− 12 (1 − Ω)−1/2 1 1 1 √ √ − =− , 2Ω 2 1+ 1−Ω Ω 1−Ω

and, using Ω=

ρm0 /a2 , ρm0 /a3 + ρΛ

we have

dΩ = −3(1 − Ω)Ω. da Putting these results into the above expression for dt/da gives dt/da = 1/(aH) (which is of course required by the definition H ≡ a/a). ˙ Finally, using Eq. (8.4) we arrive at the corrected Eq. (8.10). 9.1 Eq. (9.11) should read   x 0  1 R(x) = − 2 1 − . x A a

To arrive at this equation we begin with the nonzero metric components corresponding to Eq. (9.5). These are gxx = A, gθθ = x2 and gφφ = x2 sin2 θ. From these one finds the following nonzero Γ’s (plus those obtained by interchanging the lower indices which have equal value): Γxxx = Γxθθ = Γxφφ = Γθxθ = Γθφφ = Γφxφ = Γφθφ =

1 −1 0 1 A (A + A0 − A0 ) = A0 /A 2 2 1 −1 A (−∂x gθθ ) = −x/A 2 1 −1 A (−∂x gφφ ) = −x sin2 θ/A 2 1 −2 x ∂x gθθ = 1/x 2 1 (−∂θ gφφ ) = − sin θ cos θ 2x2 1 (∂x gφφ ) = 1/x 2x2 sin2 θ 1 (∂θ gφφ ) = cot θ. 2 2x sin2 θ

 15

We need R = Rii where Rij = Rkikj is the Ricci tensor. The nonzero components of the latter are 1 A0 xA x A0 1 = +1− 2 2A A

Rxx = Rθxθx + Rφxφx = Rθθ = Rxθxθ + Rφθφθ

Rφφ = Rxθxθ + Rφθφθ   x A0 1 +1− sin2 θ. = 2 A2 A These give R = g xx Rxx + g θθ Rθθ + g φφ Rφφ     1 A0 1 1 1 A0 1 1 1 A0 + + − + + − = x A2 2x A2 x2 Ax2 2x A2 x2 x2 A   x 0  2 . 1 − = x2 A

(3)

9.2 In the question it should be (9.12). Setting R = −6K, the corrected Eq. (9.11) gives 3Kx2 = 1 − (x/A)0 . This gives x/A = x − Kx3 − C, which gives Eq. (9.12). 9.3 In Eq. (9.14) it should be dφ2 . In a radial direction Eq. (9.14) becomes d` = dχ which means that χ is the radial distance. A circle centred at χ = 0 can be taken to have θ = π/2 and then we have d` = sin χdφ along the circle. The circumference of a circle with radius χ is therefore 2π sin χ, which is smaller than the flat-space result 2πχ. 9.4 Eqs. (9.17)–(9.20) give dw = cos χdr − r sin χdχ dz = sin χ cos θdr + r cos χ cos θdχ − r sin χ sin θdθ dx = sin χ sin θ cos φdr + r cos χ sin θ cos φdχ + r sin χ cos θ cos φdθ − r sin χ sin θ sin φdφ dy = sin χ sin θ sin φdr + r cos χ sin θ sin φdχ + r sin χ cos θ sin φdθ − r sin χ sin θ cos φdφ. Inserting these into Eq. (9.15) we find that the coefficients of dχ2 , dθ2 and dφ2 correspond to the corrected Eq. (9.14), while the coefficients of the cross terms vanish. 9.5 On a sphere of radius r we have with spherical polar angles d`2 = r2 dθ2 + r2 sin2 θdφ2 .

16 

This corresponds to gθθ = r2 , gφφ = r2 sin2 θ and gθφ = 0. The only non-zero derivative of gij is ∂θ gφφ = 2r2 sin θ cos θ. The nonzero Γs are Γθφφ = − sin θ cos θ and Γφθφ = Γφφθ = cot θ. Their nonzero derivatives are ∂θ Γθφφ = sin2 θ − cos2 θ and ∂θ Γφθφ = ∂θ Γφφθ = −1/ sin2 θ. The nonzero components of Rij are Rθθ = −1 and Rφφ = − sin2 θ, which give R = g θθ Rθθ + g φφ Rφφ   1 − sin2 θ 2 = 2 −1 + = − 2. 2 r r sin θ

(4)

(The minus sign comes from our convention for the curvature tensor. When discussing a surface, the opposite convention is generally used.) On a cylinder with radius r we have in cylindrical polar coordinates d`2 = r2 dθ2 + dz 2 , corresponding to gθθ = r2 and gzz = 1. These are independent of θ and z which means that the Γs vanish and so does the curvature tensor. The only other surface that can be made out of flat sheet of paper is conical, which means that it is the only other curved surface with zero curvature tensor. 10.1 Eq. (10.1) gives ∇ · u = H∇ · r = H∇ · (xi + yj + zk)   ∂x ∂y ∂z = H + + ∂x ∂y ∂z = 3H. Since ρ is evaluated a comoving position, ∂ρ/∂t = 0 and Eq. (3.32) gives dρ/dt = −3Hρ as required. 10.2 Since r2 = x2 + y 2 + z 2 we have ∂r/∂x = x/r. This gives ∂φ(r)/∂x = (x/r)dφ/dr and   ∂2φ 1 dφ ∂r d 1 dφ = +x ∂x2 r dr ∂x dr r dr   1 dφ x2 1 dφ 1 d2 φ = + − 2 + , r dr r r dr r dr2 giving 2

∇ φ = = =

  3 dφ 1 dφ 1 d2 φ +r − 2 + r dr r dr r dr2 2 dφ d2 φ + r dr  dr2  1 d dφ r2 . r2 dr dr

 17

A solution of ∇2 φ = 0 has dφ/dr ∝ 1/r2 which implies φ ∝ 1/r (excluding φ = const which would have no effect). Contrary to the assertion in the question, φ ∝ r2 is not a solution. 10.3 It should say, if r . 10−4 (M/M )1/3 Mpc. To demonstrate this, set the Newtonian contribution to the potential, GM/r, equal to the cosmological contribution Λr2 /6. This gives 6GM Λ



3M 4πρ0 ΩΛ

r = =

1/3



= 9.81 × 10

1/3

−5



M M ΩΛ

1/3 Mpc.

Contrary to what is stated in the question, the repulsive effect of the cosmological constant is now significant, for the relative motion of even adjacent galaxies. Of course, that was not so when galaxy clusters formed. 10.4 To verify Eq. (10.13), substitute Eqs. (10.8) and (10.11) to obtain the first-order perturbation of Eq. (3.32): ρδ ˙ + ρδ˙ = −∇ · vρ − ∇ · uρδ. Using Eq. (3.32) to evaluate the last term, and cancelling a term ρδ ˙ that then appears on both sides, this gives ρδ˙ = −∇ · vρ, and dividing by ρ gives Eq. (10.13). To derive Eq. (10.14), use Eq. (10.11) to obtain   du ˙ δ = Hv + v. dt Putting this and Eq. (10.12) into Eq. (3.33), and noting that ∇ · P = 0 for the unperturbed P , gives Eq. (10.14). To get Eq. (10.15), insert Eq; (10.12) into Eq. (3.28). 10.5 Taking the gradient of Eq. (10.18), we see that ∇ corresponds to multiplying each Fourier component by ik/a. Since vvec (k) in Eq. (10.22) is orthogonal to k, we see that the Fourier component of ∇ · v is ik · vsc /a. From Eq. (10.24) this is equal to (k/a)V (not to V as is incorrectly stated in the question). ˙ To derive this equation 10.6 In the middle term of Eq. (10.27) it should be δ. ˙ note first that Eq. (10.25) gives V = −(a/k)δ, which gives aH ˙ a ¨ V˙ = − δ − δ. k k Putting these into Eq. (10.26) with δP = 0, and using Eq. (10.28) for Φ, gives 3 δ¨ + 2H δ˙ − H 2 δ = 0. 2 10.7 To arrive at Eq. (10.36), one should subtract Eq. (10.34) from (10.35) with δPb = 0. The significance of Eq. (10.36) is already explained in the text.

18 

10.8 The derivation of the final equality in Eq. (10.42) is a standard gas theory calculation, except for a factor Cp /Cv which has been set to 1. If it held at the epoch of gas formation the minimum galaxy mass MJ would be 1.4 × 105 M (the value of MJ given in the sentence after the one containing Eq. (10.42)). 11.1 By virtue of the ergodic theorem, we can deal with just the observable Universe. If it did not hold, we would need to consider an ensemble of universes, of which the actual Universe would be just one member. 11.2 To obtain Eq. (11.8) one should replace the integral over x in Eq. (11.6) by the right hand side of Eq. (11.7). 11.3 In the absence of rotational invariance, the derivation of Eq. (11.8) still holds with now Pg depending on k. This gives σg2 ≡ hg 2 (x)i Z Z 0 1 3 = d k d2 k0 hg(k)g(k0 )ie(k+k )·x 6 (2π) Z Z 1 3 d k d2 k0 δ 3 (k + k0 )Pg (k) = (2π)3 Z 1 = d3 kPg (k). (2π)3

(5)

11.4 Using spherical polar coordinates, Eq. (11.23) becomes 3 W (kR) = 4πR3

Z

R 2

Z

dxx 2π 0

π

dθ sin θeikx cos θ .

0

Using dθ sin θ = −d(cos θ) to evaluate the second integral, this gives Z y 3 2 sin y W (y) = 3 dxx2 . 2y 0 y Differentiating this expression gives dW 3W sin y =− +3 2 . dy y y We are asked to verify that Eq. (11.24) gives the same result. Differentiating Eq. (11.24) gives   cos y sin y sin y cos y dW = 3 −3 4 + 2 +2 3 dy y3 y y y 3W sin y = − +3 2 , y y which is indeed the same result. 11.5 Because of the cosmological constant, a ¨ is initially negative as we go back in time, but it soon becomes positive because the cosmological constant soon gives a negligible contribution to the energy density.

 19

11.6 If the pressure has a perturbation δP on the slicing of uniform ρ, Eq. (11.28) becomes   ˙ t) (ρ(t) + P (t) + δP (x, t)) . ρ(t) ˙ = −3 H(t) + ζ(x, Extracting the first order perturbation from this we have ˙ t) (ρ(t) + P (t)) − 3H(t)δP (x, t), 0 = −3ζ(x, which gives ˙ t) = − H(t)δP (x, t) . ζ(x, ρ(t) + P (t) 12.1 From Eq. (12.3)(summation understood over repeated indices)   ki ki 1 Σii = − 2 + δii P Π k 3  2  k 3 = − 2+ P Π = 0.. k 3 Since we deal with the scalar mode, Σij is invariant under rotations about the k axis which means that it cannot depend on any vector or tensor other than k. The only symmetric 2 × 2 matrix that can be formed from the components of k is (up to a scalar factor) ki kj . The only symmetric 2 × 2 matrix that doesn’t involve vectors or tensors is δij . Therefore Σij is a linear combination of ki kj and δij , and requiring that it’s traceless requires that it has the form (12.9) for some scalar Π. 12.2 Using the coordinate system of Eq. (12.4), the threads have constant xi , therefore a 4-vector pointing along the threads is Aµ = (1, 0, 0, 0). The slices have constant η, therefore a vector pointing along a slice is B µ = (0, 1, 0, 0). Working to first order, the indices of a perturbation are raised and lowered with the unperturbed metric which gives Bµ = (0, 1, 0, 0) and Aµ Bµ = 0. 12.3 As stated after Eq. (12.21), we are setting δ = δγ . With w = 1/3, Eqs. (12.19) and (12.20) become ζ + Φ = δ/4 and Φ = Ψ = −(2/3)ζ. Eliminating Φ gives ζ = 3δ/4 and then the previous equation gives ζ = −2Φ. ˙ 13.1 In Eq. (13.7) the coefficient of δ˙γ should be (1/4)R/(1 + R). Also, the final expression in Eq. (13.9) should be multiplied by 1/3. To answer the question, the following are useful 1 + wf 1 − 3wf w˙ f

4ργ + 3ρb 3(ργ + ρb ) ρb = ργ + ρb  2   1 1 ρb aH ρb /ργ = − ˙= − . 3 1 + ρb /ργ ργ 3 (1 + ρb /ργ )2

=

(The final expression follows from the fact that ρb /ργ is proportional to 1/a.) To ˙ − δ˙γ ). Putting arrive at Eq. (13.7), note first that Eq. (13.5) gives Vγ = (3/4k)(4Φ this into Eq. (13.6) gives    k w˙ f 3  ˙ ργ ˙ 4Φ − δ˙γ + δγ + kΨ. Vγ = − aH(1 − 3wf ) + 1 + wf 4k 3 ρb + 34 ργ

20 

Using this expression, Eq. (13.5) gives    w˙ f ¨ ˙ − δ˙γ 4Φ δγ = aH(1 − 3wf ) + 1 + wf   2 4k 1 ργ ¨ − δγ + Ψ + 4Φ. 3 3 ρb + 43 ργ Using the expressions just given, one sees that this is equivalent to Eqs. (13.7) – (13.10). Dropping the time derivatives, Eqs. (13.7), (13.8) and (13.9) give δγ = −4(1 + R)Ψ. Ignoring the effect of the neutrinos as in Section 12.4 we have Ψ = Φ, giving δγ = −4(1 + R)Φ. 13.2 When the neutrinos are ignored we have Φ = Ψ. The tight coupling approximation ignores the neutrinos, and consists of the approximations (13.4) andP (13.6) P plus the exact equations (12.16), 13.5), (15.1), (15.2), δρ = δρi and δP = δPi . The initial conditions required to solve these equations are (12.20) – (12.22) and (11.33). 13.3 The number in Eq. (13.13) is not in fact correct. It should be 0.198. To go from Eq. (13.12) to Eq. (1.13), one can replace ρb by Ωb and similarly for the other energy densities, and then use Table 6.1 as stated after Eq. (13.13). 13.4 As a measure of the rate with which a quantity f varies, we can use f˙/f . Then we have (remembering that an overdot is denoting d/dη) A˙ R˙ a˙ da = aH. ∼ = = A R a dt A calculation of the rates of variation of B and C is beyond the level of these questions. Regarding the required slowness of the variation, the question should refer to Eq. (13.15) (not Eq. (13.14)). In that equation, A, B and C should have ˙ little variation during one period, which means |A/A|  kcs /a ∼ k/a and similarly for B and C. Since the acoustic oscillation occurs only on sub-horizon scales k > aH, this is satisfied by the rate of variation of A that we estimated. 13.5 The factor 9/16 follows from Eq. (13.4), and the definition (11.9) of the spectrum of a quantity. 13.6 To define the cross section for Thomson scattering, we can pretend that the electron has radius r and that a photon scatters whenever it hits an electron. Then σT = πr2 . If ts is the average travel time of a photon between successive Thomson scattering events, cts is the average distance travelled, which means that a tube of length cts and cross section σT should contain on average 1 electron. This requirement corresponds to cts σT ne , as stated after Eq. (13.18). 14.1 The question should say Eq. (14.7). Using it we have XX 0 ∗ 0 ∗ ha0∗ Um 0 m000 (` )Umm00 (`)ha`0 m000 a`m00 i `0 m0 a`m i = m000 m00

Using Eq. (14.8) for the right hand side, we see that m000 = m00 . Remembering that

 21

U is unitary the sum over m000 gives δmm0 and Eq. (14.7) gives C`0 δ`0 ` δm0 m = δm0 m δm0 m δ`0 ` C`0 . Setting `0 = ` and m0 = m , we conclude that C`0 = C` , corresponding to rotational invariance for C` . 14.2 The middle expression of Eq. (14.15) is h|a`m |4 i − 2h|a`m |2 iC` + C`2 . Using Eq. (14.8), this gives the final expression of Eq. (14.5). 14.3 The Gaussian probability distribution is P (g) = √ The normalisation is

1 2πσ 2

g2

e− 2σ2 .

∞

Z 1=

P (g)dg. 0

The two definite integrals giving Eq. (14.16) are Z ∞ 2 hg i = g 2 P (g)dg 0

and hg 4 i =

∞

Z

2

g 4 P (g)dg = 3hg 2 i . 0

14.4 From Eq. (13.15), the spacing of the peaks is ∆k = π/rs (ηls ), andrs (ηls ) = 145 Mpc is given after Eq. (13.19). From Eq. (13.19) this gives the spacing of the peaks of the in C` as ∆` ∼ πxls /rs (ηls ). After Eq. (14.2) is given xls = 14 Gpc and we therefore have ∆` ∼ 300. This is in remarkably good agreement with observation, as seen for instance in Figure 14.3. 14.5 This question should refer to Eq. (14.10), not Eq. (12.3). The factors 2 and 4 should not be present in Eq. (14.10), and in that equation the components are supposed to be defined in a plane orthogonal to the direction of the wave. To arrive at this equation, note that the time-averages of E 2 and B 2 in an electromagnetic plane wave are equal, and that the energy density of the wave is therefore E 2 . This is also the intensity of the wave, with c = 1 which we are adopting. Taking the trace of Eq. (14.10) we find indeed E 2 = I. 15.1 From Table 11.1 and its caption, scales outside the horizon at matter domination are roughly 1/k . 10 Mpc. As stated before Eq. (15.4), and as seen in Figure 15.1, these are the scales on which T (k) ' 1. 15.2 Using Eq. (15.4), Eq. (10.21) becomes (exhibiting the time-dependence of δ)  2 3 3 k T (k)ζ(k) = H 2 δ(k, a). 5 a 2 Instead of ‘Eq. (11.24)’ the question say ‘the definition of W given before

22 

Eq. (11.23)’. According to that definition, smoothing multiplies each Fourier component by W (kR) and so we have 2 δ(k, z, R) = W (kR) 5



k aH

2 T (k)ζ(k).

Ignoring the cosmological constant (which is a good approximation during the era discussed in Section 15.2) we have H 2 = H02 ρ/ρ0 = H02 Ωm a−3 . Putting this into the previous equation we arrive at Eq. (15.9). 15.3 Instead of ‘a/ks = 8 Mpc, given in the text for the inverse wavenumber of the Silk scale at last scattering’ the question should say ‘rs (ηls ) = 145 Mpc, given after Eq. (13.19)’. Looking at Figure 15.1, the first bump is at k ' 0.08h Mpc−1 , and the second is at k ' 0.10 Mpc−1 which gives the spacing ∆k ' 0.02h Mpc−1 . Although not stated in the Figure caption, h is in fact defined as h = H0 /100 km Mpc−1 s−1 and observation gives h = 0.7 which makes ∆k ' 0.014 Mpc−1 . From Eq. (13.15), ∆k = 2π/rs (ηls ) = 0.043 Mpc−1 which is in rough order of magnitude agreement with the previous estimate. 15.4 The expression in Eq. (15.13) should be multiplied by Ωm . Taking the energy density in Eq. (15.12) from the third line of Table A4, the numerical factor is actually 5.25 not 5.37. 15.5 In the Figure caption, 1/σ0 should be replaced by σ0 as stated at the bottom of page 119. An object forming at redshift 4 has therefore M ' 109 M which means that it is a light galaxy. 16.1 The equation after (16.5) gives, with Eq. (16.8), p = q. ˙ Putting this into Eq. (16.5), Eq. (16.8) then gives Eq. (16.9) as required. 16.2 From Eq. (16.21) we have ∂L/∂φ = −m2 φ, and ∂L/∂(∂µ φ) = −η µν ∂µ φ. (The simplest way of seeing that there is no factor 1/2 in the second expression, is to use the explicit expression for ηµν , given after Eq. (1.2).) Putting these into Eq. (16.20) gives Eq. (16.22) as required. 16.3 With Eq. (8.2), the Robertson-Walker line element p p (8.1) gives g00 = −1 and gij = δij a2 leading to g = −a6 , (−g)g 00 = −a3 and (−g)g ij = δ ij a. Using these results, Eq. (2.27) gives  1  ∂0 (a3 ∂0 φ) − δij ∂i (a∂j φ) 3 a 1 = −3H φ˙ − φ¨ − 2 δij ∂i ∂j φ. a

φ = −

Putting this into Eq. (16.26) gives Eq. (16.33). For a homogeneous field we arrive at Eq. (16.34) (not Eq. (16.33) as stated in the question). 16.4 In Eq. (16.40), ∂α ∂β φ should be ∂α φ∂β φ. In flat spacetime gµν = ηµν and we have   1 αβ µν µα νβ µν η ∂α φ∂β φ + V . T = η η ∂α φ∂β φ − η 2

 23

This gives ∂µ T

µν

=η

µα νβ

η

∂µ (∂α φ∂β φ) − η

µν



 1 αβ 0 η ∂µ (∂α ∂β φ) + ∂µ φV . 2

Using Eq. (16.26) for V 0 gives ∂µ T µν

= η µα η νβ (∂µ ∂α φ)∂β φ + η µα η νβ ∂α φ(∂µ ∂β φ) − η µν η αβ (∂µ ∂α φ)∂β φ − η µν η αβ ∂µ φ(∂α ∂β φ).

Interchanging µβ in the last term, one sees that it cancels the first term. Cyclically permuting µβα in the third term one sees that it cancels the second term. Since Dλ gµν = 0, we see that in Dµ T µν is given by the flat spacetime expression, with η µν replaced by g µν and with ∂µ replaced by Dµ . The cancellations noted in the previous solution therefore still operate giving indeed Dµ T µν = 0. 17.1 Since pˆ = qˆ˙ (the operator version of p = q, ˙ given after Eq. (16.8)), we have from Eq. (17.3)
iω pˆ = √ −e−iωt a ˆ + eiωt a ˆ† . 2ω This gives [ˆ p, qˆ] =



 i −e−iωt a ˆ + eiωt a ˆ† , eiωt a ˆ + eiωt a ˆ† . 2

This is a sum of eight terms. Four of them cancel, and the other four give i[ˆ a, a ˆ† ]. The commutator (17.1) is therefore indeed equivalent to the commutator (17.4). In Eq. (17.2), qˆ and pˆ should be interchanged, and using the above expression for pˆ one then finds indeed Eq. (17.5). The Eq. (17.4) (not (17.1)) and Eq. (17.5) give    h i   1 † † † ˆ H, a ˆ =ω a ˆ a ˆ+ ,a ˆ = ωˆ a† a ˆ, a ˆ† . 2 ˆ a Using the commutator (17.4) this gives [H, ˆ† ] = a ˆ† ω as required. ˆ a† = a ˆ + ωa† .) A state |Ei with energy E 17.2 (Eq. (17.7) should read Hˆ ˆ† H ˆ satisfies H|Ei = E|Ei. From 17.7 it therefore satisfies ˆ a† |Ei = (E + ω)ˆ Hˆ a† |Ei. Therefore a ˆ raises the energy by an amount ω. 17.3 For each term in the sum in Eq. (17.11), the proof given previously for Eq. (17.5) applies. ˆ Eq. (17.25) becomes 17.4 With Aˆ = φ, ˆ φˆ → φˆ − iZ[ˆ pz , φ]. This is the change corresponding to z → z + Z, and from Eq. (17.4) that change also corresponds to a†k → e−ikz Z a ˆ†k ' (1 − ikz Z)ˆ a†k .

24 

Comparing with Eq. (17.25) with now Aˆ = a ˆ†k , we see that [ˆ pz , a ˆ†k ] = kz a ˆ†k as required. 17.5 The right hand side of Eq. (17.22) should be multiplied by (2π)3 ), and the right hand side of Eq. (17.30) should be Λ2UV /32π 2 . The perturbation δφ can be c is then given by the right hand side of taken to have zero spatial average, and δφ Eq. (17.14). We have therefore Z 2 0 1 1 1 c p hδφ (x)i = hˆ a(k)ˆ a† (k0 )iei(k−k )·x . d3 kd3 k 0 p 6 0 (2π) 2E(k) 2E(k ) Using Eq. (17.22) and the fact that a ˆ(k) gives zero when acting on the vacuum, this gives Z 2 0 1 1 1 c p hδφ (x)i = d3 kd3 k 0 p δ 3 (k − k0 )ei(k−k )·x 0 (2π)3 2E(k) 2E(k ) Z 1 1 d3 k . (6) = (2π)3 2E(k) Imposing the cutoff on k, and taking the cutoff to be much bigger than m so that we can set E(k) ' k, this gives c (x)i = 4π hδφ (2π)3 2

Z

ΛUV

dkk/2 = 0

Λ2UV . 32π 2

18.1 Using Eq. (6.6), Eq. (8.7) becomes   4πG 3P a ¨=− aρ 1 + . 3 ρ Inflation is an early era with a ¨ > 0, which with P/ρ constant is equivalent to 3P < −ρ. 18.2 during radiation domination, a ∝ t1/2 , H = 1/(2t) and aH ∝ 1/a ∝ t−1/2 . During matter domination, a ∝ t2/3 , H = 2/3t and ρ ∝ a−3 . N is defined by Eq. (18.4). Assuming instant radiation domination after inflation, 1/4

(aH)2 ρ = 21/4 , (aH)eq ρinf where 2 indicates the end of inflation. Ignoring the cosmological constant, (aH)eq = a0 H0



a0 aeq

1/2 =

p

1 + zeq = 58.

This gives 1/4

ρ 1016 GeV N = ln 58 + ln 16inf + ln . 1/4 10 GeV ρeq

 25

The present energy density is given in Table A 4, and converting to natural units it becomes ρ0 = (1.5 × 10−3 )4 eV4 which gives 1/4

N = 63 − ln(1016 GeV/ρinf ). This is in essential agreement with Eq. (18.5), which actually corresponds to slightly different approximations than the ones that I have adopted for the solution. 18.3 This exercise is also set as part of Exercise 5 and the solution will be given at that point. 18.4 On page 86 the epoch of horizon entry for a scale k (occurring after inflation) is defined by aH = k. The epoch of horizon exit for the scale (occurring during inflation and defined explicitly on page 145) is also defined by aH = k. At any epoch, a Hubble time is defined as 1/H(t). The number of Hubble times between two epochs is therefore defined as Z t2 N (t1 , t2 ) = H(t)dt. t1

Recall that during radiation domination, a ∝ t1/2 , H = 1/(2t) and aH ∝ 1/a ∝ t−1/2 . These give N (tentry , tend ) =

1 tentry (aH)end ln = ln , 2 tend (aH)entry

where ‘entry’ denotes the epoch of horizon entry and ‘end’ denotes the end of inflation. During inflation, taking H to be constant, we have aH ∝ a and a ∝ exp(Ht) which gives aend (aH)end = = eH(tend −tex ) , (aH)ex aex and (aH)end N (tex , tend ) = H(tend − tex ) = ln , (aH)ex where ‘ex’ denotes the epoch of horizon exit. Since (aH)entry = (aH)exit we have indeed N (tend − tex ) = N (tentry − tend ). 18.5 Proof of Eq. (18.13). Eq. (18.11) gives V 0 φ˙ 3H 2 MP ' HMP while Eqs. (18.8) and (18.9) give φ˙ 2  MP2 H 2 . Evaluating φ˙ from Eqs. (18.11) and (18.10), this gives Eq. (18.13). (As we deal with a strong inequality the choice of numerical factor is unimportant, but the choice made in the definition of  is the standard one.) Proof of Eq. (18.15). Eq. (18.11) with the exact equation (16.33) ¨  3H|φ|, ˙ and then Eqs. (18.14) and (18.9) imply |V 00 |  H 2 , which (from imply |φ| the final equality of Eq. (18.15)) is equivalent to |η|  1. Proof of Eq. (18.16). Z tend Z φend H(φ) N (φ) ≡ Hdt = dφ. φ˙ t(φ) φ

26 

Using Eq. (19.11) this gives Z

φend

N (φ) = φ

3H 2 dφ. V0

Using Eq. (18.10) this gives Eq. (18.16). ˙ 2 Pδφ . Using Eq. (19.11) this gives 19.1 From Eq. (19.12), Pζ = (H/φ) Eq. (19.13), in which the right hand side is to be evaluated at the epoch aH = k. In Eq. (19.14), dk should be d(ln k). To verify Eq. (19.14), note first that the final expression of Eq. (19.14) is to be evaluated at the epoch k = aH. Since H varies slowly during inflation we have as good approximations d(aH))/dt = aH. ˙ This gives d(ln k) = Hd ln a = da/a = Hdt, where the right hand side is evaluated at the epoch aH = k. This gives " n(k) − 1 =

H 2π

2

9H 4 V 02

#−1

d H dt

"

H 2π

2

# 9H 4 . V 02

˙ + AB. ˙ we find Making repeated use of the rule (AB)˙ = AB n(k) − 1 =

˙ 00 6H˙ (V 0 )˙ 6H˙ φV − 2 = − 2 , H2 HV 0 H2 HV 0

where again the right hand sides are evaluated at the epoch aH = k. Using the approximations 3MP2 H 2 = V and 3H φ˙ = −V 0 we finally arrive at Eq. (19.14). 19.2 If V = Cφp where C is a constant, we have V 0 = Cpφp−1 and V 00 = Cp(p − 1)φp−2 . Putting these into the definitions (18.13) and (18.15) of  and η, we arrive at the required expressions. 20.1 The question should say 10−15 m (as stated on page 152), not 10−5 m. To verify this, note first that the average spacing between particles with number −1/3 −1/3 density ns is ns . As an estimate, Eq. (7.10) gives ns = T −1 . That is in natural units with kB = 1. Taking T = 100 MeV and using the first line of Table A1 we −1/3 find ns ∼ 10/(5 × 1015 ) m, which is roughly 10−15 m as required. 20.2 In the question it should be (1014 GeV)2 , not 1014 GeV. During radiation domination a ∝ t1/2 so that H = 1/2t. Working to rough order of magnitude, in natural units, the energy of a string whose length is the Hubble distance is Estring ∼ (1014 GeV)2 t. From Eq. (7.11) the energy density is of order T 4 where T is the temperature. The energy within a sphere whose diameter is the Hubble distance is therefore Esphere ∼ t3 T 4 . Using Eq. (8.16) tT 2 ∼ 1 s MeV2 . Converting this to natural units using the second line of Table A.3 this becomes tT 2 ∼ 1018 GeV which gives Esphere ∼ (1018 GeV)2 t. We have therefore Esphere /Estring ∼ 108 .