Astronomía Extragaláctica y Cosmología ObservacionalDepto. de Astronomía (UGto)

Lecture 14 Properties of the Standard Model

Redshifts

The Hubble and the Deceleration Parameters

The Critical Solution and Ω(t)

The solutions for Λ = 0

The Complete Solution and H(z)

Lookback Time

Distances Commoving Distance Angular Diameter Distance Luminosity Distance

Thermodynamics

Matter-Radiation Equality

Redshifts

Light rays travel along null geodesics (ds = 0) Consider a source emitting electromagnetic radiation at (r1, θ1, φ1, t1) and an observer at (0, 0, 0, t0). In the course of the trajectory of the light, its coordinates are [r(t), θ1, φ1, t], that is, it travels straight to the observer (direction of the source do not change) If the coordinates of source and observer are commoving, they do not change with time

this happens because the separation between the source and the observer is commoving

and, since and

∫t1→t0 c dt /a (t) = constant dt/a (t) = constantdt0/a(t0) = dt1/a(t1)

dt0/dt1= a 0/a1

dt0 = ν1 = λ0 z = λ0 – λ1 = λ0 – 1dt1 ν0 λ1 λ1 λ1

1 + z = a 0 /a 1

ds2 = c2 dt2 – a2(t) [dr2/(1 – k r2) + r2 dθ2 + r2 sin2θ dφ2] = 00 = c2 dt2 – a2(t) [dr2/(1 – k r2)]

c2 dt2 = [a2(t) dr2/(1 – k r2)]

∫t1→t0 [c /a (t)] dt = ∫r1→0 dr / (1 – k r2) = constant

The Hubble and the Deceleration Parameters

From this relation we can recover the “Hubble Law” for a small distance (dr) and time (dt):

And so we have the first fundamental “cosmological parameter”, the Hubble Parameter:

which represents the fractional rate of change of the scale factor.

Another important observable “cosmological parameter” is related to the acceleration of the expansion, ä :

which has a positive value if the expansion is slowing down (decelerating) and a negative value if it is speeding up (accelerating).

For t = 0, they assume their present-day values:• H(t0) = H0 (the Hubble Constant)• q(t0) = q0

z = a 0 /a – 1 = (1 /a ) (a + da) – 1 = 1 + (da /a ) – 1 = (å dt) / a = (å / a ) dr/c

cz = (å / a ) dr

cz = vr = H dr

H(t) ≡ (å /a )

q(t) ≡ – (a ä )/å 2 = – ä / (a H2)

The Critical Solution

The critical solution for the FLRW model represents the simplest transition between the open and closed models, and frequently is used as reference:

• flat topology: k = 0• no Cosmological Constant: Λ = 0

From it, we can define the critical density parameter:

And the matter density parameter (to simplify other equations):

Eq. (1): (å /a )2 = (8πG/3)ρ + Λ/3 – k(c/a)2

(å /a )2 = (8πG/3)ρ H2 = (8πG/3)ρ

ρcrit(t) = 3H2/(8πG)

Ω(t) = Σ Ωi = Σ ρi /ρcrit

Ωm = ρm/ρcrit = 8πG ρm /(3H2)Ωrad = ρrad/ρcrit = 8πG ρrad /(3H2)ΩΛ = ρΛ/ρcrit = 8πG ρΛ /(3H2) = Λc2/(3H2)

The Scale Parameter on the Critical Solution

We found that ρ/ρ0 = (a / a0 ) –3(1+ω), so:

For a matter dominated regime, ω = 0, thus

For a radiation dominated regime, ω = 1/3, thus

If we calculate the limit of ρ for when t = 0, we find that it is infinite!

(å /a )2 = (8πG/3) ρ0 (a 0/a)3(1+ω)

å /a = (8πG/3) ρ0 (a 0/a)3(1+ω)/2

da/a = (8πG/3) ρ0 (a 0/a)3(1+ω)/2 dt

∫ da a –1+3(1+ω)/2 = (8πG/3) ρ0 a 03(1+ω)/2 ∫ dt

2/[3(1+ω)] a 3(1+ω)/2 = (8πG/3) ρ0 a 03(1+ω)/2 t

a (t) = [3(1+ω)/2 (8πG/3) ρ0] 2/[3(1+ω)] a 0 t2/[3(1+ω)]

a (t) t2/3 ρ(t) a –3 t –2

a (t) t1/2 ρ(t) a – 4 t –2

The classical solutions for Λ = 0

The next step is to analyze the solutions for any value of k, but Λ = 0:

Eq. (1): (å /a)2 = (8πG/3)ρ + Λ/3 – k(c/a)2

H2 = (8πG/3)ρ – k(c/a)2 (: H2) 1 = (8πG/3H2) ρ – k(c/Ha)2

1 = Ω – kc2/(H2a2)

kc2 = H2a 2(Ω – 1)

A Universe with density above the critical value will be spatially closed, whereas a lower density Universe will be spatially open. If ρ =ρcrit, the Universe is Euclidean (spatially flat)

k Ω geometry Thus: +1 > 1 closed, spherical

0 = 1 flat−1 < 1 open, hyperbolic

The spatial curvature is given by the scalar curvature, that can be expressed as

and the “radius of curvature” is given by

note that for closed models (k>0) RU is just the physical radius of the three sphere

R = 6k/a2 = (6H2/c2)(Ω – 1)

RU = a/|k| = c/H|Ω – 1|

The solutions for Λ = 0

The solutions for Λ = 0

H(z) for Λ = 0

For Λ=0, in a matter dominated model, we can express H as a simple function of z:

z = a0/a – 1 dz = –(a0/a 2) da = –(1+z) da/aå/a0 = å a/a0a = å /[a (1+z)] = da /[a (1+z) dt] = –dz /[(1+z)2 dt]

kc2 = H02a0 2(Ω0 – 1)

(å /a )2 = Ω0 H02 (a 0/a)3 – H0

2a0 2(Ω0 – 1) /a 2

å 2 = Ω0 H02 (a 0

3/a) – H02a0

2 (Ω0 – 1)(å /a0)2 = Ω0 H0

2 (a 0/a) – H02 (Ω0 – 1)

(å /a0)2 = Ω0 H02 (1+z) – H0

2 (Ω0 – 1)(å /a0)2 = H0

2 [Ω0(1+z) – (Ω0 – 1)]–1/(1+z)4 (dz/dt)2 = H0

2 [Ω0 + z Ω0 – (Ω0 – 1)]dz/dt = –(1+z)2 H0 [1 + z Ω0]½

H = å /a = (å /a0)(a0/a) = –1/(1+z)2 dz/dt (1+z)

H(z) = H0 (1+z) [1 + z Ω0]½

The Complete Solution

Friedmann-Lemaître Solution

Again dividing the energy equation by H2 we have:

Conservation + state equation

Or taking the present-day values of the parameters:

2ä /a + (å /a )2 + k(c/a)2 = – 8πG (P/c2) + Λ

(å /a )2 = (8πG/3)ρ + Λc2/3 – k(c/a)2

ä /a = – 4πG/3 (ρ + 3P/c2) + Λ/3

ρ’ = –3 (a’/a) (ρ+P/c2)ρm = ρm,0 (a 0 / a )3

ρrad = ρrad,0 (a 0 / a )4

ρΛ = Λc2/(8πG) = constantρ = ρ0 (a 0 / a ) 3(1 + ω)

(å /a )2 = (8πG/3) [ρm,0 (a 0/a)3 + ρrad,0 (a 0/a)4] + Λc2/3 – k(c/a)2

(å /a )2 = Ωm H02 (a 0/a)3 + Ωrad H0

2 (a 0/a)4 + ΩΛ H02 – Ωk H0

2 (a 0/a)2

1 = (8πG/3H2)ρ + Λ/3H2 – k(c/aH)2

1 = Ω + ΩΛ – kc2/(aH)2

kc2/(aH)2 = Ω + ΩΛ – 1kc2 = H2a 2(ΣΩi – 1) H2a 2Ωk

H(z)

and now using the relation for the redshift

We may consider different models

k Λ Ω P• Einstein-de Sitter 0 0 1 0• Radiation Dominated 0 0 1 P• Open Dust -1 0 <1 0• Closed Dust +1 0 >1 0• Flat w/ Vacuum Energy 0 ΩΛ=1-Ω <1 0

Note that

(å /a )2 = Ωmat H02 (a 0/a)3 + Ωrad H0

2 (a 0/a)4 + ΩΛ H02 – Ωk H0

2 (a 0/a)2

(å /a )2 = H02 [Ωrad (1+z)4 + Ωm (1+z)3 – Ωk (1+z)2 + ΩΛ]

å /a = H(z) = H0 E(z) = H0 [Ωrad (1+z)4 + Ωm (1+z)3 – Ωk (1+z)2 + ΩΛ]½

z = a 0/a – 1 dz = (a 0/a 2) da = (1+z) da/ada/a = dz/(1+z)

H = å /a = da /(a dt) = dz/(1+z) dtdt = dz/[(1+z) H(z)]

Evolution of the Scale Factor

(EdS)

(Closed)

(Open)

(Flat Λ)

(Closed Λ)

Lookback Time

The lookback time to an object is the diference between the age t0 of the Universe now (at observation) and the age te of the Universe at the time the photons from a source were emitted

Age of the Universe (in Gyears) for different models

Ω0 Open (ΩΛ = 0) Spatially-flat (ΩΛ = 1 – Ω0)1.0 9.3 9.30.5 10.5 11.60.3 11.3 13.50.1 12.5 17.80.01 13.9 28.0

E(z) = [Ωrad (1+z)4 + Ωmat (1+z)3 – Ωk (1+z)2 + ΩΛ]½

tL = ∫te→t0 dt = 1/H0 ∫0→z dz / (1+z)E(z)

tL = 1/H0 ∫0→z dz / (1+z) [Ωrad (1+z)4 + Ωmat (1+z)3 – Ωk (1+z)2 + ΩΛ]½

Lookback Time

(EdS)

(Open)

(Flat Λ)

Distances

A small comoving distance between two nearby objects in the Universe is the distance between them which remais constant with the epoch if the two objs are moving with the Hubble flow

The angular diameter distance is defined as the ratio of an object’s physical transverse size to its angular size (in radians)

The luminosity distance is defined by the relationship between bolometric flux and bolometric luminisity

DC = ∫t→t0 [c dt /a (t)] = ∫z→0 c dz / H(z) = c/H0 ∫z→0 dz / E(z)

c/H0 Ωk-½ sinh [Ωk

½ DcH0/c] Ωk > 0DA = s/ =1/(1+z) DC Ωk = 0

c/H0 Ωk-½ sin [|Ωk|½ DcH0/c] Ωk < 0

See more on Hogg 2000 [astro-ph/9905116]

DL = (L / 4πF) = (1+z)2 DA

Angular Diameter Distance

(EdS)

(Open)

(Flat Λ)

Gravitational focusing → the matter within the beam of observation produces a gravitational lensing effect(distorts the light rays, making the image greaterat high-z)

DA = s/

Luminosity Distance

(EdS)

(Open)

(Flat Λ)

Thermodynamics

In strictest mathematical sense, it is not possible for the Universe to be in thermal equilibrium (if so it would not evolve...) but, for practical purposes, it has for much of its history been very nearly in thermal equilibrium We saw that the Friedmann equations take us to the adiabatic first law of thermodynamics. Equivalently we can say that the Universe is isentropic, since it may have had an almost constant entropy. Thus, the second law of thermodynamics states

For the radiation we have

ds = dQ = 0 dU + P dV = 0 kBT dU = – P dV

U = ε VP = (1/3) ε

d(εV) = – (1/3) ε dV(4π/3) d(ε a3) = – (1/3) (4π/3) ε d(a3)

dε a3 + ε d(a3) = – (1/3) ε d(a3)dε/ε = – (4/3) d(a3)/a3

∫ d ln ε = – (4/3) ∫ d ln (a3)ln ε = ln (a3)–(4/3)

ε = ε0 (a/a0)–4

non-viscous fluid

Thermodynamics

But the (thermal) radiation has a black body spectral energy density, that follows the Planck distribution

where h is the Planck constant, ħ = h/2π, k is the Boltzmann constant and a is the Stefan-Boltzmann constant (do not confuse with the scale factor!). Thus

εν = 8πh ν3 .

c3 ehν/kT – 1

ε = ∫0→ εν dν = π2 k4 T4 = a T4

15 ħ3 c3

a T4 = ε0 (a/a0)–4

T = (ε0/a)¼ (a/a0)–1

T = T0 (1+z)

Thermodynamics

For the matter we have

Observational test for T: the ground state transition of C I has a fine structure splitting determined by the energy density and temperature of CMBR photons – the relative strengths of the respective C I absorption lines have been observed in Lyα clouds from QSO spectra (with Keck telescope)

• Cowie et al. 1994 – Q1331+170 (z = 1.776) → TCMBR = 7.4±0.8 K (expected 7.58 K)

• Ge et al. 1997 – Q0013-004 (z = 1.9731) → TCMBR = 7.9±1.0 K (expected 8.11 K)

U = ε V = (3/2) n k T V = (3/2) N k TP = n k T (3/2) N k d(T) = –(N/V) k T dV

(3/2) dT = –(1/V) T dVdT/T = –(2/3) d(a3)/a3

∫ d ln T = –(2/3) ∫ d ln (a3)ln T = ln (a3)–2/3

T = T0 (a/a0)–2 = T0 (1+z)2

ideal monoatomic gas

mean energy

Matter-Radiation Equality

We have found that the evolution of the main cosmological parameters differs for (non-relativistic) matter and (relativistic) radiation

One can realize that there is a time in the past when the densities of matter and radiation were equal (consider a0 = 1)

The CMBR temperature (2.725±0.002) give us a good estimate of the present density parameter for radiation (4.1510–5 h–2); considering the density parameter for matter to be 0.3 and h to be 0.75 we have

ω a ρ Tmatter 0 t2/3 ρm,0 (1+z)3 T0 (1+z)2

radiation (γ,ν) ⅓ t1/2 ρr,0 (1+z)4 T0 (1+z)

Ωm = ρm,0 / ρcrit = ρm / ρcrit (a /a0)3 ρm = Ωm ρcrit / a3 Ωr = ρr,0 / ρcrit = ρr / ρcrit (a /a0)4 ρr = Ωr ρcrit / a4

Ωm ρcrit / aeq3 = Ωr ρcrit / aeq

4

aeq = Ωr / Ωm

aeq ≈ 0.00025 zeq ≈ 4000

Matter-density Equality

The matter-radiation equality divides the thermal history of the Universe in a radiation domination era and a matter domination era When the matter and radiation are coupled (by Thomson scattering), in the radiation era, it is the Trad a –1 that dominates, because the number density of photons is very much greater than that of nucleons (nγ / np ~ 109)

Further readings

Papers:

P.J.E. Peebles 1998, PASP 111, 274 (astro-ph/9806201) – The Standard Cosmological Model A.R. Liddle 1999, astro-ph/9901124 – An Introduction to Cosmological Inflation S.M. Carroll 2000, astro-ph/0004075 – The Cosmological Constant M.S. Turner 2002, astro-ph/0202007 – The New Cosmology O. Lahav & Y. Suto 2004 (NED: level 5) – Living Reviews (in Relativity)