Imperial College QFFF, 2018-19

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CosmologyLecture notes

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Toby Wiseman

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Toby Wiseman; Huxley 507, email: t.wiseman@imperial.ac.uk

Books

This course is not based directly on any one book. Appropriate reading forthe course is;

• Weinberg, ”Cosmology”also an older book ”Gravitation and cosmology”

• Kolb and Turner, ”The early universe”(somewhat dated now!)

• Liddle and Lyth, ”Cosmological inflation and large scale structure”

• Peacock, ”Cosmological physics”

• Mukhanov, ”Physical foundations of cosmology”

• Dodelson, ”Modern cosmology”

For issues of GR, my favourite book is obviously Wald, ”General Relativity”.

Mathematica

I have written some mathematica notebooks to illustrate certain calculations.These can be downloaded from my PWP. In order to run them you will needMathematica. Instructions for downloading this are at;www3.imperial.ac.uk/physics/staff/computing/software

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Conventions

I will use the following conventions; c = 1 and (�+++) signature.

For geometry I will use;

�cab(x) ⌘

1

2gcd✓@gdb@xa

+@gda@xb

� @gab@xd

◆

so,

ravb ⌘ @av

b + �bacv

c

And,

[r↵,r�]vµ = R �↵�µ v�

so,

R �↵�µ = @��

�↵µ � @↵�

��µ + �⌫

↵µ���⌫ � �⌫

�µ��↵⌫

3

Brief history of universe

10�43 s, 1019 GeV

Planck scale - quantum gravity - GUTs???

⇠ 10(�25,�40) s, ⇠ 1010�18 GeV

Inflation - quantum perturbations created, reheating

Hot big bang begins...

⇠ 10(�25,�38) s, ⇠ 1010�16 GeV

Baryogenesis

10�10 s, 1016 K, 100 GeV

Electroweak scale - dark matter annihilates to give relics??

10�4 s, 1013 K, 100 MeV

Quarks condense to hadrons

10 s, 1011 K, 1 MeV

Neutrino decoupling and e+e� annihilation (e+ + e� $ ⌫⌫)

Observation era begins...

102 s, 1010 K, 0.1 MeV

nucleosynthesis (n+ ⌫ $ p+ e�)

105 yrs, 104�5 K

Matter domination (Radiation-Matter equality)

3⇥ 105 yrs, 103�4 K, 1 eV

Photon decoupling (p+ e $ H), CMB formed

4

108 yrs

Structure formation; First objects - universe no longer close to FRW

5⇥ 109 yrs

Dark energy comes to dominate

13.8⇥ 109 yrs, 2.7 K (e↵ective temperature of photons)

Now

Plan

1. FRW; symmetries, Friedmann equation, cosmologies for perfect fluidsand scalar fields, observables, ⇤CDM model

2. Hot matter; stat. mech. description of matter in thermal equibilir-ium, out of equilibrium description (Boltzmann equation), relics (darkmatter)

3. Thermal history; hot radiation era, nucleosynthesis, recombination

4. Inflation; cosmological puzzles (flatness, horizon, monopole), scalarfield cosmologies, slow roll inflation, generation of fluctuations duringinflation

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1 FRW

1.1 Cosmological principle

The universe is statistically the same at every location in it, and in everydirection, ie. there is nothing special about where we live.

On suitably large scales (essentially beyond scales which are gravitationallybound - galaxy cluster scales ⇠ Mpc) and after a suitable averaging, theuniverse should be homogeneous and isotropic for some suitably chosenset of freely falling observers.

These observers define a foliation of spacetime into spatial slices. For anyspacetime filled with observers we may (locally at least) write the metric as,

ds2 = �dt2 + hij(t, x)dxidxj (1)

This is simply a coordinate choice, where the time coordinate is the propertime as measured for a freely falling observer sitting at the constant spatiallocation x - (ie. x =const is a geodesic).

Technically we take homogeneity and isotropy to mean we may write themetric of spacetime as,

ds2 = �dt2 + d⌃2 , d⌃2 = hij(t, x)dxidxj (2)

where at each constant time t, then d⌃ is the metric on a space which ishomogeneous and isotropic.

1.1.1 Homogeneity and isotropy spatial geometries

Now at a fixed time t then ⌃ - the constant time slice - is simply a spatialgeometry. Let us now suppress time in this discussion and simply considerhomogeneous isotropic spatial geometries.

Consider first isotropy. Then at any point in ⌃ all directions must appearthe same. Recall that any Riemannian geometry locally is flat. Consider apoint, and then locally about that point we may write,

d⌃2 = hijdxidxj = A(⇢)d⇢2 +B(⇢)d⌦2

(2)

(3)

6

where d⌦2

(2)

= d✓2 + sin2 ✓d�2 is the round unit 2-sphere.

An important point is that A and B do not depend on ✓ and � as we wish tohave isotropy, and also there are no o↵ diagonal ⇢✓ or ⇢� terms in the metricfor the same reason.

[More precisely the metric has isometry group SO(3), generated by the 3Killing vectors of the 2-sphere.]

Let us chose a convenient coordinate r2 = B(⇢), and then,

d⌃2 = A(r)

✓2rdr

B0(⇢)

◆2

+ r2d⌦2

(2)

= S(r)dr2 + r2d⌦2

(2)

(4)

Now since any space is locally flat, as r ! 0 we must have limr!0

S(r) = 1.

Now consider homogeneity. A necessary, but not su�cient condition forhomogeneity is that the Ricci scalar is constant - ie. it doesn’t depend on xi.Now,

R⌃

=2

r2

✓1� 1

S+

rS 0(r)

S2

◆(5)

where R⌃

is the Ricci scalar of the metric on ⌃. Let us set this equal to aconstant k. Then choose the normalization of the constant so,

R⌃

= 6k (6)

Then one may solve the resulting equation, and with the condition that thespace is locally flat at r ⇠ 0 we have,

S(r) =1

1� kr2(7)

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One then finds that the Ricci tensor is;

(R⌃

)ij = 2khij (8)

and the metric is,

d⌃2 =dr2

1� kr2+ r2d⌦2

(2)

(9)

We see the Ricci tensor is proportional to the metric - this is a special con-dition called an Einstein space.

It is convenient to perform a coordinate transformation;

r ! r0 =1

ar (10)

for a constant a so then,

d⌃2 = a2✓

dr02

1� ka2 r02+ r02d⌦2

(2)

◆(11)

By performing this scaling we may always write the metric as;

d⌃2 = a2✓

dr02

1� k0 r02+ r02d⌦2

(2)

◆= a2d⌃02 (12)

where now k0 = 0,±1, and a is a constant. Note now, R⌃

0 = 6 k0.

The 3 cases of k0 are distinct, and a simply sets the size or ‘scale’ of thegeometry. Let us now drop the ’primes’ on r, k and consider the 3 casesk = 0,±1.

Case k = 0 - flatThen clearly k = 0 is the case of Euclidean 3-d flat space;

d⌃2 = dr2 + r2d⌦2

(2)

= dx2 + dy2 + dz2 (13)

The homogeneity and isotropy are apparent in the cartesian and sphericalcoordinates; 3 translations and 3 rotations respectively.

8

Case k = 1 - 3-sphereFor k = 1 then ⌃ is a unit radius (round) 3-sphere. Let us see why. Consider4-dimensional Euclidean space with coordinates xA = (z, xi). Then we maywrite the metric as,

ds2(4d) = dz2 + dr2 + r2d⌦2

(2)

= �ABdXAdXB (14)

Now consider embedding a unit 3-sphere ⌃ into this space,

z2 + r2 = 1 =) rdr = �zdz (15)

Hence the induced metric is,

d⌃2 =

✓dz

dr

◆2

dr2 + dr2 + r2d⌦2

(2)

=

✓1

1� r2

◆dr2 + r2d⌦2

(2)

(16)

using,

1 +

✓dz

dr

◆2

= 1 +r2

z2=

z2 + r2

z2=

1

1� r2(17)

Homogeneity and isotropy; We may see the isometries of the 3-sphere by not-ing that the full isometry group SO(4) acts on the 3-sphere in the obviousway in the Cartesian coordinates XA in the 4-d Euclidean space. One cantranslate this action into the z, r, ✓,� coordinates although it is complicated.The action of homogeneity are the 3 rotations of a point to another point,and of isotropy are the 3 rotations about an axis.

Case k < 0 - 3-hyperboloidThe k = �1 case is similar to the sphere. Now instead of embedding a surfacein Euclidean space, we instead embed in 4d Minkowski spacetime,

ds2 = �dz2 + dr2 + r2d⌦2

(2)

= �dz2 + �ijdXidXj (18)

where now the extra coordinate z is a ’time’ coordinate. Now a hyperboloid,⌃, with unit radius is embedded as,

z2 � r2 = =) rdr = zdz (19)

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and its induced metric is,

d⌃2 = �✓dz

dr

◆2

dr2 + dr2 + r2d⌦2

(2)

=

✓1

1 + r2

◆dr2 + r2d⌦2

(2)

(20)

now using,

1�✓dz

dr

◆2

= 1� r2

z2=

z2 � r2

z2=

1

1 + r2(21)

Homogeneity and isotropy; We see the embedding and metric of the hyper-boloid are invariant under a Lorentz transformation of the 4-d embeddingMinkowski spacetime, ie. under SO(1, 3). Hence the isometry group of the3-hyperboloid ⌃ is SO(1, 3). The action on the Minkowski coordinates z,X i

is straightforward, but is complicated in the coordinates r, ✓,�. The actionof homogeneity and isotropy are generated by combinations of the 3 boostsand 3 rotations.

1.1.2 The FRW spacetime

Now let us return to the full spacetime, rather than a constant t slice, andconsider the time dependence.

We have seen that the geometry of a constant t slice can be written as,

d⌃2 = a2✓

1

1� kr2dr2 + r2d⌦2

(2)

◆(22)

for a constant a setting the scale of the space, and k = 0,±1 setting itscharacter.

However we must recall that a was a constant of integration, and may bedi↵erent depending on the spatial slice we pick. We can use the above coor-dinates on each time slice provided we let a = a(t). Now the full spacetimemetric takes the form,

ds2 = �dt2 + a(t)2d⌃2

(k) , d⌃2

(k) =1

1� kr2dr2 + r2d⌦2

(2)

, k = 0,±1 (23)

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and the function a(t) is the ‘scale factor’ controlling how the size of the ho-mogeneous isotropic spatial slices changes in time.

[ Note that one might wonder why k 6= k(t)? Such topology change is notpossible in a smooth manner. ]

We use the terminology;

1. k = 0 is a flat universe

2. k = 1 is a closed universe (sphere spatial sections)

3. k = �1 is an open universe (hyperbolic spatial sections)

Note that in the flat and open cases the geometry of the spatial slices is infi-nite - the spatial volume is infinite. However, for the closed case the universehas finite volume.

1.1.3 Conformal time

We may choose a new time coordinate, ⌧ called conformal time such that;

ds2 = a(⌧)2��d⌧ 2 + d⌃2

(k)

�(24)

defined by dt = a(⌧)d⌧ so that d⌧ = dt/a(t) and hence,

⌧(t) =

Z t

t0

dt0

a(t0)(25)

One simplification of using conformal time is that null geodesics in this space-time follow the same curves as in the spacetime ds2 = �d⌧ 2+d⌃2

(k). However⌧ does not have an interpretation as a proper time.

1.1.4 Another coordinate chart

We have seen that we may write,

d⌃2

(k) =1

1� kr2dr2 + r2d⌦2

(2)

(26)

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in isotropic coordinates. There is another convenient chart for computingthe Christo↵el symbol �(h)i

jk,

d⌃2

(k) = hijdxidxj =

✓�ij +

k xixj

1� k|x|2

◆dxidxj , |x|2 = xnxm�nm (27)

and in this chart one can conveniently write,

�(h)ijk = k xi hjk (28)

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1.2 Properties of FRW

Writing FRW in the general form earlier,

ds2 = �dt2 + a(t)2⌃2

(0,±1)

= �dt2 + a(t)2hij(x)dxidxj (29)

then one finds the metric and inverse metric in coordinates xµ = (t, xi) is,

gµ⌫ =

✓�1 00 a2(t)hij

◆, gµ⌫ =

✓�1 00 1

a2(t)hij

◆(30)

where hij is the inverse metric to hij.The non-vanishing Christo↵el components are then,

�tij = a ahij

�itj =

a

a�ij

�ijk = �(h)i

jk (31)

where �(h)ijk is the Christo↵el symbol of the 3-d spatial geometry hij(x).

The components, �tti = �i

tt = 0 by isotropy, and the components�t

tt = 0 happen to vanish using this proper time coordinate t.Then one finds the Ricci tensor components;

Rtt = �3a

aRti = 0

Rij = R(h)ij +

�2a2 + aa

�hij (32)

where R(h)ij is the Ricci tensor of hij and we recall that R(h)

ij = 2khij andhence,

Rij =�2k + 2a2 + aa

�hij

(33)

Again Rti = 0 due to isotropy.

This yields an Einstein tensor Gµ⌫ = Rµ⌫ � 1

2

gµ⌫R as,

Gtt =3k

a2+

3a2

a2Gti = 0

Gij =��k � a2 � 2aa

�hij (34)

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1.2.1 Special geodesics of FRW

By symmetry, xi =constant is a timelike geodesic - with proper time t. Theseare the worldlines our preferred observers. These observers are called comov-ing observers as they free fall with the homogeneous, isotropic frame. Theyhave 4-velocity vµ = dxµ/d⌧ = dxµ/dt = (1, 0, 0, 0).

A simple set of geodesics are the null curves passing through r = 0 isisotropic coordinates,

ds2 = �dt2 + a2(t)

✓1

1� kr2dr2 + r2d⌦2

(2)

◆(35)

By symmetry ✓,� =constant for these null geodesics, so the metric on theirworld line is,

ds2curve = �dt2 +a2(t)

1� kr2dr2 = 0 (36)

which must vanish for a null curve so,

1

adt = ± 1p

1� kr2dr (37)

for the curve. Consider a null geodesic arriving at an observer at r = 0at time t = to, having previously passed through radius R at time T (withto > T ). Thus r is decreasing with t and so we require the negative signabove. Then integrating (and flipping the limits in

Rdr) we obtain,

Z to

T

1

a(t)dt =

Z R

0

1p1� kr2

dr =

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

⌘ fk(R) (38)

1.2.2 General geodesics of FRW

Consider a geodesic xµ = (T (�), R(�),⇥(�),�(�)) with a�ne parameter �.In order to obtain the geodesic equations we may vary the action,

S =

Zd�L , L = gµ⌫

dxµ

d�

dx⌫

d�= �T 2 + a2(T )

R2

1� kR2

+R2

⇣⇥2 + sin2 ⇥�2

⌘!(39)

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where =d/d� (not d/dt).

Consider a geodesic passing through r = 0. Then by symmetry again itshould have ⇥,� =constant. More generally symmetry implies there shouldbe a class of geodesics with only radial motion, i.e. with ⇥,� =constant.

Check: � and ⇥ equations are (respectively;

a2R2 sin2 ⇥� = J ,d

d�

⇣a2R2⇥

⌘= a2R2 sin⇥ cos⇥�2 (40)

for constant of integration J . These are indeed satisfied for ⇥,� =constant,and J = 0.

Note: We might also consider geodesic motion which is not radial. For thiswe would have to solve the above complicated system. However, we knowfrom homogeneity that we can consider any point as r = 0, so other raysnot going through r = 0 are always related to ones through the origin byhomogeneity.

Now we are left with varying the line element along the curve,

L = �T 2 +a2(T )

1� kR2

R2 (41)

We have already solved this in the null case above. Consider now the timelikecase. Using,

@L

@R=

2a2R

1� kR2

,@L

@R=

2ka2RR2

(1� kR2)2(42)

this yields the Euler-Lagrange equation, d/d�(@L/@R) = @L/@R;

d

d�

a2R

1� kR2

!=

ka2RR2

(1� kR2)2(43)

Now dividing by a2R/(1� kR2) gives,

d

d�

ln

a2R

1� kR2

!!=

kRR

1� kR2

= �1

2

d

d�

�ln�1� kR2

��(44)

15

So then,

R =A

a2p1� kR2 (45)

where A is a constant of integration.Now using L = µ ⌘ 0,±1 we have,

T 2 =a2(T )

1� kR2

R2 � µ =A2 � µa2

a2(46)

so that,

T =

pA2 � µa2

a(47)

Hence the curve obeys,

dR

dT=

R

T=

Ap1� kR2

ap

A2 � µa2(48)

Consider a ray through r = 0 at t = to and previously through r = R att = T with to > T . Hence during the interval T < t < t

0

then dR/dT < 0and so A < 0. Then (flipping the limits in R);

Z to

T

|A|dTa(T )

pA2 � µ a2(T )

=

Z R

0

dRp1� kR2

=

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

(49)

and so we may calculate R = R(T ). This obviously agrees with the caseµ = 0 from before.

Aside on photons/wave

Consider a high frequency photon - so its wavelength is much shorter thanthe local curvature scales. Then in an LIF we may write that,

Aµ = eikµxµ

eµ (50)

for constant polarisation eµ (obeying constraints) and constant wave vectorkµ, obeying k2 = 0. The 4-momentum of the photon is,

pµ = ~kµ (51)

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The fact that kµ is constant may be written covariantly as,

kµrµk⌫ = 0 (52)

ie. the wave crests follow null curves. Now more generally in a curvedspacetime, a high frequency photon may be written as,

Aµ = ei�(t,x)eµ(t, x) (53)

where the phase � is rapidly varying, and now the field kµ = @µ� again inthe short wavelength limit, obeys the geodesic equation,

kµrµk⌫ = 0 (54)

and the 4-momentum is again pµ = ~kµ. This is what is known as thegeometric optics approximation.

1.2.3 Gravitational redshift

Consider a photon travelling along a null geodesic (µ = 0). Then its 4-momentum is pµ = ~kµ with kµ obeying the null geodesic condition kµrµk⌫ =0 with kµkµ = 0. Hence we may write,

kµ =dxµ

d�(55)

for some a�ne parameter �. Now (for ~ = 1, c = 1)

pµ =dxµ

d�=⇣T , R, 0, 0

⌘=

✓A

a(T ),

A

a2(T )

p1� kR2, 0, 0

◆(56)

Consider a comoving observer with 4-velocity vµ = (1, 0, 0, 0). They measurethe photon to have an energy Ecomove = �pµvµ, so,

Ecomove = +pt =A

a(T )(57)

Hence the energy of the photon measured by observers in the cosmologicalframe redshifts as E ⇠ 1/a(T ).

17

A unit norm spatial vector (tangent to constant t slices) for such anobserver, in the direction of the photon, is nµ = (0, 1/

pgrr, 0, 0), so that

nµnµ = 1 and nµvµ = 0. Then the magnitude of momentum |p|comove thatthe observer measures the photon to have in the direction nµ is,

|p|comove = |pµnµ| = |grrprnr| = |pgrrpr| = ap

1� kR2

|A|a(T )2

p1� kR2

=|A|a(T )

(58)

Thus we have, a(T )|p|comove =constant. In the case k = 0, this resultis seen as a conservation law using Cartestian spatial coordinates due totranslation invariance in space. More generally it is a conservation law dueto the isometry of homogeneity. The fact that Ecomove ⇠ 1/a is simply aresult of the fact that in any LIF a photon has E = |p|.

We have derived the famous gravitational redshift result, namely thatfor a photon emitted at time te with comoving energy Ee (and momentum|p|e = Ee), and received at to, comoving energy Eo (and momentum |p|o =Eo), then,

Eo

Ee

=a(te)

a(to)(59)

Hence a photon is redshifted by an expanding universe - in a sense it istunnelling out of a gravitational potential well. Since frequency of a photonE = ~⌫, one also has,

⌫o⌫e

=a(te)

a(to)(60)

It is conventional to define the redshift Z as,

1 + Z ⌘ ⌫e⌫o

=a(to)

a(te)(61)

where now to should be taken as the time today. Thus redshift Z(te) vanishesfor photons emitted today (ie. very close by) and is Z > 0 for photons emittedin our past. It is important as it provides a very physical measure of time,Z = Z(t) by,

Z(t) =a(to)

a(t)� 1 (62)

18

where we think of t < to as the time in the past when a photon which wereceive today was originally emitted. Inverting this relation, t = t(Z) givesan elegant relation between comoving proper time and redshift which is di-rectly observed. However, this relation clearly depends on the detail ofa(t), and hence the cosmological model.

A more pedestrian derivation:Consider a null ray emitted at r = R at time t = te and propagating to theorigin at time t = to. Then we have previously found,

Z to

te

dt

a(t)=

Z R

0

drp1� kr2

=

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

(63)

Now consider a second ray emitted slightly later at t = te + �te and receivedat t = to + �to. Then,

Z to+�t0

te+�te

dt

a(t)=

Z R

0

drp1� kr2

(64)

Hence, di↵erencing these relations we obtain,

0 =

Z to+�t0

te+�te

dt

a(t)�Z to

te

dt

a(t)

=�tea(te)

� �toa(to)

(65)

Now since the relation between the emitted and observed comoving frequen-cies is ⌫o

⌫e= �te

�tothen we obtain as before;

1 + Z ⌘ ⌫e⌫o

=a(to)

a(te)(66)

Massive geodesicsLet us continue our discussion of geodesics and conclude with a massiveparticle (µ = �1 and � = ⌧ 6= t). Then it has 4-momentum pµ = mvµ,

pµ = mdxµ

d�= m

⇣T , R, 0, 0

⌘=

mpA2 + a2(T )

a(T ),mA

a2(T )

p1� kR2, 0, 0

!(67)

19

Now a comoving observer measures Ecomove and |p|comove as;

Ecomove = pt = m

s

1 +A2

a2(T )= m� (68)

where � is the usual Lorentz factor � = 1/p1� v2 for observed velocity v

(so v2 = 1� 1/�2), and so,

v2 =A2

a2 + A2

(69)

Then note that,

�v =|A|p

a2 + A2

s

1 +A2

a2(T )=

|A|a(T )

(70)

And for the momentum,

|p|comove = |pµnµ| =ap

1� kR2

m|A|a(T )2

p1� kR2

=m|A|a(T )

= m�v (71)

with observed velocity v. Thus E2

comove = |p|2comove +m2 as usual, but with ared shifting momentum, |p|comove ⇠ 1/a(T ).

Note that as for the photon, we have a|p|comove =constant, which followsfrom the conservation law due to spatial homogeneity. Recall this is easilyseen computing geodesics for the flat k = 0 case in Cartesian coordinateswhere homogeneity is manifest.

Thus more generally the redshift is really related to observed and emittedcomoving 3-momentum;

1 + Z =a(te)

a(to)=

|p|o|p|e

(72)

for any free falling object - null or timelike. However, for photons E = |p|.Note in terms of observed energy for a massive particle we have,

1 + Z =|p|o|p|e

=

sE2

o �m2

E2

e �m2

(73)

20

1.3 Observables

Let us now consider some important quantities which are related to observ-ables in cosmology.

1.3.1 Comoving distance

If we are at r = 0 and a distant object is a radial coordinate position Rat time t then we say it has comoving distance R. Note that if we and theobject are comoving, then this distance remains constant in time.

1.3.2 Proper distance

The foliation of spacetime by the family of cosmological observers allowsus to define the proper distance between two objects at a given time t.Suppose we consider an observer at time t at the origin r = 0, and an objectat that time at radius r = R. Then the proper distance d(t, R) is the geodesicdistance within the spatial section ⌃. The spatial slice ⌃ at constant t is;

d⌃2 = a(t)2✓

dr2

1� kr2+ r2d⌦2

(2)

◆(74)

and by symmetry the geodesic in ⌃ between the observer and the objecttravels on the radial line ✓,� =constant. Hence its proper distance is,

d(t, R) = a(t)

Z R

0

drp1� kr2

= a(t)

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

(75)

Note that this is not a geodesic in the full spacetime, only within the geometryof the constant spatial slice defined by our cosmological observers.

Note also that for a null geodesic emitted at time t = te from radiusr = R and reaching the origin r = 0 at time t = to we have,

Z to

te

dt

a(t)=

Z R

0

drp1� kr2

(76)

Hence, we have for a null ray,

d(to, R)

a(to)=

d(te, R)

a(te)=

Z to

te

dt

a(t)(77)

21

1.3.3 Hubble function

A very important quantity is the Hubble function,

H(t) =a0(t)

a(t)(78)

and its value today, ie. t = t0

, is called the Hubble constant H0

= H(t0

).Consider a comoving observer at r = 0, and a comoving object at r =

R. Then the Hubble function determines the rate of change of the properdistance between these objects.

d

dtd(t, R) = a0(t)

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

= H(t)d(t, R) (79)

Note that d/d is independent of R.This can be interpreted as a version of Hubble’s law, namely that objects

in an expanding universe recede from us with a velocity proportional to theirdistance. This movement of objects away from us is called the Hubble flow.

Note however that the ‘velocity’ d is not a very physical one. Let usconsider now a more physical version of Hubble’s law.

1.3.4 Nearby sources

Consider a source nearby to us at r = 0, t = to, emitting at time te and atradius r = R, so �t = to � te is small.

Now using our previous relation for the proper distance travelled by anull ray,

d(to, R)

a(to)=

Z to

te

dt

a(t)⇠ �t

a(to)(80)

then we see,

d(to, R) ⇠ �t (81)

This simply says the proper distance is approximately the light travel time.

22

Then we may also expand,

a(te) = a(to) + a0(to)(te � to) + . . .

= a(to) (1��t ·Ho + . . .) (82)

Hence,

1 + Z =a(to)

a(te)= 1 +�tHo + . . . =) Z ' �tHo (83)

Thus the Hubble constant determines the redshift of local sources.Putting these results together we obtain Hubble’s experimental result fornearby sources;

Z ' Ho · d(to, R) (84)

1.3.5 Value of Hubble constant

The Hubble constant Ho has a value, usually quoted in peculiar units;

H ' 70 km s�1 Mpc�1 (85)

What funny units! Recall a parsec; 1pc =distance at which 1 AU subtendsan angle of 1 arc second! So pc = 3.2 light yr = 3.1 ⇥ 1016m. In terms ofastronomical scales;

• Nearest star (Proxima Centauri) ⇠ 1 pc

• Milky way (our galaxy) ⇠ 1 Kpc

• Galaxy cluster ⇠ 1-10 Mpc

• Hubble horizon (approx size of observable universe) ⇠ 3000 Mpc

These units are useful in the sense of d = Hd so that an object 1Mpc awaylooks as though it is receding at a velocity 70 km s�1.

For reference, note that our peculiar motion, ie. the motion relative to thecosmological frame, is ⇠ 400 km s�1 (which is typical value). Thus on largescales (> cluster size) peculiar motion become negligible compared to theHubble flow. On small scales it is an important e↵ect, dominating the motiondue to cosmological expansion.

23

1.3.6 Angular diameter and Luminosity distance

So far we have the relation R(t) and Z(t) for a null ray provided we knowthe cosmological expansion a(t). Suppose we observe light from an objectand measure its redshift, then we know the time of emission t and the radiusR the light was emitted from.

Now since we also know the proper distance relation d(t, R) to the objectthen if we have an independent measure of distance then we can check thisagrees. This will be a check of our cosmological model.

Turned around, if we know the distance d and redshift Z of many observedobjects, we can hope to reconstruct the function a(t) and deduce the correctcosmological expansion rate (and then using Einstein’s equations understandthe matter driving the expansion).

However while Z is something directly and accurately observed, the properdistance d is not something that is easy to deduce simply looking at a dis-tance source. Hence it is useful to have two more practical distance measures;angular diameter distance dA and luminosity distance dL.

Angular diameter distance dALet us be comoving observers at r = 0. Consider a comoving object at radiusr = R with proper size � (for example a distant galaxy). Assume we knowthis size - for example we have a good model for galaxies. Suppose it emitslight at time t = te, and we observe this at time t = to. Let us see the objectto have an angular diameter ⇥ on the sky.

Then the angular diameter distance is defined to be the distance the ob-ject appears to be from us ie. assuming a simple Euclidean spatialgeometry and infinite speed of light. So,

dA =�

⇥(86)

Now from the spatial geometry at time te,

ds2space = a(te)2

✓dr2

1� kr2+ r2

�d✓2 + sin2 ✓d�2

�◆(87)

we have,

� = a(te)R⇥ (88)

24

and thus we see,

dA = a(te)R =1

1 + Za(t

0

)R (89)

Recall that,

d(t, R) = a(t)

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

,d(to, R)

a(to)=

d(te, R)

a(te)(90)

Hence we can relate dA to the proper distance today d(to, R). For exampleif k = 0,

dA = a(te)R = d(te, R) =a(te)

a(to)d(to, R)

=1

(1 + Z)d(to, R) (91)

Thus if we know the size of an object � we can measure dA and its redshiftZ directly, and then deduce its proper distance d from these.

Luminosity distance dLSuppose we have a similar source, but now instead of knowing the size � ofthe source, we instead know its intrinsic luminosity L = energy emitted(isotropically) per time. When we observe the light we see an apparentluminosity ` =energy at receiver per area per time.

We define the luminosity distance which is the distance the object ap-pears to be from us;

` =L

4⇡d2L(92)

From the spatial geometry at time to, there light from the source is spreadover a 2-sphere with area,

A(S2

) = 4⇡ (a(to)R)2 (93)

The apparent luminosity is given by;

` =1

(1 + Z)2L

A(S2

)(94)

where the two factors of 1/(1 + Z) are due to;

25

• The energy of each photon is redshifted by ⌫o/⌫e = 1/(1 + Z)

• The rate of receiving photons is reduced relative to emission frequencyby the same factor. Recall our previous calculation of gravitationalredshift precisely showed in section 1.2.3 (see equation 65) that �to =(1 + Z)�e.

Hence we see, that;

` =L

4⇡d2L=

1

(1 + Z)2L

A(S2

)=) dL = (1 + Z)a(to)R (95)

Recalling that dA = a(te)R, we see,

dL = (1 + Z)2dA (96)

Again we can relate dL to the proper distance today d(to, R). For example ifk = 0,

dL = (1 + Z)a(to)R = (1 + Z)d(to, R) (97)

ExperimentWe measure a distant source, and determine its redshift Z. Either knowingits size �, or its luminosity L we then determine its distance dA or dL.

Recall,

dA = a(te)R =1

(1 + Z)a(to)R , dL = (1 + Z)a(to)R (98)

and so our measurements give us directly Z and a(to)R.

Suppose we understand a(t) and wish to test our cosmological model. Thenour measurement for a source gives both Z and R. But the photon relations,

1 + Z =a(to)

a(te),

Z to

te

dt

a(t)=

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

(99)

relates these as we know R = R(te) and Z = Z(te) so we know R = R(Z).Thus we can check the agreement.

In practice we use lots of sources to build up the relation between Z and R,and then constrain parameters in our model to fit this.

26

1.4 Horizons and the big bang

The inverse of the Hubble function, H�1, defines a distance scale, denotedthe Hubble scale. This is sometimes loosely referred to as the ’horizon size’.Generally it doesn’t refer to any horizon, but is an important physical scalethat essentially determines the scale beyond which the spacetime doesn’t lookflat for causal physics. However in certain situations there are two types ofhorizon - particle and event horizons.

Suppose we have an expanding universe which started at t = tBB at a BigBang, so that a(tBB) = 0. Using the null relation for a ray reaching r = 0 attime t starting at a radius r = RH at the Big Bang;

Z t

tBB

dt0

a(t0)=

8<

:

RH k = 0sin�1(RH) k = 1sinh�1(RH) k = �1

(100)

Thus this gives a relation RH = RH(t). The Particle horizon size dH(t) attime t is the proper radius of the comoving volume with this radius RH(t).Recall the proper distance,

d(t, R) = a(t)

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

(101)

and hence we see,

dH(t) = d(t, RH(t)) = a(t)

Z t

tBB

dt0

a(t0)(102)

Existence of a particle horizonNote that this integral may or may not converge depending on the be-haviour of a(t) at the big bang. We shall discuss this later. For usual matterand radiation this does converge (with a ⇠ (t � tBB)1/2 for hot radiation).However, for exotic inflaton matter, this does not.

Event horizonIf the integral

Z 1

t

dt0

a(t0)=

8<

:

RE k = 0sin�1(RE) k = 1sinh�1(RE) k = �1

(103)

27

converges, it means that an event at r = 0 and time t can only influencecomoving observers within a radius r RE(t). An observer outside theradius RE will never see the event, however long they wait. This comovingsize is known as the event horizon for the event at r = 0, time t, and itsproper size is,

dE(t) = d(t, RE(t)) = a(t)

Z 1

t

dt0

a(t0)(104)

Note that if the universe recollapses then t may have a maximum value tmax.In this case the event horizon is defined in the obvious way as,

Z tmax

t

dt0

a(t0)=

8<

:

RE k = 0sin�1(RE) k = 1sinh�1(RE) k = �1

(105)

Comment: The existence of dark energy leads to an event horizon, wheredE ⇠ H�1

0

. Hence as the universe expands, objects which are not gravita-tionally bound to us move further away (d = Hd). Once they move beyonda proper distance dE they can never influence us again.

Note that while the terminology horizon is used, these are not proper causalhorizons in the sense of black holes. Their definition is dependent on thechoice of an observer. The region associated to one observers horizon willnot coincide with that of another.

1.5 Perfect fluids

Consider the stress tensor for a single perfect fluid;

Tµ⌫ = (⇢+ P ) uµu⌫ + Pgµ⌫ (106)

where ⇢, P are the energy density and pressure, and uµ is the local fluid 4-velocity (so u2 = �1). The relation P = P (⇢) is the equation of state ofthe fluid. Recall that for such a fluid the dynamics is simply determined bythe conservation of stress-energy,

rµTµ⌫ = 0 (107)

which yields relativistic hydrodynamics.

28

Important: We may have several such fluids in a spacetime and providedthey don’t exchange energy-momentum with other matter, each one willindividually obey the above equations.

The equations of hydro are best described by projecting into the uµ andorthogonal directions. Firstly project the conservation equation into thedirection uµ;

uµr⌫Tµ⌫ = 0 =) u⌫@⌫⇢+ (⇢+ P )r⌫u

⌫ = 0 (108)

recalling u2 = �1. This gives an evolution equation for the density.

Let us now project onto an orthogonal direction nµ to the motion uµ, so thatuµnµ = 0. Note that nµ must be space like (since vµ is timelike). Then,

nµr⌫Tµ⌫ = 0 =) nµ ((⇢+ P ) u⌫r⌫u

µ + gµ⌫@⌫P ) = 0 (109)

and this is equivalent to,

(⇢+ P ) u⌫r⌫uµ +��⌫µ + uµu

⌫�@⌫P = 0 (110)

again recalling that u2 = �1. The quantity ?⌫µ = �⌫µ + uµu⌫ is a projector

into the directions orthogonal to uµ. It has the property that,

?⌫µu

µ = 0 , ?⌫µn

µ = n⌫ (111)

for any nµ orthogonal to uµ. Then,

(⇢+ P ) u⌫r⌫uµ = �?⌫µ@⌫P (112)

gives an evolution equation for the velocity.

We will derive this later, but taking P = w⇢ for a constant w, then,

• Dust / cold matter: w = 0, so P ' 0

• Radiation / hot matter: w = 1

3

, so P ' 1

3

⇢

• Vacuum energy: w = �1

29

Note in the case of vacuum energy it isn’t really a fluid;

Tµ⌫ = �⇢gµ⌫ (113)

However, the condition rµT µ⌫ = 0 then implies @µ⇢ = 0, and hence theenergy density is not dynamical, but constant in time and space. It is naturalto define the cosmological constant

⇤ ⌘ 8⇡GN⇢ (114)

so we may write,

Tµ⌫ = � 1

8⇡GN

⇤gµ⌫ (115)

and then we move this term to the LHS of the Einstein equations,

Rµ⌫ �1

2gµ⌫R + ⇤gµ⌫ = 8⇡GN Tµ⌫ (116)

where Tµ⌫ is the stress tensor due to other matter - not a cosmological term.Now ⇤ is an inverse length squared, where the length gives the radius ofcurvature associated to the cosmological constant.

1.6 The cosmological stress tensor

Now consider cosmological matter in FRW,

ds2 = gµ⌫dxµdx⌫ = �dt2 + a(t)2hij(x)dx

idxj (117)

so that the stress tensor is homogeneous and isotropic. Then since there is nopreferred direction we must have Tti = 0, and Tij / hij. Due to homogeneitywe must have Ttt is a function of time only, and Tij = f(t)hij for a function oftime f . Then we define the total density and total pressure of cosmologicalmatter in FRW to be,

Ttt ⌘ ⇢tot(t) , Tij ⌘ a2(t)Ptot(t)hij(x) (118)

where Tµ⌫ is the sum of all matter components. Why? Consider a perfectfluid which is homogeneous and isotropic. Then uµ = (1, 0, 0, 0) and ⇢ = ⇢(t),P = P (t). Thus,

Ttt = (⇢+ P ) utut + Pgtt = ⇢

Tti = 0

Tij = Pgij = a2Phij (119)

30

and thus the density and pressure agree.Note however, that while any matter with cosmological symmetry has

a stress tensor with the form above, it does not mean it behaves like asingle perfect fluid with a simple local equation of state Ptot(t) = F (⇢tot(t)).In general matter with have a complicated non-local (in time) equation ofstate.

Let us consider the conservation equation rµTµ⌫ = 0 for a stress tensorsharing the cosmological symmetry. Recall that this applies to the total stresstensor (as a result of the Bianchi identities), but also applies separately tothe stress tensor of any matter component that doesn’t interact with othermatter. The time component of the conservation equation implies;

0 = rµTµt = @tTtt � �µµ↵T

↵t � �µ

t↵T↵µ

= ⇢� �iitT

tt � �j

tiTij

= ⇢� 3a

a(�⇢) +

a

a�jiP �ij

= ⇢+ 3a

a(⇢+ P ) (120)

This applies to both the total stress tensor, but also separately to any com-ponent that doesn’t interact with other components. The only assumptionis that the matter has the cosmological symmetry.

1.7 Cosmological perfect fluids

Consider now the behaviour of a single perfect fluid with cosmological sym-metry that doesn’t interact with other matter. Let it have equation of stateP = w⇢ for constant w. Then conservation implies,

⇢+ 3a

a⇢ (1 + w) = 0 =) ⇢ =

k

a3(1+w)

(121)

for a constant of integration k. We usually write this constant in terms ofthe scale factor today, ao, and the value of the fluid density today, ⇢o, as,

⇢ = ⇢0

⇣a0

a

⌘3(1+w)

(122)

Consider the important cases;

• Dust / cold matter: w = 0; ⇢ ⇠ 1

a3, so the matter simply dilutes.

31

• Radiation / hot matter: w = 1

3

; ⇢ ⇠ 1

a4, so the radiation both

dilutes and its constituents redshift.

• Vacuum energy: w = �1, as we saw before we have ⇢ =constant.

1.8 Dynamics and the Friedmann equation

We have previously states the components of the Einstein tensor. Recall,

ds2 = �dt2 + a(t)2hij(x)dxidxj , R(h)

ij = 2khij

Gtt =3k

a2+

3a2

a2, Gij =

��k � a2 � 2aa

�hij (123)

Then the Einstein equations are Gµ⌫ = 8⇡GNTµ⌫ where we emphasise thatTµ⌫ is the total stress tensor, and hence is conserved and in our FRW ap-proximation must have cosmological symmetry. Using equations (119) wethen arrive at two important relations. Firstly the tt component directlygives;

a2

a2+

k

a2=

8⇡GN

3⇢tot =) H2 =

8⇡GN

3⇢tot �

k

a2(124)

This is the Friedmann equation and directly determines the Hubble func-tion H = a/a. Note that we also have the matter conservation equation;

⇢+ 3a

a(⇢tot + Ptot) = 0 (125)

Suppose we have a single fluid for matter. Recall the fluid equationsof motion are simply given by stress energy conservation. Then given theuniverse at a time ti with scale factor a(ti) and density ⇢(ti), with knownequation of state P = P (⇢) then the Friedmann and conservation equationsallow us to integrate forward in time to determine a(t) and ⇢(t).

For a combination of perfect fluids (with cosmological symmetry), thenthe total energy and pressure will be the sum of the various components

⇢tot =X

i

⇢i , Ptot =X

i

Pi (126)

and each will be conserved separately;

⇢i + 3a

a(⇢i + Pi) = 0 (127)

32

(Obviously this implies the total stress tensor is conserved). Again the Fried-mann equation, the individual conservation equations and equations of statePi = Pi(⇢i), allow one to integrate a(t) and ⇢i(t) forward in time.

More generally, for any matter with cosmological symmetry, the Fried-mann equation together with the matter equations of motion allow integra-tion forward in time. For example, for an (interacting) scalar field � theequation of motion is,

r2� = rµrµ� = V 0(�) (128)

and the stress tensor is,

Tµ⌫ = @µ�@⌫�� gµ⌫

✓1

2(@↵�)

2 + V (�)

◆(129)

and it is easy to show the matter equation of motion implies the stress ten-sor is conserved. For a mass, then V (�) = m2

2

�2. Then for cosmologicalsymmetry the scalar equation becomes;

�+ 3a

a� = �V 0(�) (130)

and so knowing a,� and � at a time ti then the Friedmann equation andscalar equation allow integration forward in time. In this case,

⇢tot =1

2�2 + V (�) , Ptot =

1

2�2 � V (�) (131)

Note that the case V (�) = 0 is P = ⇢ (so w = 1 - a ’sti↵’ equation of state).However, for general V (�) this is not of a simple P = P (⇢) form.

What about the spatial components of the Einstein equations? The spatialij-components may be combined with the tt component to eliminate termsinvolving H2 to give another equation which is useful;

a

a= �4⇡GN

3(⇢+ 3P ) (132)

Note that this equation is independent of k. An important point is that(due to the Bianchi identities) this equation is equivalent to the Friedmannequation and total matter conservation equation. Thus it contains no new

33

information, but nicely summarises how a behaves.

Acceleration and expansionCosmological expansion is characterised by a > 0. However, acceleration ischaracterised by a > 0. We see for this to hold we require;

⇢+ 3P < 0 (133)

where we note this is the total energy density and pressure. Suppose thestress tensor is dominated by a perfect fluid with equation of state P = w⇢.In this case acceleration implies,

w < �1

3(134)

Hence we see cold or hot matter (w = 0, 13

) give a decelerated expansion.However dark energy gives acceleration.

Note that if ⇢ > 0 then if a > 0 at some time, then the universe remainsexpanding unless k = 1 (the closed case). In the case k = 0,�1 this isdecelerated or accelerated expansion depending on the sign of a. In theclosed case k = 1, if we have a < 0 then if the scale factor reaches a radiussuch that a = 0 so,

8⇡G

3⇢ =

1

a2(135)

then after this point a < 0, and the universe will recollapse to a big crunch.

1.9 Simple cosmological solutions

Consider the flat case k = 0 with a single perfect fluid P = w⇢ for constantw with w > �1. Then from conservation we already saw;

⇢ = ⇢o⇣aoa

⌘3(1+w)

(136)

Putting this into the Friedmann equation;

a2

a2=

8⇡GN⇢o3

⇣aoa

⌘3(1+w)

= H2

o

⇣aoa

⌘3(1+w)

(137)

34

Hence (taking the positive root for an expanding universe);

a1+3w

2 a = Hoa3(1+w)

2o (138)

and integrating,

2

3(1 + w)a

3(1+w)2 = Hoa

3(1+w)2

o (t� k) (139)

for a constant of integration k = 0. We now see why we restricted to w > �1.We conventionally fix this by taking the big bang a = 0 to be at time t = 0,so,

a = ao

✓3(1 + w)

2Hot

◆ 23(1+w)

(140)

Then the density goes as,

⇢

⇢o=

✓a

ao

◆�3(1+w)

=

✓3(1 + w)

2Hot

◆ 23(1+w)

!�3(1+w)

=

✓2

3(1 + w)

1

Hot

◆2

(141)

Note then that for all values of w we have ⇢ ⇠ t�2.

Hubble functionThe Hubble function is;

H =a

a=

2

3(1 + w)

1

t(142)

Null geodesicsNull geodesics are governed by the equation (recall k = 0);

R =

Z to

te

dt

a(t)=

1

ao

✓3(1 + w)

2Ho

◆� 23(1+w)

Z to

te

dt t�2

3(1+w) (143)

where,Z to

te

dt t�2

3(1+w) =3(1 + w)

1 + 3w

ht

1+3w3(1+w)

itote

(144)

35

Lower limit: For the lower limit to converge for te ! 0 we require;

0 <1 + 3w

3(1 + w)=) w > �1

3or w < �1 (145)

Thus for non-accelerating matter �1

3

< w precisely gives a finite lower limit.For accelerating matter in the range �1 w < �1

3

it does not. (The rangew < �1 is unlikely to be physically relevant).Upper limit: For the upper limit to converge for to ! 1 we require theopposite;

1 + 3w

3(1 + w)< 0 =) �1 < w < �1

3(146)

Thus this requires accelerated expansion.

Particle horizonIn such a model the particle horizon size (recall k = 0) at time t = to is;

dH(to) = a(to)

Z to

0

dt

a(t)=

✓3(1 + w)

2Ho

◆� 23(1+w)

Z to

0

dt t�2

3(1+w)

(147)

Hence it is finite for non-accelerating matter with �1

3

< w such as dust w = 0and radiation w = 1/3. Then the lower limit gives zero, and so,

dH(to) =

✓3(1 + w)

2Ho

◆� 23(1+w) 3(1 + w)

1 + 3w(to)

1+3w3(1+w)

=3(1 + w)

1 + 3w

✓3(1 + w)

2Ho

◆�1

=2

1 + 3w

1

Ho

(148)

recalling that Ho =2

3(1+w)

1

to.

Event horizonIn this model the event horizon size (recall k = 0) at time t = to is;

dE(t) = a(to)

Z 1

to

dt

a(t)=

✓3(1 + w)

2Ho

◆� 23(1+w)

Z 1

to

dt t�2

3(1+w)

(149)

36

Thus we see for �1 < w < �1

3

that the upper limit gives zero and so up toa sign we get the same as for the particle horizon above,

dE(t) =

✓3(1 + w)

2Ho

◆� 23(1+w) 3(1 + w)

1 + 3w

ht

1+3w3(1+w)

i1to

= �3(1 + w)

1 + 3w

✓3(1 + w)

2Ho

◆�1

= � 2

1 + 3w

1

Ho

(150)

where the quantity is positive due to the range of w.

Cold matter and k = 0Consider our special case of cold matter. Note that this case is called theEinstein-de Sitter model (although it has nothing to do with de Sitterspacetime). Then,

⇢ = ⇢o⇣aoa

⌘3

(151)

Then we have decelerated expansion with;

a

ao=

✓3

2Hot

◆ 23

, H =2

3t(152)

Recall (as for all w) then ⇢ ⇠ t�2. The particle horizon size is;

dH(to) =2

Ho

(153)

but there is no event horizon.

Hot matter/Radiation w = 1/3 and k = 0For radiation we have,

⇢ = ⇢o⇣aoa

⌘4

(154)

Again this yields ⇢ ⇠ t�2. As for dust we have decelerated expansion;

a

ao= (2Hot)

12 , H =

1

2t(155)

37

and the particle horizon is;

dH(to) =1

Ho

(156)

and again there is no event horizon without acceleration.

Special case: Cosmological constant w = �1 and k = 0This is an example of de Sitter spacetime. Then as we saw earlier the energydensity is constant,

⇢ = ⇢o ⌘⇤

8⇡GN

(157)

with ⇤ a constant inverse length squared. Let us assume that ⇤ > 0 - apositive cosmological constant, so that the energy density ⇢ = �P > 0.This negative pressure leads to accelerated expansion of the spacetime. Thenthe Friedmann equation gives;

H2 =a2

a2=

8⇡GN⇢o3

=⇤

3(158)

so (taking the positive root);

a = a0

ep

⇤3 (t�to) , H = Ho =

r⇤

3(159)

where the Hubble function is constant. An important point is that nowa ! 0 as t ! �1. So the singularity or ‘big bang’ is not at t = 0, but isan infinite proper time in the past. For this reason there is no particlehorizon;

Z to

�1

dt

a(t)=

Z to

�1dte�

p⇤3 (t�to) =

1

�q

⇤

3

he�

p⇤3 (t�to)

ito�1

! 1

(160)

Another important feature is that we see exponential expansion. So the

38

rate of acceleration is very quick. This acceleration leads to an event horizon;

dE(t) = a(to)

Z 1

to

dt

a(t)=

Z 1

to

dt e�p

⇤3 (t�to)

=1

�q

⇤

3

he�

p⇤3 (t�to)

i1to

=

r3

⇤(161)

with the upper limit now giving zero.

1.10 Some other cosmological solutions

More generally the de Sitter solutions are simple to write down for k = ±1, 0and ⇤ > 0. They solve the equation,

Rµ⌫ �1

2gµ⌫R + ⇤gµ⌫ = 0 =) Rµ⌫ = ⇤gµ⌫ (162)

Let us define;

Ho =

r⇤

3(163)

as above.

Flat slicing of de Sitter, k = 0: we have already seen this;

ds2 = �dt2 + e2Hot�ijdxidxj (164)

where a ! 0 as t ! �1.

Global de Sitter, k = +1:

ds2 = �dt2 +1

H2

o

coshHot d⌃2

k=+1

(165)

Note that in this case we have no a = 0 ’big bang’, and for t < 0 we have acontracting universe!

39

Hyperbolic slicing of de Sitter, k = �1:

ds2 = �dt2 +1

H2

o

sinhHot d⌃2

k=�1

(166)

where a ! 0 as t ! 0.

In fact all these are coordinates on the same spacetime, with the globalk = +1 case covering the whole of de Sitter, and the others only coveringportions of it. Thus in fact for k = 0 and k = �1, then a ! 0 isn’t reallythe big bang at all but rather just a coordinate singularity.

Matter plus radiation (k = 0): Consider now again the case k = 0 withboth a matter and radiation perfect fluid. Now,

a2

a2=

8⇡GN

3

✓⇢m⇣a

0

a

⌘3

+ ⇢r⇣a

0

a

⌘4

◆(167)

and thus,

a2 =A

a+

B

a2, A =

8⇡GN

3⇢ma

3

o , B =8⇡GN

3⇢ra

4

o (168)

Then,

apAa+B

da = dt =) t� tk =2

3pA

✓a� 2

B

A

◆ra+

B

A(169)

so,

(t� tk) =

r⇢ma3o6⇡G

✓a(t)� 2

⇢ra0⇢m

◆ra(t) +

⇢ra0⇢m

(170)

Note we may set the big bang to t = 0 by choosing an appropriate tk;

p6⇡Gt =

p⇢ma3o

✓a(t)� 2

⇢ra0⇢m

◆ra(t) +

⇢ra0⇢m

+ 2⇢3/2r a3o⇢m

(171)

Note that this is not straightforward to invert to obtain a = a(t). Note thatsetting ⇢r = 0 we recover t ⇠ a3/2 (i.e.. a ⇠ t2/3). Note that taking ⇢m ! 0is possible - a divergence must be absorbed into tk - and then one is left with

40

t ⇠ a2 (ie. a ⇠ t1/2).

Matter plus ⇤ (k = 0): One can also solve for matter plus a cosmologicalconstant. Then,

a2

a2=

8⇡GN

3

✓⇢⇤

+ ⇢m⇣a

0

a

⌘3

◆(172)

and so,

8⇡GN

3dt =

ra

⇢⇤

a3 + ⇢ma3oda

=2

3p⇢⇤

lnha

32⇢

⇤

+p⇢⇤

pa3⇢

⇤

+ ⇢ma3o

i(173)

1.11 General cosmologies

Let us use the notation that to is the time today, with Ho and ⇢o being thevalues of H(to) and ⇢tot(to) today. We then define the critical density;

⇢crit =3

8⇡GN

H2

o (174)

This is the matter density a flat (k = 0) would have in order to reproducethe expansion rate Ho. Experimentally we find,

⇢crit ' 10�26 h2 kg m�3 ⇠ 1 proton/m�3 (175)

We may write the Friedmann equation today as;

⇢crit = ⇢o �3

8⇡GN

k

a2(176)

Hence we see;

• By construction for a flat universe ⇢o = ⇢crit

• A closed universe is over dense: ⇢o > ⇢crit

• An open universe is under dense: ⇢o < ⇢crit

41

Now consider an FRW universe with matter given by a cosmological con-stant ⇤, cold matter ⇢M and hot matter/radiation ⇢R, all of which are notinteracting with the other components. Then we write,

⇢⇤

=⇤

8⇡GN

⇢M = ⇢M,o

⇣a0

a

⌘3

⇢R = ⇢R,o

⇣a0

a

⌘4

(177)

for matter and radiation densities ⇢M,o, ⇢R,o today. Then we have,

H2 +k

a2=

8⇡GN

3

✓⇢⇤

+ ⇢M,o

⇣a0

a

⌘3

+ ⇢R,o

⇣a0

a

⌘4

◆(178)

Now let us use the critical density to rewrite this expression by defining;

⌦i =⇢i,o⇢crit

,8⇡GN

3⇢crit = H2

o (179)

and we also define,

8⇡GN

3⇢crit⌦k = H2

o⌦k ⌘ � k

a2o(180)

These ⌦i for i = ⇤,M,R, k give the fraction of the energy density at t = toin a component compared to the energy density ⇢crit. Note that for k = 0then ⌦i give the fraction in a component compared to the total energy den-sity. For k 6= 0 we should think of ⇢crit as the e↵ective total energy densityincluding thinking of the spatial curvature as a matter component.

Then we may write,

H2 =8⇡GN

3⇢crit

✓⌦

⇤

+ ⌦k

⇣a0

a

⌘2

+ ⌦M

⇣a0

a

⌘3

+ ⌦R

⇣a0

a

⌘4

◆(181)

and so the Friedmann equation (note we have also used all the fluid equationsto get here) takes the simple form;

✓H

Ho

◆2

= ⌦⇤

+ ⌦k

⇣a0

a

⌘2

+ ⌦M

⇣a0

a

⌘3

+ ⌦R

⇣a0

a

⌘4

(182)

42

and this relation completely determines the dynamics of the scale factor.Note that for t = to we obtain the important relation;

1 =X

i

⌦i = ⌦⇤

+ ⌦k + ⌦M + ⌦R (183)

An important comment; clearly for very small a, radiation (if present) alwaysdominates. At late times if a ! 1 then ⇤ (if present) dominates. In thatcase then;

H !p

⌦⇤

H0

(184)

Note that for our universe we believe ⌦⇤

= 0.7. Hence H, which today isHo, is very close to its final value, so H ! 0.8H

0

.

To obtain the dynamics we take the positive root (for an expanding universe)and integrate;

t =1

Ho

Z a(t)

0

da

aq⌦

⇤

+ . . .+ ⌦R

�a0a

�4

(185)

where we assume that we have a big bang a = 0 at t = 0. A more physicalparameterization is in terms of redshift. Define;

x(t) ⌘ a(t)

ao=

1

1 + Z(t)(186)

Note then that;

t = 0 , a = 0 =) x = 0

t = to , a = ao =) x = 1 (187)

Then,

t =1

Ho

Z 11+Z(t)

0

dx

xp⌦

⇤

+ ⌦kx�2 + ⌦Mx�3 + ⌦Rx�4

(188)

and the age of the universe today is;

to =1

Ho

Z1

0

dx

xp⌦

⇤

+ ⌦kx�2 + ⌦Mx�3 + ⌦Rx�4

(189)

43

Consider a null geodesic received at t = to and r = 0, emitted at time t andradius R. Recall our usual relation;

Z to

t

dt0

a(t0)= Fk(R) , Fk(R) =

8<

:

R k = 0sin�1(R) k = 1sinh�1(R) k = �1

(190)

Now,

Fk(R) =

Z ao

a(t)

1

a

dt

dada =

Z ao

a(t)

da

a2 H(a)

=1

aoHo

Z1

11+Z

dx

x2

p⌦

⇤

+ ⌦kx�2 + ⌦Mx�3 + ⌦Rx�4

(191)

A nice way to write this is;

aoR(Z) =1

Ho

p⌦k

sinh

"p⌦k

Z1

11+Z

dx

x2

p⌦

⇤

+ ⌦kx�2 + ⌦Mx�3 + ⌦Rx�4

#

(192)

where we recall ⌦k = �k/(a2oH2

o ).

1.12 The ⇤CDM model

We now consider the ⇤CDM universe which is our ’standard model’ of cos-mology. It is a universe that is started by inflation, goes through a period ofhot big bang, and at late times is dominated by Cold Dark Matter (CDM)together with conventional matter, and a cosmological constant.

At late times in the ⇤CDM universe we assume that the universe has ex-panded to the point where radiation is irrelevant, so ⌦r ⌧ 1. In fact as wediscuss later for our universe ⌦r ⇠ 10�5. Thus we neglect it.

We also assume that |⌦k| ⌧ 1. Note that this does not strictly require k = 0,but just that the size of the spatial sections are su�ciently large that anyspatial curvature is irrelevant. This is justified by the theory of inflation wewill discuss later.

44

Now recalling that we have the constraint (setting ⌦k = ⌦r = 0);

1 = ⌦⇤

+ ⌦m (193)

then at late times in the universe (ie. now) there are just two parameters tomeasure. Firstly Ho, and secondly say ⌦m, the fraction of energy density inmatter relative to the cosmological constant.

Recall the angular diameter and luminosity distances for an object at redshiftZ are;

dA =1

(1 + Z)ao R(Z) , dL = (1 + Z)ao R(Z) (194)

The most elegant determination of the ⇤CDM parameters is from super-novae. Their redshift is directly measured. Certain (Type IIA) supernovaare believed to be ’standard candles’. Once their light curves have beenmeasured, we believe we can infer their total luminosity L. Then we maydetermine their luminosity distance from observation of apparent luminosity`, as ` = L/(4⇡d2L). Then,

dL(Z) = (1 + Z)ao R(Z)

=(1 + Z)

Ho

Z1

11+Z

dx

x2

p⌦

⇤

+ ⌦Mx�3

(195)

give the relation between measured luminosity distance and redshift.

Note that this integral can be computed in closed form - it is given in termsof an elliptic integral. However the analytic expression is not terribly useful.

After observing lots of supernovae, one fits the data to this relation to deter-mine the best fit of the parameters Ho and ⌦m. One famously finds,

h ' 0.7 , Ho = 100h kms�1Mpc�1

⌦m ' 0.3 =) ⌦⇤

' 0.7 (196)

Hence there is a positive cosmological constant (dark energy).

45

Current state of the art measurements are from Planck satellite from a fullCMB analysis

h ' 0.67± 0.015

⌦⇤

' 0.686± 0.02 (197)

As you can see these agree nicely!

It is a great mystery why the cosmological constant is so small (according toany QFT calculation it should be enormous!), and yet is of order the energydensity in matter today. No one knows the answers to these questions. Theidea that dark energy may be dynamic (so called quintessence) rather thana cosmological constant may try to explain this, but the real issues I believeshould be addressed by quantum gravity.

There are two important epochs;

Acceleration: Recall that,

a

a= �4⇡G

3(⇢+ 3P )

= �4⇡G

3

��2⇢

⇤

+ ⇢m,0 (1 + Z)3�

(198)

Hence the redshift Zacc at which acceleration began in the late universe was,

2⌦⇤

= ⌦m (1 + Zacc)3 =) Zacc = 0.67 (199)

which is around 6.1Gyr ago.

⇤-matter equality: Defined as the time or redshift Z⇤�m when the energy

densities in the two components (which drive the Friedmann equation) areequal. Hence.

⌦⇤

= ⌦m (1 + Z⇤�meq)

3 =) Z⇤�meq = 0.33 (200)

which is around 3.6Gyr ago.

46

1.13 Dark matter

The ⌦m = 0.3 matter component is the total cold matter. This is the usualbaryonic matter - stars and gas - but also dark matter. There are varioustechniques in astronomy to measure the amount of baryonic matter in theuniverse. However there are also methods to measure the total mass thatgravitates in a system - for example weak lensing. The mass of galaxies as-sessed by directly accounting for the baryonic matter, or looking at the totalgravitating mass is considerably di↵erent. Likewise the galactic dynamicsthat would follow only form the observed baryonic matter doesn’t reproduceobserved dynamics at all. The implication is that cold matter is actuallycomposed not only of baryons but also a non-luminous dark matter.

If we split the cold matter into a dark matter and baryonic component, so,

⇢m = ⇢DM + ⇢B (201)

then di↵erent methods agree that,

⌦B

⌦m

⇠ 0.15 (202)

ie. around only 15% of cold matter is baryonic. The rest is dark matter.Since ⌦m = 0.32 (Planck) then,

⌦B ⇠ 0.05 , ⌦DM ⇠ 0.27 (203)

47