Cosmology 2004-20051 Thermal History Prof. Guido Chincarini Here I give a brief summary of the most important steps in the thermal evolution of the Universe

Cosmology 2004-2005 1

Thermal History

Prof. Guido Chincarini

Here I give a brief summary of the most important steps in the thermal evolution of the Universe. The student should try to compute the various parameters and check the similarities with other branches of Astrophysics.

After this we will deal with the coupling of matter and radiation and the formation of cosmic structures.


The cosmological epochs

• The present Universe– T=t0 z=0 Estimate of the Cosmological Parameters and of the

distribution of Matter.

• The epoch of recombination– Protons and electron combine to form Hydrogen

• The epoch of equivalence– The density of radiation equal the density of matter.

• The Nucleosynthesis– Deuterium and Helium

• The Planck Time– The Frontiers of physics


Recombination

0 0

3

25 1e2 kT

e H21

eTot 1 2 2 e

Tot

2 2 2e e2

e Tot1 1 Tot e

23 3 30 ,c 00 0 0

Tot 0p p p

Saha equation : 0.038 H 72

2 m kTNN e ; 13.6 eV ; k 8.62 10 eV deg

N h

NN N N x and for Hydrogen N N

N

N NN xN N

N N N N 1 x

3HN z N 1 z 1 z 1 z 1

m m 8 G m

3

erecombination

Tot

z

Nt t x 0.5

N


Zrecombination


A Play Approach

• We consider a mixture of photons and particles (protons and electrons) and assume thermal equilibrium and photoionization as a function of Temperature (same as time and redshift).

• I follow the equations as discussed in a photoionization equilibrium and I use the coefficients as given in Osterbrock, see however also Cox Allen’s Atrophyiscal Quantities.

• A more detailed approach using the parameters as a function of the Temperature will be done later on.

• The solution of the equilibrium equation must be done by numerical integration.

• The Recombination Temperature is defined as the Temperature for which we have: Ne = Np=Nho=0.50

b,0 h2 =0.02 H0=72


The Equations

0 00

0

0

H 0 e p 0 H p

3

0 e p 0 e Tot 0 30

2 230 0

0 b,0 e b,03p 0 p

230

0 b,0 030 p

N N a H d N N H ,T ; N N

TN a H d N N H ,T ; N 0.5 N T 0.5 N

T

for T I use Radiation Temperature

3H 3HTN ; N 0.5

8 Gm T 8 Gm

3HTN a H d 0.5 H ,T

T 8 Gm


0 0

0 0

318 0

0

33

218 0

0 h

kT

18 30

2 h h

kT kT

2330

b,0 030 p

for a H 6.3 10

2h4 cI Part N a H d 6.3 10 dh

e 1

2* 4 * 6.3 10 * d dB*

ce 1 e 1

3HTII Part 0.5 H ,T C*T

T 8 Gm


3200 3400 3600 3800 4000 4200 4400Temperature

-23

-22

-21

-20

-19

-18

-17

noitcnuF

Recombination Temperature


The Agreement is excellent


Time of equivalence

r 4 3r m eq r eq m eq2

3 40 0

m eq m,0 r eq r ,03 4eq eq

0m,0eq

r ,0 eq

2300

m,0 c ,0 m,0

434 3 5r ,0

r ,0 r ,0 220

3

eq

tt a t t a t t t t

c

a t a tt t

a t a t

a t1 z

a t

3H0.3 2.9110

8 G

T4.46 10 g cm ; 4.6 10

3Hc8 G

2.91101 z

0

346540

4.46 10


The need of Nucleosynthesis

• I assume that the Luminosity of the Galaxy has been the same over the Hubble time and due to the conversion of H into He.

• To get the observed Luminosity I need only to convert 1% of the nucleons and that is in disagreement with the observed Helium abundance which is of about 25%.

• The time approximation is rough but reasonable because most of the time elapsed between the galaxy formation and the present time [see the relation t=t(z)].

• To assume galaxies 100 time more luminous would be somewhat in contradiction with the observed mean Luminosity of a galaxy.

• Obviously the following estimate is extremely coarse and could be easily done in more details.


* 10 33 10 7

73G 12

11 1 1

11 33 10 773

12

L 2.3110 2 10 1.36 10 3.15 10L in eV over Hubble time 1.24 10

1.6 10

or in a different way

LL2 10 ; 10 ; 0.1 ; 2 erg s gr

L

0.2 2 10 2 10 1.36 10 3.15 10L( in eV ) 0.2 HubbleTime 2.14 10

1.6 10

# nucleons

MM

M M

M

11 3368

24

735

68

2 10 2 102.5 10

mp 1.6 10

L 2 10Em per Nucleon 0.8 10 eV 0.08 MeV

# nucleons 2.5 10

The reaction H He produces 6 MeV so that

0.08I need only 1.3% nucleons to react

6

M


Temperature and Cosmic Time

2

1 1r t2 2

0 0

1 122 2

4r

112 42

10

1 GMm dr 2GMmv 0

2 r dt r

1dr r 2GM dt t accurately slide 16

6 G

3 3 ct

32 G 32 G T

3 cT t

32 G

for t 1 s T 1.52 10 K Nuclear Reaction are possible


Gamow 1948

The reasoning by Gamow was rather simple. To form heavy elements I must start from elementary particles and in particular I should be able to form Deuterium from protons and neutrons according to the reaction:

n + p => d + A reaction that need a Temperature of about 109 degrees Kelvin. If I

have the photons at higher Temperature (T>109) these dissociate the Deuterium as soon as it forms. The Temperature is therefore very critical if I want to accumulate Deuterium as a first step in the formation of heavy elements.

The Density is critical as well. The density must allow a reasonable number of reaction. However it must not be too high since at the end I also have a constraint due to the amount of Hydrogen I observe and indeed I need protons to form hydrogen.

Finally since the Temperature is a function of time the time of reaction in order the reaction to occur must be smaller than the time of expansion of the Universe.


Gamow

During the time t I have :Number of encounters = nvtn = Density at the time t = cross sectionv = thermal velocity of particlesOr I have one enclounter in 1/nvt seconds and if the temperature is

correct I will have the reaction. As we have said therefore the reaction time must by fast, in other words I must have:

1/ nv t < texp or nv texp >1. Here we obviously have a radiation

dominated Universe and =1 (E=0 in the simple minded Newtonian model. Let play quickly:

n

V (t=1s)

r2 cross section

2 2020

402 20 4

020

2 22 0

0 0 20

Ha 8 G*

a 3 H

a8 G Ha a

a 3 H

aa1 H

a a


R R R c R

RR

a da H a dt egrating

a ta H a t a H a t t

H a t

a H Ht t t

a t G t G H

a T ct

Gt c G Ga

20 0

1

22 2 2 2

0 0 0 0 0 0 00 0 0

412 24 2 22

0 0 0,0 ,4

0 0 0

14 22

2 2

int

1 1 1; ; ;

2 2 2

3 31

8 8 2

3 3 3;

32 32 32 7.

R

R

for T sT

a T

a T

1

29

15 4

4 404 4

,0 0

10 230.556 10

Given by Physicsv derived from Temp.t from Cosmologyn(t) to be determined

From slide 15: n v t = 1n (t) = 1018 nucleons cm-3


Look at this

ee e e

ee

e

Deuterium Deuterium

e

H km s Mpc

a t tRadiation t

a t t Ht

a t tDust t

a t t H

a tt slide t

a t z

a t a t a tT t T

a t a t

0

1

2

1702

3

00 0 0

31.5

217 11

00

00

65 / /

1

24.75 10 ;

2 1

3

110 4.75 10 8.98 10

1

230.5

e Deuterium e

e

t t

a t t t

Kelvin

1 2

2 3

0 0

2111 32

11 17

230.5 8.98 102.5

8.98 10 4.75 10


And

Deuterium e Deuterium e

e e

H

a tn t n

a t

a t a t t t

a t a t t t

cm m

cm

3

00

31 232 3

18 18

0 0

32111 32

8 311 17

8 3

230.5

10 10

230.5 8.98 101.47 10 *

8.98 10 4.75 10

1.47 10 1.67 10

B B B

B

g g cm Barions

Nowadays

Hh

G

hg cm

G

24 32 3

22 0

22 5231 3

8

2.4 10 /

:

30.02

8

3 0.02 0.65 100 10 / 3.09 10^ 243 1001.58 10

8 8 6.67 10


See Gamow and MWB

Following the brief description of Gamow reasoning I introduce here the Microwave background and the discussion on the discovery.

The discovery by Penzias and Wilson and related history.

The details of the distribution with the point measured by Penzias and Wilson.

The demonstration that a Blackbody remains a Blackbody during the expansion.

The explanation of the observed dipole as the motion of the observer respect the background touching also upon the Mach principle.

Some example and computation before going to the next nucleosynthesis slides.

The analysis of the WMAP observations and the related fluctuations will be eventually discussed later also in relation with the density perturbation and the formation of galaxies and large scale structure. The Horizon and The Power spectrum and Clusters of galaxies.


The main reactions

210 9em c

for T 10 T 6 10k

p n

p n

e e and when T decrea

e

e

e eses only

That is at some point after the temperature decreases under a critical value I will not produce pairs from radiation but I still will produce radiation by annihilation of positrons electrons pairs.

That is at this lower temperature the reaction above, proton + electron and neutron + positron do not occur any more and the number of protons and neutron remain frozen.


Boltzman Equation

• At this point we have protons and neutrons which could react to form deuterium and start the formation of light elements. The temperature must be high enough to get the reaction but not too high otherwise the particles would pass by too fast and the nuclear force have no time to react.

• The equation are always Boltzman equilibrium equations.

6 122n p

16 10

2 2p n

1.294 10 1.6 10 to ergm m c

1.38 10 10neutrons kT

protons

e

1 Neutron1

every5

5 Pr oto

m c 938.2592 MeV ; m c 939.5527 MeV ; 1.2935 MeV

ne e 0.22

n

Neutron could decay n p e

However it takes 15 mi

n

nutes, too

s

l

ong !!


2i i

2 2d d d d

2 2n p n p n pn n pp d d

3m c2

i dkTi i n p n p d3

ii

Tot

3

23

3 m c m cd2d kT kT

d d d33 3

3 3

2 2

m m cm m m m m cd dkT k

3 3

B

2 m kT 2gn g e ; g g 2 ;

h 3n

Xn

Tm k

2 m kT 2 2n g e g e

h 2

T Tm k m k

2 23 e 3 e

T


2n n

2p p

d

3

2

m cnn kT

n n 3Tot Tot

3

2

m cpp kT

p p 3Tot Tot

3 3B2 2

3 k Td d dTot

n p p n n p

Tm k

n 1 2X g e

n n

Tm k

n 1 2X g e

n n

X g m k Tn e

X X g g m m 2

Plot as a function of T


Comment

• As it will be clear from the following Figure in the temperature range 1 – 2 10^9 the configuration moves sharply toward an high Deuterium abundance, from free neutrons to deuterons.

• Now we should compute the probability of reaction to estimate whether it is really true that most of the free neutrons are cooked up into deuterium.

• Xd changes only weakly with B h2

• For T > 5 10^9 Xd is very small since the high Temperature would favor photo-dissociation of the Deuterium.


0 1109 2109 3109 4109 5109

Temperature

-10

-8

-6

-4

-2

0

2

4goLdX

pXnX

Deuterium Equilibrium Temperature


Helium


More


Compare to Observations

Abu

ndan

ce R

ela

tive

to

Hyd

roge

n

B h2

The current estimates are:

From the D/H Ratio in Quasars Abs. Lines:

B h2 = 0.0214 0.002

From the Power Spectrum of the CMB:

B h2 = 0.0224 0.001


After Deuterium

3

3

3 4

3 4

d d He n p p n

d d H p p n n Tritium

d H He n p p n n

d He He p p p n n


Probability of Reaction

• I assume also that at the time of these reaction each neutron collides and reacts with 1 proton. Indeed the Probability for that reaction at this Temperature is shown to be, even with a rough approximation, very high.

Number of collision per second= r2 v n

I assume an high probability of Collision so that each neutron

Collides with a proton.Probability Q is very high so that it

Is reasonable to assume that all electronsReact.

n

V (t=1s)

r2 cross section

2

3

9 13 8 1crit b ,0

n 0

5

Q r n v t # of collision Pr obability of collision

kT TT 10 ; t 231 s; r 10 ; v Sqrt 2.9 10 cm s ; n n

m T

Q 3.2110 1 see however det ailed computation


Finally

He TotTot He n

p p

npHe He He He n

Tot Tot p Tot Tot

1 neutron 5 protons1 He every 10 protons

2 neutrons 10 protons

n n 1 10.2 0.17 Accurate computation 0.12

pp n p 61n

m 14 n n n

m m 2

1n 4mn m n 4 n2Y * 2 0.24

m n n n p

M

MM M


Neutrinos• 1930 Wolfgang Pauli (1945 Nobel) assumes the existence of a third particle to

save the principle of the conservation of Energy in the reactions below. Because of the extremely low mass Fermi called it neutrino.

• The neutrino is detected by Clyde Cowan and Fred Reines in 1955 using the reaction below and to them is assigned the Nobel Prize.

• The Muon neutrinos have been detected in 1962 by L. Lederman, M. Schwartz, and J. Steinberg. These received the Nobel Prize in 1988.

• We will show that the density of the neutrinos in the Cosmo is about the density of the photons.

• The temperature of the neutrinos is about 1.4 smaller than the temperature of the photons. And this is the consequence of the fact that by decreasing temperature I stop the creation of pairs from radiation and howver I keep annihilating positrons and electrons adding energy to the photon field.

9.7e

12.1

13.3

Leprons Neutral Mass Temperature Fermions

e 15eV 10 Massless ?

.17MeV 10 Move at speed of light

24MeV 10 Follow geodet ics


Recent results• It has been demonstrated by recent experiments [Super

Kamiokande collaboration in Japan] that the neutrinos oscillate. For an early theoretical discussion see Pontecorvo paper.

• The experiment carried out for various arrival anles and distances travelled by the Neutrinos is in very good agreement with the prediction with neutrino oscillations and in disagreement with neutrinos without oscillations.

• The oscillations imply a mass so that finally it has been demonstrated that the neutrinos are massive particles.

• The mass is however very small. Indeed the average mass we can consider is of 0.05 eV.

• The small mass, as we will see later, is of no interest for the closure of the Universe.

• On the other hand it is an important element of the Universe and the total mass is of the order of the baryonic mass.

• www => neutrino.kek.jp // hep.bu.edu/~superk


2i iq q3 3 3kT kT

343 4

p x0

33 3

q3 3 x 40kT

The distribution function for Fermions is :

g 4 g 41 1 1f q n q q dq ; q E h for photons

h c he 1 e 1

and for the density of Energy

Bx 21 2 5.62 p

e 1 p 8

4 1 x 1q dq 1 2 4 4

c h e 1e 1

4

0

416

4 15 4 43 310 27

5 44 4 15

3 3

5.68

4 5.68 1.38 10 7T 3.3 10 T T

163 10 6.625 10

7 k 7T T ; 7.56 10 ; Density

c 30 h 16

Bernoulli Number

Riemann Zeta

Gamma

Here I use g = 1


Conventions

I is the chemical potential + for Fermions and – for Bosons• gi Number of spin states

– Neutrinos and antineutrinos g=1– Photons, electrons, muons, nucleons etc g=2

e- = e+ = 2 =7/8 T4 (I use in the previous slide ge = 2)• Neutrinos have no electric charge and are not directly coupled to photons.

They do not interact much with baryons either due to the low density of baryons.

• At high temperatures ~ 1011-1012 the equilibrium is kept through the reactions

• • Later at lower temperature we have electrons and photons in equilibrium ad

neutrinos are not coupled anymore• At 5 109 we have the difference before and after as shown in the next slides.

e ee e; ;


See Weinberg Page 533

Assuming the particles are in thermal equilibrium it is possible to derive an equation stating the constancy of the entropy in a volume a3 and expressing it as:

At some point during the expansion of the Universe and the decrease of the Temperature we will create, as stated earlier, radiation (Gamma rays), and however the will not be enough Energy to create electron pairs.

We are indeed warming up the photon gas while the neutrino gas remains at a lower Temperature because we are not pumping in ant Energy.

eq eq

aS a T T p T

T

33 ,


Entropy

3

3 3

e e e e

2 2e

6 129 4

16 e e

9

At some Temperature we have only e e in thermal equilibrium

as a Volume a p p p &

T1

p c Relativistic Re gime kT m c3

0.5 10 1.6 10 7electrons are relativistic T 5.9 10 ; T

1.

T 5

38 10 8

s a

10

3 3 33

Tot Tot Tot e e

334 4 4

333

9

This quantity wil

a 1 a 4 a 4

T 3 T 3 T 3

a 4 7 7 4 11T T T aT

T 3 8 8 4 3

e e warms up photon field we are left with photons

a 4 4s a aT

l be conserved

T 5 10

This quantity will3 3

bT

e conserved


s a3

Temperature

Time

T ~ 5 109

11/3 (aT)3 4/3 (aT)3

Conserve Entropy

9

9

9

1

33 33 T 10

T 10

1

1 1

3 3

,0 ,0

T 10

R,

aT11 4 11s a aT aT

3 3 aT 4

while neutrinos & antineutrinos are not warmed up T T a

T 11 41.401 T T 1.9 K

T 4 11

Density of Radiation due to neutrinos

3 spec

4

34 4

4

7 7 4ies neutrinos * 2 neut &antineut * * T 6 T

16 16 11

0.68 T


2p q 30

32 3 3iq3 3 3 3kT

3,0

2 30 230 3 3 30

0 ,c 12

3 31q dq for p 0

e 1 p

g 4 1 1 1 4n q q dq N 3 3 k T N 108.6 cm 326 cm

h c c he 1

n 420 cm

3H 9.72 10 c9.72 10 g cm eVcm 5460 eVcm

8 G 1.6 10

5460closure with 16.7 eV m

Obs326

erved

0.05eV


Time

PhotonsNeutrinos


Planck Time

• We define this time and all the related variables starting from the indetermination principle. See however Zeldovich and Novikov for discussion and inflation theory.

5 2

3 3 p2 2 2p p p p p p p2

p

43p 5

593 3

p 2 233pp p 3

33 2

p3 5 3 98 3p p p p p

p

52 19

p p p

E t

c t1E t m c t c t c t c t c t

G t G

Gt 10 s

c

1 c4 10 g cmG

G t Gl c t 1.7 10 cmc

c cm l 2.5 10 g n l 10 cm

G m G

c EE m c 1.2 10 GeV TG

5p 1 32c

k 1.4 10 Kk G


Curiosity – Schwarzschild Radius

• It is of the order of magnitude of the radius that should have a body in order to have Mass Rest Energy = Gravitational Energy.

• And the photons are trapped because the escape velocity is equal to the velocity of light.

22

s 2s

2s

s s 3s

s p3 5p

G m2 G m m crrc

rc G m 2 G mt time to cross r

2 r c c

2 G c Gc t 2 2 twith m m c G cG


The Compton time

• I define the Compton time as the time during which I can violate the conservation of Energy E = mc2 t=t. I use the indetermination principle.

• During this time I create a pait of particles tc = / m c.

• In essence it is the same definition as the Planck time for m = mp.

Compton C p

1

C p2 2 5

cl c t and for m m

m c G

c Gt t

m c c G c


Preliminaries

1 Mole: Amount in grams equal to the molecular weight express in amu.

1 amu = 1/12 of the weight of the C12 atom = 1.66 10-24 g.

NA = Number of Atoms in 1 Mole.

24

23 1

1 1

1 6610 1

6 0210

1

A

A

A

Mass Mole M gramsN

. * M mass molecule or atom in amu amu

N . mole N Density; Molecular weight in amu

NN amu N

amu amu


From Thermodynamics

1 1

19 1 1

2

8 31

5 1910

1

1

AA

v pV P

v

P V

T

. J mole KN k T RP NkT T R N k Gas cons tant

. eV mole K

dQ dE PdV

dQ dE dQ dE dVc c P

dT dT dT dT dT

E c T

lnPlnTP P R

c c Ideal Gas abovelnPT TlnT

By definit

1

1P

V VV

c R RTion ; c ; E c T

c


Gas + Black Body

The two systems gas and photons coexist. We neglect any interaction between the two constituents of the mixture except for what is needed to keep thermal equilibrium. They can be considered as two independent systems. For the Black Body (See Cox and Giuli for instance):

P=1/3 a’ T4 andE = a’ T4 V = a’ T4 / per unit mass

So that I finally have, if I consider a mass :

42

4

1 42 2

1

1

31 1

1

Total

Gas Rad

R T EE a T Density

c

P n k T a T

n m n k T a Tc c


T as a function of a

The total pressure of a system is given by:

And the density of Energy isSee Cox and Giuli Page 217:

We conserve the number of particles:

And Conserve Energy as an adiabatic expansion.

41

3P n k T a T

1 42 2

1 11n m n k T a T

c c

3 30 0n a n a

3 2 2 3 2 2

0

3 3

dQ dE PdV dE PdV

dd a c P a da a c P a

da


13 2 2 3 4 3 2 2 4 22 2

3 2 3 20 0

1 13 30 0

4 3 4 2 3 3

13 3 20 0

1 11 3

0

1 1

3 4

4 1 3 4

dn m a c c n k T a a T a c n k Ta a T a

da c c

d dn m a c n m a c

da dad dT

n k T a n a kda da

d dTa T a a T a a a T Put together

da dadT

a a T n a k n k Ta ada

4 2

13 3 3 3 20 0

3

3 31

33

3

4 1 3 4

3

4 1 41 1

3 3 3

474

3deg rees

T a

dT Ta a T n a k n k a a T a

da a

Divide by n k a

dT a T T a T

da n k a n k

Ta T; k photon entropy per gas particle

n k n cm


3 31

1

3 1

1

1

4 1 41 1

3 3 3

11 1

3

1

3 1 3 1

11

1 13

1

dT a T T a T

da n k a n k

a dT

T da

a dT dT da;

T da T aa dT

T a adiabatic exp ansion ideal gasT da

a dTT a

T da

Documents

Cosmology 2004-20051 Thermal History Prof. Guido Chincarini Here I give a brief summary of the most important steps in the thermal evolution of the Universe