ASSIGNMENTS Week 10 (A. Achterberg) (NWI-NM026C) 1 Cold...

ASSIGNMENTS Week 10 (A. Achterberg)(NWI-NM026C)

1 Cold Dark Matter perturbations and driven baryonicfluctuations

Cold Dark Matter (CDM) is the prime candidate for the material that dominates the mattercontent of the Universe. In that picture ordinary (baryonic) matter is only a minorityspecies: ΩCDM ' 0.20, Ωba ' 0.04. The term ’cold’ in CDM means that these particlesdecoupled from the rest of the universe when they where non-relativistic, that is at atemperature Tdec such that kbTdec mcdmc2. One thinks that CDM consists of WIMPS:Weakly Interacting Massive Particles. Several experiments are being conducted to detectsuch particles.

In this assignment we look at small fluctuations in the CDM density in the early uni-verse to find out how they grow. We will describe these in terms of the fractional densityperturbation,

∆cdm(x , t) =

(δρ(x , t)

ρ

)cdm

. (1.1)

Here x is the comoving coordinate, related to the physical position vector r by r = a(t) x.

In the limit where the contribution of baryons to the gravitational potential can beneglected the cold dark matter density fluctuations satisfy

[d2

dt2 + 2( aa

) ddt− 4πG ρcdm(a)

]∆cdm(x , t) = 0 . (1.2)

Here a = da/dt = Ha is the time derivative of the scale factor. The sound speed of CDMmay be taken to be so small that it does not matter. The unperturbed CDM density scalesin the usual manner:

ρcdm(a) ∝ a−3 . (1.3)

We will also look at how the gravitational forces associated with CDM density fluctuationsdrive density fluctuations in the gas of baryons.

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a. Consider a variable change from t to a(t) so that one uses the scale factor a(t) ratherthan time to track the expansion of the universe. This leads to:

ddt

= ad

da. (1.4)

Show that

d2

dt2 = a2 d2

da2 + ad

da, (1.5)

where a ≡ d2a/dt2.

b. Friedmann’s equation for a flat (k = 0), matter-dominated universe leads to:

a2 =8πGρ

3a2 , a = −

4πGρ3

a . (1.6)

Show that the CDM wave equation (1.2) can be rewritten as:

(23 a2 d2

da2 + adda−ρcdm

ρ

)∆cdm(a) = 0 . (1.7)

c. It is a reasonable approximation in the matter-dominated phase of the Universe toassume ρ ' ρcdm so that the last term inside the brackets of Eqn. (1.7) can bereplaced by unity. Show that in this case solutions of the form

∆cdm(a) ∝ ∆0

(aa0

)s

(1.8)

solve Eqn. (1.7), and find the two possible values for the exponent s.

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Baryons (as a minority species) feel the perturbation in gravity caused by the CDM den-sity perturbations. This leads to baryonic density perturbations that are driven by theCDM fluctuations. In the limit ρba ρcdm and for scales much larger than the baryonicjeans length, the baryonic density fluctuations satisfy:

[d2

dt2 + 2( aa

) ddt

]∆ba(t) = 4πGρcdm ∆cdm(t) . (1.9)

Here ∆ba(t) = δρba/ρba is the fractional baryonic density perturbation. The term on theright-hand side of this equation is the driving term: the effect of the gravitational acceler-ation ∝ ∆cdm(t) felt by the baryons due to the gravity induced by CDM fluctuations. Theself-gravity of the baryons has been neglected as we assumed ρba ρcdm.

d. Now show that for the growing solution obtained in c for ∆cdm, the solution withpositive exponent s, the following expression for ∆ba solves Eqn. (1.9) for the initialcondition ∆ba(a0) = 0:

∆ba(a) = ∆0

(aa0− 1

). (1.10)

Hint: first convert the equation for the baryons to the form analogous to (1.7) usingρ ' ρcdm! That should not be any problem now.

e. What does this mean for the relative magnitude of CDM and baryonic mass fluctu-ations δρcdm and δρba at late times after substantial growth, that is for a a0?

You will need the current densities as derived from WMAP, conveniently expressedin terms of the Omega-parameters: Ωcdm0 = 0.20, Ωba0 = 0.04.

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2 Radiation drag: a covariant approach

In the following, I largely adopt the notation of Mihalas & Mihalas, 1984: Foundationsof Radiation Hydrodynamics, Oxford University Press. Let χ(x , t ; n , ν) be the opacityof a medium at rest for photons with momentum

pph =hνc

n (2.11)

and frequency ν, photon energy E = hν. The quantity η(x , t ; n , ν) is the emissivitycoefficient of the medium. The standard radiation transfer equation, the transport equationfor photons in the medium in terms of the radiation intensity I(x , t ; n , ν), reads:

(1c∂

∂t+ n · ∇

)I(x , t ; n , ν) = η(x , t ; n , ν) − χ(x , t ; n , ν) I(x , t ; n , ν) . (2.12)

The two terms on the right-hand side of this equation are respectively the emission of newor scattered photons in the direction defined by n and with frequency ν, and the secondterm is the absorption/scattering of existing photons by the medium.

If one wants to think in terms of a photon distribution function fph(x , t , pph), so that

fph(x , t , pph) d3 pph (2.13)

is the number of photons in the momentum inteval pph, pph + dpph at space-time event(x , t), one has:

I(x , t ; n , ν) =

(h4ν3

c2

)fph(x , t , pph) , (2.14)

see Mihalas & Mihalas Ch. 6 for more details. From this point on I will suppress thespace-time dependence in all quantities, but keep the dependence on ν and n as it iscrucial in the derivations.

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The Minkowski force density on the medium due to the interaction with the radiation is

Gµ =(G0 , G

), (2.15)

with

G0 =1c

∫ ∞

0dν

∮dΩ

χ(n , ν) I(n , ν) − η(n , ν)

, (2.16)

and

G =1c

∫ ∞

0dν

∮dΩ

χ(n , ν) I(n , ν) − η(n , ν)

n . (2.17)

The integration is over photon frequencies and solid angle. G0 corresponds to the netenergy per unit time gained (or lost) per unit volume by the medium, divided by c fordimensional grounds, while G is the net momentum transfer per unit time and unit volumefrom photons to the medium.

a. One can define the photon four-momentum by writing

pph =hνc

(1 , n) ≡hνc

nµ (2.18)

Show that nµ is a null-vector (as it should!), and the show that the Minkowski forcedensity can be written covariantly (in terms of four-vectors) as

Gµ =1c

∫ ∞

0dν

∮dΩ

χ(n , ν) I(n , ν) − η(n , ν)

nµ . (2.19)

b. The energy-momentum tensor of the radiation field, which we will call Rµν ratherthan the usual T µν, is related to the intensity I(x , t ; n , ν) by:

Rµν =1c

∫ ∞

0dν

∮dΩ I(n , ν) nµnν =

ε F/c

F/c P

. (2.20)

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Here ε is the radiation energy density, F is the radiative energy flux (so that F/cequals the photon momentum flux M) and P is the radiative stress tensor in three-space. Show that the following relations hold:

ε =1c

∫ ∞

0dν

∮dΩ I(n , ν) ,

Fc

=1c

∫ ∞

0dν

∮dΩ I(n , ν) n , (2.21)

P =1c

∫ ∞

0dν

∮dΩ I(n , ν) n⊗n .

Here n⊗n is a tensor in three-space with components (n⊗n)i j = ni n j.

c. Let us now concentrate on simple Thomson scattering by electrons at rest, withχ(n , ν) ∼ σTne with ne the electron proper density and σT the Thomson crosssection. We use [1] the fact that the scattering is symmetric1 so that the we can usethe average scattering coefficient, and [2] the fact that there is no net energy transferand that the scattered photons on average do not carry away any momentum so thatthe momentum term involving η(n , ν) integrates to zero. For simple scattering likethis one has

η(n , ν) = neσT

∮dΩ′

4πI(n′ , ν) , (2.22)

and the scattered radiation is isotropic so that η does not depend on the direction n.

Now show that this implies the following for the Minkowski force in the rest frameof the electron fluid:

G0 = 0 , G =neσT

c

∫ ∞

0dν

∮dΩ I(n , ν) n . (2.23)

1It contains a factor (1 + cos2 θs) with θs the scattering angle, assuming unpolarized incident radiation.

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d. We have assumed at the outset that the medium (the electron fluid) is at rest, withfour-velocity

Uµ = (1 , 0 , 0 , 0) (2.24)

Show that the Minkowski radiation force due to Thomson scattering on an electronfluid in the electron rest frame as derived in c can be represented covariantly interms of the radiation energy-momentum tensor Rµν as:

Gµ = neσT hµτ Uσ Rστ .

(2.25)

Summation over double indices is implied, and I define the projection tensor

hαβ = δαβ − UαUβ , (2.26)

which projects onto the plane perpendicular to the four-velocity as hαβUβ = Uαhαβ =

0. These definitions assume that the Minkowski metric is ηµν = diag(1, −1, −1, −1)and uses ηµν = δ

µν.

The crucial point of this derivation, and result (2.25), is the following:Expression (2.25) contains only four-vectors and four-tensors and scalars like the properdensity ne and σT. The expression is therefore fully covariant. Consequently, it can beapplied in any reference frame, even though the derivation leading to this result assumedthat the electrons are at rest! We will apply that principle in what follows.

Question continues on next page!

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e. Consider slow (that is: non-relativistic) motion of the electron fluid with a four-velocity2

Uµ ≈(1 + 1

2β2 , β

), Uµ ≈

(1 + 1

2β2 , −β

)(2.27)

with |β| = |v|/c 1. The electrons moves through an isotropic radiation field wherethe radiation energy-momentum tensor takes the simple form

Rµν =

ε 0 0 0

0 13ε 0 0

0 0 13ε 0

0 0 0 13ε

. (2.28)

This expression for Rµν uses the fact in this case F = 0, and that the pressure three-tensor satisfies P = 1

3ε I, with I = diag(1 , 1 , 1) the unit tensor in three-space.It is formally the same as the energy-momentum tensor for an ideal fluid at rest,T µν = diag(ρc2 , P , P , P) with ρc2 the energy density and P the pressure, as forisotropic radiation we have P = ε/3.

Now show that the Minkowski force density excerted by the radiation on the elec-tron fluid equals, neglecting all terms of order β3 or higher:

Gµ = −4neσT ε

3

(β2, β

).

(2.29)

Note: the largest term ∝ β is a frictional force (radiation drag) as it is in the oppositedirection of the velocity of the electron fluid. The much smaller G0 ∝ β2 term correspondsto the energy transfer from electrons to the radiation due to Compton scattering in thelimit β 1.In the early Universe the photons are in thermal equilibrium so that ε = ar T 4 with ar theradiation constant and T the temperature.

2The β2 terms must be included to preserve the normalization UσUσ = 1 to order β2.

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3 Thinking in terms of redshift

The number of cosmological parameters that is directly observable are few: the (current)Hubble constant H0, the Ω-parameters and redshift z come to mind. H0 and z are relatedto the scale factor of the universe by

H0 =

(1a

dadt

)t=t0

, 1 + z =a(t0)a(t)

, (3.1)

with t the time of photon emission for a hypothetical photon that arrives at the observerat the current epoch t0 with redshift z. It is therefore not surprising that most astronomersthink in terms of redshift, and use redshift as a measure of distance, and (implicitly) ofthe time when the photons they observe were emitted. So: ”at redshift z” means ”at adistance/time corresponding to a present photon redshift z”. Given a certain redshift z, thescale factor is a(t) = a(t0)/(1 + z) and one can calculate t from a(t) provided one knowsthe type of universe one is living in! Remember in all this that ”large z” is equivalent to”distant” and ”far into the past”.

In this assignment we follow that approach, and calculate a number of important quan-tities in terms of redshift (rather than time). We will also use a scaling where a(t0) = 1,which is always possible. This means that the scale factor at time of emission is related tothe photon redshift z at time of observation by

a(t) =1

1 + z. (3.2)

We will now convert many of the fundamental relationships in cosmology from a formu-lation in terms of scale factor to a formulation in terms of redshift. In order to make directcontact with other ’observables’ this formulation will use the Omega-parameters (ratio ofdensity over critical density, ρco = 3H2

0/8πG) for matter, radiation and vacuum energy:

Ωm0 =8πG ρm0

3H20

, Ωr0 =8πG ρr0

3H20

, ΩΛ =Λ

3H20

=8πG ρv

3H20

(3.3)

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3.1 Densities

Show that the matter and radiation density at time t ≤ t0 can be respectively written interms of photon redshift of the photons originating from that time as

ρm = ρm0 (1 + z)3 , ρr = ρr0 (1 + z)4 . (3.4)

3.2 Friedmann’s equation in z-language

Show that the Hubble parameter can be written as

H(t) = −1

1 + zdzdt. (3.5)

What sign does dz/dt have when you (correctly) interpret t as the time of emission?

Now show that Friedmann’s equation can be written as

(dzdt

)2

= H20(1 + z)2

[Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2

], (3.6)

where

Ωk0 ≡ Ωm0 + Ωr0 + ΩΛ − 1 ≡ Ω0 − 1 . (3.7)

is a measure for the curvature of the universe at time t0.

Hint: Get rid of the curvature constant k by looking at the Friedmann equation at time t0

when a(t0) = 1 and H = H0.

3.3 Hubble parameter

What is the value of the Hubble parameter H(z) at the time corresponding with redshift z?

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3.4 Friedmann equation in a Lemaitre Universe

Now assume that the radiation term can be neglected, in effect putting Ωr0 = 0. Show thatFriedmann’s equation in this case can be written as

(dzdt

)2

= H20(1 + z)2

[(Ωm0 z + 1) (1 + z)2 −ΩΛz (2 + z)

]. (3.8)

This corresponds to Eqn. 7.64 on p. 217 of Longair’s book.

(Hint: you may need the special product (1 + z)3 = 1 + 3z + 3z2 + z3.)

3.5 Source distance as a function of observed redshift z0

In the lectures we calculated the distance to a source at the moment of photon reception,which is the distance always quoted in research papers, as

D0 =

∫ t0

t

c dta(t)

, (3.9)

assuming a(t0) = 1 as before. Now show that this distance can be written through avariable change from t to z as:

D0 =c

H0

∫ z0

0

dz√Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2

. (3.10)

Here z0 is the observed redshift of a photon from that source, which started the journey attime t.

Hint: think carefully about the way to go from an integral over time to an integral overredshift, and what this means for the integration boundaries!

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3.5.1 The small-z limit

Show that for z0 1 the above integral for D0 always yields the result

D0 =cz0

H0+ O(z2

0) . (3.11)

3.6 Photon travel time as a function of observed redshift z0

Using a similar approach, converting an integral over time into one over redshift, showthat photon travel time ∆t ≡ t0 − t of a photon with observed redshift z0 equals

∆t =1

H0

∫ z0

0

dz

(1 + z)√

Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2. (3.12)

3.7 Horizon distance

The horizon distance corresponds to the distance a photon can travel during the entire ageof the universe. Answer the following questions:

• Assuming a(0) = 0, what is the redshift z0 at the moment t0 of photon reception;

• What then is the horizon distance Dhor according to (13.71)?

3.8 Distances in a flat universe

I now consider a flat universe with Ωk0 = 0 and with a negligible radiation density so thatwe can put Ωr0 = 0. In this case we have:

Ωm0 + ΩΛ = 1 , (3.13)

and result (13.71) reduces to

D0 =c

H0

∫ z0

0

dz√Ωm0 (1 + z)3 + ΩΛ

. (3.14)

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Calculate D0 for two cases:

1. A flat, matter-dominated universe with Ωm0 = 1 and ΩΛ = 0;

2. A flat De-Sitter universe with Ωm0 = 0 and ΩΛ = 1.

Can you now immediately write down the horizon distance Dhor for the first case?

3.9 Omega parameter and deviation from flatness

The Ω- parameter of some universal constituent i with density ρi is defined in the usualfashion as

Ωi =8πG ρi

3H2 . (3.15)

The deviation of flatness is

∣∣∣∣∣∣∣ ∑

all constituents

Ωi

− 1

∣∣∣∣∣∣∣ ≡ |Ωtot − 1| . (3.16)

Consider a universe filled with a mixture of cold matter (”dust”), radiation and vacuumdensity ρv = Λ/8πG. Use the results of the previous questions to show that the followingrelation holds, with Ω0 ≡ Ωtot(t0) = Ωm0 + Ωr0 + ΩΛ:

Ω(z) − 1 =(Ω0 − 1) (1 + z)2

Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ − (Ω0 − 1) (1 + z)2 (3.17)

Use this result to show the following/answer the following:

1. Far in the past, the deviation from flatness (assuming of course Ω0 , 1) alwaysbecomes very small regardless the values of Ωm0, Ωr0 and ΩΛ.

2. The derivative d |Ω − 1| /dz is always negative (for Ω0 , 1) when ΩΛ = 0. Whatdoes this mean for the sign of the time derivative d |Ω − 1| /dt ?

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3. If ΩΛ , 0 and Ω0 , 1 the derivative d |Ω − 1| /dz changes sign, from negative atlarge z to positive at small z, when the equality

Ωr0 (1 + z)4 + 12Ωm0 (1 + z)3 = ΩΛ (3.18)

can be satisfied for z ≥ 0. Use this result to calculate the redshift z∗ when our Uni-verse started to become flatter again (if it ever deviated significantly from flatnessat all!), using Ωmo ' 0.27, Ωr0 ' 0 and ΩΛ ' 0.7.

4 Angular distance, early horizon size and related issues

In this assignment we continue to look at various observable quantities, and how we cancalculate them using the properties of the current universe such as redshift, the Omegaparameters and the Hubble constant H0.

Whenever possible we will convert time-integrals into redshift integrals in the samemanner as was done in the first assignment. That means that we will keep using integralconversions of the sort

∫ t2

t1dt · · · =

∫ z1

z2

dz

(1 + z)√

Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ0 −Ωk0 (1 + z)2· · · . (4.1)

Here z1 (z2) is the redshift of a photon observed at some fixed time t0 if it was emittedat some earlier time t1 (t2). Note that we do not necessarily assume that t2 = to (photonobservation time = current age of the universe) as we did earlier. That does not influencein any way the validity of the expression, which is solely based on a variable change,the definition of redshift z in terms of the scale factor a(t) for the choice a(t0) = 1 andFriedmann’s equation.

Also do not forget that z2 < z1 if t2 > t1: photons emitted later are observed with a smallerredshift.

The ultimate aim of this assignment is to have you calculate the angular size in the skyof the horizon distance as it was at the moment that the Cosmic Microwave Background(CMWB) was born. This moment in the evolution of our universe corresponds with aredshift z = zCMWB ' 1100, as explained in the Lectures, or in Chapter 9 of Longair’sbook.

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a. If we put ourselves at the origin r = 0 of a Robertson-Walker metric, all photonorbits that reach us are purely radial orbits that have (θ , φ) constant. Also, in aRobertson-Walker metric the way angles are measured does not explicitly dependon curvature.

Use this information to prove that a distant object of fixed linear size d that is muchsmaller than its distance is seen with an angular size ∆θ on the sky that equals

∆θ = 2 arctan(

d2D

)'

dD

(assuming d D, ∆θ 1) , (4.2)

with D the distance of the source at the moment of emission of the photons. (It mayhelp to draw a figure showing the situation at photon emission, at photon receptionas well as the photon orbits.) We will use the small angle approximation (∆θ 1)from this point onwards.

b. If the source is observed at t0 with a redshift z0, then show that the so-called angu-lar distance Da ' d/∆θ is related to the source distance D0 at the moment of theobservation by

Da =d

∆θ=

D0

1 + z0. (4.3)

c. Use one of the results of this week’s first assignment to show that

Da =c

H0 (1 + z0)

∫ z0

0

dz√Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2

. (4.4)

NO COMPLICATED CALCULATION SHOULD BE REQUIRED!

(Continues on next page!)

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d. At any time t > 0, not just t0, photons emitted at the Big Bang arrive at an observerwith redshift z = ∞. That means that the horizon size at some 0 < t ≤ t0 is

Dhor(t) = a(t)r = a(t)∫ t

0

c dt′

a(t′)(4.5)

Show that this allows you to obtain the horizon size ”at a redshift z” from

Dhor(z) =c

H0(1 + z)

∫ ∞

z

dz′√Ωm0(1 + z′)3 + Ωr0 (1 + z′)4 + ΩΛ −Ωk0 (1 + z′)2

.

(4.6)

e. The angle ∆θ under which an object of size d is observed equals for small anglesaccording to result b

∆θ(z0) =d

Da= (1 + z0)

dD0

. (4.7)

Calculate ∆θ(z0) for the simple case of a flat universe dominated by cold matter,that is for Ωm0 = 1, Ωr0 = ΩΛ0 = Ωk0 = 0.

Make a rough graph of ∆θ(z0) and show that the angular size, given d, first de-creases with increasing redshift (distance), but then increases for large z0 (i.e. largedistances!). Show that the minimum angular size occurs in this universe at a redshift

zmin = 1.25 , (4.8)

that this minimum angular size equals

∆θmin =278

H0dc

= 3.375H0d

c, (4.9)

and that it occurs for a source that currently resides at a distance

D0 = D(z = 0) =23

cH0

=Dhor

3. (4.10)

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Here Dhor = 2c/H0 is the horizon size for this flat, matter-dominated universe attime t0.

Could you give a physical explanation for the fact that the every-day intuitive prin-ciple ”the more distant it is, the smaller it looks” does not necessarily always applyin an expanding universe?

f. We now consider the birth of the Cosmic Microwave Background (CMWB) at z =

zCMWB ' 1100. At that (and larger) redshifts it is a fair assumption to approximatethe universe as a radiation-dominated, flat universe. That means that we only needto take the radiation term ∝ Ωr0 into account in integrals for z ≥ 1100 as that termdominates over the other terms, and can put Ωk0 = 0 as we are dealing with aflat universe. Show that the horizon size at the birth of the CMWB can then becalculated from (4.6) as

Dhor(zCMWB) 'c

√Ωr0 H0

1(1 + zCMWB)2 ' 10−6 c

√Ωr0 H0

. (4.11)

g. The distance associated with redshift z = zCMWB ' 1100 defines a fictitious spheri-cal surface around each observer that corresponds to the last scattering surface, thesurface where the CMWB photons experienced (on average) their last scattering andbeyond which (i.e. at larger redshifts) the universe is opaque to visible light. Thecorresponding distance is DLS . Using a model with Ωr = 0 (a fair approximationfor z < 1100) to calculate DLS with zCMWB ' 1100 one finds roughly:

DLS 'c

H0

∫ zCMWB

0

dz√Ωm0 (1 + z)3 + ΩΛ0

'3cH0

. (4.12)

The integration uses the observed values for Ωm0 = 0.27 and ΩΛ0 ' 0.73. Thecurrent radiation density corresponds to

Ωr0 ' 10−4 . (4.13)

Use this information to calculate the angular size in the sky of the horizon distanceat the moment the CMWB was born, first in radians and then in degrees!


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h. As a last application we consider the past light cone. For simplicity I will againassume a flat universe. Photon orbits follow from the equation for null geodesics

ds2 = c2dt2 − a2(t) dr2 = 0 , (4.14)

which gives the comoving distance ∆r traveled in the cosmic time interval ∆t =

t2 − t1 as

∆rphot(t1|t2) =

∫ t2

t1

c dta(t)

. (4.15)

The past light cone is usually defined as the physical distance at the moment ofemission (te) that corresponds with the comoving distance ∆rphot(te|t0) traveled by aphoton in the time interval t0 − te if the photon is received at t = t0. This gives

DPLC = a(te) ∆rphot(te|t0) = a(te)∫ t0

te

c dta(t)

. (4.16)

The reason why this particular definition is useful is that one can in principle seehow many objects are within the past light cone when the light was emitted. Bythe time the photon is received (t0), this distance has been stretched by the Hubbleexpansion by a factor a(t0)/a(te) = 1 + z0. Assuming a scaling where a(t0) = 1 thisimplies3 D0(PLC) = ∆rphot.

Now show that for photons received with redshift z0 the past light cone size DPLC atthat redshift coincides with the angular distance, i.e.

DPLC = Da =c

H0 (1 + z0)

∫ z0

0

dz√Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2

.

(4.17)


3This is the basis of the remark that the choice a(t0) = 1 means that the current physical distance andthe comoving distance coincide.

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i. Can you now exploit the fact that DPLC = Da in the following way for a matter-dominated universe with for Ωm0 = 1, Ωr0 = ΩΛ0 = Ωk0 = 0, see assignment ewhere we did something similar for the angular size:

1. Demonstrate that DPLC increases for small z, but decreases at large z;

2. Give the redshift where DPLC has a maximum, and calculate DPLC at that point.

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5 Hand-in Assignment:Entropy conservation and temperature jumps

In Big Bang cosmology the Universe is self-contained, with no ‘outside’. That means thatthere is nothing that can add heat to, or extract heat from the universe at-large. Thereforethe entropy of any comoving volume in the universe is conserved. The entropy density(entropy per unit volume) for a thermal gas of relativistic particles with a chemical po-tential µ = 0 follows from the fundamental thermodynamic relation

s =ρ + P

T, (5.1)

with ρ the energy density (in natural units) and P the pressure. For an isotropic gas ofrelativistic particles we have P = ρ/3 and therefore

s = 43

ρ

T. (5.2)

Table 1 below gives the properties for a thermal gas of relativistic bosons or fermions

Table 1: Thermal gas of relativistic particles

Property (with c = ~ = kb = 1) symbol Bosons Fermions(per degree of freedom)

Number density: n(T )ζ(3)π2 T 3 3

4ζ(3)π2 T 3

Energy density: ρ(T )π2

30T 4 7

8π2

30T 4

Pressure: P(T ) = 13 ρ(T )

π2

90T 4 7

8π2

90T 4

Entropy density: s(T ) = 43ρ(T )

T2π2

45T 3 7π2

180T 3

(assuming chemical potential µ = 0)

20

In this table ζ(3) ' 1.202 is the third zeta-function.

Since volumes in an expanding universe increase as V ∝ a3, with a(t) the scale factor,entropy conservation means:

2π2

45

∑bosons

gi (Ti a)3 +78

∑fermions

gi (Ti a)3

= constant.

(5.3)

Here the gi are the statistical weights of each species that give the number of degrees offreedom. Distinguishable anti-particles are counted separately in this sum.

If there is only a single relativistic particle species, for instance photons that have g = 2,this relation reduces to Ta = constant, with T the radiation temperature, and we recoverthe result from redshift: the radiation temperature scales as T ∝ a−1.

In more complicated situations, where we have a mixture of species that interact fre-quently through collisions thereby maintaining thermal equilibrium, one should userelation (5.3), which is more fundamental. Species that are decoupled (meaning: the timebetween collisions Hubble time) should not be included in the sum over species. Theydo not interact with the rest of the Universe!

a. In the Standard Model of particle physics all three neutrino species are mass-less.Several experiments have shown that this is not true, but unfortunately only upperlimits to the neutrino masses exist as most (but not all!) of these experiments deter-mine quadratic mass differences ∆

(m2

)= |m2

i − m2j | rather than individual masses.

Nevertheless, the simplest models aimed to explain these neutrino oscillation ex-periments estimate the mass of the heaviest neutrino in the range m ∼ 0.05 eV.One can also obtain a limit on the mass of neutrinos from the data on the CosmicMicrowave Background. In 2013 the Planck collaboration reported a limit

∑three known species

mν < 0.66 eV (95% confidence limit). (5.4)

21

Since the known cross section for high-energy electron-neutrino scattering predictsthat the neutrino’s decouple at a temperature T ∼ 3 MeV (in energy units) we canprobably safely assume that neutrinos are relativistic at decoupling. Since (afterdecoupling) neutrinos are simply redshifted (just like photons) and diluted (so thatnν ∝ a−3) we have a neutrino temperature that scales as Tν ∝ a−1, just like thephoton temperature, as long as they remain relativistic. Since the temperature ofthe radiation- and neutrino gas were equal at decoupling, they remain equal afterdecoupling as long as nothing more happens.

But something does happen: at T ∼ 1 MeV electrons and positrons annihilate(e+ + e− → 2γ), a process that adds extra photons and energy to the pool ofphotons, thereby raising the photon temperature and entropy.

Use the entropy law for a coupled thermal gas of photons, positrons and electrons,all at the same temperature T , to show that the entropy conservation law (5.3) dic-tates that the photon temperature after annihilation is

Tγ(after) =

(114

)1/3

T . (5.5)

You may assume that the annihilation proceeds very fast, much faster than the Hub-ble time. You will also need the following statistical weights: ge± = 2 and gγ = 2.

b. After the electron-positron annihilation the usual redshift law Tγ ∝ a−1 once againapplies to photons. Use this to calculate the ratio of the number density in neutri-nos (all species) and photons after annihilation. You may use that there are threeneutrino species (and three corresponding anti-neutrino species), each with gν = 1,in the Standard Model. Does this ratio change as the Universe expands?

c. Calculate the number density nν0 and temperature Tν0 of each individual neutrinospecies in the Cosmic Neutrino Background of the present Universe, using that thephotons of the Cosmic Microwave Background have (in normal cgs units)

nγ0 ' 420 cm−3 , Tγ0 ' 2.7 K . (5.6)

You may assume neutrinos are mass-less in this calculation.

22

d. Neutrinos at small energies are very weakly interacting. If the interaction crosssection is σν, the neutrino velocity in units of lightspeed is βν and their densityis nν, the interaction rate (event rate) R in a detector containing in total Nt targetparticles is

R = σν βνc nν Nt s−1 . (5.7)

For incoherent neutrino-nucleon (nucleon = proton or neutron) scattering we havefor neutrinos with energy Eν (including rest energy):

σν ' 2 × 10−56( Eν

1 eV

)2

cm2 . (5.8)

In this calculation you may assume that neutrinos are massive, with mass mν '

0.1 eV in energy units, and with βν ' 10−3: neutrinos move around the galaxy witha speed roughly equal to orbital speed of the Solar System. Using the answer for nνfrom b (which is still valid!) calculate the following:

– The number of target particles (nucleons) needed to get a reaction rate R equalto once a year; Remember: 1 yr ' 3 × 107 s;

– The number of liters of water you need if you use a water detector (like high-energy neutrino experiments like ANTARES do), where the mass per nucleonis roughly the proton mass, mp ' 1.67 × 10−27 kg.

How large (in kilometers) would the linear size of a cubic volume of waterneed to be to achieve this event rate of once a year?

e. Repeat the same calculation, but now assuming again that the neutrinos are mass-less, with an energy Eν = kbTν0. You may use the answer obtained in c. for thecurrent neutrino temperature Tν0.

23


6 Simple inflation model: the massive real scalar field

In this assignment we will consider the simplest incarnation for an inflation model, whichemploys a massive (mass m), spatially uniform, one-component scalar field with La-grangian

L (ϕ , ϕ) = 12 ϕ

2 − V(ϕ) = 12 ϕ

2 − 12m2 ϕ2 . (6.1)

Here

ϕ ≡∂ϕ

∂t, (6.2)

employing natural units with c = ~ = kb = 1 and G = 1/m2pl, with mpl =

√~c/G =

1.2 × 1019 GeV the Planck mass. Density and pressure associated with this field are:

ρϕ = 12 ϕ

2 + 12m2 ϕ2 , Pϕ = 1

2 ϕ2 − 1

2m2 ϕ2 . (6.3)

Field dynamics in an expanding universe with Hubble parameter H(t) is described by the’equation of motion’. Introducing the notation ϕ ≡ ∂2ϕ/∂t2:

ϕ + 3H ϕ + m2 ϕ = 0 . (6.4)

The universe as a whole is described by Friedmann’s equation, again in natural units:

(Htpl

)2=

4π3g∗45

(T

mpl

)4

+8π3

ρϕ

m4pl

− κ

(T

mpl

)2

. (6.5)

The three terms on the right-hand side respectively correspond to the contribution of theenergy density of radiation, the energy density of the scalar field and the curvature term.In these units we measure the temperature T , mass m and field amplitude ϕ all in GeV, andtime in units of GeV−1. The natural curvature parameter κ may be taken to be a constant.

24

a. In order to gain some understanding about the behavior of a scalar field in an ex-panding universe, consider the (admittedly) not very realistic case of H = constant.Try a quasi-oscillatory complex solution4 of the form

ϕ(t) = ϕ0 e−iωt . (6.6)

For which ω does this solution indeed solve (6.4)?

(Hint: the well-known formula for the solutions of a quadratic equation ax2 + bx +

c = 0 also works if the three coefficients (a , b , c) are complex numbers.)

b. Show that the solutions you found in a always decay exponentially, i.e. |ϕ(t)| ≡√ϕ(t)ϕ∗(t) ∝ exp(−λt). Give all the possible values for λ = Im(ω) for m > 3H/2

and for m < 3H/2, and find the slowest decaying solution for m 3H/2.

The morale of this calculation: The expansion of the universe leads to the decay of theamplitude |ϕ| of the scalar field. This remains true even if H depends on time.

c. In the slow roll approximation one assumes

|ϕ| (H |ϕ| , m2 |ϕ|

), 1

2 ϕ2 V(φ) = 1

2m2 ϕ2 . (6.7)

Use this approximation to show that a variable change, from ϕ(t) to ϕ(a(t)) witha(t) the scale factor, leads to an equation for the scalar field of the form

3H2(d lnϕd ln a

)+ m2 = 0 . (6.8)

(Hint: you will need the definition of the Hubble parameter H(t)).Also show that the general solution of this equation is a quasi-exponential decay:

ϕ(a) = ϕ(a0) exp(−

m2

3

∫ a

a0

da′

a′ H2(a′)

)(with H(t) → H(a(t))) . (6.9)

4You can always take the real part of a complex solution to get a real field since we are dealing with alinear equation for ϕ(t), in effect putting ϕ(t) = Re

(φ0 e−iωt

)after finding the complex solution!

25

d. We now do a consistency check on the solution to see if it satisfies the slow-rollassumption under which it was derived. Show that the second (energy) conditionfor the slow-roll approximation can only be satisfied if we additionally assume that

m 32 H . (6.10)

e. Since the temperature in an expanding universe drops as T ∝ a−1 and the ra-diation density as ρr ∝ T 4 ∝ a−4 it is likely that at some point the energy densityof the scalar field starts to dominate over the effects of radiation and curvature.Friedmann’s equation then simplifies in the slow-roll approximation for ϕ(t) to

H2t2pl =

H2

m2pl

=8π3

V(ϕ)m4

pl

=4π3

m2ϕ2

m4pl

. (6.11)

Show that a simple re-interpretation of the first result of question c then leads to:

d ln adϕ

= −4π ϕm2

pl

. (6.12)

Also show that the condition derived in d is justified as long as the scalar fieldamplitude remains sufficiently large:

|ϕ|

mpl

1√

3π' 0.326 . (6.13)

f. What is the solution of Eqn. (6.12) for a as a function of φ if a = a0 for φ = φ0 atsome time t0?


26

g. In the scenario of Chaotic Inflation (as proposed by Andei Linde) Inflation startsimmediately after the Big Bang at the Planck time tpl ' 10−43 s. At that timequantum fluctuations are large, allowing the scalar field to carry an energy densityroughly equal to the Planck energy density ρpl ' m4

pl:

V(φ) = 12m2ϕ2

0 ' m4pl ⇐⇒

|ϕ0|

mpl'√

2mpl

m(6.14)

Assuming that inflation ends when ϕ ' 0, give the typical inflation factor, the factorby which the scale factor increases over the inflation period:

Zinfl =a(ϕ ' 0)

a(ϕ0)=

a(ϕ ' 0)a0

. (6.15)

Calculate Zinfl for the (not unreasonable) case where the mass of the scalar fieldequals m = 0.01 mpl.

27

7 A universe filled with a free scalar field

We consider a universe whose dominant component is a uniform and free scalar field ϕ(t)with density ρϕ. ”Free” implies that the scalar field potential vanishes: V(ϕ) = 0. In thatcase the Friedmann’s equation and the field dynamics are described by:

(1a

dadt

)2

=8πG ρϕ

3,

∂2ϕ

∂t2 + 3H(t)∂ϕ

∂t= 0 , (7.1)

ρϕ = 12

(∂ϕ

∂t

)2

.

In this problem you are asked to analyze the dynamics of such a universe.We employ natural units with ~ = c = kb = 1 and an energy scale of 1 GeV through-out this question!

a. Show that ϕ ≡ ∂ϕ/∂t satisfies

ϕ(t) = ϕ0

(a(t)a0

)−3

. (7.2)

Here ϕ0 and a0 are reference values at some arbitrary reference time t0.

b. Solve Friedmann’s equation using result a and show that the solution can be writtenas:

a(t)a0

=

(tt0

)α. (7.3)

Give the value of the exponent α and the expression for t0 in terms of ρϕ0 ≡12 ϕ

20.

Question continues on next page!

28

Now consider the situation where the universe also contains radiation with density ρrad,so that the total density is

ρ = ρϕ + ρrad , (7.4)

both functions of the scale factor a(t). The radiation is thermal radiation, with ρrad ∝ T 4.

c. Show that in a universe that initially starts with ρϕ ρrad radiation will start todominate after some time teq, where

a(teq)a0

=

√ρϕ0

ρrad0. (7.5)

(Hint: you will need the correct dependence of ρrad on a!)

We now choose t0 = teq and a0 = a(teq) so that when a = a0 we have ρϕ0 = ρrad0 ≡ ρeq,equality between the scalar field density and radiation density. We also define

a(t) ≡a(t)a0

, Heq ≡

√8πG ρeq

3. (7.6)

d. Show that Friedmann’s equation can be manipulated into

dadt

= Heq

√1 + a2

a2 . (7.7)

(question continues on next page!)

29

e. Show that the solution of Eqn. (7.7) can be represented parametrically through

a√

1 + a2 − sinh−1 a = 2Heq t . (7.8)

Here sinh−1 is the inverse hyperbolic sine.

Check also that this reproduces the correct behavior for a(t) as found in b for thescalar field-dominated phase, when a 1 and ρϕ ρrad, using well-known ap-proximations for sinh(y) and

√1 + y2 valid for positive y 1.

How does a(t) behave in the radiation-dominated phase, ρϕ ρrad, where a 1?

Hint:

The proof of (7.8) involves an integration where the substitution a = sinh x may beuseful, as well as the identities

a2

√1 + a2

=√

1 + a2 −1

√1 + a2

,

cosh2 x − sinh2 x = 1 , (7.9)∫sinh2 x dx = 1

2 (sinh x cosh x − x) .

30

8 From radiation domination to inflation

Consider a flat universe filled with radiation and a constant vacuum density. In naturalunits (with kb = ~ = c = 1 and G = 1/m2

pl) Friedmann’s equation reads:

H2 =

(1a

dadt

)2

=4π3g∗45m2

pl

T 4 +8π

3m2pl

ρvac . (8.1)

Note that in natural units the dimensions are as follows: [T ] = GeV, [t] = GeV−1 and[ρvac] = GeV4. The radiation temperature T falls in the usual fashion due to redshift:

T ∝ a−1 . (8.2)

a. Let T∗ be the temperature at which the radiation density and the vacuum density areexactly equal. The corresponding scale factor is a∗. Now show that by defining

R(t) =a(t)a∗

=T∗

T (t), (8.3)

Friedmann’s equation can be written as

(1R

dRdt

)2

= H2vac

(R−4 + 1

). (8.4)

Give the expression for the constant Hvac.

(Question continues on next page!)

31

b. Show that Eqn. (8.4) can be rewritten in terms of two new variables,

τ = Hvact and Z = R2 , (8.5)

as

dZdτ

= 2√

1 +Z2 . (8.6)

c. Find the solution forZ(τ), and then for R(τ), using the primitive5:

∫dx

√1 + x2

= sinh−1(x) . (8.7)

Assume an initial condition R(0) = 0.

d. Show that the time where the radiation density and the vacuum density are exactlyequal is given by:

teq =sinh−1(1)

2Hvac. (8.8)

e. Show that the solution correctly approaches the de Sitter solution for large R andlarge t:

R(t) ∝ exp (Hvact) . (8.9)

5The notation ”sinh−1” denotes the inverse hyperbolic sine.

32

9 HAND-IN ASSIGNMENT Friedmann models:flow in the Ω −Ω plane

Friedmanns equation for a universe filled with matter with density ρm, and with vacuumenergy ρvac = Λ/8πG (Λ is Einstein’s Cosmological Constant) reads:

H2 =

(1a

dadt

)2

=8πG

3(ρm + ρvac) −

ka2 . (9.10)

If matter is cold, its mass density drops as

ρ ∝ a−3 (9.11)

as a result of the increase in each physical volume (V ∝ a3) due to the universal expan-sion. The Omega-parameters for cold matter and vacuum energy are defined as

Ωm =8πG ρm

3H2 , ΩΛ =8πG ρvac

3H2 =Λ

3H2 . (9.12)

a. Show that differentiation of (a2×(Friedmann’s equation)) leads to:

d2adt2 = −

4πGρm

3a +

8πG ρvac

3a . (9.13)

b. Now show that the following three relations hold (you will need the first for theremaining two relations!):

dHdt

=1a

d2adt2 − H2 ,

(9.14)dΩm

dt= H Ωm (Ωm − 2ΩΛ − 1) ,

dΩΛ

dt= H ΩΛ (Ωm − 2ΩΛ + 2) .

33

c. Now consider the Omega-Omega plane of Figure 1 on the next page. The x-axiscorresponds to Ωm and the y-axis to ΩΛ. Defining

Ωm =dΩm

dt, ΩΛ =

dΩΛ

dt(9.15)

a particular universe (as it evolves) follows a trajectory in this plane with velocity

V = (vx , vy) =(Ωm , ΩΛ

). (9.16)

These trajectories define how the universe behaves.

Now answer/do the following:

1. What is the significance of the two long-dashed lines, ΩΛ = 12 (Ωm + 2) and

ΩΛ = 12 (Ωm − 1), in this figure?

2. Sketch the direction of the velocity V (using little arrows, as in a phase dia-gram in Classical Mechanics that tracks position and momentum) in differentparts of the diagram;

3. Show that the three points (Ωm ,ΩΛ) = (0 , 1), (1 , 0) and (0 , 0) are fixedpoints; In the Figure they are indicated with black dots.

4. Show that only the fixed point (0 , 1) is a stable fixed point in the sense thatUniverses are attracted (as opposed to repelled) by it. Show also that the othertwo fixed point are unstable.

34

Figure 1: The Omega-Omega plane can serve as a phase diagram for all possible cold Universemodels. The short dashed line Ωm + ΩΛ = 1 corresponds to a spatially flat universe. In this casewe allow negative values for ΩΛ.

35


10 The re-heating temperature

At the end of inflation with a total expansion factor Zinfl, the particle density has beenreduced by a factorZ3

infl, and the temperature is reduced by a factorZinfl. Given thatZinfl

is a large number, ranging fromZinfl ∼ 1032 in classical inflation models toZinfl ∼ 101010

in some chaotic inflation models, this means that the Universe is practically empty andmuch too cold. This problem is solved by the Little Bang, where the energy in the scalarfield driving inflation is converted into a new population of fresh particles. Here we willuse general arguments to prove that the temperature of the gas after the Little Bang is notto different from the temperature at the beginning of the inflationary period.

Natural units with kb = ~ = c = 1 and G = 1/m2pl are used throughout!

a. Inflation happens so early in the Universe that all known particles are relativistic,and behave as radiation. Adding a homogeneous real scalar field (the simplestpossible model) gives the energy density of the Universe as

ρ = ρrad + ρϕ =π2

30g∗T 4 + 1

2

(∂ϕ

∂t

)2

+ V(ϕ , T ) (10.1)

Here V(ϕ , T ) is the effective potential of the scalar field, and g∗ counts the numberof relativistic degrees of freedom in bosons and fermions. In the standard model athigh temperatures one has g∗ = 106, 75.

Before inflation the scalar field expectation value is at ϕ ' 0, the minimum of thehigh-temperature effective potential. This corresponds to a potential energy densityV(ϕ = 0) ≡ V0. At what temperature will the scalar field start to dominate if youassume the slow-roll approximation (small ∂ϕ/∂t) is valid?

36

b. If one appeals to some form of symmetry breaking that happens at an energy scaleEsb, the fact that in natural units the dimension of the energy density is [energy]4,one often assumes that

V0 ∼ E4sb . (10.2)

This essentially assumes that the value of V0 only depends on the symmetry break-ing scale, and not on some additional (arbitrary) small or large parameter. In otherwords: V0 = α E4

sb with α = O(1).

Show that answer a implies that inflation starts once the temperature has fallen to

T ∼ 0.41 Esb ≡ Tcr . (10.3)

You may assume the standard model is valid.

c. The equation of motion for the scalar field in a universe with Hubble parameter H(t)is

∂2ϕ

∂t2 + 3H∂ϕ

∂t= −

∂V∂φ

. (10.4)

Show that this equation of motion leads to the following equation for the energydensity of the scalar field:

∂ ρϕ

∂t= −3H

(∂ϕ

∂t

)2

. (10.5)

d. If the slow-roll approximation applies the right-hand side of the equation derivedin c is a small quantity, and the energy density of the scalar field is approximatelyconserved. What is its value if inflation starts at ϕ = 0 with (∂ϕ/∂t)0 ' 0?

e. Now show that, if all the above is true, instantaneous conversion of all vacuumenergy ρφ into ‘radiation’ would fill the universe with a radiation gas with exactlythe same temperature (and exactly the same denity!) Tcr as the temperature wherescalar field energy starts to dominate.(In practice, you do loose some energy as result c shows, and the reheating tem-perature is somewhat lower than Tcr.)

37

11 To slow roll or not to slow roll...

If we adopt the dot-notation for time derivatives,

ϕ =∂ϕ

∂t, ϕ =

∂2ϕ

∂t2 , (11.1)

and the notation

V ′ =∂V∂ϕ

, V ′′ =∂2V∂ϕ2 (11.2)

for the derivatives of the effective potential, the equations for the inflationary universe are:

H2 =8πG

3

(12 ϕ

2 + V(ϕ))

(Friedmanns equation, neglecting curvature),

(11.3)ϕ + 3H ϕ + V ′ = 0 (Eqn. of motion of uniform scalar field ϕ(t).)

a. Use the two fundamental equations (11.3) to show (by time-differentiation of Fried-mann’s equation) that

H = −4πG ϕ2 . (11.4)

b. For a slow roll approximation to apply one should demand that the Hubble constantchanges little over one Hubble time tH = 1/H. Show that this condition (|∆H| H)is equivalent with:

|H| H2 , (11.5)

and that this is only possible if

ϕ2 |V(ϕ)| . (11.6)

38

This is the first slow roll condition.

c. One usually also assumes the second slow roll condition holds: the change of ϕ ina Hubble time tH = 1/H is small compared with ϕ, that is |∆ϕ| |ϕ|. Show thatthis condition is satisfied if

|ϕ| H |ϕ| . (11.7)

d. Now assume that both slow roll conditions are satisfied. Give the approximatesolution for H and for ϕ in this case, and then show using result a that

|H|H2 =

116πG

(V ′(ϕ)V(ϕ)

)2

. (11.8)

What condition on the ratio |V ′(ϕ)/V(ϕ)|, a property of the potential, should oneimpose in order for the slow-roll approximation to be internally consistent?

e. Show also that if both slow roll conditions are satisfied the following relations hold:

H ϕ = −V ′(ϕ)/3 , ϕ =V ′

24πG

V ′′

V− 1

2

(V ′

V

)2 . (11.9)

(Hint: differentiate the approximate solution for ϕ in order to obtain the secondrelation!)

f. Show that the second slow roll condition (11.7) can only be satisfied consistently ifthe shape of the potential is such that it satisfies the condition

∣∣∣∣∣V ′′(ϕ)V(ϕ)

∣∣∣∣∣ 24πG (11.10)

is satisfied. You will have to use the answer obtained in d in comparing the magni-tude of the contribution from the two terms in the expression for ϕ!

39

g. From the definition of the Hubble parameter H(t) one sees immediately that theexpansion factorZ between cosmic time t1 and cosmic time t2 equals

Z =a(t2)a(t1)

= exp(∫ t2

t1dt H(t)

)≡ eN . (11.11)

Show that in the slow-roll approximation in an inflationary universe we have

N = −

∫ ϕ2

ϕ1

dϕ(8πG V(ϕ)

V ′(ϕ)

), (11.12)

with ϕ1 ≡ ϕ(t1) and ϕ2 ≡ ϕ(t2).

h. Show that result d, which puts a limit on |V ′(ϕ)/V(ϕ)| in order for the slow-rollapproximation to apply, implies that one gets a large inflation factor (i.e. N 1)provided

φ1 − φ2 ≥1√

4πG=

mpl√

4π(in natural units with G = 1/m2

pl) . (11.13)

40

12 Hand-in assignment: GUT inflation

One of the earliest inflation models, proposed around 1981 by US physicist Alan Guth6,employed a Higgs-like scalar field associated with the unification of the stong nuclearforce with the electro-weak force. The energy density of this field acts in a brief period asa cosmological constant.

Even though we know now that this particular model does not work, it is a useful exerciseto retrace some of Guth’s calculations to get a feel for the typical numbers. That is whatwe will do in this assignment. Natural units with ~ = kb = c = 1 and an energy scale of 1GeV will be employed throughout. The Friedmann equation can then be written as

H2 ≡

(1a

dadt

)2

=4π3g∗

45

T 4

m2pl

+8π3

ρϕ

m2pl

− κ T 2 . (12.1)

As in the previous assignments we have an energy density of the scalar field ϕ equal to

ρϕ = 12 ϕ

2 + V(ϕ) . (12.2)

For practical calculations it is useful to know the dimensions (in natural units) of thevarious quantities:

[t] = GeV−1 , [H] = [T ] = [m] = [ϕ] = GeV , [ρϕ] = GeV4 . (12.3)

The quantities a(t), g∗ and κ are dimensionless numbers. This may be handy in checkingyour algebra. The Planck mass equals

mpl = 1.22 × 1019 GeV . (12.4)

Finally: GUT unification is expected to occur on a mass/temperature scale (energy scale)

mGUT = TGUT = 1015 GeV . (12.5)

6Guth, A.H., Phys. Rev. D 23, 347, 1981.

41

a. It is usually assumed that after the Big Bang the universe is dominated by radiation.One also often assumes that the potential energy (really an energy density!) beforesymmetry breaking takes the ’natural’ value

V(ϕ ' 0) ' m4GUT . (12.6)

In the symmetric state, the simplest unified theories have a number of extra particles(gauge bosons), leading to

g∗ = Nbosons +78

Nfermions =675

4' 169 . (12.7)

Now calculate the typical temperature T∗ where the energy density of the scalarfield starts to dominate the expansion of the universe, assuming that the curvatureterm is still smaller than the radiation term7.

b. As explained in the Lectures, inflation models employ scalar field potentials thatare modeled after the potentials used for the Higgs mechanism. A simple modelpotential has the form

V(ϕ) = V0 −12m2 ϕ2 + λ ϕ4 . (12.8)

Here V0 is a constant with [V0] = GeV4, [m] = GeV and λ is dimensionless. How-ever, what matters classically is the effective potential that governs the behavior ofthe expectation value of ϕ. It takes the form (where ϕ now is the expectation value!)

Veff(ϕ , T ) ' V(ϕ) +λ

2T 2 ϕ2 = V0 −

12m2 ϕ2 + λ ϕ4 +

λ

2T 2 ϕ2 . (12.9)

The extra term ∝ T 2 is due to the coupling of the scalar field to the heath bath ofthermal particles. Now show the following:

1. For T > Tcr ≡ m/√λ the effective potential has a single minimum at φ = 0;

2. For T < Tcr = m/√λ the effective potential two minima at φ = ±φm with

φm > 0. Calculate φm in the limit T Tcr. φ = 0 now corresponds to amaximum!

7Otherwise, the Universe would already have started to collapse!

42

c. You have just found the critical temperature of this model, below which the symme-try will be broken as the expectation value ϕ always evolves towards the minimum.In this case that implies that ϕ ' 0 at high temperatures, and ϕ = ±ϕm at low tem-peratures. Make a sketch of the shape of the potential Veff(ϕ , T ) for T > Tcr andfor T < Tcr.

d. What value should V0 take if you require that, long after symmetry breaking whenT Tcr and |φ| ' φm = constant, there is no large cosmological constant due to thescalar field? You may simply assume that effectively Λ = 0, and use the fact thatthe thermal term ∝ T 2 in expression (12.9) for Veff may be neglected if T Tcr.

Show also that this implies that the effective potential has the form

Veff(ϕ , T ) = λ

[ϕ2 −

m2

4λ

]2

+λ

2T 2 ϕ2 . (12.10)

This simple requirement reduces the number of independent free parameters in Veff(φ,T )from three (V0, m and λ) to just two (m and λ)!

e. We now use an argument of ’naturalness’, which states that one expects that theparameters of the correct theory should be such that they do not introduce an extraenergy scale that is too different from the fundamental unification scale, mGUT ∼

1015 GeV. In this particular case this means that one should demand:

Veff(ϕ = 0) ' m4GUT , Tcr ' TGUT = mGUT (12.11)

Show that this can only be satisfied if the parameter λ is of order unity (and not, forinstance, 102 or 10−2), and that this also implies that m ' mGUT.

43

ANSWERS TO SELECTED ASSIGNMENTS (A.Achterberg) (NWI-NM026C)

13 Week 12

1a. The required answer follows directly from:

d2

dt2 =ddt

(a

dda

)=

dadt

dda

+ ad2

dt da(13.12)

= ad

da+ a2 d2

da2 .

1b. The first step is use the results from a to rewrite the terms involving time-derivatives,specifically:

d2

dt2 + 2( aa

) ddt

= a2 d2

da2 +

(a + 2

a2

a

)d

da. (13.13)

Then you must use the two versions of Friedmann’s equation to eliminate a and a. Inparticular, the term in front of d/da is

a + 2a2

a= −

4πGρ3

a +16πGρ

3a = 4πGρ a . (13.14)

This yields:

d2

dt2 + 2( aa

) ddt

= 4πGρ(

23 a2 d2

da2 + ad

da

). (13.15)

Substituting this into (1.2) and dividing the resulting expression by 4πGρ gives relation(1.7). Remember: generally ρcdm < ρ as there is also radiation and baryons in the uni-verse that contribute to the effective mass density.

44

1c. In the matter-dominate phase we can put ρ ' ρcdm. This was the case up to the pointwhere the universe reaches 75% of its current age, and the cosmological constant becameimportant. In that case relation (1.7) becomes:

(23 a2 d2

da2 + ad

da− 1

)∆cdm(a) = 0 . (13.16)

A solution ∆cdm(a) ∝ as solves this differential equation provided s satisfies the indicativeequation:

23 s(s − 1) + s − 1 = (2

3 s + 1)(s − 1) = 0 , s = 1 or s = −3/2 . (13.17)

This is a growing solution (positive s = 1, as a increases with time) and a decayingsolution (negative s = −3/2). The growing solution, needed in d, is

∆cdm(a) = ∆0

(aa0

). (13.18)

1d. If one substitutes the growing solution for ∆cdm(a) into (1.9) one gets:

[d2

dt2 + 2( aa

) ddt

]∆ba(t) = 4πGρcdm ∆0

(aa0

). (13.19)

Converting from time-derivatives to a-derivatives with ρ ' ρcdm yields, analogously towhat we found in b:

(23 a2 d2

da2 + adda

)∆ba(a) = ∆0

(aa0

). (13.20)

This equation has a particular (’driven’) linear8 solution with ∆ba(a0) = 0:

∆ba(a) = ∆0

(aa0− 1

), (13.21)

as direct substitution proves.8So that d2∆ba/da2 = 0.

45

1e. For a a0 we have:

∆ba(a) ' ∆cdm(a) = ∆0

(aa0

). (13.22)

Since ∆ba = δρba/ρba and ∆cdm = δρcdm/ρcdm this (approximate) equality implies:

δρba

δρcdm=

ρba

ρcdm

∆ba

∆cdm'

ρba

ρcdm=

Ωba0

Ωcdm0=

0.040.20

= 0.2 . (13.23)

We can use the current values of the Omega-parameters for baryons and CDM since bothdensities scale as ρ ∝ a−3, so the density ratio remains constant as the universe expands.

The exact solution is also easily obtained from result c/d:

δρba

δρcdm=

Ωba0

Ωcdm0

(1 −

a0

a

). (13.24)

2a. Since n is a unit vector we have

nµ nµ = 1 − n · n = 1 − 1 = 0 . (13.25)

Also:

Gµ = (G0 , G) =1c

∫ ∞

0dν

∮dΩ

χ(n , ν) I(n , ν) − η(n , ν)

(1 , n)

(13.26)

= =1c

∫ ∞

0dν

∮dΩ

χ(n , ν) I(n , ν) − η(n , ν)

nµ ,

which is the required answer.

2b. A trivial exercise in substitution, using the second equality in the definition of Rµν andthe definition of the four-vector nµ.

46

2c. The point here is that the emission is isotropic, that is it it does not depend on thedirection of n. This means the following for the solid-angle integrations of the η(n , ν)that appear in the definition for G0 and G:

∮dΩ η(n , ν) = 4π

(neσT

∮dΩ′

4πI(n′ , ν)

)= neσT

∮dΩ

4πI(n , ν) ,

(13.27)∮dΩ η(n , ν) n =

(neσT

∮dΩ′

4πI(n′ , ν)

) ∮dΩ n = 0 .

That means that the absorption term in G0 is exactly canceled by the emission term (nonet energy transfer between photons and a medium at rest), and that the emssion term inG is identically zero.

2d. The crucial point is that for Uµ = (1 , 0 , 0 , 0) one has

hµν = diag(0 , 1 , 1 , 1) , (13.28)

which guarantees that G0 = 0. Also,

Uσ Rστ = R0τ = (ε , F/c) =1c

∫ ∞

0dν

∮dΩ I(n , ν) nτ . (13.29)

and

hµτ nτ = (0 , n) . (13.30)

The relation

neσT hµτ Uσ Rστ =neσT

c

∫ ∞

0dν

∮dΩ I(n , ν) (0 , n) = Gµ (13.31)

is then immediately obvious.

47

2e. This needs a little work: First of all, with Rµν in the form (2.28):

Uσ Rστ ≡ Wτ =((1 + 1

2β2) ε , −1

3β ε). (13.32)

Secondly

hµτWτ = Wµ − Uµ (UτWτ)

= Wµ −((1 + 1

2β2) W0 − β ·W

)Uµ (13.33)

= Wµ −(1 + 4

3 β2)ε Uµ .

Here I have neglected a term of order β4. Substituting for Wµ and Uµ and neglecting allterms of order β3 and higher you get:

hµτWτ = −4

3 ε(β2 , β

), (13.34)

and the answer follows immediately from Gµ = neσT hµτ Uσ Rστ = neσT hµτ Wτ.

3.1. With the scaling a(t0) = 1, the known density laws for (cold) matter and radiationand the redshift definition 1 + z = a−1 you have:

ρm = ρm0 a−3 = ρm0 (1 + z)3 , ρr = ρr0 a−4 = ρr0 (1 + z)4 . (13.35)

3.2. Just use the definition for the Hubble parameter and a(t) = 1/(1 + z):

H ≡(1a

dadt

)= (1 + z)

d(1 + z)−1

dt= −

11 + z

dzdt. (13.36)

Here t stands for the emission time. If it increases (photon is emitted later) it arrives witha smaller redshift; If it is emitted earlier (smaller t) the redshift is larger. This impliesdz/dt < 0.

48

The second part, deriving the Friedmann equation in terms of z and dz/dt, uses the fol-lowing trick: for matter the term on the right-hand side of the Friedmann equation is

8πGρm

3=

8πGρm0 (1 + z)3

3= H2

08πGρm0

3H20

(1 + z)3 = H20 Ωm0 (1 + z)3 . (13.37)

The radiation term can be treated similarly, and for the cosmological constant term youwrite Λ/3 = H2

0(Λ/3H20) = H2

0 ΩΛ0. The curvature term requires some thought. Fried-mann’s equation at t = t0 can be written as:

H20 = H2

0 (Ωm0 + Ωr0 + ΩΛ0) − k (13.38)

since a(t0) = 1 by definition. Solving for k we get:

k = H20 (Ωm0 + Ωr0 + ΩΛ0 − 1) ≡ H2

0 Ωk0 . (13.39)

Friedmann’s equation,

H2 =8πG

3(ρm + ρr) +

Λ

3−

ka2 , (13.40)

can now be rewritten as:

(1

1 + zdzdt

)2

= H20

[Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2

](13.41)

Multiplying both sides with (1 + z)2 gives the relation asked for in this part of the assign-ment.

49

3.3. This is essentially solved by working out the powers of 1+z for Ωr0 = 0. In particular:

Ωm0(1 + z)3 + ΩΛ −Ωk0 (1 + z)2 =

Ωm0

(1 + 3z + 3z2 + z3

)+ ΩΛ0 − (Ωm0 + ΩΛ0 − 1)

(1 + 2z + z2

)=

(13.42)Ωm0

(z + 2z2 + z3

)−ΩΛ0 (2z + z2) + (1 + z)2 =

(Ωm0z + 1) (1 + z)2 −ΩΛ0z (2 + z) .

In the last step I have used z + 2z2 + z3 = z (1 + z)2.

3.4. To convert integrals over cosmic time t into integrals over z you use:

dz =

(dzdt

)dt ⇐⇒ dt =

dtz

with z ≡dzdt. (13.43)

The form of the Friedmann equation derived in 3.3 gives us z (note the sign!):

z = −H0(1 + z)√

Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2 (13.44)

so that

dt = −1

H0

dz

(1 + z)√

Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2. (13.45)

The perhaps unexpected minus sign will sort itself out when we use this in actual integrals,and is a simple consequence of z < 0 if t is the time of emission (rather than reception) asis the case here. This relation will be used again and again in what follows.

50

We can immediately apply this to the source distance D0. What you have to realize isthat a photon emitted by the source at time t has an observed redshift zo at t0, while a(hypothetical) photon emitted at t0 and observed at t0 has no redshift at all: z = 0. Thismeans in general that

∫ t0

tdt · · · =

∫ 0

zo

dzz· · · = −

∫ zo

0

dzz· · · , (13.46)

where reversing the order of integration in the last equality introduces an extra minus signthat actually cancels pesky minus sign in z in Eqn. (13.44) (check this out!). For thedistance integral that gives us D0 this procedure, together with a(t) = (1+ z)−1 and (13.45)immediately yields:

D0 =

∫ t0

t

c dta(t)

= c∫ zo

0

dzz

(1 + z)

(13.47)

=c

H0

∫ zo

0

dz√Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2

.

For zo 1 one can use

Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2 ' 1 + Ω z (13.48)

with

Ω ≡ 2 + Ωm0 + 2Ωr0 − 2ΩΛ0 (13.49)

This means that the integrand in (13.47) can be approximated by

(1 + Ω z

)−1/2≈ 1 − 1

2Ω z + O(z2) , (13.50)

and that the distance integral for small z0 is

D0 ≈c

H0

∫ zo

0dz

(1 − 1

2Ω z)

=czo

H0+ O(z2

o) . (13.51)

51

3.5. One can calculate photon travel times by employing the almost trivial relation

∆t = t0 − t =

∫ t0

tdt , (13.52)

and replacing the time integral by an integral over redshift using (13.45) and (13.46). Thisapproach immediately gives the answer.

3.6. Photons that are produced at the Big Bang, when t = 0 and (formally) a = 0, arrivewith a redshift zo = ∞. This sets the upper limit in the integral for the distance to thesource, so the horizon distance is defined by

Dhor ≡c

H0

∫ ∞

0

dz√Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2

. (13.53)

Note that this integral always converges! At different observation times one gets a differ-ent answer since the Omega parameters change when one chooses t0 differently.

3.7. Equation (3.14) applied to the two limiting cases Ωm0 ≡ Ω0 = 1, ΩΛ0 ≡ ΩΛ = 0 andΩm0 ≡ Ω0 = 0, ΩΛ0 ≡ ΩΛ = 1 yield the integrals

D0 =

c

H0

∫ zo

0

dz(1 + z)3/2 =

2cH0

(1 −

1√

1 + zo

)case 1: (Ω0 , ΩΛ) = (1 , 0)

cH0

∫ zo

0dz =

czo

H0case 2: (Ω0 , ΩΛ) = (0 , 1)

.

(13.54)

In case 1 the horizon distance follows from putting zo = ∞ and one finds:

Dhor =2cH0

. (13.55)

In case 2 this procedure formally yields Dhor = ∞. This may seem strange, but is putinto perspective by looking at what this distance is at the moment of emission t. In a flatuniverse we have for a(t0) = 1 that D(t) = D0/(1 + zo), so the distance to the source whenthe photon was emitted is

52

D(t) =D0

1 + zo=

cH0 (1 + zo)

∫ z0

0

dz√Ω0 (1 + z)3 + ΩΛ

. (13.56)

In case 2, where H0 = HΛ =√

Λ/3 (from Friedmann!), this gives:

D =czo

H0 (1 + zo)=

czo

HΛ (1 + zo)(13.57)

In the limit zo → ∞ this yields a finite answer:

D(zo → ∞) =c

HΛ

. (13.58)

As we have seen in one of the earlier assignments this is exactly the distance that separatesphotons that effectively move towards the observer and ultimately reach him (for D <c/HΛ) and those that will never reach the observer (for D > c/HΛ) in a De-Sitter universe.Those photons starting exactly at D = c/HΛ take an infinite amount of time to reach theobserver.

3.8 The aim is to evaluate the integral

t0 =1

H0

∫ ∞

0

dz

(1 + z)√

Ω0 (1 + z)3 + ΩΛ

(13.59)

that gives us the age of the universe, given the (in principle observable) current values ofthe two Omega parameters Ω0 ≡ Ωm0 and ΩΛ ≡ ΩΛ0 (Longairs notation!) and H0. Thesuggested substitution is aimed at simplifying the square root9:

1 + z =

(ΩΛ

Ω0

)1/3

tan2/3 θ ⇐⇒√

Ω0 (1 + z)3 + ΩΛ =

√ΩΛ

cos θ. (13.60)

We also have

dz =23

(ΩΛ

Ω0

)1/3 dθcos2 θ tan1/3 θ

. (13.61)

9This is a standard procedure for integrals such as this.

53

The integral in (13.59) then becomes

t0 =2

3√

ΩΛH0

∫ π/2

θ0

dθsin θ

. (13.62)

Here the angle θ0 corresponds to z = 0, and follows from:

(ΩΛ

Ω0

)1/3

tan2/3 θ0 = 1 ⇐⇒ tan θ0 =

√Ω0

ΩΛ

' 0.608 (from observed values) . (13.63)

At θ = π/2 tan θ goes to infinity so this corresponds to z = ∞. Using the primitive for theintegral over θ given in the Assignment one finds:

t0 =2

3√

ΩΛH0ln

(1 + cos θ0

sin θ0

). (13.64)

What remains is expressing this result in terms of the Omega parameters using the def-inition of θ0. To that end you can use the relations that follow straightforwardly fromtan = sin / cos and sin2 + cos2 = 1. In symbolic notation:

sin =tan

√1 + tan2

, cos =1

√1 + tan2

. (13.65)

We then get

1 + cossin

=

√1 + tan2 + 1

tan=

√1 + tan2 + 1√

1 + tan2 − 1

1/2

. (13.66)

In the last equality I have used tan2 = (√

1 + tan2 + 1)(√

1 + tan2 − 1) to get a moresymmetric-looking equation. Putting in tan θ0 =

√Ω0/ΩΛ and using Ω0 + ΩΛ = 1 then

yields after some straightforward algebra:

1 + cos θ0

sin θ0=

(1 +√

ΩΛ

1 −√

ΩΛ

)1/2

. (13.67)

54

The expression for t0 that follows from substituting this into (13.64) agrees with our earlierresult (??) for the age of a Friedmann-Lemaitre universe:

t0 =2

3H0√

ΩΛ0

ln

(1 +√

ΩΛ0

1 −√

ΩΛ0

)1/2 . (13.68)

4a. By drawing a picture, it is easily seen that photons that hit the observer from thetwo ends of the object should leave with the correct direction so that they ultimately bothhit the observer. That direction is set by the distance at photon emission, see the figurebelow. That the source moves further away from the observer as the photons propagate isimmaterial.

Figure 2: The situation that applies to the determination of the angular distance Da. Two lightrays from either end of the object come in with an angle 1

2∆θ with respect to the normal to theobject. The figure is for a flat universe where distances scale as D(t) = r a(t), with r the fixedcomoving coordinate. However, the result derived in this assignment is generally true.

55

Elementary trigonometry tells you that

tan(

12∆θ

)=

d2Da

⇐⇒ ∆θ = 2 arctan(

d2Da

)= 2 arctan

(d

2a(te) r

)(13.69)

4b. For small angles we have tan(12∆θ) ' 1

2∆θ and we find:

Da = a(tem) r =

(a(te)a(t0)

)a(t0)r =

D0

1 + z'

d∆θ

. (13.70)

4c. One simply uses relation (13.71) for D0 and Da = D0/(1 + z0):

Da =c

H0 (1 + z0)

∫ z0

0

dz√Ωm0(1 + z)3 + Ωr0 (1 + z)4 + ΩΛ −Ωk0 (1 + z)2

. (13.71)

4d. The only differences with earlier calculations of this type are

1. The integration boundaries that now run from redshift z to z = ∞ (moment of theBig Bang);

2. The fact that the (physical) distance is evaluated not at t = t0 (current epoch) whena(t0) = 1 by definition, but at the time corresponding to redshift z. The scale factorat that time is a(t) = 1/(1 + z), which accounts for the extra factor 1/(1 + z) in thefinal answer (4.6).

4e. It is easiest if one first calculates D0, the source distance at photon reception:

D0 =c

H0

∫ z0

0

dz(1 + z)3/2 =

2cH0

(1 −

1√

1 + z0

). (13.72)

The angular distance then follows immediately:

56

Da =D0

1 + z0=

2cH0 (1 + z0)3/2

( √1 + z0 − 1

). (13.73)

The angular size (in the small angle approximation) then is

∆θ =d

Da=

H0d2c

(1 + z0)3/2

√1 + z0 − 1

. (13.74)

For small z0 this scales as

∆θ ' H0d/cz0 ∝ z−10 . (13.75)

Since D0 ' cz0/H0 for z0 1 we have the intuitively obvious case where the angular sizedeclines as 1/distance. For large z0 on the other hand we have

∆θ ∝ z0 (13.76)

and the angular size increases in a counter-intuitive manner with increasing redshift.

This automatically implies that ∆θ has a minimum. The redshift for which that that mini-mum angular diameter occurs follows from

ddz0

((1 + z0)3/2

√1 + z0 − 1

)=

1 + z0 −32

√1 + z0(√

1 + z0 − 1)2 = 0 . (13.77)

That minimum occurs at a redshift z0 = z∗ such that

√1 + z∗ = 3

2 ⇐⇒ z∗ = 54 . (13.78)

The corresponding minimum angular size, angular distance and current source distanceare:

∆θmin =278

H0dc

, Da =8

27c

H0, D0 = (1 + z∗) Da =

23

cH0

. (13.79)

57

The (somewhat far-fetched) intuitive explanation for the increasing angular size at highredshift for a universe such as this is gravitational lensing by the gravitational attractionof the matter in the universe as a whole. All models with Ωm0 , 0 produce such anincrease, see the discussion and figures in Section 7.4.4. of Longairs book. The largerΩm0, the stronger the lensing and the smaller the redshift where the angular diameter hasa minimum.

4f. In a radiation-dominated flat universe with z > zCMWB ' 140 we can approximateresult (4.6) for the horizon at redshift zCMWB by

Dhor(zCMWB) =c

H0(1 + zCMWB)

∫ ∞

zCMWB

dz′√Ωr0 (1 + z′)4

(13.80)

=c

H0√

Ωr0 (1 + zCMWB)

∫ ∞

zCMWB

dz′

(1 + z′)2 =c

H0√

Ωr0

1(1 + zCMWB)2 .

If we put in numbers, specifically Ωr0 ' 4−4 and zCMWB ' 140, one finds Dhor(zCMWB) '4−4 (c/H0).

4g. With the distance to the surface of last scattering being DLS ' 3 (c/H0), and the cor-responding angular distance being DLS/(1 + zCMWB), the angular size of the then-horizonat the moment of CMWB creation is

∆θCMWB ' (1 + zCMWB)Dhor(zCMWB)

3 (c/H0)=

13√

Ωr0 (1 + zCMWB)' 3 × 4−2 radian . (13.81)

Since one radian corresponds to ∼ 57.3 degrees, the angular size of the horizon is about1.8 degrees on the sky. This means that regions separated by more than this angular dis-tance were outside each other’s horizon when the CMWB was created, and could notcommunicate in any way about the precise conditions, such as density or radiation tem-perature. There is therefore no obvious reason why these regions should look very similar(as observed, with deviations smaller than one part in 45 in radiation temperature, forinstance). This is an example of the horizon problem in standard Friedmann cosmology.

58

4h/i. This is just a straightforward application of the recipe (4.1) with t1 = te and t2 = t0

(current epoch) as well as

z(te) = z0 , z(t0) = 0 , a(te) =1

1 + z0, a(t) =

11 + z

. (13.82)

That immediately leads to result (4.17), and a comparison with expression (4.4) for theangular distance Da show that they are indeed identical.

In e we proved that ∆θ = d/Da has a minimum at z = z∗ for a flat, matter-dominateduniverse. That means that Da = DPLC must have a maximum at that redshift, which equals(see above)

DPLC(z = z∗ = 54 ) =

827

cH0

. (13.83)

59

14 Week 14

5a. Before annihilation we had a total statistical weight due to photons and electron-positron pairs

gbefore = 2 + 2 ×78

=224

(14.84)

After the pairs have annihilated only the photons are left with gafter = gγ = 2. Entropyconservation then requires:

gbefore T 3before = gafter T 3

after , (14.85)

neglecting the change in the scale factor a during annihilation. So the photon temperaturejumps according to:

Tafter

Tbefore=

(gbefore

gafter

)3

=

(114

)1/3

. (14.86)

The neutrinos no longer interact with the rest of the universe through collisions, and aretherefore unaffected by pair annihilation.

5b. The table gives the number density per degree of freedom, so generally:

nfoton(T ) =2ζ(3)π2 T 3

γ , nν(T ) = 2 × 3 ×(

34

ζ(3)π2 T 3

ν

)(14.87)

This uses that there are 3 neutrinos, 3 anti-neutrinos each with a weight gν = 1, or equiv-alently gν = 2 per species. So we get:

nνnγ

=(9/2) T 3

ν

2 T 3γ

=94×

411

=9

11. (14.88)

Since both the photon density (after pair annihilation!) and the neutrino density fall of asa−3 this ratio does not change as the Universe expands.

60

5c. Using the numbers:

nν0 = 344 cm−3 , Tν0 = 1.93 K . (14.89)

5d. I NEGLECTED TO GIVE THE SOLAR SYSTEM SPEED: ∼ 220 km/s. We will useβν ' 10−3. A rate of one event per year corresponds to

R ' 3 × 10−8 s−1 , (14.90)

and using the cross section, Eν ∼ mν = 0.1 eV c ' 3 × 1010 cm/s, βν ' 10−3 and nν ' 300cm−3 we have:

Nt =R

σνβνc nν' 2 × 1040 . (14.91)

The specific weight of water is ∼ 1 g/cm3 = kg per liter. One liter if water contains1/1.67× 10−27 ∼ 6× 1026 nucleons. That means we need (2× 1040)/(6× 1026) ∼ 3× 1013

liters of water. This corresponds of a cube with linear size 3 × 104 dm = 300 kilometer.

61

15 Week 14

6a. Substituting the trial solution into Eqn.(6.4) yields the following relation, assumingH =constant:

(−ω2 − 3i Hω + m2

)ϕ0e−iωt = 0 . (15.1)

This relation can only be satisfied for all t if the term inside the brackets vanishes:

ω2 + 3i Hω − m2 = 0 . (15.2)

Solving for ω:

ω = ω± ≡ −32 iH ±

√m2 − 9

4 H2 (15.3)

6b. All solutions have λ ≡ −Im(ω) > 0 (Note the minus sign!). Defining Re(ω) ≡ ωr andwriting

ω = ωr − iλ , (15.4)

the exponential factor becomes

e−iωt = e−λt e−iωrt . (15.5)

This is a combination of sinusoidal behavior in time, and exponential decay in time. Asa result, the field amplitude decays as

√ϕϕ∗ ∝ e−λt. The following possibilities present

themselves for λ:

λ+ = λ− = 32 H if m > 3

2 H;(15.6)

λ± = 32 H ∓

√94 H2 − m2 if m < 3

2 H;

62

In the limit m 3H/2 the slowest decaying mode (longest surviving solution) occurs forλ+:

λ+ = 32 H −

√94 H2 − m2

= 32 H

1 −√

1 −4m2

9H2

(15.7)

' 32 H

(1 −

[1 −

2m2

9H2

])=

m2

3H.

For comparison purposes: in the same limit the other solution has λ− ' 3H so the ratio ofthe two decay rates is λ+/λ− = m2/9H2 1.

6c. In the slow-roll approximation the field dynamics can be approximated by:

3H ϕ + m2ϕ = 0 . (15.8)

Using the scale factor a(t) rather than time to track the expansion of the universe, in effectmaking the variable change ϕ(t) → ϕ(a(t)), one can use:

ϕ =∂ϕ

∂t=

dadt

∂ϕ

∂a. (15.9)

The definition of the Hubble parameter yields

dadt

= H(t) a(t) (15.10)

so that

ϕ = Ha∂ϕ

∂a= H

dϕd ln a

. (15.11)

In the last step I have replaced the partial derivative by an ordinary derivative since ϕ isassumed to be uniform: it only depends on time or (in this formulation) on a(t).

63

Substituting this into (15.8) we get:

3H2 dϕd ln a

+ m2 ϕ = ϕ

(3H2 d lnϕ

d ln a+ m2

)= 0 . (15.12)

The required relation follows immediately since ϕ , 0.

The formal solution follows from rewriting the equation for ϕ as

d lnϕ = −m2

3H2(a)daa, (15.13)

using d ln a = da/a. Note that I consider the Hubble parameter now as a function of arather than time, the same procedure as used for ϕ. Formally integrating,

∫ ϕ

ϕ0

d lnϕ′ = −

∫ a

a0

m2

3H2(a′)da′

a′(15.14)

with ϕ0 ≡ ϕ(a0), yields the required solution (6.9) as the integral on the left-hand sidegives lnϕ− lnϕ0 = ln(ϕ/ϕ0). It then becomes a simple matter of exponentiation to get theanswer.

6d. The field will decay slowly if the factor in the exponent changes only slightly whenexpands by a factor ∼ 2, that is when ∆a ∼ a. This condition gives

m2

3H2 1 ⇐⇒ m √

3 H ' 1.7 H . (15.15)

This is very close to the condition m 32 H, which has a more physical interpretation in

view of what we found in b.

Note also that if we write

da′ =da′

dt′dt′ = H(a′)a′ dt′ (15.16)

the quasi-exponential solution (6.9) for ϕ involves the exponent

64

m2

3

∫ a

a0

da′

a′ H2(a′)=

m2

3

∫ t

t0

dt′

H(a′)≡

∫ t

t0dt′λ(t′) , (15.17)

with

λ(t) ≡m2

3H(t), (15.18)

formally equal to the decay rate λ+ that we found as the slowest decaying mode in thecase of constant H. The only difference is that now λ depends on time, so we must use anintegral: λ+t −→

∫λ(t) dt.

This incidentally shows that the slowest decaying solution is the one that in fact satisfiesthe slow-roll approximation, while the other solution does not.

6e. If we look at the scale factor a as a function of ϕ rather than the other way around, theinverse relation, we can use

dadϕ

dϕda

= 1 ⇐⇒dadϕ

=

(dϕda

)−1

. (15.19)

Relation (6.8) then gives:

d ln adϕ

=1a

dadϕ

=1a

(dϕda

)−1

=1ϕ

(d lnϕd ln a

)−1

= −3H2am2ϕ

. (15.20)

If the scalar field dominates over the other two terms on the right-hand side of Friedmann’sequation (6.5) we have (using tpl = 1/mpl in natural units)

H2 '4π3

m2ϕ2

m2pl

. (15.21)

Substituting this into (15.20) we get the required result:

d ln adϕ

= −4π ϕm2

pl

. (15.22)

65

The slow-roll condition (6.10) with (15.21) for the Hubble parameter gives

m2 94

H2 = 3πm2ϕ2

m2pl

⇐⇒ϕ

mpl

1√

3π. (15.23)

6f. The solution follows from writing

d ln a = −4πϕm2

pl

dϕ = −2πm2

pl

d(ϕ2

)(15.24)

and formally integrating:

ln a − ln a0 = ln(

aa0

)= −

2πm2

pl

(ϕ2 − ϕ2

0

). (15.25)

Here a = a(ϕ) and a0 ≡ a(ϕ0). Equivalently:

a(ϕ) = a0 exp

2πm2

pl

(ϕ2

0 − ϕ2) . (15.26)

If ϕ decays, a(ϕ) grows quasi-exponentially. This is in fact equivalent to what we foundin e: if a grows ϕ decays quasi-exponentially. Both ways of looking at this problem hasit’s uses.

6g. If we use the maximum value allowed for ϕ0 on energetic grounds at the Planck time,that is ϕ0 =

√2 (m2

pl/m), and let the field decay to ϕ ϕ0 (i.e. ϕ ' 0) we get:

a(ϕ ' 0)a(ϕ0)

' exp

2π ϕ20

m2pl

= exp

4π m2pl

m2

≡ Zinfl . (15.27)

If we assume m = 0.01 mpl, the inflation factor equals:

Zinfl = e1.26×125= 125.5×124

. (15.28)

66

Such an enormous expansion factor would solve both the flatness problem and the horizonproblem of standard cosmology.

7a. If we define u ≡ ϕ the equation of motion for the scalar field reads

u + 3(1a

dadt

)u = 0 , (15.29)

where I have substituted the definition for the Hubble parameter. Changing the indepen-dent variable from t to a(t) so that

u =dadt

duda

, (15.30)

the above equation becomes

dadt

(duda

+ 3au)

= 0 , (15.31)

or equivalently:

d ln ud ln a

= −3 ⇐⇒ u = ϕ ∝ a−3 . (15.32)

The energy density in the scalar field decays as

ρϕ = 12u2 ∝

(aa0

)−6

. (15.33)

7b. If we neglect all contributions to the density except that of the scalar field we havefrom Friedmann’s equation:

(1a

dadt

)2

=8πG ρϕ

3=

4πG u2

3=

4πG ϕ20

3

(aa0

)−6

. (15.34)

67

Assuming power-law behaviour a(t) ∝ tα we have H = α/t and this equation becomes:

α2

t2 =4πG ϕ2

0

3

(tt0

)−6α

. (15.35)

This must be true at all t, so one must have 6α = 2 (same power of time on both sides) sothat α = 1/3. Applying the relation at t = t0 with α = 1/3 yields:

19t2

0

=8πG ρϕ0

3⇐⇒ t0 =

1√24πG ρϕ0

. (15.36)

7c. Since ρrad always decays as ρrad ∝ T 4 ∝ a−4, it will always dominate for sufficientlylarge values of a/a0 Equality of the two contributions to the density requires

ρrad0

(aa0

)−4

= ρϕ0

(aa0

)−6

(15.37)

This implies:

a(teq)a0

=

√ρϕ0

ρrad0. (15.38)

7d. With the choice a0 = a(teq) we have:

ρrad = ρeq a−4 , ρϕ = ρeq a−6 , (15.39)

and Friedmann’s equation for a universe filled with a free scalar field and radiation can bewritten as

(1a

dadt

)2

=8πG ρeq

3

(a−4 + a−6

). (15.40)

Multiplying both sides by a2 and taking the square root of the resulting equation yields:

68

dadt

=

√8πG ρeq

3

√1 + a2

a2 ≡ Heq

√1 + a2

a2 . (15.41)

7e. The result in d can be reduced to quadrature by separating variables:

Heq dt =a2 da√

1 + a2. (15.42)

Then we can use the trick

a2

√1 + a2

=a2 + 1 − 1√

1 + a2(15.43)

to write:

Heq dt =√

1 + a2 da −da

√1 + a2

. (15.44)

Upon integrating this equation

∫Heq dt = Heqt =

∫√

1 + a2 da −∫

da√

1 + a2(15.45)

and using the standard integrals one finds the result.

69

For a 1 one can use:

√1 + a2 ' 1 + 1

2 a2 , sinh−1 a ' a − 16 a2 . (15.46)

The parametric relation

a√

1 + a2 − sinh−1 a = 2Heqt (15.47)

then reduces in leading order to:

a(1 + 1

2 a2)−

(a − 1

6 a2)

= 23 a3 = 2Heqt . (15.48)

Therefore:

a(t) =(3Heqt

)1/3. (15.49)

For a 1 we have10

a√

1 + a2 ' a2 , sinh−1 a ' ln 2a (15.50)

so that we find to leading order:

a2 = 2Heqt ⇐⇒ a(t) =(2Heqt

)1/2. (15.51)

You will (hopefully) recognize the standard behavior a ∝√

t of a flat, radiation-dominateduniverse.

10See a standard book like Abramovitz & Stegun, Handbook of Mathematical Functions, where you willfind a relation like sinh−1 x = ln(x +

√1 + x2), which is useful here.

70

8a. Using the definition of R we have

ρrad = ρvac R−4 , H =

1R

dRdt

. (15.52)

Friedmanns equation for a flat universe filled with dark energy and radiation then is

(1R

dRdt

)2

=8πG ρvac

3

(R−4 + 1

)≡ H2

vac

(R−4 + 1

). (15.53)

8b. With τ = Hvact andZ = R2 we have:

H = 12

(1Z

dZdt

)= Hvac

(1

2ZdZdτ

),√R−4 + 1 =

√1 +Z2

Z. (15.54)

As a consequence, Friedmann’s equation can be written as

(1

2ZdZdτ

)2

=1 +Z2

Z2 . (15.55)

Taking the square root on both sides gives the answer.

8c. Another example of separation of variables: we have

2 dτ =dZ

1 +Z2 . (15.56)

Integration:

2τ = sinh−1Z + constant . (15.57)

If we impose the initial condition R(0) = 0 (and therefore Z(0) = 0) the integrationconstant vanishes. So the solution is:

Z(τ) = R2(τ) = sinh(2τ) ⇐⇒ R(t) = sinh1/2(2Hvact) . (15.58)

71

8d. Equality of the two densities occurs as R = 1 (by definition!) and therefore atZ = 1.So we must have

sinh(2Hvacteq) = 1 ⇐⇒ teq =sinh−1(1)

2Hvac=

ln(1 +√

2)2Hvac

'0.4407

Hvac. (15.59)

8e. For large R (and even largerZ) we may use that also Heqt 1 so that

sinh(2Hvact) 'exp(2Hvact)

2. (15.60)

Putting this asymptotic result into the expression for R one finds

R(t) '1√

2exp(Hvact) , (15.61)

which corresponds to de Sitter expansion.

72

9a. The starting point is a2× Friedmann, which can be arranged into:

(dadt

)2

−8πG

3

(ρm a2 + ρvac a2

)= −k (15.62)

We differentiate this equation11.

We will need:

ddt

(dadt

)2

= 2(dadt

)d2adt2

ddt

(ρm a2

)=

dadt

(2a ρm + a2 dρm

da

)= −ρm a

(dadt

)as ρm ∝ a−3 , (15.63)

ddt

(ρvac a2

)= 2a ρvac

(dadt

)

Remember that k is a constant! The result is

2(dadt

) [d2adt2 +

4πGρm

3a −

8πG ρvac

3a]

= 0 . (15.64)

Excluding the static solution, where da/dt = 0, the term inside the square brackets mustvanish, which leads to the required result.

11We take this route, rather than taking the square root on both sides, and differentiating the resultingequation for da/dt. That road leads to all kinds of trouble, even though formally there is nothing wrongwith it.

73

9b. The equation for dH/dt simply uses

dHdt

=ddt

(1a

dadt

)=

1a

d2adt2 −

1a2

(dadt

)2

(15.65)

=1a

d2adt2 − H2 .

The remaining two equations need some work, for instance from Ωm = 8πGρm/3H2:

dΩm

dt=

8πG3

(1

H2

dρm

dt−

2ρm

H3

dHdt

)

=8πG

3

[1

H2(−3H ρm) −

2ρm

H3

(1a

d2adt2 − H2

) ](15.66)

Here I have used that ρm ∝ a−3 which allows you to write dρm/dt = −3H ρm (check this!).We can now use Friedmann’s equation in the form derived in a,

d2adt2 = −

4πG ρm

3a +

8πG ρvac

3a , (15.67)

to get rid of the d2a/dt2 term. This leads to

dΩm

dt=

8πG ρm

3H

[8πG ρm

3H2 −16πG ρvac

3H2 − 1]. (15.68)

If we then use the definitions for the two Omega parameters Ωm and ΩΛ again we getthe required result for dΩm/dt. The equation for the time-derivative of ΩΛ is derived inexactly the same manner.

9c. The line ΩΛ = 12 (Ωm +2) is the line where Ωm = 0, and the line ΩΛ = 1

2 (Ωm−1) is theline on which ΩΛ = 0. This allows you to divide the Ω − Ω plane in sections accordingto the sign of Ωm and of ΩΛ, see Figure 2 below. From that figure one can infer how theflow in the Omega-Omega plane is, see Figure 3.

74

From the second figure one can also see that of the three fixed points, the points whereΩm = ΩΛ = 0, only the point Ωm = 0, ΩΛ = 1 is a stable fixed point: the flow inits immediate vicinity is towards that point. The other two fixed points have the flowpointing away from it (the (1,0) point) or avoiding it (the (0,0) point: an empty universe!):if you ‘perturb’ the Universe away from these last two fixed points, it will start movingaway from them rather than back towards the fixed point.Note also: the flat solution, the line Ωm + ΩΛ = 1, is a trajectory since flat solutions withk = 0 stay flat! The arrows lie along that line towards the stable fixed point Ωm = 0,ΩΛ = 1.

Figure 3: The division of the Ω −Ω plane according to the signs of Ωm and ΩΛ.

75

Figure 4: Schematic representation of the flow in the Ω −Ω plane.

76

16 Week 15

10a. The field starts to dominate if

V(ϕ , T ) ' V0 >π2

30g∗T 4 ⇐⇒ T <

(30V0

π2g∗

)1/4

≡ Tcr . (16.69)

10b. With V0 ' E4sb answer a yields with g∗ = 106.75:

Tcr =

(30π2g∗

)1/4

Esb = (2.85 × 10−2)1/4 Esb = 0.41 Esb; . (16.70)

10c. The way to do this is to multiply the whole equation of motion with ∂ϕ/∂t and use:

∂ϕ

∂t∂2ϕ

∂t2 = 12

∂

∂t

(∂ϕ

∂t

)2

,∂ϕ

∂t∂V∂ϕ

=∂V∂t

. (16.71)

Note that inflationary models always assume that the scalar field is spatially uniform, sowe could have replaced ∂/∂t with d/dt.

10d. If you assume the slow-roll approximation holds and start inflation at ϕ ' 0 one hasρϕ ' V0.

10e. In the same approximation ρϕ is approximately conserved, so if all the scalar fieldenergy is released as radiation we have

π2

30g∗T 4

rh ' V0 ⇐⇒ Trh '

(30V0

π2g∗

)1/4

= Tcr . (16.72)

The re-heating temperature Trh roughly equals the critical temperature at which the infla-tion started.

77

11a. Differentiation of Friedmann’s equation leads to

2H H =8πG

3(ϕϕ + ϕ V ′

)(16.73)

Using the equation of motion for the scalar field in the form

ϕ = −3H ϕ − V ′ (16.74)

to eliminate the ϕ term inside the brackets immediately yields

2H H = −8πH (ϕ)2⇐⇒ H = −4πG ϕ2 . (16.75)

11b. The condition is, with ∆t ' tH ' 1/H:

|∆H| = |H ∆t| ''|H|H H ⇐⇒ |H| H2 . (16.76)

Using the expression for H derived in a:

4πG ϕ2 H2 ⇐⇒ |ϕ| H√

4πG. (16.77)

But is this is true we also have

8πG3

(12 ϕ

2) H2 , (16.78)

and Friedmann’s equation can only be satisfied if the V(ϕ) term is much larger than the12 ϕ

2 term.

11c. We must demand

|∆ϕ| = |ϕ| ∆t '|ϕ|

H |ϕ| ⇐⇒ |ϕ| H |ϕ| . (16.79)

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11d. In the slow-roll approximation we have:

H '

√8πG

3V(ϕ) , ϕ ' −

∂V/∂ϕ3H

= −V ′(ϕ)3H

= −V ′(ϕ)√

24πG V(ϕ). (16.80)

Then from a we have:

|H|H2 =

4πG ϕ2

H2 = 4πG(

38πG V

) V ′(ϕ)√24πG V(ϕ)

2

=1

16πG

(V ′(ϕ)V(ϕ)

)2

. (16.81)

The slow-roll approximation corresponds to |H| H2, so we must impose:

116πG

(V ′(ϕ)V(ϕ)

)2

1 ⇐⇒

∣∣∣∣∣V ′(ϕ)V(ϕ)

∣∣∣∣∣ √16πG . (16.82)

11e. The first relation has already been used, and follows from the equation of motion forthe scalar field. The second relation follows by using this result, and a time differentiation:

ϕ =∂

∂t

− V ′(ϕ)√24πG V(ϕ)

= −1

√24πG

∂

∂t

(V ′√

V

). (16.83)

Since we have

∂

∂t

(V ′√

V

)= ϕ

(V ′′√

V− 1

2

V ′2

V3/2

)(16.84)

and the relation ϕ = −V ′/√

24πGV the required answer follows immediately.

11f. One can show that the term involving (V/V)2 is already small compared to the otherterm because of the first slow-roll condition (answer d). The condition |ϕ| H |ϕ| =

|V ′|/3| can only be satisfied if in addition the V ′′ term in result e is sufficiently small:

79

|V ′ V ′′|24πGV

|V ′|3⇐⇒

∣∣∣∣∣V ′′V ′

∣∣∣∣∣ 8πG . (16.85)

(The answer given in the original assignment is a bit sloppy, it misses the factor 3!)

11g. The only thing you have to do is to change the integration variable and use theslow-roll approximation:

dt =dϕϕ

= −3Hdϕ

V ′(ϕ). (16.86)

Then

∫dt H(t) = −

∫3H2 dϕ

V ′(ϕ)= −

∫dϕ

8πG V(ϕ)V ′(ϕ)

≡ N . (16.87)

11h. In d we derived that one of the conditions for slow-roll is

∣∣∣∣∣V ′(ϕ)V(ϕ)

∣∣∣∣∣ √16πG . (16.88)

This implies that

8πG V(ϕ)V ′(ϕ)

√

4πG , (16.89)

and the integral for N (the number of e-foldings) satisfies:

calN √

4πG |ϕ1 − ϕ2| (16.90)

One gets sufficient inflation if N 1, which is guaranteed if one puts

|ϕ1 − ϕ2| >1√

4πG=

mpl√

4π. (16.91)

80

12a. Since the temperature of radiation falls as T ∝ a−1 and the energy of radiation asa−4 there inevitable comes a moment where the energy density of the scalar field starts todominate if it does not fall rapidly. The threshold condition is

ρrad =π2g∗30

T 4 ' ρϕ ' m4GUT , (16.92)

which corresponds to a temperature (with g∗ ' 169):

T '(

30π2g∗

)1/4

mGUT ' 0.37 mGUT = 3.7 × 1214 GeV(= 4.2 × 1224 K) (16.93)

12b. The answer is most easily seen by writing (12.9) in the form:

Veff(ϕ , T ) = V0 −12µ

2 ϕ2 + λ ϕ4 , (16.94)

with

µ2 ≡ m2 − λ T 2 . (16.95)

The extrema of the effective potential follow from

∂Veff

∂φ= 4λ ϕ3 − µ2 ϕ = 4λ ϕ

(ϕ2 −

µ2

4λ

)= 0 (16.96)

There is always a minimum or a maximum at ϕ = 0. Which of the two is determined bythe sign of the second derivative at ϕ = 0,

(∂2Veff

∂φ2

)ϕ=0

= −µ2 = λ T 2 − m2 . (16.97)

If the derivative is positive, so for µ2 < 0 and T > m/√λ (high temperatures) the

extremum at φ = 0 is a minimum. In that case there are no other minima or maxima.

81

At lower temperatures, for T < m/√λ, the second derivative of the potential at ϕ = 0

changes its sign. Now ϕ = 0 corresponds to a maximum. In that case µ2 > 0 andcondition (16.96) can also be satisfied at

φ = ±µ

2√λ≡ ±ϕm . (16.98)

It is easily checked that these both points correspond to a minimum. At very low temper-ature (that is for T m/

√λ) we have µ2 ≈ m2 and ϕm ' m/2

√λ.

12c. For T > m/√λ the shape is quasi-parabolic, with the ”wings” of the parabola

getting steeper and steeper as temperature increases. For T < m/√λ you get a ’mexican

hat profile’, look it up on Internet!

12d. If you want no large cosmological constant when the scalar field ”sits” in one of thetwo minima at ϕ = ±ϕM you should demand V(ϕ = ±ϕm) ' 0 when T m/

√λ. In that

case you can neglect the temperature term ∝ T 2 in Veff and put µ2 ' m2 and ϕm = m/2√λ:

V0 −12m2 ϕ2

m + λ ϕ4m = V0 −

m4

16λ= 0 ⇐⇒ V0 =

m4

16λ. (16.99)

If you substitute this back into definition (12.9) you find (after some rearrangement ofterms) answer (??).

12e. The naturalness argument says

V0 =m4

16λ' m4

GUT , Tcr ' mGUT . (16.100)

The second condition is almost automatically satisfied if the first condition is satisfied.The first condition is satisfied when m ' mGUT and λ ∼ 1. That also implies

φm =m

2√λ' 1

2mGUT , (16.101)

up to factors of order unity. This last condition states that φm also has a ’natural’ value.

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