Cosmological solutions of the Einstein-Friedmann equations ...

Cosmology 

Cosmological solutions of the Einstein-Friedmann equations 

Summary of Friedmann’s equations 

Ingredients are the Einstein equations of GRT: 

Rµν − 1 

2 R gµν − Λ gµν = κ Tµν , 

c○ 2009, F. Jegerlehner ≪❘ Lect. 7 ❘≫ 439

Cosmology 




Rµν − 1 


with the cosmic ideal fluid energy-momentum tensor 

T µν = (ρ + p) (t) u µ u ν − p(t) g µν . 


Cosmology 




Rµν − 1 




The cosmological constant term Λ gµν nowadays is usually included as part of Tµν 

as an additional contribution to the energy density 


Cosmology 




Rµν − 1 




The cosmological constant term Λ gµν nowadays is usually included as part of Tµν 

as an additional contribution to the energy density 

ρ(t) → ρ(t) + ρΛ ; ρΛ = Λ 

κ , 


Cosmology 

claimed to represent the vacuum energy density. 


Cosmology 

claimed to represent the vacuum energy density. The ideal fluid tensor structure 

then requires the vacuum energy, now usually called dark energy to satisfy the 

exotic equation of state 

pΛ = −ρΛ . 


Cosmology 




pΛ = −ρΛ . 

In the following we discuss the resulting set of Friedmann equations 


Cosmology 




pΛ = −ρΛ . 


˙S 2 

c2 = κ 

3 ρ S 2 − k Friedmann equation 

 

d 3 

dS ρ S = −3 p S 2 energy equation 

p = p(ρ) equation of state 


Cosmology 




pΛ = −ρΛ . 


˙S 2 

c2 = κ 


 

d 3 



with ρ and p now assumed to include the cosmological constant term. 


Cosmology 




pΛ = −ρΛ . 


˙S 2 

c2 = κ 


 

d 3 



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 need 

3 equations to uniquely determine them in terms of some initial values. 


Cosmology 

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


Cosmology 


In spite of the fact that such models are unrealistic, they give some insight into the 

geometrical and kinematic possibilities in cosmological model building. In this 

limiting case we have the following basic equations: 


Cosmology 





˙S 2 

c 2 = Λ 

3 S 2 − k ; 

¨S 

c 2 = Λ 

3 S 


Cosmology 





or 

˙S 2 

c 2 = Λ 

3 S 2 − k ; 

¨S 

c 2 = Λ 

3 S 

Λ = −3 H2 (t) 

c 2 q(t) ; K(t) = − H2 (t) 

c 2 (q(t) + 1) . 


Cosmology 

The general solution is 

S (t) = a ct + b ; Λ = 0 , a 2 = −k , 

S (t) = 1 

2 

a e αct + b e −αct ; Λ 0 , α 2 = Λ 

3 

; Λ 

3 

ab = k = (1, 0, −1) . 


Cosmology 


S (t) = a ct + b ; Λ = 0 , a 2 = −k , 

S (t) = 1 

2 


3 

Because of time reversal symmetry or anti-symmetry: 

when b 0: b = ±a with a > 0. 

; Λ 

3 

ab = k = (1, 0, −1) . 


Cosmology 


S (t) = a ct + b ; Λ = 0 , a 2 = −k , 

S (t) = 1 

2 


3 



Cases: 

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

; Λ 

3 

ab = k = (1, 0, −1) . 


Cosmology 


S (t) = a ct + b ; Λ = 0 , a 2 = −k , 

S (t) = 1 

2 


3 



Cases: 


; Λ 

3 

(b) Λ = 0 , k = −1: S = ct ; H = 1/t , q = 0 ; z = t0/t1 − 1 , 

Milne’s Universe 

ab = k = (1, 0, −1) . 


Cosmology 


S (t) = a ct + b ; Λ = 0 , a 2 = −k , 

S (t) = 1 

2 


3 



Cases: 


; Λ 

3 

(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 

ct 

a . 

ab = k = (1, 0, −1) . 


Cosmology 

(d) Λ > 0 , k = 0: S = exp 

ct 

c 

a = exp (Ht) ; H ≡ a = κ ρvac c2 /3 

de Sitter Universe: here H(t) = constant and q(t) ≡ −1, this universe has infinite 

age! 


Cosmology 

(d) Λ > 0 , k = 0: S = exp 

ct 

c 



age! Today’s pure dark energy universe! 

(e) Λ > 0 , k = −1: S = a sinh 

ct 

a . 


Cosmology 

(d) Λ > 0 , k = 0: S = exp 

ct 

c 




(e) Λ > 0 , k = −1: S = a sinh 

ct 

a . 

(f) Λ < 0 , k = −1: S = a sin 

ct 

a . 


Cosmology 

(d) Λ > 0 , k = 0: S = exp 

ct 

c 




(e) Λ > 0 , k = −1: S = a sinh 

ct 

a . 

(f) Λ < 0 , k = −1: S = a sin 

ct 

a . 

where in all cases 


Cosmology 

(d) Λ > 0 , k = 0: S = exp 

ct 

c 




(e) Λ > 0 , k = −1: S = a sinh 

ct 

a . 

(f) Λ < 0 , k = −1: S = a sin 

ct 

a . 


a = |3/Λ| . 


Cosmology 

(d) Λ > 0 , k = 0: S = exp 

ct 

c 




(e) Λ > 0 , k = −1: S = a sinh 

ct 

a . 

(f) Λ < 0 , k = −1: S = a sin 

ct 

a . 


a = |3/Λ| . 

Models (b), (d), (e) and (f) some times have been called “incomplete” as they 

exhibit more space than “substrate”. 


Cosmology 

Terminology: Λ > 0 , k = 0, ±1 (ρ = p = 0) de Sitter spaces 


Cosmology 


Λ < 0 , k = −1 (ρ = p = 0) Anti-de Sitter spaces 


Cosmology 



Isometric embedding: ds 2 = dz 0 2 

− dz 1 2 

− dz 2 2 

− dz 3 2 

− k dz 4 2 


Cosmology 




− dz 1 2 

− dz 2 2 

− dz 3 2 

− k dz 4 2 


Cosmology 




− dz 1 2 

− dz 2 2 

− dz 3 2 

− k dz 4 2 


Cosmology 




S − 4 

✻ 

z 0 

− dz 1 2 

 

 

 

 

 

 

 

 

 

✲ 

✒ 

z3 de Sitter 

z 4 

S + 4 

− dz 2 2 

− dz 3 2 

z 0 

✻ 

− k dz 4 2 

 

 

 

 

 

 

 

 

 

 

✒ 

Anti de Sitter 

c○ 2009, F. Jegerlehner ≪❘ Lect. 7 ❘≫ 444 

z 4 

z 3 

✲

Cosmology 

B) Matter dominated: ρ > 0 , p = 0 


Cosmology 


Λ = 0: preferable class of models 

satisfies Einstein’s boundary criterion: empty space = flat space 

Base equations: ρ S 3 = ρ0 S 3 

0 

= constant 

˙S 2 

c2 = κ 

3 ρ S 2 − k = C 

S 

− k ; C κ 

3 ρ S 3 = constant 

which can be integrated elementary: dS/ √ C/S − k = ±c dt 


Cosmology 





0 

= constant 

˙S 2 

c2 = κ 

3 ρ S 2 − k = C 

S 

− k ; C κ 



❖ k = 0 : Einstein-de Sitter universe 

S (t) = 9 

4 C 1/3 

(ct) 2/3 


Cosmology 





0 

= constant 

˙S 2 

c2 = κ 

3 ρ S 2 − k = C 

S 

− k ; C κ 



❖ k = 0 : Einstein-de Sitter universe 

S (t) = 9 

4 C 1/3 

(ct) 2/3 

❖ k = 1 : Friedmann-Einstein universe 


Cosmology 

ct = C arcsin √ X − √ X − X 2 ; X = S (t)/C 


Cosmology 


Parameter representation: 


Cosmology 



S (t) = 1 

2 C (1 − cos η) 

1 ct = 2 C (η − sin η) 

 

cycloid 


Cosmology 



❖ k = −1 : 

S (t) = 1 

2 C (1 − cos η) 

1 ct = 2 C (η − sin η) 

ct = C √ X + X 2 − arcsinh √ X ; X = S (t)/C 

 

cycloid 


Cosmology 



❖ k = −1 : 

S (t) = 1 

2 C (1 − cos η) 

1 ct = 2 C (η − sin η) 



 

cycloid 


Cosmology 



❖ k = −1 : 

S (t) = 1 

2 C (1 − cos η) 

1 ct = 2 C (η − sin η) 



 

S (t) = 1 

2 C (cosh η − 1) 

1 ct = 2 C (sinh η − η) 

cycloid 


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 


Cosmology 



S (t) 9 

4 C 1/3 

(ct) 2/3 ; t → 0 


Cosmology 



S (t) 9 

4 C 1/3 

For the observables we have the relationships 

(ct) 2/3 ; t → 0 


Cosmology 



S (t) 9 

4 C 1/3 


κ 

2 ρ(t) = 3 H2 (t) 

K(t) = 

(ct) 2/3 ; t → 0 

c 2 q(t) ✄ q(t) > 0! 

H 2 (t) 

c 2 (2 q(t) − 1) ✄ q(t) = 

⎧ 

⎪⎨ 

⎪⎩ 

> 1 

2 

= 1 

2 

< 1 

2 

; k = 1 

; k = 0 

; k = −1 


Cosmology 



S (t) 9 

4 C 1/3 


κ 

2 ρ(t) = 3 H2 (t) 

K(t) = 

(ct) 2/3 ; t → 0 

c 2 q(t) ✄ q(t) > 0! 

H 2 (t) 

c 2 (2 q(t) − 1) ✄ q(t) = 

⎧ 

⎪⎨ 

⎪⎩ 

> 1 

2 

= 1 

2 

< 1 

2 

; k = 1 

; k = 0 

; k = −1 

The following graphical representation is shown for C = 1 and c = 1, i.e, S (t) as 

well as ct are represented in units of C. 


Cosmology 

0.16 

0.21 

0.25 

0.30 

1.0 

2.0 

7.5 

0.5 

∞ 

↑ 

1 

2 ctc 

q = 0.10 

↑ 

ctc = π 


Cosmology 

Einstein-de Sitter universe 

The flat case k = 0 is of particular interest as a critical universe in between positive 

and negative curvature. In this case many things are fixed: 


Cosmology 




First 

K(t) ≡ 0 ✄ qEdS(t) ≡ 1 

2 and ρEdS(t) = 3 

κ 

H 2 (t) 

c 2 


.

Cosmology 




First 

K(t) ≡ 0 ✄ qEdS(t) ≡ 1 

2 and ρEdS(t) = 3 

κ 

In addition, using S 3 = 9 

4 C(ct)2 , we obtain 

C = κ 

3 ρ S 3 = κ 

3 

3 H 

· 

κ 

2 (t) 

c2 where C cancels and we get 1 = 9 

4 H2 (t) (t) 2 or 

· S 3 = 9 

4 C H2 (t) 

c 2 

H 2 (t) 

c 2 

(ct) 2 


.

Cosmology 

t = 2 

3 H−1 (t) (EdS) 


Cosmology 

t = 2 

3 H−1 (t) (EdS) 

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


Cosmology 

t = 2 

3 H−1 (t) (EdS) 


Present values: 

H0 0.83 × 10 −28 cm −1 × c 

ρ0 EdS 1.2 × 10 −29 gr/cm 3 × c 2 

t0 EdS = 2 

3 H−1 

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


Cosmology 

t = 2 

3 H−1 (t) (EdS) 


Present values: 

H0 0.83 × 10 −28 cm −1 × c 

ρ0 EdS 1.2 × 10 −29 gr/cm 3 × c 2 

t0 EdS = 2 

3 H−1 

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

where we used 

κ = 8πGN 

c 2 

= 1.86637 × 10 −27 cm/gr . 


Cosmology 

S(t) 

t0 

tH = H −1 

0 

Einstein-de Sitter universe: a typical expansion pattern (for closed universes only 

for t ≪ trecontraction) 


t

Cosmology 

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


Cosmology 

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

C = κ 

3 ρ(t) S (t)3 = 2 H2 (t) 

q(t) S (t)3 

c2 c○ 2009, F. Jegerlehner ≪❘ Lect. 7 ❘≫ 453

Cosmology 


thus, with 

C = κ 

3 ρ(t) S (t)3 = 2 H2 (t) 

q(t) S (t)3 

c2 k S (t) −2 = H2 (t) 

c 2 

(2 q(t) − 1) , 


Cosmology 


thus, with 

we may write 

C = κ 

3 ρ(t) S (t)3 = 2 H2 (t) 

q(t) S (t)3 

c2 k S (t) −2 = H2 (t) 

c 2 

S (t) = c 

H(t) 

(2 q(t) − 1) , 

√ 1 

|2 q(t)−1| 


Cosmology 

in terms of the observables H(t) and q(t). 


Cosmology 

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

C = 2 c 

H(t) 

q(t) 

; q(t) ≥ 0 ! 

3/2 

(|2 q(t) − 1|) 


Cosmology 


Also, with 

˙S 2 

C 

= 

c2 S 

C = 2 c 

H(t) 

− k , 

q(t) 

; q(t) ≥ 0 ! 

3/2 

(|2 q(t) − 1|) 

¨S C 

= − 

c2 2S 2 ; q = − ¨S S 1 

= 

˙S 2 2 

C 

C − kS 


Cosmology 


Also, with 

we obtain 

˙S 2 

C 

= 

c2 S 

C = 2 c 

H(t) 

− k , 

q(t) 

; q(t) ≥ 0 ! 

3/2 

(|2 q(t) − 1|) 

¨S C 

= − 

c2 2S 2 ; q = − ¨S S 1 

= 

˙S 2 2 

C 

C − kS 


Cosmology 


Also, with 

we obtain 7 

˙S 2 

C 

= 

c2 S 

C = 2 c 

H(t) 

− k , 

q(t) 

; q(t) ≥ 0 ! 

3/2 

(|2 q(t) − 1|) 

¨S C 

= − 

c2 2S 2 ; q = − ¨S S 1 

= 

˙S 2 2 

C 

C − kS 


Cosmology 


Also, with 

we obtain 7 

˙S 2 

C 

= 

c2 S 

C = 2 c 

H(t) 

− k , 

q(t) 

; q(t) ≥ 0 ! 

3/2 

(|2 q(t) − 1|) 

¨S C 

= − 

c2 2S 2 ; q = − ¨S S 1 

= 

˙S 2 2 

k S (t) = C 

2 q(t) − 1 

2 q(t) 

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

, 

C 

C − kS 


Cosmology 


Also, with 

we obtain 7 

˙S 2 

C 

= 

c2 S 

C = 2 c 

H(t) 

− k , 

q(t) 

; q(t) ≥ 0 ! 

3/2 

(|2 q(t) − 1|) 

¨S C 

= − 

c2 2S 2 ; q = − ¨S S 1 

= 

˙S 2 2 

k S (t) = C 

2 q(t) − 1 

2 q(t) 

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

7 We also may write 

ct k=1 

= C 

 

arcsin 

 

1 − 2q −1 − 

√ 

2q−1 

2q 

; ct k=−1 

√ 

1−2q 

= C 2q − arcsinh 

, 

 

2q−1 

− 1 

C 

C − kS 


Cosmology 

The conclusion of the above discussion: provided Λ = 0 and p = 0, we may 

determine the age and the curvature of the universe by evaluating the observable 

relations 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 the 

critical density. 


Cosmology 





critical density. Generally, 

such that 

ρ(t) = ρEdS(t) · 2 q(t) 


Cosmology 






such that 

ρ(t) = ρEdS(t) · 2 q(t) 

ρ0 > ρ0 EdS ✄ k = 1 

ρ0 < ρ0 EdS ✄ k = −1 


Cosmology 






ρ(t) = ρEdS(t) · 2 q(t) 

such that 

ρ0 > ρ0 EdS ✄ k = 1 

Thus: 

ρ0 < ρ0 EdS ✄ k = −1 

● Lot of matter – space closes under gravity, gravity wins. 


Cosmology 






ρ(t) = ρEdS(t) · 2 q(t) 

such that 

ρ0 > ρ0 EdS ✄ k = 1 

Thus: 

ρ0 < ρ0 EdS ✄ k = −1 


● Little matter – space is open and matter spreads forever. 


Cosmology 






ρ(t) = ρEdS(t) · 2 q(t) 

such that 

ρ0 > ρ0 EdS ✄ k = 1 

Thus: 

ρ0 < ρ0 EdS ✄ k = −1 


● 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 by 

the normal baryonic matter which stars, planets, dust and interstellar gas are 

made of. There are many indications that baryonic matter is only a fraction of the 

total gravitating matter, which is mainly Dark Matter (DM). Dark matter, which only 

manifests to us by gravitational interaction, is seen in galaxies (velocity profiles), in 

relative motion of the objects in clusters of galaxies (Fritz Zwicky 1933, applying 

the virial theorem to Coma Cluster) as well as in the universe as a whole (CMB 

fluctuations). A very important “tool” in dark matter search is gravitational lensing, 

which convincingly supports the other findings, and in addition provides important 

information concerning the distribution of DM. We will discussed this in detail later, 

together with Observational findings. 


Cosmology 

C) Radiation dominated: p = ρ/3 


Cosmology 


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


Cosmology 






Cosmology 





Base equations: ρ S 4 = ρ0 S 4 0 

= constant 

˙S 2 

c 2 = κ 

3 ρ S 2 − k = ¯C 

S 2 − k ; ¯C κ 


which can be integrated elementary. 


Cosmology 





Base equations: ρ S 4 = ρ0 S 4 0 

= constant 

˙S 2 

c 2 = κ 

3 ρ S 2 − k = ¯C 

S 2 − k ; ¯C κ 


which can be integrated elementary. We actually have 

√ 

S dS = ±cdt 

¯C−kS 2 


Cosmology 

❖ k = 0 : 

S (t) = 4 ¯C 1/4 

(ct) 1/2 


Cosmology 

❖ k = 0 : 

S (t) = 4 ¯C 1/4 

❖ k = 1 : 

S (t) = 

(ct) 1/2 

¯C − √ ¯C − ct 2 1/2 

in 0 < ct < 2 √ ¯C with cyclic extension ! 


Cosmology 

❖ k = 0 : 

S (t) = 4 ¯C 1/4 

❖ k = 1 : 

S (t) = 

(ct) 1/2 

¯C − √ ¯C − ct 2 1/2 

in 0 < ct < 2 √ ¯C with cyclic extension ! 

❖ k = −1 : 

S (t) = 

√ ¯C + ct 2 

1/2 − ¯C 


Cosmology 

In each case we have S (t) → 0 (t → 0) i.e., all of them are Big-Bang 

cosmologies. 


Cosmology 


cosmologies. As in the matter dominated scenario, for small t the ¯C term is 

dominating, such that, independently of k, 

S (t) 4 ¯C 1/4 

(ct) 1/2 ; t → 0 


Cosmology 




S (t) 4 ¯C 1/4 

The behavior for large t is easily obtained: 

k=0 : S (t) = 4 ¯C 1/4 (ct) 1/2 

k=-1 

k=1 

: 

: 

S (t) ct 

cyclic, cycle time ctc = 2 √ ¯C 

(ct) 1/2 ; t → 0 


Cosmology 




S (t) 4 ¯C 1/4 


k=0 : S (t) = 4 ¯C 1/4 (ct) 1/2 

k=-1 

k=1 

: 

: 

S (t) ct 


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

c . 

(ct) 1/2 ; t → 0 


Cosmology 




S (t) 4 ¯C 1/4 


k=0 : S (t) = 4 ¯C 1/4 (ct) 1/2 

k=-1 

k=1 

: 

: 

S (t) ct 


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

c . 


(ct) 1/2 ; t → 0 


Cosmology 

κ ρ(t) = 3 H2 (t) 

K(t) = 

c 2 q(t) ✄ q(t) ≥ 0! 

H 2 (t) 

c 2 (q(t) − 1) ✄ q(t) = 

⎧ 

⎪⎨ 

⎪⎩ 

> 1 ; k = 1 

= 1 ; k = 0 

< 1 ; k = −1 


Cosmology 

κ ρ(t) = 3 H2 (t) 

K(t) = 

c 2 q(t) ✄ q(t) ≥ 0! 

H 2 (t) 

c 2 (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) as 

well as ct are represented in units of √ ¯C. 


Cosmology 

0.5 

0.3 

0.2 

3.0 

1.0 

ր 6.0 ∞ 

↑ 

1 

2 ctc 

q = 0.10 

↑ 

ctc = 2 √ C 


Cosmology 

Actually, in the radiation dominated era the situation is qualitatively very similar to 

the matter dominated p = 0 scenario. 


Cosmology 



Exercise: Express q(t) in terms of S and C (matter dominated) or ¯C (radiation 

dominated) scenario, respectively. 


Cosmology 





For radiation in thermal equilibrium (black body radiation) the Stefan-Boltzmann 

law holds: 

ρradiation = a T 4 c 2 ; a = π2 k 4 B 

15 3 c 2 = 8.418 × 10 −36 gr/cm 3 ◦ K −4 


Cosmology 





For radiation in thermal equilibrium (black body radiation) the Stefan-Boltzmann 

law holds: 

ρradiation = a T 4 c 2 ; a = π2 k 4 B 

15 3 c 2 = 8.418 × 10 −36 gr/cm 3 ◦ K −4 

and the spectral distribution is given by the Planck-distribution 


Cosmology 

8π hν ργ(ν) dν = 3 dν 

exp 

hν 

k B Tγ 


 

−1

Cosmology 


which characterizes the perfect radiator. 

exp 

hν 

k B Tγ 


 

−1

Cosmology 


which characterizes the perfect radiator. 

exp 

hν 

k B Tγ 


 

−1

Cosmology 

Since ρ(t)S (t) 4 = ρ0S 4 0 

= constant we have 

ρ(t) ∝ T 4 (t) ∝ 1 

S (t) 4 


Cosmology 

Since ρ(t)S (t) 4 = ρ0S 4 0 

implying 


ρ(t) ∝ T 4 (t) ∝ 1 

S (t) 4 

T(t) = T0 S 0 

S (t) 


Cosmology 

Since ρ(t)S (t) 4 = ρ0S 4 0 

implying 


ρ(t) ∝ T 4 (t) ∝ 1 

S (t) 4 

T(t) = T0 S 0 

S (t) 

In fact, the radiation cools down by the expansion so much that it decouples from 

matter. This was predicted by Gamow and Alpher and Herman in 1948, 


Cosmology 


Cosmology 

and Dicke and Peebles at Princeton were searching for it when it was actually 

discovered “by accident” in 1965 by Penzias and Wilson at Bell Laboratories. 


Cosmology 

and Dicke and Peebles at Princeton were searching for it when it was actually 

discovered “by accident” in 1965 by Penzias and Wilson at Bell Laboratories. For 

more history read:✲≫✲≫ 


Cosmology 

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

temperature 


Cosmology 

T0 = (2.725 ± 0.002) ◦ K 


Cosmology 

T0 = (2.725 ± 0.002) ◦ K 

not only clearly favored Big-Bang cosmologies and essentially ruled out 

steady-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. 


Cosmology 

T0 = (2.725 ± 0.002) ◦ K 




What does it mean? First we have 

ρ0γ = a T 4 0 c2 ✄ ρ0γ 4.64 × 10 −34 gr/cm 3 × c 2 , 


Cosmology 

T0 = (2.725 ± 0.002) ◦ K 




What does it mean? First we have 

ρ0γ = a T 4 0 c2 ✄ ρ0γ 4.64 × 10 −34 gr/cm 3 × c 2 , n0γ 410 photons/cm 3 

which confronts with the present baryonic matter density 

ρ0,mat 3 × 10 −31 gr/cm 3 × c 2 . 


Cosmology 

However, the evolution of the two densities are different 


Cosmology 


ρrad(t) = ρ0,rad 

4 S 0 

S (t) 

whereas ρmat = ρ0,mat 

3 S 0 

S (t) 


,

Cosmology 


ρrad(t) = ρ0,rad 

4 S 0 

S (t) 

whereas ρmat = ρ0,mat 

3 S 0 

S (t) 

and since S (t) → 0 (t → 0) in a Big-Bang cosmology radiation always dominates 

in the early universe! 


,

Cosmology 

ρ(t) 

Big Bang 

radiation 

era 

ρrad(t) 

ρrad = ρmat 

matter 

era 

ρmat(t) 

curvature ? 


present 

t

Cosmology 

ρ(t) 

Big Bang 

radiation 

era 

ρrad(t) 

ρrad = ρmat 

matter 

era 

ρmat(t) 

curvature ? 

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


present 

t

Cosmology 

ρ(t) 

Big Bang 

radiation 

era 

ρrad(t) 

ρrad = ρmat 

matter 

era 

ρmat(t) 

curvature ? 

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 

discuss later. 


present 

t

Cosmology 

The critical energy density and the flatness problem 


Cosmology 


In general we have some mixture of vacuum energy, relativistic and non-relativistic 

matter. We have seen that the Einstein-de Sitter universe with k=0 plays a special 

role. In the present matter dominated era ρEdS plays the boundary case between 

expansion forever or recontraction. 


Cosmology 





expansion forever or recontraction. We therefore define 

ρ0,crit = ρEdS = 3H2 0 

8πGN = 1.878 × 10−29 h 2 gr/cm 3 , 


Cosmology 







8πGN = 1.878 × 10−29 h 2 gr/cm 3 , 

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. 


Cosmology 







8πGN = 1.878 × 10−29 h 2 gr/cm 3 , 


Mpc −1 , and express the energy density in units of ρ0,crit. Thus the present density 

ρ0 is represented by 

Ω0 = ρ0/ρ0,crit 


Cosmology 







8πGN = 1.878 × 10−29 h 2 gr/cm 3 , 


Mpc −1 , and express the energy density in units of ρ0,crit. Thus the present density 

ρ0 is represented by 

and 

Ω0 = ρ0/ρ0,crit 


Cosmology 

Ω0 = 1 


Cosmology 

Ω0 = 1 

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


Cosmology 

Ω0 = 1 


Ω0 > 1 


Cosmology 

Ω0 = 1 


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 heat 

death! 


Cosmology 

Ω0 = 1 


Ω0 > 1 



death! 

Ω0 < 1 


Cosmology 

Ω0 = 1 


Ω0 > 1 



death! 

Ω0 < 1 

the case of little matter, gravity not sufficient to stop expansion, universe expands 

forever, space open, ends in freezing to death! 


Cosmology 

Ω0 = 1 


Ω0 > 1 



death! 

Ω0 < 1 

the case of little matter, gravity not sufficient to stop expansion, universe expands 

forever, space open, ends in freezing to death! 

If we include the cosmological constant as a vacuum energy density in the total 

density ρ, we may write the Friedmann equations in the form 


Cosmology 

˙S 2 

c 2 + k = κ 

3 ρ S 2 and 3 ¨S 

c 2 S 

= −κ 

2 

(3 p + ρ) 


Cosmology 

˙S 2 

c 2 + k = κ 

3 ρ S 2 and 3 ¨S 

c 2 S 

and the energy-momentum conservation as 

= −κ 

2 

(3 p + ρ) 


Cosmology 

˙S 2 

c 2 + k = κ 

3 ρ S 2 and 3 ¨S 

c 2 S 


˙ρ = −3 ˙S 

S 

(ρ + p) . 

= −κ 

2 

(3 p + ρ) 


Cosmology 

˙S 2 

c 2 + k = κ 

3 ρ S 2 and 3 ¨S 

c 2 S 


˙ρ = −3 ˙S 

S 

(ρ + p) . 

= −κ 

2 

(3 p + ρ) 

If the mixture of radiation and matter dominates over the vacuum energy 3p + ρ is 

always positive and thus we have ¨S /S ≤ 0. This means that the expansion must 

have started with S = 0 in the past and the present age must be lower than the 

Hubble age: t0 < H−1 0 (see figure in EdS case above). For k = 1 this also implies 

the recontraction to S = 0. 


Cosmology 

˙S 2 

c 2 + k = κ 

3 ρ S 2 and 3 ¨S 

c 2 S 


˙ρ = −3 ˙S 

S 

(ρ + p) . 

= −κ 

2 

(3 p + ρ) 

If the mixture of radiation and matter dominates over the vacuum energy 3p + ρ is 

always positive and thus we have ¨S /S ≤ 0. This means that the expansion must 

have started with S = 0 in the past and the present age must be lower than the 

Hubble age: t0 < H−1 0 (see figure in EdS case above). For k = 1 this also implies 

the recontraction to S = 0. 

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

0 can be written as 


Cosmology 

q0 = κ (ρ0+3 p0) 

6 H 2 0 

= ρ0+3 p0 

2 ρ 0,crit 


Cosmology 

q0 = κ (ρ0+3 p0) 

6 H 2 0 

= ρ0+3 p0 

2 ρ 0,crit 

This equation provides a simple explanation for the different values q0 takes 

depending 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 = 1 

2 , 

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

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


i.e.

Cosmology 




2 Ωmat q0 = 1 

2 , 



Whatever the mix of energies the total energy density 


i.e.

Cosmology 




2 Ωmat q0 = 1 

2 , 




ρ(t) = ρ0,crit 

⎧ 

⎪⎨ 

⎪⎩ ΩΛ + Ω0,mat 

3 S 0 

S (t) 

+ Ω0,rad 

⎫ 

4 

S 0 ⎪⎬ 

S (t) ⎪⎭ 


i.e.

Cosmology 




2 Ωmat q0 = 1 

2 , 





⎧ 

⎪⎨ 


3 S 0 

S (t) 

+ Ω0,rad 

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

⎫ 

4 

S 0 ⎪⎬ 

S (t) ⎪⎭ 


i.e.

Cosmology 




2 Ωmat q0 = 1 

2 , 





⎧ 

⎪⎨ 


3 S 0 

S (t) 

+ Ω0,rad 

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

ρtot ρ0,crit Ω0,rad 

4 S 0 

S (t) 

⎫ 

4 

S 0 ⎪⎬ 

S (t) ⎪⎭ 


, 

i.e.

Cosmology 

in which case, independent of k! 


Cosmology 


S (t) 4 ¯C 1/4 

(ct) 1/2 ; ¯C = κ 

3 ρ0,rad S 4 0 i.e. 

S 4 0 

S 4 (t) = 

3 

4 κ (ct) 2 ρ0,rad 


,

Cosmology 


such that 

S (t) 4 ¯C 1/4 

is universal. 

(ct) 1/2 ; ¯C = κ 

3 ρ0,rad S 4 0 i.e. 

ρtot(t) 3 

4 κ (ct) 2 

S 4 0 

S 4 (t) = 

3 

4 κ (ct) 2 ρ0,rad 


,

Cosmology 


such that 

S (t) 4 ¯C 1/4 

is universal. 

(ct) 1/2 ; ¯C = κ 

3 ρ0,rad S 4 0 i.e. 

ρtot(t) 3 

4 κ (ct) 2 

S 4 0 

S 4 (t) = 

Now, in the radiation dominated early phase after the Big Bang 

3 

4 κ (ct) 2 ρ0,rad 


,

Cosmology 


such that 

S (t) 4 ¯C 1/4 

is universal. 

(ct) 1/2 ; ¯C = κ 

3 ρ0,rad S 4 0 i.e. 

ρtot(t) 3 

4 κ (ct) 2 

S 4 0 

S 4 (t) = 


3 

4 κ (ct) 2 ρ0,rad 

˙S 2 

c 2 = κ 

3 ρ S 2 − k = ¯C 

S 2 − k ; ¯C κ 



,

Cosmology 


such that 

S (t) 4 ¯C 1/4 

is universal. 

(ct) 1/2 ; ¯C = κ 

3 ρ0,rad S 4 0 i.e. 

ρtot(t) 3 

4 κ (ct) 2 

S 4 0 

S 4 (t) = 


3 

4 κ (ct) 2 ρ0,rad 

˙S 2 

c 2 = κ 

3 ρ S 2 − k = ¯C 

S 2 − k ; ¯C κ 


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


,

Cosmology 

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

In fact the observed energy density at present is close to the critical one. This we 

will discuss later. Even if one counts the normal baryonic matter only one is not 

much more than a factor of 10 off from the critical value Ω = 1 (atoms ∼ 4%) . The 


Cosmology 


˙S 2 

¯C 

→ 

c2 S 

2 or 

˙S 2 

S 2 → c2 ¯C 

S 4 = c2 κρ 

3 





Cosmology 


i.e. 

˙S 2 

¯C 

→ 

c2 S 

2 or 

˙S 2 

S 2 → c2 ¯C 

S 4 = c2 κρ 

3 

ρ(t) = ρEdS(t) = 3 

κ 

H 2 (t) 

c 2 . 





Cosmology 


i.e. 

˙S 2 

¯C 

→ 

c2 S 

2 or 

˙S 2 

S 2 → c2 ¯C 

S 4 = c2 κρ 

3 

ρ(t) = ρEdS(t) = 3 

κ 

H 2 (t) 

c 2 . 

It is truly remarkable that at that early times, precisely when we would expect 

curvature the be most important, the evolution automatically picks the flat solution. 

This does not mean, however, that at later times when the other energy density 

components come in to play and even dominate the scene we have to expect a flat 

space one. 





Cosmology 

missing part making Ω = 1 complete is dark matter (cosmological constant 73%) 

and cold dark matter (what is it ? 23%). 


Cosmology 


and cold dark matter (what is it ? 23%). The strongly time and scenario 

dependent energy density ρ(t) easily deviates by 60 orders of magnitude from 

ρ0,crit today, given the evolution during the enormous time span of the present age 

of the universe. In other words, the flat solution Ω = 1 is highly unstable. How is it 

possible then that we apparently live in a universe close to that unstable point? 


Cosmology 







This is the so called flatness problem (Dicke 1969). The solution: either 

cosmological fine-tuning assuming we really started with an accuracy of 62 or 

more digits with Ω = 1 or some dynamical mechanism transmutes this unstable 

point into an stable attractor, this is what inflation theory, to be discussed later, can 

do for us. 


Cosmology 







This is the so called flatness problem (Dicke 1969). The solution: either 

cosmological fine-tuning assuming we really started with an accuracy of 62 or 

more digits with Ω = 1 or some dynamical mechanism transmutes this unstable 

point into an stable attractor, this is what inflation theory, to be discussed later, can 

do for us. 

Exercise: confirm the necessary fine tuning in the cosmic evolution by explicit 

calculation. 


Cosmology 


Cosmology 

Appendix: Λ 0 

The issue about the true value of the cosmological constant, which likely is 

non-vanishing as mentioned earlier, will be reconsidered later. In any case we 

have to discuss the solution of the cosmological equations for the case on 

non-zero cosmological constant. The consequences for the empty world case we 

already discussed: flat “background space” gets replaced by de Sitter or anti-de 

Sitter space. The generalization to the matter dominated scenario is almost trivial. 

A glimpse at the Friedmann equation shows that one can treat the cosmological 

constant as a contribution to the energy density: 

κρmat + Λ → κρtot, 

where ρtot = ρtot + ρΛ with ρΛ = Λ 

κ . Provided p = 0, all solutions remain unchanged, 

except for a different interpretation of the energy density, which in this case is not 

identical with the “normal” mass density of baryonic plus dark matter. Note that 

the cosmological constant in general enters the equation of state in a not a priori 


Cosmology 

known way. However, this seems not to matter as long as we have p 0. 

That a non-vanishing Λ spoils the geometry ⇔ matter duality, because the 

Einstein equation remains true no matter on which side of the equation we write 

the cosmological term. Many believe it has some thing to do with quantum 

vacuum fluctuations, but no answer can be given why it is so small. In any case if 

we take the energy momentum tensor as given on the quantum level by the 

Standard Model (SM) of strong and electroweak interactions we would expect the 

Higgs field vacuum expectation value (Bose condensate) as well as the quark 

condensates to contribute to the cosmological constant. Again, the value obtained 

is about 50 orders of magnitude to big! What tames the cosmological constant to 

a small non-vanishing value? 

Exercise: estimate the cosmological constant as induced by the Higgs mechanism 

and by spontaneous symmetry breaking of chiral symmetry in QCD. 

A cosmological constant of course affects local gravitation in particular it might 

affect black holes by changing the Schwarzschild radius etc. In order to study the 


Cosmology 

physical effect of a non-vanishing Λ we consider the spherically symmetric mass 

distribution in outer space: 

➊ Λ = 0 : Schwarzschild 

ds 2 = 

Gµν ≡ 0 ✄ Gµν = Λ gµν 

 

1 − r0 

 

r 

(cdt) 2 − 1 

1 − r0 

r 

dr 2 − r 2 dΩ 2 

which is unique, static with boundary condition: flat as r → ∞. It has the Newto- 

nian approximation with potential: ϕ − m 

r 

with m r0 

2 and where g00 1 + 2ϕ 

c 2 


Cosmology 

➋ Λ 0 : 

α = 1 − r0 

r 

→ 1 − r0 

r 

ds 2 = 

 

1 − r0 

r 

Λ − 3 R2 . Denoting K = Λ 

3 

− Kr2 

 

(cdt) 2 − 

1 

1 − r0 

r − Kr2 dr2 − r 2 dΩ 2 

which is unique, static with boundary condition: Schwarzschild as Λ → 0. In the 

Newtonian approximation: ϕ −m 1 

r − 2kr2 Observation of planetary motions yields: |Λ| ≪ 10−42 cm−2 ! 

In particular: in empty space (no matter) m = r0 

2 

Λ > 0. The gravity potential is 

ϕ − 1 

2 Kr2 repulsive linear force F ∝ Kr 

= 0 we have de Sitter space if 


Cosmology 

and the metric 

ds 2 = 1 − Kr 2 (cdt) 2 − 

1 

1 − Kr 2 dr2 − r 2 dΩ 2 

which is spherical symmetric with respect to any point and is regular. For K > 0 

there is the de Sitter coordinate singularity K r 2 = 1. While for m 0 we have a 

true singularity at r = 0, such a singularity is absent for m = 0. 

Mappings 


Cosmology 


ds 2 = 1 − Kr 2 (cdt) 2 − 

1 

1 − Kr 2 dr2 − r 2 dΩ 2 




Mappings 

Λ > 0: de Sitter space dS 4 


Cosmology 


ds 2 = 1 − Kr 2 (cdt) 2 − 

1 

1 − Kr 2 dr2 − r 2 dΩ 2 




Mappings 


embedding into M 5 = R 1,4 : (cT) 2 − X 2 − Y 2 − Z 2 − W 2 = −a 2 


Cosmology 


ds 2 = 1 − Kr 2 (cdt) 2 − 

1 

1 − Kr 2 dr2 − r 2 dΩ 2 




Mappings 


embedding into M 5 = R 1,4 : (cT) 2 − X 2 − Y 2 − Z 2 − W 2 = −a 2 

ds 2 = (cdT) 2 − dX 2 + dY 2 + dZ 2 + dW 2 


Cosmology 

With a 2 1 

K 

= 3 

Λ 

; a > 0 the mapping reads: 


Cosmology 

With a 2 1 

K 

= 3 

Λ 


X = r sin θ sin ϕ 

Y = r sin θ cos ϕ 

Z = r cos θ 

 

W = a 1 − r2 

a2 cosh ct 

 

a 

T = a 1 − r2 

a2 sinh ct 

a 

W = a 

T = a 

r 2 

r 2 

a2 − 1 sinh ct 

a 

a 2 − 1 cosh ct 

a 

⎫ 

⎪⎬ 

r < a 

⎪⎭ 

⎫ 

⎪⎬ 

r > a . 

⎪⎭ 


Cosmology 

With a 2 1 

K 

= 3 

Λ 




Z = r cos θ 

 

W = a 1 − r2 

a2 cosh ct 

 

a 

T = a 1 − r2 

a2 sinh ct 

a 

W = a 

T = a 

r 2 

r 2 

a2 − 1 sinh ct 

a 

a 2 − 1 cosh ct 

a 

⎫ 

⎪⎬ 

r < a 

⎪⎭ 

⎫ 

⎪⎬ 

r > a . 

⎪⎭ 

Note: z = constant is the hyperboloid: (cT) 2 − W2 = constant. Homogeneity and 

Isotropy follow from the symmetry: S − 4 is invariant under 5-dimensional Lorentz 

Λ r. 

transformations. The force is linear repulsive F ∝ 1 

3 


Cosmology 

Λ < 0: Anti-de Sitter space AdS 4 


Cosmology 


embedding into R 2,3 : (cT) 2 − X 2 − Y 2 − Z 2 + W 2 = a 2 


Cosmology 





Cosmology 



With a 2 1 

K 

= − 3 

Λ 




Cosmology 



With a 2 1 

K 

= − 3 

Λ 





Z = r cos θ 

 

W = a 1 + r2 ct 

cos 

a2 a 

 

T = a 1 + r2 ct 

sin 

a2 a 


Cosmology 

Note: z = constant is the circle: (cT) 2 + W2 = constant. Homogeneity and Isotropy 

follow from the symmetry: S + 4 is invariant under S O(2, 3). The force is linear 

attractive F ∝ −1 3 Λ r. Causality problem: periodicity in time ct → ct + 2 π a. The 

circle in the (cT, W) plane is acausal. May be cured by mapping the circle to the 

real line: which is the universal cover of the AdS space. 

Exercise: find the model Einstein proposed: an eternal static solution. Show that 

Einstein’s GRT with vanishing cosmological constant has no such solution. 

Exercise: the observation of planetary motions constrains the cosmological 

constant to |Λ| ≪ 10 −42 cm −2 !. Show that this is compatible with a dark energy 

density, which has been determined to be ΩΛ ∼ 0.74 ± 0.03. 

Previous ≪❘ , next ❘≫ lecture. 


Cosmology 

Note: z = constant is the circle: (cT) 2 + W2 = constant. Homogeneity and Isotropy 

follow from the symmetry: S + 4 is invariant under S O(2, 3). The force is linear 

attractive F ∝ −1 3 Λ r. Causality problem: periodicity in time ct → ct + 2 π a. The 

circle in the (cT, W) plane is acausal. May be cured by mapping the circle to the 

real line: which is the universal cover of the AdS space. 

For the cosmological solutions of the Einstein-Friedmann equations the RW-metric 

is only affected as far as S (t) solves a different dynamical equation. 

Exercise: find the model Einstein proposed: an eternal static solution. Show that 

Einstein’s GRT with vanishing cosmological constant has no such solution. 

Exercise: the observation of planetary motions constrains the cosmological 

constant to |Λ| ≪ 10 −42 cm −2 !. Show that this is compatible with a dark energy 

density, which has been determined to be ΩΛ ∼ 0.74 ± 0.03. 

Previous ≪❘ , next ❘≫ lecture.

Cosmological solutions of the Einstein-Friedmann equations ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?