Cosmological solutions of the Einstein-Friedmann equations ...fjeger/Cosmolect1-7.pdf ·...

Cosmological solutions of the Einstein-Friedmann equations

Summary of Friedmann’s equationsIngredients are the Einstein equations of GRT:

Rµν −12

R gµν − Λ gµν = κ Tµν ,

with the cosmic ideal fluid energy-momentum tensor

T µν = (ρ + p) (t) uµuν − p(t) gµν .

The cosmological constant term Λ gµν nowadays is usually included as part of Tµνas an additional contribution to the energy density

ρ(t)→ ρ(t) + ρΛ ; ρΛ =Λ

κ,

c© 2009, F. Jegerlehner ≪x Lect. 7 x≫

http://www-com.physik.hu-berlin.de/~fjeger/Cosmolect1-2.pdf


Cosmology

claimed to represent the vacuum energy density. The ideal fluid tensor structurethen requires the vacuum energy, now usually called dark energy to satisfy theexotic equation of state

pΛ = −ρΛ .

In the following we discuss the resulting set of Friedmann equations

S 2

c2 = κ3 ρ S 2 − k Friedmann equation

ddS

(ρ S 3

)= −3 p S 2 energy equation

p = p(ρ) equation of state

with ρ and p now assumed to include the cosmological constant term.Note, we have 3 functions S (t), ρ(t) and p(t) to be determined, for which we need3 equations to uniquely determine them in terms of some initial values.

c© 2009, F. Jegerlehner ≪x Lect. 7 x≫ 1



Cosmology

A) Empty Worlds, vacuum energy at most: ρ = p = 0 ,Λ = κ ρvac

In spite of the fact that such models are unrealistic, they give some insight into thegeometrical and kinematic possibilities in cosmological model building. In thislimiting case we have the following basic equations:

S 2

c2 = Λ3 S 2 − k ; S

c2 = Λ3 S

or

Λ = −3 H2(t)c2 q(t) ; K(t) = −

H2(t)c2 (q(t) + 1) .

The general solution is

S (t) = a ct + b ; Λ = 0 , a2 = −k ,

S (t) =12

(a eαct + b e−αct

); Λ , 0 , α2 =

Λ

3;

Λ

3ab = k = (1, 0,−1) .




Cosmology

Because of time reversal symmetry or anti-symmetry:when b , 0: b = ±a with a > 0.

Cases:

(a) Λ = 0 , k = 0: S = S 0 = constant ; H = 0 , q = 0 .

(b) Λ = 0 , k = −1: S = ct ; H = 1/t , q = 0 ; z = t0/t1 − 1 ,Milne’s Universe

(c) Λ > 0 , k = 1: S = a cosh(

cta

).

(d) Λ > 0 , k = 0: S = exp(

cta

)= exp (Ht) ; H ≡ c

a =√κ ρvac c2/3

de Sitter Universe: here H(t) = constant and q(t) ≡ −1, this universe has infiniteage! Today’s pure dark energy universe!

(e) Λ > 0 , k = −1: S = a sinh(

cta

).




Cosmology

(f) Λ < 0 , k = −1: S = a sin(

cta

).

where in all cases

a =√|3/Λ| .

Models (b), (d), (e) and (f) some times have been called “incomplete” as theyexhibit more space than “substrate”.




Cosmology

Terminology: Λ > 0 , k = 0,±1 (ρ = p = 0) de Sitter spacesΛ < 0 , k = −1 (ρ = p = 0) Anti-de Sitter spaces

Isometric embedding: ds2 =(dz0

)2−

(dz1

)2−

(dz2

)2−

(dz3

)2− k

(dz4

)2

-

6

z0

z3

z4S −4

de Sitter

-

6

z0

z3

z4S +4

Anti de Sitter




Cosmology

B) Matter dominated: ρ > 0 , p = 0

Λ = 0: preferable class of modelssatisfies Einstein’s boundary criterion: empty space = flat space

Base equations: ρ S 3 = ρ0 S 30 = constant

S 2

c2 = κ3 ρ S 2 − k = C

S − k ; C κ3 ρ S 3 = constant

which can be integrated elementary: dS/√

C/S − k = ±c dt

v k = 0 : Einstein-de Sitter universe

S (t) =(

94 C

)1/3(ct)2/3

v k = 1 : Friedmann-Einstein universe




Cosmology

ct = C(arcsin

√X −√

X − X2)

; X = S (t)/C

Parameter representation:

S (t) = 12 C (1 − cos η)

ct = 12 C (η − sin η)

cycloid

v k = −1 :

ct = C(√

X + X2 − arcsinh√

X)

; X = S (t)/C

Parameter representation:

S (t) = 12 C (cosh η − 1)

ct = 12 C (sinh η − η)




Cosmology




Cosmology

In each case we have S (t)→ 0 (t → 0) i.e., all of them are Big-Bang cosmologies.For small t the C term is dominating, such that

S (t) '(

94 C

)1/3(ct)2/3 ; t → 0

For the observables we have the relationshipsκ2 ρ(t) = 3 H2(t)

c2 q(t) q(t) > 0!

K(t) =H2(t)

c2 (2 q(t) − 1) q(t) =

> 1

2 ; k = 1= 1

2 ; k = 0< 1

2 ; k = −1

The following graphical representation is shown for C = 1 and c = 1, i.e, S (t) aswell as ct are represented in units of C.




Cosmology

∞

q = 0.10

0.16

0.210.25

0.30

0.51.0 2.0

7.5

↑↑ctc = π1

2ctc




Cosmology

Einstein-de Sitter universeThe flat case k = 0 is of particular interest as a critical universe in between positiveand negative curvature. In this case many things are fixed:

First

K(t) ≡ 0 qEdS(t) ≡12

and ρEdS(t) =3κ

H2(t)c2 .

In addition, using S 3 = 94C(ct)2, we obtain

C =κ

3ρ S 3 =

κ

3·

3κ

H2(t)c2 · S 3 =

94

CH2(t)

c2 (ct)2

where C cancels and we get 1 = 94 H2(t) (t)2 or

t = 23 H−1(t) (EdS)




Cosmology

Hence in the EdS universe everything is fixed for given H0:

Present values:

H0 ' 0.83 × 10−28 cm−1 × cρ0 EdS ' 1.2 × 10−29 gr/cm3 × c2

t0 EdS = 23H−1

0 ' 8.1 × 109 years (EdS age of the universe)

where we usedκ =

8πGNc2 = 1.86637 × 10−27 cm/gr .




Cosmology

a(t)

tt0

tH = H−10

Einstein-de Sitter universe: a typical expansion pattern (for closed universes onlyfor t trecontraction)




Cosmology

Back to the non-flat geometries k = ±1: here we have

C =κ

3ρ(t) S (t)3 = 2

H2(t)c2 q(t) S (t)3

thus, with

k S (t)−2 =H2(t)

c2 (2 q(t) − 1) ,

we may write

S (t) = cH(t)

1√|2 q(t)−1|

in terms of the observables H(t) and q(t). Furthermore, we have

C = 2c

H(t)q(t)

(|2 q(t) − 1|)3/2 ; q(t) ≥ 0 !




Cosmology

Also, with

S 2

c2 =CS− k ,

Sc2 = −

C2S 2 ; q = −

S SS 2

=12

CC − kS

we obtain 1

k S (t) = C2 q(t) − 1

2 q(t),

which means that kS (t) is determined by C and q(t).

1We also may write

ct k=1= C

arcsin

√1 − 2q−1 −

√2q−12q

; ct k=−1

= C √

1−2q2q − arcsinh

√2q−1 − 1




Cosmology

The conclusion of the above discussion: provided Λ = 0 and p = 0, we maydetermine the age and the curvature of the universe by evaluating the observablerelations at t = t0. Since, in principle, we can determine H0 and ρ0 independently,we can find whether k = 0,+1 or − 1. The density ρEdS for k = 0, shows up as thecritical density. Generally,

ρ(t) = ρEdS(t) · 2 q(t)

such thatρ0 > ρ0 EdS k = 1ρ0 < ρ0 EdS k = −1

Thus:l Lot of matter – space closes under gravity, gravity wins.l Little matter – space is open and matter spreads forever.

i.e, either confinement or asymptotic freedom




Cosmology

In any case the matter dominated energy density ρmat is known not to be given bythe normal baryonic matter which stars, planets, dust and interstellar gas aremade of. There are many indications that baryonic matter is only a fraction of thetotal gravitating matter, which is mainly Dark Matter (DM). Dark matter, which onlymanifests to us by gravitational interaction, is seen in galaxies (velocity profiles), inrelative motion of the objects in clusters of galaxies (Fritz Zwicky 1933, applyingthe virial theorem to Coma Cluster) as well as in the universe as a whole (CMBfluctuations). A very important “tool” in dark matter search is gravitational lensing,which convincingly supports the other findings, and in addition provides importantinformation concerning the distribution of DM. We will discussed this in detail later,together with Observational findings.




Cosmology

C) Radiation dominated: p = ρ/3

Realistic for the early stage of a Big-Bang cosmology.

Λ = 0: preferable class of modelssatisfies Einstein’s boundary criterion: empty space = flat space

Base equations: ρ S 4 = ρ0 S 40 = constant

S 2

c2 = κ3 ρ S 2 − k = C

S 2 − k ; C κ3 ρ S 4 = constant

which can be integrated elementary. We actually have

S dS√

C−kS 2 = ±cdt

v k = 0 :




Cosmology

S (t) =(4 C

)1/4(ct)1/2

v k = 1 :

S (t) =

(C −

(√C − ct

)2)1/2

in 0 < ct < 2√

C with cyclic extension !

v k = −1 :

S (t) =

((√C + ct

)2− C

)1/2

In each case we have S (t)→ 0 (t → 0) i.e., all of them are Big-Bang cosmologies.As in the matter dominated scenario, for small t the C term is dominating, such




Cosmology

that, independently of k,

S (t) '(4 C

)1/4(ct)1/2 ; t → 0

The behavior for large t is easily obtained:k=0 : S (t) =

(4 C

)1/4(ct)1/2

k=-1 : S (t) ' ctk=1 : cyclic, cycle time ctc = 2

√C

For k = 1 cyclicity is t → t − t0n ; t0n = n · 2√

Cc .

For the observables we have the relationships




Cosmology

κ ρ(t) = 3 H2(t)c2 q(t) q(t) ≥ 0!

K(t) =H2(t)

c2 (q(t) − 1) q(t) =

> 1 ; k = 1= 1 ; k = 0< 1 ; k = −1

The following graphical representation is shown for C = 1 and c = 1, i.e, S (t) aswell as ct are represented in units of

√C.




Cosmology

q = 0.10

0.2

0.3

0.5

1.03.0ր 6.0 ∞

↑↑ctc = 2

√C1

2ctc




Cosmology

Actually, in the radiation dominated era the situation is qualitatively very similar tothe matter dominated p = 0 scenario.

Exercise: Express q(t) in terms of S and C (matter dominated) or C (radiationdominated) scenario, respectively.

For radiation in thermal equilibrium (black body radiation) the Stefan-Boltzmannlaw holds:

ρradiation = a T 4 c2 ; a =π2 k4

B15 ~3 c2 = 8.418 × 10−36 gr/cm3 K−4

and the spectral distribution is given by the Planck-distribution




Cosmology

ργ(ν) dν = 8π hν3dν(exp

(hν

kBTγ

)−1

)

which characterizes the perfect radiator.




Cosmology

Since ρ(t)S (t)4 = ρ0S 40 = constant we have

ρ(t) ∝ T 4(t) ∝1

S (t)4

implying

T (t) = T0S 0

S (t)

In fact, the radiation cools down by the expansion so much that it decouples frommatter. This was predicted by Gamow and Alpher and Herman in 1948,




Cosmology

and Dicke and Peebles at Princeton were searching for it when it was actuallydiscovered “by accident” in 1965 by Penzias and Wilson at Bell Laboratories.For more history read:R≫R≫


http://www.aip.org/history/cosmology/ideas/bigbang.htm#gamow

http://www.dfi.uem.br/~macedane/history_of_2.7k.html



Cosmology

The discovery of the isotropic Cosmic Microwave Background (CMB) oftemperature




Cosmology

T0 = (2.725 ± 0.002) K

not only clearly favored Big-Bang cosmologies and essentially ruled outsteady-state theory, it by now provides the best test of isotropy at the level of 0.1%over 360 of the sky! We will come to that later.

What does it mean? First we have

ρ0γ = a T 40 c2 ρ0γ ' 4.64 × 10−34 gr/cm3 × c2 , n0γ ' 410 photons/cm3

which confronts with the present baryonic matter density

ρ0,mat ' 3 × 10−31 gr/cm3 × c2 .

However, the evolution of the two densities are different

ρrad(t) = ρ0,rad

(S 0

S (t)

)4

whereas ρmat = ρ0,mat

(S 0

S (t)

)3

,




Cosmology

and since S (t)→ 0 (t → 0) in a Big-Bang cosmology radiation always dominatesin the early universe!

tcurvature ?

ρ(t)ρrad(t)

ρmat(t)

ρrad = ρmat

radiationera

matterera

Big

Bang

present

Radiation dominates the Big Bang, at present it is a cold relict the CMB

Besides photons, a similar sea of cold neutrinos fills our universe. This we will




Cosmology

discuss later.




Cosmology

The critical energy density and the flatness problem

In general we have some mixture of vacuum energy, relativistic and non-relativisticmatter. We have seen that the Einstein-de Sitter universe with k=0 plays a specialrole. In the present matter dominated era ρEdS plays the boundary case betweenexpansion forever or recontraction. We therefore define

ρ0,crit = ρEdS =3H2

08πGN

= 1.878 × 10−29 h2 gr/cm3,

where H0 is the present Hubble constant, and h its value in units of 100 km s−1

Mpc−1, and express the energy density in units of ρ0,crit. Thus the presentdensity ρ0 is represented by

Ω0 = ρ0/ρ0,crit

and




Cosmology

Ω0 = 1

is the critical density for which the universe just remains open, infinite and flat.Ω0 > 1

the case of much matter, space closes due to gravitational attraction of mass.Gravity stops the expansion after finite time, the universe collapses, ends in heatdeath!

Ω0 < 1

the case of little matter, gravity not sufficient to stop expansion, universe expandsforever, space open, ends in freezing to death!

If we include the cosmological constant as a vacuum energy density in the totaldensity ρ, we may write the Friedmann equations in the form

S 2

c2 + k = κ3 ρ S 2 and 3 S

c2S= −κ2 (3 p + ρ)

and the energy-momentum conservation as




Cosmology

ρ = −3 SS (ρ + p) .

If the mixture of radiation and matter dominates over the vacuum energy 3p + ρ isalways positive and thus we have S /S ≤ 0. This means that the expansion musthave started with S = 0 in the past and the present age must be lower than theHubble age: t0 < H−1

0 (see figure in EdS case above). For k = 1 this also impliesthe recontraction to S = 0.

The present deceleration parameter q0 = −S 0/(S 0 H2

0

)can be written as

q0 =κ (ρ0+3 p0)

6 H20

=ρ0+3 p02 ρ0,crit

This equation provides a simple explanation for the different values q0 takesdepending on the form of the equation of state:




Cosmology

Form of energy equation of state q0 q0(k = 0)a) vacuum energy: pΛ = −ρΛ q0 = −ΩΛ q0 = −1,b) non-relativistic matter: pmat = 0 q0 = 1

2 Ωmat q0 = 12,

c) relativistic matter, radiation: prad = ρrad/3 q0 = Ωrad q0 = 1,

i.e.

the different forms of matter contribute with different signs and weights to q0.

Whatever the mix of energies the total energy density

ρ(t) = ρ0,crit

ΩΛ + Ω0,mat

(S 0

S (t)

)3

+ Ω0,rad

(S 0

S (t)

)4

for t → 0 is dominated in any case by the radiation part:

ρtot ' ρ0,crit Ω0,rad

(S 0

S (t)

)4

,




Cosmology

in which case, independent of k!

S (t) '(4C

)1/4(ct)1/2 ; C =

κ

3ρ0,rad S 4

0 i.e.S 4

0

S 4(t)=

34 κ (ct)2 ρ0,rad

,

such that

ρtot(t) ' 34 κ (ct)2

is universal.

Now, in the radiation dominated early phase after the Big BangS 2

c2 = κ3 ρ S 2 − k = C

S 2 − k ; C κ3 ρ S 4 = constant

shows that the curvature term proportional to k is subleading and may be dropped




Cosmology

(in accord with the universal behavior just mentioned before):

S 2

c2 →CS 2 or

S 2

S 2 →c2 CS 4 =

c2 κρ

3

i.e.

ρ(t) = ρEdS(t) = 3κ

H2(t)c2 .

It is truly remarkable that at that early times, precisely when we would expectcurvature the be most important, the evolution automatically picks the flat solution.This does not mean, however, that at later times when the other energy densitycomponents come in to play and even dominate the scene we have to expect a flatspace one.

In fact the observed energy density at present is close to the critical one. This wewill discuss later. Even if one counts the normal baryonic matter only one is notmuch more than a factor of 10 off from the critical value Ω = 1 (atoms ∼ 4%) . The




Cosmology

missing part making Ω = 1 complete is dark matter (cosmological constant 73%)and cold dark matter (what is it ? 23%). The strongly time and scenariodependent energy density ρ(t) easily deviates by 60 orders of magnitude fromρ0,crit today, given the evolution during the enormous time span of the present ageof the universe. In other words, the flat solution Ω = 1 is highly unstable. How is itpossible then that we apparently live in a universe close to that unstable point?This is the so called flatness problem (Dicke 1969). The solution: eithercosmological fine-tuning assuming we really started with an accuracy of 62 ormore digits with Ω = 1 or some dynamical mechanism transmutes this unstablepoint into an stable attractor, this is what inflation theory, to be discussed later, cando for us.

Exercise: confirm the necessary fine tuning in the cosmic evolution by explicitcalculation.




Cosmology




Cosmology

Appendix: Λ , 0The issue about the true value of the cosmological constant, which likely isnon-vanishing as mentioned earlier, will be reconsidered later. In any case wehave to discuss the solution of the cosmological equations for the case onnon-zero cosmological constant. The consequences for the empty world case wealready discussed: flat “background space” gets replaced by de Sitter or anti-deSitter space. The generalization to the matter dominated scenario is almost trivial.A glimpse at the Friedmann equation shows that one can treat the cosmologicalconstant as a contribution to the energy density:

κρmat + Λ→ κρtot,

where ρtot = ρtot + ρΛ with ρΛ = Λκ. Provided p = 0, all solutions remain unchanged,

except for a different interpretation of the energy density, which in this case is notidentical with the “normal” mass density of baryonic plus dark matter. Note thatthe cosmological constant in general enters the equation of state in a not a prioriknown way. However, this seems not to matter as long as we have p ' 0.




Cosmology

That a non-vanishing Λ spoils the geometry⇔ matter duality, because theEinstein equation remains true no matter on which side of the equation we writethe cosmological term. Many believe it has some thing to do with quantumvacuum fluctuations, but no answer can be given why it is so small. In any case ifwe take the energy momentum tensor as given on the quantum level by theStandard Model (SM) of strong and electroweak interactions we would expect theHiggs field vacuum expectation value (Bose condensate) as well as the quarkcondensates to contribute to the cosmological constant. Again, the value obtainedis about 50 orders of magnitude to big! What tames the cosmological constant toa small non-vanishing value?

Exercise: estimate the cosmological constant as induced by the Higgs mechanismand by spontaneous symmetry breaking of chiral symmetry in QCD.

A cosmological constant of course affects local gravitation in particular it mightaffect black holes by changing the Schwarzschild radius etc. In order to study thephysical effect of a non-vanishing Λ we consider the spherically symmetric mass




Cosmology

distribution in outer space:

Gµν ≡ 0 Gµν = Λ gµν

Ê Λ = 0 : Schwarzschild

ds2 =

(1 −

r0

r

)(cdt)2 −

11 − r0

r

dr2 − r2 dΩ2

which is unique, static with boundary condition: flat as r → ∞. It has the Newto-nian approximation with potential: ϕ ' −m

r with m r02 and where g00 ' 1 +

2ϕc2

Ë Λ , 0 :




Cosmology

α =(1 − r0

r

)→ 1 − r0

r −Λ3 R2. Denoting K = Λ

3

ds2 =

(1 −

r0

r− Kr2

)(cdt)2 −

11 − r0

r − Kr2dr2 − r2 dΩ2

which is unique, static with boundary condition: Schwarzschild as Λ→ 0. In theNewtonian approximation: ϕ ' −m

r −12kr2

Observation of planetary motions yields: |Λ| 10−42 cm−2 !

In particular: in empty space (no matter) m =r02 = 0 we have de Sitter space if

Λ > 0. The gravity potential is

ϕ ' −12

Kr2 repulsive linear force F ∝ Kr




Cosmology

and the metric

ds2 =(1 − Kr2

)(cdt)2 −

11 − Kr2 dr2 − r2 dΩ2

which is spherical symmetric with respect to any point and is regular. For K > 0there is the de Sitter coordinate singularity K r2 = 1. While for m , 0 we have atrue singularity at r = 0, such a singularity is absent for m = 0.

Mappings

Λ > 0: de Sitter space dS 4

embedding into M5 = R1,4 : (cT )2 − X2 − Y2 − Z2 −W2 = −a2

ds2 = (cdT )2 −(dX2 + dY2 + dZ2 + dW2

)c© 2009, F. Jegerlehner ≪x Lect. 7 x≫ 43



Cosmology

With a2 1K = 3

Λ; a > 0 the mapping reads:

X = r sin θ sinϕ

Y = r sin θ cosϕ

Z = r cos θ

W = a√

1 − r2

a2 cosh cta

T = a√

1 − r2

a2 sinh cta

r < a

W = a√

r2

a2 − 1 sinh cta

T = a√

r2

a2 − 1 cosh cta

r > a .

Note: z = constant is the hyperboloid: (cT )2 −W2 = constant. Homogeneity andIsotropy follow from the symmetry: S −4 is invariant under 5-dimensional Lorentztransformations. The force is linear repulsive F ∝ 1

3 Λ r.




Cosmology

Λ < 0: Anti-de Sitter space AdS 4

embedding into R2,3 : (cT )2 − X2 − Y2 − Z2 + W2 = a2

ds2 = (cdT )2 −(dX2 + dY2 + dZ2

)+ dW2

With a2 1K = − 3

Λ; a > 0 the mapping reads:

X = r sin θ sinϕ

Y = r sin θ cosϕ

Z = r cos θ

W = a

√1 +

r2

a2 coscta

T = a

√1 +

r2

a2 sincta




Cosmology

Note: z = constant is the circle: (cT )2 + W2 = constant. Homogeneity and Isotropyfollow from the symmetry: S +

4 is invariant under S O(2, 3). The force is linearattractive F ∝ −1

3 Λ r. Causality problem: periodicity in time ct → ct + 2 π a. Thecircle in the (cT,W) plane is acausal. May be cured by mapping the circle to thereal line: which is the universal cover of the AdS space.

For the cosmological solutions of the Einstein-Friedmann equations the RW-metricis only affected as far as S (t) solves a different dynamical equation.

Exercise: find the model Einstein proposed: an eternal static solution. Show thatEinstein’s GRT with vanishing cosmological constant has no such solution.

Exercise: the observation of planetary motions constrains the cosmologicalconstant to |Λ| 10−42 cm−2 !. Show that this is compatible with a dark energydensity, which has been determined to be ΩΛ ∼ 0.74 ± 0.03.

Previous ≪x , next x≫ lecture.






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