Transcript Diapositiva 1 - Universidad de Guanajuato

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
Depto. de Astronomía (UGto)
Astronomía Extragaláctica y Cosmología Observacional
 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
ds2 = c2 dt2 – a2(t) [dr2/(1 – k r2) + r2 dθ2 + r2 sin2θ dφ2] = 0
0 = 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
this happens because the separation between the source and the observer is commoving
∫t1→t0
and, since
c dt /a (t) = constant  dt/a (t) = constant
dt0/a(t0) = dt1/a(t1)
dt0/dt1= a 0/a1
dt0 = ν1 = λ0
dt1
ν 0 λ1
and
1 + z = a 0 /a 1
z = λ0 – λ1 = λ0 – 1
λ1
λ1
 The Hubble and the Deceleration Parameters
 From this relation we can recover the “Hubble Law” for a small distance (dr) and time (dt):
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
 And so we have the first fundamental “cosmological parameter”, the Hubble Parameter:
H(t) ≡ (å /a )
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 :
q(t) ≡ – (a ä )/å 2 = – ä / (a H2)
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
 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
Eq. (1):
(å /a )2 = (8πG/3)ρ + Λ/3 – k(c/a)2
(å /a )2 = (8πG/3)ρ
H2 = (8πG/3)ρ
 From it, we can define the critical density parameter:
ρcrit(t) = 3H2/(8πG)
 And the matter density parameter (to simplify other equations):
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:

(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)]
 For a matter dominated regime, ω = 0, thus
a (t)  t2/3
ρ(t)  a –3  t –2
 For a radiation dominated regime, ω = 1/3, thus
a (t)  t1/2
ρ(t)  a – 4  t –2
 If we calculate the limit of ρ for when t = 0, we find that it is infinite!
 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
1 = (8πG/3H2) ρ – k(c/Ha)2
1 = Ω – kc2/(H2a2)
(: H2)
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
R = 6k/a2 = (6H2/c2)(Ω – 1)
 and the “radius of curvature” is given by
RU = a/|k| = c/H|Ω – 1|
note that for closed models (k>0) RU is just the physical radius of the three sphere
 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/a
å/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 – H02a0 2(Ω0 – 1) /a 2
å 2 = Ω0 H02 (a 03/a) – H02a02 (Ω0 – 1)
(å /a0)2 = Ω0 H02 (a 0/a) – H02 (Ω0 – 1)
(å /a0)2 = Ω0 H02 (1+z) – H02 (Ω0 – 1)
(å /a0)2 = H02 [Ω0(1+z) – (Ω0 – 1)]
–1/(1+z)4 (dz/dt)2 = H02 [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
(å /a )2 = (8πG/3)ρ + Λc2/3 – k(c/a)2
2ä /a + (å /a )2 + k(c/a)2 = – 8πG (P/c2) + Λ
ä /a = – 4πG/3 (ρ + 3P/c2) + Λ/3
 Again dividing the energy equation by H2 we have:
1 = (8πG/3H2)ρ + Λ/3H2 – k(c/aH)2
1 = Ω + ΩΛ – kc2/(aH)2
kc2/(aH)2 = Ω + ΩΛ – 1
kc2 = H2a 2(ΣΩi – 1)  H2a 2Ωk
 Conservation + state equation
ρ’ = –3 (a’/a)
(ρ+P/c2)
ρ = ρ0 (a 0 / a ) 3(1 + ω)
ρm = ρm,0 (a 0 / a )3
ρrad = ρrad,0 (a 0 / a )4
ρΛ = Λc2/(8πG) = constant
 Or taking the present-day values of the parameters:
(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 H02 (a 0/a)4 + ΩΛ H02 – Ωk H02 (a 0/a)2
 H(z)
(å /a )2 = Ωmat H02 (a 0/a)3 + Ωrad H02 (a 0/a)4 + ΩΛ H02 – Ωk H02 (a 0/a)2
and now using the relation for the redshift
(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
 We may consider different models
• Einstein-de Sitter
• Radiation Dominated
• Open Dust
• Closed Dust
• Flat w/ Vacuum Energy
 Note that
k
0
0
-1
+1
0
Λ
0
0
0
0
ΩΛ=1-Ω
Ω
1
1
<1
>1
<1
z = a 0/a – 1  dz = (a 0/a 2) da = (1+z) da/a
da/a = dz/(1+z)
H = å /a = da /(a dt) = dz/(1+z) dt
dt = dz/[(1+z) H(z)]
P
0
P
0
0
0
 Evolution of the Scale Factor
(Closed Λ)
(Flat Λ)
(Open)
(EdS)
(Closed)
 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
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
 Age of the Universe (in Gyears) for different models
Ω0
1.0
0.5
0.3
0.1
0.01
Open (ΩΛ = 0)
9.3
10.5
11.3
12.5
13.9
Spatially-flat (ΩΛ = 1 – Ω0)
9.3
11.6
13.5
17.8
28.0
 Lookback Time
(Flat Λ)
(Open)
(EdS)
 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
DC = ∫t→t0 [c dt /a (t)] =
∫z→0 c dz / H(z)
= c/H0 ∫z→0 dz / E(z)
 The angular diameter distance is defined as the ratio of an object’s physical transverse
size to its angular size (in radians)
c/H0 Ωk-½ sinh [Ωk½ DcH0/c] Ωk > 0
DA = s/ =1/(1+z) DC
c/H0 Ωk-½ sin [|Ωk|½ DcH0/c] Ωk < 0
Ωk = 0
 The luminosity distance is defined by the relationship between bolometric flux and
bolometric luminisity
DL = (L / 4πF) = (1+z)2 DA
See more on Hogg 2000 [astro-ph/9905116]
 Angular Diameter Distance
(Open)
(Flat Λ)
DA = s/
(EdS)
Gravitational focusing →
the matter within the beam
of observation produces a
gravitational lensing effect
(distorts the light rays,
making the image greater
at high-z)
 Luminosity Distance
(Open)
(Flat Λ)
(EdS)
 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
ds = dQ = 0
kBT

dU + P dV = 0
dU = – P dV
 For the radiation we have
U=εV
P = (1/3) ε
non-viscous fluid
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
 Thermodynamics
 But the (thermal) radiation has a black body spectral energy density, that follows the Planck
distribution
εν = 8πh
ν3
c3 ehν/kT – 1
.
ε = ∫0→ εν dν = π2 k4 T4 = a T4
15 ħ3 c3
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
a T4 = ε0 (a/a0)–4
T = (ε0/a)¼ (a/a0)–1
T = T0 (1+z)
 Thermodynamics
 For the matter we have
U = ε V = (3/2) n k T V = (3/2) N k T
P=nkT
mean energy
ideal monoatomic gas
(3/2) N k d(T) = –(N/V) k T dV
(3/2) dT = –(1/V) T dV
dT/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
 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)
 Matter-Radiation Equality
 We have found that the evolution of the main cosmological parameters differs for
(non-relativistic) matter and (relativistic) radiation
matter
radiation (γ,ν)
ω
0
⅓
ρ
ρm,0 (1+z)3
ρr,0 (1+z)4
a
 t2/3
 t1/2
T
T0 (1+z)2
T0 (1+z)
 One can realize that there is a time in the past when the densities of matter and radiation
were equal (consider a0 = 1)


Ωm = ρm,0 / ρcrit = ρm / ρcrit (a /a0)3
Ωr = ρr,0 / ρcrit = ρr / ρcrit (a /a0)4
ρm = Ωm ρcrit / a3
ρr = Ωr ρcrit / a4
Ωm ρcrit / aeq3 = Ωr ρcrit / aeq4
aeq = Ωr / Ωm
 The CMBR temperature (2.725±0.002) give us a good estimate of the present density
parameter for radiation (4.1510–5 h–2); considering the density parameter for matter to
be 0.3 and h to be 0.75 we have
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)