Transcript Slide 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.
Cosmology 2004-2005
1
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
Cosmology 2004-2005
2
Recombination
Saha equation : 0  0.038 H 0  72
3
2

N2
 2 me kT   kT
5
1
Ne  
e
;


13.6
eV
;
k

8.62
10
eV
deg
H

2
N1
h


N
NTot  N1  N 2 x  e and for Hydrogen N 2  N e
NTot
Ne2
Ne2
N2
x2
Ne 


NTot
N1
N1 NTot  N e 1  x
NTot  z   N0  1  z  
3
0
mp
1  z 
3

0 ,c 0
mp
1  z 
3
3H 02 0
3

1

z
 
8 G m p


N
trecombination  t  x  e  0.5 
NTot


Cosmology 2004-2005
3
Zrecombination
Cosmology 2004-2005
4
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
Cosmology 2004-2005
5
The Equations

N H 0  N a  H 0  d  N e N p  H 0 ,T  ; N H 0  N p
0
T3
0 N a  H 0  d  Ne N p  H 0 ,T  ; N e  0.5 NTot T   0.5 N0 T03

for T I use Radiation Temperature
N0  b ,0
3H 02
3H 02
T3
; N e  0.5 3 b ,0
8 Gm p
T0
8 Gm p
3H 02
T3
0 N a  H 0  d  0.5 T03 b ,0 8 Gm p   H 0 ,T 

Cosmology 2004-2005
6
18   0 
for a  H 0   6.3 10  
 
3

2h
3
c 6.3 10 18   0  d 
 
h
0
0
 
kT
e 1

2* 4 * 6.3 10 18 *  03 
d
d


B*
0  h 
0  h 
c2
  e kT  1 
  e kT  1 




3H 02
T3
II Part  0.5 3 b ,0
  H 0 ,T   C* T 3
T0
8 Gm p

I Part   N a  H 0  d  

4
h
Cosmology 2004-2005
2
3
7
Recombination
Temperature
-17
-18
Function
-19
-20
-21
-22
-23
3200
3400
3600
3800
4000
Temperature
Cosmology 2004-2005
4200
4400
8
The Agreement is excellent
Cosmology 2004-2005
9
Time of equivalence
r  t  
 r t 
c2
 a 4  t   m  t   a 3  t  teq   r  teq    m  teq 
 m  teq    m,0
a 3  t0 
a
3
t 
  r  teq    r ,0
eq
a 4  t0 
a 4  teq 
 m,0 a  t0 

 1  zeq
 r ,0 a  teq 
 m,0   c ,0  m,0
 r ,0 
T 4
c
2
 4.46 10
34
3H 02

0.3  2.91 10 30
8 G
g cm ;  r ,0 
3
 r ,0
2
0
3H
8 G
 4.6 10 5
2.91 10 30
1  zeq 
 6540
34
4.46 10
Cosmology 2004-2005
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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.
Cosmology 2004-2005
11
LG  in eV over Hubble time 
L   2.31 10


*
10
2 10 331.36 10 10 3.15 107
1.6 10 12
or in a different way
L
M
L
11
M  2 10 ;
 10 ;
 0.1 ;
 2 erg s 1 gr 1
L
M
M
 1.24 1073
0.2 2 10 112 10 331.36 10 10 3.15 107
L( in eV )  0.2 M HubbleTime 
 2.14 1073
12
1.6 10
M 2 10 112 10 33
# nucleons 

 2.5 10 68
24
mp
1.6 10
L
2 1073
Em per Nucleon 

 0.8 10 5 eV  0.08 MeV
68
# nucleons 2.5 10
The reaction H  He produces 6 MeV so that
0.08
I need only
 1.3% nucleons to react
6
Cosmology 2004-2005
12
Temperature and Cosmic Time
1 2 GMm
dr
2GM
mv 
0

2
r
dt
r

r
0
1
2
1
2
1
dr r    2GM  dt  t 
t
6 G
0
1
2
accurately  slide 16 




3
3c
t
 
4 
32

G

32

G

T



r 
2
1
2
1
4
 3 c 2   21
T 
 t
 32  G  
for t  1 s T  1.52 10 10 K Nuclear Re action are possible
Cosmology 2004-2005
13
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.
Cosmology 2004-2005
14
Gamow
During the time t I have :
Number of encounters = nvt
n = Density at the time t
 = cross section
v = thermal velocity of particles
Or I have one enclounter in
1/nvt 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/ nv t < texp or nv texp >1.
Here we obviously have a radiation
dominated Universe and  =1 (E=0
in the simple minded Newtonian
model. Let play quickly:
Cosmology 2004-2005
 r2 cross section
n
V (t=1s)
H 02
 a  8 G 
* 2
  
3
H0
a
2
a04
2
8 G   0 4 2
a
a H0

 
3
H 02
a
2
2
a
2 a0
   0   1 H 0 2
a
 a0 
15
a da  H 0 a02 dt
int egrating
1 2
1
1
a t 
a  H 0 a02 t ;  a02  H 0 a02t0 ; t0 
;
 
2
2
2 H0
a0  t0 
1
2
4
 R   R ,0
1
2
2


4
2
2
2






a0
3
H
3
H
t
t
t
0
0
  R ,c  R   1     





4
  t0  
a
8 G  t0 
8 G  2 H 0 


1
2
1
2




3
aT 4
3
3c 2
  for T  10 9  230.5 s
R 
 2 ; t 
 
2
15
4
 32 Ga  7.56 10
32 Gt
c
T 
 32 G  R 

 R a04 T 4
 4  4
 R ,0 a T0

 Given by Physics
v derived from Temp.
t from Cosmology
n(t) to be determined



From slide 15: n  v t = 1
n (t) = 1018 nucleons cm-3
Cosmology 2004-2005
16
Look at this
H 0  65 km / s / Mpc

1 
 te 
2 H e 
17
 t0  4.75 10 ;
2

a t   t  3
2 1

Dust
    t0 
a  t0   t0 
3 H0

a t 
t 
Radiation
 
a  te   te 
1
2
3
2
 a  te  
 1 
te   slide 10   
t0  


 a t  
1

z
0 
e 


T  t0   T  230.5 
a  t Deuterium 
a  t0 
1
2

1.5
4.75 10 17  8.98 10 11
a  t Deuterium  a  te 
1
2
2
3
t
 t 
  Deuterium   e  
a  t0   t e   t 0 
a  te 
2
3
 230.5   8.98 10 

 2.5 Kelvin
11  
17 
8.98
10
4.75
10

 

11
Cosmology 2004-2005
17
And
 a t  
n  t0   n  230.5  

 a  t  
0 

3
 a  t Deuterium  a  te  
 10 
 a  t  a  t  
e
0 

18
3
3
1
2


2 
3



t
t
 10 18   Deuterium   e   


  t e   t0  


3
2
1

11 3 
2
 230.5   8.98 10  
 
 1.47 10 8 cm 3  * mH



11
17
  8.98 10   4.75 10  




1.47 10 8 cm 3 1.67 10 24 g  2.4 10 32 g / cm 3 Barions
Nowadays :
3 H 02
B  B

8 G
 B h  0.02
2


3 B h 100 3 0.02  0.65  100 10 /  3.09 10 ^ 24 

8 G
8  6.67 10 8
2
2
5
Cosmology 2004-2005

2
 1.58 10 31 g cm 3
18
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.
Cosmology 2004-2005
19
The main reactions
for T
e  e
p  e
me c 2
k
n 
p 
e  n
1010
T
6 10 9
 and when T decreases e  e   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.
Cosmology 2004-2005
20
Boltzman Equation
m p c 2  938.2592 MeV ; mn c 2  939.5527 MeV ;   1.2935 MeV
nneutrons
e
n protons
mn  m p  c 2


kT
e

1.294 106 1.6 10 12  to erg 
1.38 10 16 1010
 0.22
1 Neutron
1
every
5
5 Pr otons
Neutron could  decay n  p  e  e
However it takes 15 min utes, too long !!
•
•
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.
Cosmology 2004-2005
21
ni  gi
 2 mi kT 
3
2
e
h3
 mi c 2  i
kT
; gn  g p 
Xi 
3
2
2g d
 2 ; n   p  d
3
ni
nTot
3
2
c 
2 md kT   2   m kT

nd  g d
e
3
h3
 2 
3
2
d
d
T 

m
k
 d
  md c2  d
2 
 gd 
e kT 
3
3
2
3
2
T 
T 


2
m
k
m
k

m

m

m

m

m
c






n
p
n
p
d
n
p
 d

 d
  mn  m p c2  Bd  n   p
2 
2 

kT
kT
3
e

3
e
3
3
Cosmology 2004-2005
22
3
2
T 

m
k
 mn c 2   n


n
nn
1
2 

kT
Xn 

gn
e
3
nTot nTot
3
2
T 

mp k
 m p c2   p


np
1
2 
kT
Xp 

gp 
e
3
nTot nTot
3
2
3
2
Bd


Xd
gd
md
3k T 
kT
 nTot
e




XnX p
g p g n  mn m p 
2



Plot as a function of T
Cosmology 2004-2005

23
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.
Cosmology 2004-2005
24
Deuterium
Equilibrium
Temperature
4
2
Log Xd XnXp
0
-2
-4
-6
-8
-10
0
1 10
9
9
2 10
3 10
Temperature
Cosmology 2004-2005
9
4 10
9
5 10
9
25
Helium
Cosmology 2004-2005
26
More
Cosmology 2004-2005
27
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
Abundance Relative to Hydrogen
Compare to Observations
B h2
Cosmology 2004-2005
28
After Deuterium
d  d  He  n
 p pn
d d  H3  p
 p  n  n Tritium
3
d  H  He  n  p  p  n  n
3
4
d  He3  He4  p  p  p  n  n
Cosmology 2004-2005
29
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
2 cross section

r
n
Collision so that each neutron
Collides with a proton.
Probability Q is very high so that it
V (t=1s)
Is reasonable to assume that all electrons
React.
2
Q   r n v t  # of collision  Pr obability of collision
 kT 
T 
T  10 ;  t  231 s; r 10 ; v  Sqrt    2.9 10 8 cm s 1 ; n  ncrit  b ,0  
 mn 
 T0 
Q  3.21 10 5 1  see however det ailed computation 
9
3
13
Cosmology 2004-2005
30
Finally
1 neutron  5 protons 

  1 He every 10 protons
2 neutrons  10 protons 
n
n
1
1
 0.2

  0.17  Accurate computation 0.12
p
n  p 1 p 6
n
m He
M Tot
1
4
 nTot nHe  nn
mp
mp
2
M He
Y
M Tot
1
nHe mHe m p nHe 4 2 nn 4
nn

*


2
 0.24
M Tot m p
nTot
nTot
n p
Cosmology 2004-2005
31
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.
Leprons
Neutral
Mass
e
e


 15eV


 .17MeV
 24MeV
Temperature 
Fermions

10 9.7
Massless ?


1012.1
 Move at speed of light
 Follow geo det ics
1013.3
Cosmology 2004-2005
32
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
Cosmology 2004-2005
33
The distribution function for Fermions is :
g 4 1
g 4 1
1
2
f q  i 3 3 q
 n q   i 3
q
dq ; q  E  h for photons
q
h c kT
h
e 1
e kT  1
and for the density of Energy
Bernoulli Number


x3

3  2  B4

1

2

  p  8  5.62 p 4
Here I use g = 1

0 e p x  1


Gamma



4  1
x3
1
3
  3 3  q
q dq     x
 4  1  2 3    4    4   5.68  4
0 e
c h 0 kT
1 

e 1
Riemann Zeta

16 4
 4 5.68 1.38 10  T 4  3.3 10 15 T 4  7  T 4

10 3
27 3
16
  3 10   6.625 10 
7 5 k 4 4 7
 3
T   T 4 ;   7.56 10 15 ; Density
3
c 30 h
16
Cosmology 2004-2005
34
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
e  
   e ;    
 e  e ;
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.
Cosmology 2004-2005
35
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:
3
a
S a 3 ,T  eq T   peq T 
T


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.
Cosmology 2004-2005
36
Entropy
At some Temperature we have only e  e   in thermal equilibrium


a3
s a Volume  a  
e  e    pe  pe  p &
T
1
p   c 2 Re lativistic Re gime kT  mec 2 T  5 10 9
3
0.5 106 1.6 10 12
7
9
electrons are relativistic T 

5.9
10
;




T 4


16
e
e
1.38 10
8
3
3
3
a3 
1
a3 4
 a 4
s a   Tot  Tot  
Tot 
 e   e   
T 
3
T 3
 T 3
a3 4 7
7
4
11
3
4
4
4

  T   T   T     aT  This quantity will be conserved
T 3 8
8
4
 3

3

T  5 10 9 e   e    warms up photon field  we are left with photons
a3 4
4
3
sa 
    aT  This quantity will be conserved
T 3
3
3
Cosmology 2004-2005
37
Conserve Entropy
s a3
11/3 (aT)3
4/3 (aT)3
Time
T ~ 5 109
s a3 
Temperature
 aT T 109
11
4
3
3
 11 
  aT      aT   
 
3
3
 aT T 109  4 
1
3
while neutrinos & antineutrinos are not warmed up T  T  a 1
1
3
1
3
 T 
 11 
4


1.401

T

 
 ,0
 
  T ,0  1.9 K
 11 
 T T 109  4 
Density of Radiation due to neutrinos
 R ,  3  species neutrinos * 2  neut & antineut *
4
3
7
7 4 
*  T4  6    T4 
16
16  11 
 0.68  T4
Cosmology 2004-2005
38
3  3 
1
2
0 e p q  1 q dq  p 3 for p  0
1 4
3
q 2 dq  N 3 3 3   3  k T   N 108.6 cm 3  326 cm 3
c h

n q  
gi 4 1
h3 c 3
1
e
q
kT
1
n ,0  420 cm 3
0 ,c
3H 0 2
9.72 10 30 c 2
30
3

 9.72 10 g cm 
eVcm 3  5460 eVcm 3
12
8 G
1.6 10
5460
 16.7 eV   m 
326
Observed  0.05 eV
closure with
Cosmology 2004-2005
39
Time
Cosmology 2004-2005
40
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.
 E t
 E t  m p c t p   p  c t p 
2
tp 
lp  c tp 
m p   p l p3 
3
c 5t 2p
3 2
1
c tp 
c t p  c t p  G 
G t 2p
2
G
c5
G
 1.7 10 33 cm
3
c
c
 2.5 10 5 g
G
 c5 
19
E p  mpc  
  1.2 10 GeV
 G 
2
10 43 s
1
c5
p 
 2  4 10 93 g cm 3
2
G tp G
n p  l p3 
p
3
2
c 
98
3

  10 cm
mp  G 
3
 c 5  1
32
Tp 
 
 k  1.4 10 K
k
 G 
Ep
Cosmology 2004-2005
41
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.
2
G
m
m c2 
rs
2Gm
rs 
c2
c2 G m

2
rs
with m  m p 
ts  time to cross rs  
c
G
ts 
2G
c3
c
2
G
Cosmology 2004-2005
rs 2 G m

c
c3
G
 2 tp
5
c
42
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.
lCompton  c tC 
mc

tC 
 2 
2
mc
c 
and for m  m p 
c
G
1
c
 
G 
Cosmology 2004-2005
G
 tp
5
c
43
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.
Mass 1 Mole  M  grams 
1
NA 

1.66 1024 * M  mass 1 molecule or atom in amu  amu
N A  6.02 1023 mole 1
N  Density;   Molecular weight in amu
  N  amu  N 

 amu

Cosmology 2004-2005
1  N A

amu 

44
From Thermodynamics
P  NkT 
N A k T


R

 8.31 J mole 1 K 1
R  N A k Gas cons tan t 
19
1 1
5.19 10 eV mole K
dQ  dE  PdV
T
dE
 dQ 
cv  


 dT V dT
 lnP

P   lnT
cP  cV 
T  lnP

  lnT
dE
dV
 dQ 
cp  


P

dT
 dT  P dT
E  cv T
2

 1
P
R

  Ideal Gas above 

T 

 1
T
c
R
RT
 By definition    P ; cV   1  ; E  cV T 
cV

    1
Cosmology 2004-2005
45
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 and
E = a’ T4 V = a’ T4 /  per unit mass
So that I finally have, if I consider a mass :
E  aT 
4
R T

 1

E
Density  2
c
1
PTotal  n k T  aT 4
3
1
1
1
Gas  Rad  n m  2    1 n k T  2 aT 4
c
c
Cosmology 2004-2005
46
T as a function of a
1
4
The total pressure of a system is given by: P  n k T  aT
3
And the density of Energy is
  n m
1
1
1
4



1
n
k
T

a
T


c2
c2
See Cox and Giuli Page 217:
We conserve the number of particles:
n a3  n0 a03
And Conserve Energy as an adiabatic expansion.
dQ  0  dE  PdV
 dE   PdV
d
3 2
2
d   a c   3 P a da 
 a 3 c 2   3 P a 2

da
Cosmology 2004-2005
47
d 
1
1
1 2
3 2
3
4 3 2

n
m
a
c



1
c
n
k
T
a

a
T
a c   3 n k Ta 2  a T 4 a 2


2
2

da 
c
c

d
d
 n m a 3c 2  
n0 m a03c 2   0

da
da
d 
dT
1
1
  1 n k T a 3      1 n0 a03k


da 
da
d
dT
 aT 4 a 3   aT 4 3 a 2  a a 3 4 T 3
Put together
da
da
dT 
1
a a 3 4 T     1 n0 a03k   3 n k Ta 2  4 aT 4 a 2

da 
dT 
T
1
a a 3 4 T     1 n0 a03k    3 n k a 3  4 aT 3 a 2 
 a
da 
Divide by 3 n k a 3
dT
da
 4 a T 3 1
T  4 aT 3 
1 
 3 n k  3    1    a 1  3 n k 




3
Tdeg rees 
4 a T
   74
;  k  photon entropy per gas particle
3 nk
n cm 3 
3
Cosmology 2004-2005
48
dT
da


a dT
 
T da


 4 a T 3 1
T  4 aT 3 
1 
 3 n k  3    1    a 1  3 n k 




a dT 
1
1 




1
  1   




T da 
3

  1

 a dT
dT
da

  T da  3    1 ; T   3    1 a

1 
3  1
T

a
adiabatic exp ansion ideal gas


1
1
    1  
  1
3


a dT
1


1
T

a

T da

Cosmology 2004-2005
49