Thermodynamics of the Early Universe
Fundamental Interactions
Interaction Rates
i
Rudiments of Equilibrium Thermodynamics
Entropy
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 1
Fundamental Interactions { the Standard Model
Introduction to the Standard Model:Experimental constraints
Gauge symmetry:SU(3)
C
SU(2)
L
U(1)
Y
L
1
4
F
a
F
a

{z
}
SU(3)
C
1
4
W
i
W
i

{z
}
SU(2)
L
1
4
B
B

{z
}
U(1)
Y
+ +
G
a
j
a=1;:::8
W
;Z
;A
The Higgs sector:
The minimal choice H =
G
+
(h +iG
0
)=
p
2
necessary for SU(2)
L
U(1)
Y
!
U(1)
EM
.
L (D
H)
y
D
H V (H)
for D
@
+igW
i
T
i
+ig
0
1
2
Y B
and V (H) =
2
jHj
2
+jHj
4
with Y
H
=
1
2
If
2
< 0 then h0jjHj
2
j0i =
1
2
2
v
2
2
(spontaneous symmetry breaking,the
origin of mass)
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 2
Boson masses:m
h
=
p
2v,m
W
=
1
2
gv and m
Z
= m
W
=c
W
,for c
W
cos
W
=
g=(g
2
+g
0 2
)
1=2
Fermions
fermion
T
T
3
1
2
Y
Q
i L
1
2
+
1
2
1
2
0
l
i L
1
2
1
2
1
2
1
u
i L
1
2
+
1
2
1
6
2
3
d
i L
1
2
1
2
1
6
1
3
l
i R
0
0
1
1
u
i R
0
0
2
3
2
3
d
i R
0
0
1
3
1
3
i R
0
0
0
0
i = 1;:::;N
f
= 3;
L;R
1
2
(1
5
) (parity violation),Q = T
3
+
1
2
Y
Neutrino masses:
Dirac mass:f
ij
L
i L
j R
~
H +H.c.for
~
H i
2
H
?
Majorana mass:
1
2
M
ij
i R
C
j R
+H.c.
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 3
Gauge transformations: (x)!exp
igT
i
i
(x) ig
0
1
2
Y (x)
(x)
Gauge interactions:
L
X
i
D
for D
@
+igW
i
T
i
+ig
0
1
2
Y B
Yukawa interactions:
L
3
X
i;j=1
~
ij
u
i R
~
H
y
Q
j L
+
ij
d
i R
H
y
Q
j L
+H.c.
+
if hHi 6= 0 then m
q
6= 0
L
q mass
=
3
X
i;j=1
u
i R
M
u
ij
u
j L
+
d
i R
M
d
ij
d
j L
+H.c.
for
M
u
ij
=
v
p
2
~
ij
M
d
ij
=
v
p
2
ij
) no FCNC for one Higgs boson doublet
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 4
0
@
u
1
u
2
u
3
1
A
L;R
= U
L;R
0
@
u
c
t
1
A
L;R
0
@
d
1
d
2
d
3
1
A
L;R
= D
L;R
0
@
d
s
b
1
A
L;R
U
y
R
M
u
U
L
= diag(m
u
;m
c
;m
t
) D
y
R
M
d
D
L
= diag(m
d
;m
s
;m
b
)
+
~
; diagonal (g
f
=
p
2
m
f
v
) ) no FCNC
charged currents:
P
u
i L
d
i L
= (u;c;
t)
L
U
y
L
D
L

{z
}
U
CKM
0
@
d
s
b
1
A
L
neutral currents:
P
u
i L
u
i L
,
P
d
i L
d
i L
remain unchanged upon U
L;R
,D
L;R
transformations
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 5
U
CKM
:
unitary complex N N matrix,q
i L
!e
i
i
q
i L
)
1
2
(N 1)(N 2) phases in
U
CKM
N 3 ) CP violation in charged currents
Masses in the SM:m
V
/gv m
h
/
1=2
v m
f
/g
f
v
Leptons:
m
e
<
3 eV m
<
0:2 MeV m
<
18 MeV
m
e
= 0:5 MeV m
= 105:5 MeV m
= 1:78 GeV
Quarks:
m
u
'2 MeV m
c
'1:2 GeV m
t
'174 GeV
m
d
= 5 MeV m
s
= 0:1 GeV m
b
= 4:3 GeV
Bosons:
m
W
= 80:4 GeV m
Z
= 91:2 GeV m
h
115 GeV
+
Fine tuning:
m
e
m
t
<
0:5 10
11
)
g
e
g
t
<
0:5 10
11
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 6
Introduction to the Standard Model:Experimental constraints
Perfect agreement with the existing data
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 7
The scalar sector weakly constrained
{ Higgsboson representation:
m
2
W
m
2
Z
cos
2
W
;SM ) = 1 +O()
for general Higgs multiplets: =
P
i
[
T
i
(T
i
+1)T
2
i 3
]
v
2
i
P
i
2T
2
i 3
v
2
i
data: = 1:0002
+0:0024
0:0009
) T =
1
2
(doublets are favored)
{ Higgsboson interactions:no direct tests of the scalar potential
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 8
Outstanding problems of the SM
GaugeHiggs sector:
) m
2
h
= m
(tree) 2
h
c
2
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 9
The hierarchy problem
if m
h
'100 GeV and c'
g
2
t
4
2
'
1
40
then
1 =
10m
(tree)
h
1 TeV
!
2
1:6
1 TeV
2
If 1 TeV then a ne tuned cancellation is needed to obtain m
h
'100 GeV,
e.g.for = M
Pl
= 10
19
GeV one has
1 =
10m
(tree)
h
1 TeV
!
2
2:510
30
) to avoid ne tuning
<
5m
h
<
1 TeV
+
The New Physics is expected at E'1 TeV
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 10
Why is there only one Higgs boson?
{ The Higgs eld was introduced just to make the model renormalizable (unitary)
{ There exist many fermions and vector bosons,so why only one scalar?Why,for
instance,not a dedicated scalar for each fermion?
The strong CP problem:
{ symmetries of the SM allow for
tr
F
~
F
1
2
tr (F
F
)
P
!tr
F
~
F
{ odd under CP
L
=
g
2
s
32
2
F
a
~
F
a
) neutron EDM D
n
'2:7 10
16
e cm
+
data:D
n
<
1:1 10
25
e cm )
<
3 10
10
The strong CP problem:why is so small?
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 11
The avor sector:
parity violation:
W
+
u
i
(1
5
)d
j
P
!W
+
u
i
(1+
5
)d
j
Maximal parity violation,why?
Charge quantization,why q
u
=
2
3
,q
d
=
1
3
and q
l
= 1?
Number of generations,why N = 3?
Why is the top quark so heavy (m
t
'174 GeV while m
b
'4:3 GeV)?
m
t
'v = h0jHj0i'246 GeV
+
top quark is very dierent (possibly sensitive to the spontaneous symmetry
breaking)
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 12
Mixing angles and fermion masses:
L
3
X
i;j=1
~
ij
u
i R
~
H
y
Q
j L
+
ij
d
i R
H
y
Q
j L
+H.c.
+
L
q mass
=
3
X
i;j=1
u
i R
M
u
ij
u
j L
+
d
i R
M
d
ij
d
j L
+H.c.
for M
u
ij
=
v
p
2
~
ij
;M
d
ij
=
v
p
2
ij
0
@
u
1
u
2
u
3
1
A
L;R
= U
L;R
0
@
u
c
t
1
A
L;R
0
@
d
1
d
2
d
3
1
A
L;R
= D
L;R
0
@
d
s
b
1
A
L;R
U
y
R
M
u
U
L
= diag(m
u
;m
c
;m
t
) D
y
R
M
d
D
L
= diag(m
d
;m
s
;m
b
)
+
X
u
i L
d
i L
= (u;c;
t)
L
U
y
L
D
L

{z
}
U
CKM
0
@
d
s
b
1
A
L
It is natural to expect that U
CKM
= U
CKM
(m
q
=m
0
q
).
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 13
Parameters of the SM:
m
e
m
m
m
u
m
c
m
t
m
e
m
m
m
d
m
s
m
b
g;g
0

{z
}
(
QED
;sin
W
)
;g
s
{z}
(
QCD
)
;m
h
;

{z
}
(;)
;U
CKM

{z
}
1;2;3
;
CP
21 parameters!
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 14
Cosmology:
Dark matter and dark energy
i
i
c
for
c
=
3H
2
0
8G
N
= 11h
2
m
p
=m
3
for h'0:7
data )
=
3H
2
0
'70%,
DM
'27% and
B
'3%
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 15
SM has no candidate for dark matter
=
c
'0:7 )
'10
120
M
4
Pl
= (10
3
eV)
4
while typical scale of the
SM is O(100 GeV)!Fine tuning again!
In ation:period of fast expansion of the very early Universe,R(t)'exp
q
3
t
Again the SM has no means to explain the in ation (no in aton in the SM).For a
typical in aton m
'
10
13
GeV and 10
13
,so the SM Higgs boson is not an
in aton.
Baryogenesis and SM CP violation
n
b
n
b
n
'
n
b
n
'6 10
10
The Sakharov's necessary conditions for baryogenesis:
{ B number violation
{ C and CP violation
{ Departure from thermal equilibrium
SM:
{ B number violation:OK
{ C and CP violation:too weak CP violation/ImQ,for Q U
ud
U
cb
U
?
ub
U
?
cd
(rephasing invariant)
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 16
{ Departure from thermal equilibrium:no electroweak phase transition for m
h
>
73 GeV
Conclusion:the SM doesn't explain the baryogenesis
Why is gravity so weak?Or,why M
Pl
= 10
19
GeV v = 246 GeV?
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 17
Possible extensions of the SM  a subjective view
Extra Higgs bosons
SM single Higgs doublets quite unnatural,why only one?
extra sources of CP violation from the scalar sector (needed for baryogenesis)
hope for an explanation of weak mixing angles through horizontal symmetries
L
N
H
X
=1
3
X
i;j=1
~
ij
u
i R
~
H
y
Q
j L
+
ij
d
i R
H
y
Q
j L
+H.c.
H
!H
H
;u
i R
!U
j
i
u
j R
;d
i R
!D
j
i
u
j R
;Q
i L
!Q
j
i
Q
j L
+
constraints on fermion massmatrices:
M
u
ij
=
N
H
X
=1
~
ij
v
p
2
;M
d
ij
=
N
H
X
=1
ij
v
p
2
U
y
R
M
u
U
L
= diag(m
u
;m
c
;m
t
) D
y
R
M
d
D
L
= diag(m
d
;m
s
;m
b
)
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 18
If M
u;d
suciently constraints,then U
CKM
U
y
L
D
L
= U
CKM
(m
q
=m
q
0
)
Multidoublet models favored by the measurement
An example of extra Higgs boson scenario:the 2 Higgs Doublet Model:
V (
1
;
2
) = m
2
1
j
1
j
2
+m
2
2
j
2
j
2
+m
2
3
(e
i
3
y
1
2
+e
i
3
y
2
1
)+
+
1
(
y
1
1
)
2
+
2
(
y
2
2
)
2
+
3
(
y
1
1
)(
y
2
2
)+
+
4
(
y
1
2
)(
y
2
1
) +
5
h
e
i
5
(
y
1
2
)
2
+H.c.
i
+
+
6
(
y
1
1
)
h
e
i
6
y
1
2
+H.c.
i
+
7
(
y
2
2
)
h
e
i
7
y
1
2
+H.c.
i
where m
2
i
and
i
real
under CP:
i
(t;~x)
CP
!e
i
i
?
i
(t;~x) for i = 1;2
explicit CP violation:
i
6= 0
y
1
2
CP
!e
i(
2
1
)
y
2
1
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 19
spontaneous CP violation (
i
= 0)
h
1
i =
0
v
1
p
2
h
2
i =
0
v
1
e
i
p
2
!
cos =
2m
2
3
6
v
2
1
7
v
2
2
4
5
v
1
v
2
) SCPV if 6= 0;
Diculties of extraHiggsboson scenarios:
{ many new parameters (m
2
i
;
i
;
i
)
{ treelevel FCNC to be suppressed
M
u
ij
=
N
H
X
=1
~
ij
v
p
2
;M
d
ij
=
N
H
X
=1
ij
v
p
2
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 20
Extra gauge symmetries
GUTs,e.g.SU(5):unication of gauge couplings,...
LR symmetry,SU(2)
L
SU(2)
R
U(1):spontaneous parity violation
SU(2)
L
U(1) U(1)
0
:just extra Z
0
Extra dimensions (more special dimensions)
Motivations:
Unication of gravity and gauge interactions in g
AB
(Kaluza & Klein)
Quantization of gravity (strings)
Solution (amelioration) of the hierarchy problem
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 21
The interaction rates
i
Denition of the CrossSection:
The transition matrix element w
i!f
gives the probability for the transition to occur:
P
i!f
= jw
i!f
j
2
= jhfjiij
2
The translational invariance allows to write the matrix element as
w
i!f
=
if
+i(2)
4
4
(p
f
p
i
)T
i!f
The above formula denes the transition matrix T.
Let's consider the following scattering process
a +b!c
1
+c
2
+ +c
n
We assume that b is at rest,and the velocity of a is v = j~p
a
j=E
a
.The number
of particles b per target volume is (that denes the normalization of plane waves):
2E
b
= 2m
b
as b is at rest.The incident ux is the velocity of a times their
number density 2E
a
,so 2j~p
a
j.If the reaction volume is V and the reaction takes
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 22
place during the time T,then the crosssection is dened such that the transition
probability per unit time and unit volume equals the target density the incident
ux the crosssection ,that is,2m
b
2j~p
a
j .On the other hand it is equal to
jw
i!f
j
2
=(V T).Hence summing over all available momenta for the nal state we get
(a +b!c
1
+c
2
+ +c
n
) =
=
1
4m
b
j~p
a
j
Z
n
Y
j=1
d
3
p
j
2E
j
(2)
3
(2)
4
4
(p
a
+p
b
p
1
p
n
)j
~
Tj
2
where for unpolarized initial state we have
j
~
Tj
2
=
1
S
1
(2s
a
+1)(2s
b
+1)
X
nal spins
jT
i!f
j
2
The spins of initial states are denoted by s
a
and s
b
.The symmetry factor S appears
because in quantum mechanics we can't distinguish between two nal states which
dier only by the exchange of identical particles,in general,if there are k groups of
n
i
(i = 1;2;:::;k) identical particles in the nal state,one has S = n
1
!n
2
!:::n
k
!.
In order to have the crosssection in a Lorentz invariant form one has to replace
m
b
j~p
a
j!
(p
a
p
b
)
2
m
2
a
m
2
b
1=2
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 23
For decays
a!c
1
+c
2
+ +c
n
we get instead of the crosssection the decay width
(a!c
1
+c
2
+ +c
n
) =
=
1
4m
a
Z
n
Y
j=1
d
3
p
j
2E
j
(2)
3
(2)
4
4
(p
a
p
1
p
n
)j
~
Tj
2
for
j
~
Tj
2
=
1
S
1
(2s
a
+1)
X
nal spins
jT
i!f
j
2
Summing over all nal states we get the total width
tot
=
X
nal states f
(a!f)
Then the life time is given by
=
1
tot
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 24
while the branching ratio reads
BR(a!f) =
(a!f)
tot
(a)
Strong and Electroweak Transitions:
Estimates of crosssections:
em
(e
+
e
!
+
)
e
2
4
2
1
s
for s (p
e
+ +p
e
)
2
m
2
e
where
e
2
4
QED
'
1
128
,for
p
s'100 GeV.
strong
(qq!qq)
g
2
QCD
4
!
2
1
s
for s m
2
q
where
g
2
QCD
4
QCD
'10
1
,
weak
(
e
+e
+
!
+
+
)
g
2
weak
4
2
s
(s m
2
W
)
2
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 25
where
g
2
weak
4
=
e
2
4 sin
2
W
=
QED
0:23
The Interaction Rate:
If interactions between species are fast enough they could be in local equilibrium
(state of maximal entropy).The reaction rate responsible for establishing equilibrium
can be characterized by the collision time:
t
c
1
nv
where is the crosssection,n is the number density of the target particles and v is
their relative velocity.In order to maintain the equilibrium this time must be much
shorter than the Universe age t
H
H
1
:
t
c
t
H
(1)
Then the local equilibrium is reached before the expansion becomes relevant.
Let's consider T
>
500 GeV,then the crosssection for strong and electroweak
interactions could be estimated applying just dimensional analysis for typical energy
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 26
momentum p T (masses are irrelevant at that energy)
2
T
2
where is the ne structure constant for strong or electroweak interactions '
10
1
10
2
.Taking into account that the number density of relativistic species
behaves (see next section for details) as n R
3
T
3
we obtain
t
c
1
2
T
If the Universe is dominated by relativistic species then we have (see next section for
details)
t
H
1
H
1
(
rad
=M
2
Pl
)
1=2
M
Pl
T
2
Hence we can see that the collision (reaction) time t
c
decreases slower than the
Hubble time t
H
,so if T is too large then (1) can not be satised.Note that since
rad
T
4
during the radiation dominated epoch we have H T
2
=M
Pl
(see next
section for details).Therefore at temperatures T
2
M
Pl
'10
15
10
17
GeV,we
obtain t
c
't
H
.So for T
<
10
15
10
17
GeV but above few hundred GeV (where
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 27
2
T
2
) the inequality (1) is satised and the Universe made of quarks,leptons,gauge
bosons and Higgses remain in equilibrium.Above 10
17
GeV the interaction that we
know are too slow too keep the universe in equilibrium.Below 100 GeV,the masses of
gauge bosons W
(m
W
= 80:4030:029 GeV) and Z (m
Z
= 91:18760:0021 GeV)
become relevant and the crosssections scale as
2
weak
T
2
=m
4
W
,so
t
c
1
2
weak
m
W
T
4
1
T
for T
<
100 GeV.Therefore the interactions become too slow to maintain the
equilibrium,as a consequence,e.g.neutrinos decouple at T'1 MeV (more on that
later).
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 28
Rudiments of Equilibrium Thermodynamics
Assumptions
The Universe is a dilute and weakly interacting gas.
If rates of interactions between constituents of the Universe are large enough,then
we assume the Universe is in local equilibrium (so the state of maximal entropy,
see Mukhanov for detailed discussion).
Then the number density n
i
,the energy density
i
,and the pressure for particles
with g
i
internal degrees of freedom (massless gauge boson has g=2,massive gauge
boson has g=3,massless fermion has g = 1,massive fermion has g = 2,the same for
antifermions) is given by the following integrals of the expected number density of
particles in states with energy E
i
(phase space distribution or occupancy functions)
f
i
(~p;T):
n
i
(T) = g
i
Z
f
i
(~p;T)
d
3
p
(2)
3
(2)
i
(T) = g
i
Z
E
i
(~p)f
i
(~p;T)
d
3
p
(2)
3
for E
i
(~p) = (j~pj
2
+m
2
i
)
1=2
(3)
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 29
p
i
(T) = g
i
Z
j~pj
2
3E
i
(~p)
f
i
(~p;T)
d
3
p
(2)
3
(4)
The phase space distribution (the expected number of particles in an energy state) is
given by the FermiDirac (for fermions,+ sign below) or BoseEinstein (for bosons,
sign below) distributions
f
i
(~p;T) =
1
e
[E
i
(~p)
i
]=T
1
where
i
is the chemical potential of the species,for our unit choice k
B
= 1.It
will be usually assumed that can be neglected in the early Universe.Performing
the angular integrations and changing variables from j~pj to E = (j~pj
2
+m
2
)
1=2
,so
j~pjdj~pj = EdE,so that d
3
p!4(E
2
m
2
)
1=2
EdE and we obtain
n(T) =
g
2
2
Z
1
m
(E
2
m
2
)
1=2
exp[E(~p) ]=T 1
EdE
(T) =
g
2
2
Z
1
m
(E
2
m
2
)
1=2
exp[E(~p) ]=T 1
E
2
dE
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 30
p(T) =
g
6
2
Z
1
m
(E
2
m
2
)
3=2
exp[E(~p) ]=T 1
dE
In the relativistic limit (T m) we get
n(T) =
(
(3)
2
gT
3
for bosons
3
4
(3)
2
gT
3
for fermions
(T) =
(
2
30
gT
4
for bosons
7
8
2
30
gT
4
for fermions
p(T) =
(T)
3
(5)
where (3) = 1:202:::is the Riemann zeta function of 3.
In the nonrelativistic limit (T m) there is no dierence between fermions and
bosons
n(T) = g
mT
2
3=2
exp(m=T) (T) = mn(T) p(T) = n(T)T (T) (6)
For nonrelativistic species the average energy per particle reads
hEi
n
=
(
4
30(3)
T'2:701 T for bosons
7
4
180(3)
T'3:151 T for fermions
(7)
For the rhs of Friedmann equations we need the total contribution to the energy
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 31
density and the pressure,that is
tot
= T
4
X
i
T
i
T
4
g
i
2
2
Z
1
x
i
(y
2
x
2
i
)
1=2
y
2
dy
exp(y) 1
(8)
p
tot
= T
4
X
i
T
i
T
4
g
i
2
2
Z
1
x
i
(y
2
x
2
i
)
3=2
y
2
dy
exp(y) 1
(9)
where x
i
m
i
=T and y = E=T,and it has been taken into account that some
species may have decoupled (remaining in equilibrium) so that they may have dierent
temperature T
i
.
Note that at a given temperature the ratio of the energy density for nonrelativistic
species to relativistic one reads
nrel
rel
/
m
T
5=2
e
m=T
For the species to be nonrelativistic one needs m T so the e
m=T
is a strong
suppression factor,so that we will neglect contributions from nonrelativistic species
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 32
while calculating total energy density.In that case we get
tot
(T) =
2
30
g
?
T
4
and p(T) =
(T)
3
=
2
90
g
?
T
4
(10)
where g
?
counts only massless (m
i
T) degrees of freedom:
g
?
=
X
bosons
g
i
T
i
T
4
+
7
8
X
fermions
g
i
T
i
T
4
(11)
Note that g
?
= g
?
(T) is a function of temperature.An exact form of g
?
(T) could
be easily obtained from (8) and (9).For T 100 MeV g
?
= 106
3
4
,for T 1 MeV
g
?
= 3:36,while for 100 MeV
>
T
>
1 MeV one gets g
?
= 10
3
4
(see class).
During the radiation dominated epoch (t
<
4 10
10
s see class for this number),
tot
=
rad
hence inserting (10) into the Friedmann equation one gets the very
important formula for the physics of early Universe:
H =
8G
3
tot
(T)
1=2
=
8G
3
2
30
g
?
T
4
1=2
= 1:66
g
1=2
?
T
2
M
Pl
For the radiation dominated Universe we have obtained earlier the following time
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 33
dependence of the scale factor
R(t)/t
1=2
So,for the radiation domination one has
H
_
R
R
=
1
2t
Hence the time { temperature relation could be obtained
t = 0:30
M
Pl
g
1=2
?
T
2
1 MeV
T
2
s
The above is a useful formula to memorize as T'1 MeV is a very important
temperature in the evolution of the early Universe.
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 34
Entropy
Let's dene the entropy through its dierential
TdS(V;T) d[(T)V ] +p(T)dV = d[( +p)V ] V dp (12)
In general we have
dS(V;T) =
@S(V;T)
@V
dV +
@S(V;T)
@T
dT
So,we get
@S(V;T)
@V
=
1
T
[(T) +p(T)] and
@S(V;T)
@T
=
V
T
d(T)
dT
The integrability condition tells us that
@
2
S(V;T)
@T@V
=
@
2
S(V;T)
@V @T
=)
@
@T
1
T
[(T) +p(T)]
=
@
@V
V
T
d(T)
dT
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 35
+
dp(T)
dT
=
1
T
[(T) +p(T)] () dp(T) =
(T) +p(T)
T
dT (13)
The above equation could be derived (see class) from
(T) =
g
i
2
2
Z
1
m
(E
2
m
2
)
1=2
exp[E(~p) ]=T 1
E
2
dE;p(T) =
g
6
2
Z
1
m
(E
2
m
2
)
3=2
exp[E(~p) ]=T 1
dE
Inserting (13) into (12) we get
dS =
1
T
d[( +p)V ] V [(T) +p(T)]
dT
T
2
= d
V
T
[(T) +p(T)] +const.
So the entropy,up to an integration constant is given by
S(V;T) =
V
T
[(T) +p(T)]
Recall now the"rst law of thermodynamics"(equivalently T
;
= 0)
R
3
dp(T)
dt
=
d
dt
R
3
[(T) +p(T)]
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 36
Combining with (13) we get
R
3
1
T
dT
dt

{z
}
T
d
dt
(
1
T
)
[(T) +p(T)] =
d
dt
R
3
[(T) +p(T)]
Hence
d
dt
R
3
T
[(T) +p(T)]
= 0
Therefore,identifying volume with R
3
we can conclude that the entropy in the volume
V is conserved.It proves useful to dene the entropy density
s(T)
S(T)
V
=
(T) +p(T)
T
Since for the relativistic particles both (T) and p(T) are dominated by relativistic
species,the same happens for the entropy density.Using (5) one gets:
s =
2
2
45
g
?S
T
3
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 37
where
g
?S
=
X
bosons
g
i
T
i
T
3
+
7
8
X
fermions
g
i
T
i
T
3
(14)
Since n
/T
3
:
n
=
2(3)
2
T
3
therefore one can derive the following relation
s =
4
45(3)
g
?S
n
'1:8g
?S
n
Note that the entropy conservation implies that g
?S
T
3
R
3
= const.,therefore in
the early Universe (R 0) the temperature was maximal (roughly T/R
1
),
consequently all species can be treated as highly relativistic.
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 38
Let's now illustrate the possibility of some species having dierent temperatures by
the decoupling of neutrinos at about T 1 MeV.For weak interactions we had
weak
(e
+
+e
!
i
+
i
)
g
2
weak
4
2
s
(s m
2
W
)
2
sm
2
W
'
g
2
weak
4
2
s
m
4
W
So,since hEi 3T therefore at T m
W
we get
weak
(e
+
+e
!
i
+
i
)'
g
2
weak
4
2
T
2
m
4
W
Since the interaction rate
int
t
1
c
= nv therefore we get for n T
3
and v = 1
int
'
2
weak
T
5
m
4
W
'G
2
F
T
5
where G
F
= 1:1664 10
5
GeV
2
is the Fermi constant.Let's compare the
interaction rate with the expansion rate H g
1=2
?
T
2
=M
Pl
int
H
'
G
2
F
T
5
g
1=2
?
T
2
=M
Pl
'
G
2
F
T
5
T
2
=M
Pl
'
T
0:7 MeV
3
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 39
So,at T
<
1 MeV the interactions are too slow to provide an equilibrium between
leptons and neutrinos.Neutrinos decouple ("the freezeout") from the SM and evolve
separately,so the possibility for neutrinos to have dierent temperature appears.So,
their energy (temperature) is being redshifted the same way as for photons
T
= T
dec
R
dec
R
1
R
The entropy is separately conserved for each decoupled species,so
g
?S
(RT)
3
= const.=) T (g
?S
)
1=3
1
R
Hence we can see that neutrino distribution will be the same as if it was still
in thermal equilibrium with photons as long as (g
?S
) does not change.However
around the same temperature electrons become nonrelativistic m
e
'0:5 MeV so they
annihilate e
+
e
! (the inverse process is being suppressed as the averaged energy
decreases roughly below 2m
e
).So,the number of relativistic degrees of freedom (rdf)
drops down:
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 40
for T
>
2m
e
'1 MeV:
g
?S
=
X
bosons
g
i
T
i
T
3
+
7
8
X
fermions
g
i
T
i
T
3
= 2 +
7
8
4 =
11
2
for T 2m
e
:
g
?S
= 2
From continuity of the entropy we get the following condition
g
?S
(RT)
3
before
=
g
?S
(RT)
3
after
which implies
11
2
(RT)
3
before
= 2(RT)
3
after
=) T
before
=
4
11
1=3
T
after
For the temperature"before",the neutrinos even though they decoupled a bit earlier,
have the same temperature at photons,however at T 2m
e
photons are heated
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 41
up by e
+
e
! as the entropy is transfered (since it is continues function of
T) from e
+
e
to photons.The already decoupled neutrinos do not benet from
that reheating,since for them (RT
)
3
before
= (RT
)
3
after
(in other words the entropy
of neutrinos is conserved separately after the decoupling).Consequently there is a
dierence in temperatures for neutrinos and photons after e
+
e
freezeout:
T
=
4
11
1=3
T
Strictly speaking the photon temperature does not jump at T = 2m
e
,but rather start
to decrease slower already at temperatures slightly above T = 2m
e
(in reality the
freezeout process is smooth and starts already before T = 2m
e
).
For CMB photons T
= 2:73 K,so there should be also cosmic neutrino background
with the temperature T
= 1:95 K.
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 42
Let's now determine the present energy density,number density and entropy density
for CMB photons and neutrinos assuming T
0
= 2:75 K.
g
?
=
P
b
g
i
T
i
T
4
+
7
8
P
f
g
i
T
i
T
4
2
7
8
2 3
4
11
4=3
= 1:36
g
?S
=
P
b
g
i
T
i
T
3
+
7
8
P
f
g
i
T
i
T
3
2
7
8
2 3
4
11
= 1:91
=
2
30
g
?
T
4
4:64 10
34
g cm
3
3:16 10
34
g cm
3
n =
2(3)
2
T
3
410 cm
3
149 cm
3
s =
2
2
45
g
?S
T
3
1478 cm
3
1412 cm
3
h
2
=
8G
3(H
0
=h)
2
2:47 10
5
1:68 10
5
Where I used the following conversion factors:1 K = 4:3668 cm
1
= 8:6170
10
14
GeV = 1:5361 10
37
g,1 Mpc = 1:5637 10
38
GeV
1
,G = 6:7065
10
39
GeV
2
and H
0
= h 2:1317 10
42
GeV.
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 43
(plot from Kolb&Turner)
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 44
There exists also a possibility for another kind of radiation present as a relic of the
early Universe,this is the graviton,the massless quantum excitation of a uctuation of
the gravitational eld.The reaction responsible for maintaining the equilibrium would
be e.g.
$ hh,where h is the graviton.Gravitons h
interact with ordinary
matter through the standard Lagrangian/1=M
Pl
h
T
,where T
is the energy
momentum tensor,therefore the reaction width (the inverse of the reaction rate) is
grav
T
5
M
4
Pl
Since at the early Universe H g
1=2
?
T
2
=M
Pl
therefore we get (g
1=2
?
10 for the SM
at T
>
100 GeV)
grav
H
1
10
T
M
Pl
3
So the gravitons freezeout roughly at the Planck temperature T 2M
Pl
10
19
GeV.
Using the continuity of entropy at the moment of graviton freezeout and all the SM
thresholds we get the relation between graviton temperature and the CMB photon
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 45
temperature at the present moment (see class for the discussion):
T
grav
=
g
now
?S
g
Planck
?S
1=3
T
0
'1 K
where we have approximated g
Planck
?S
by its SM value for T
>
100 GeV,i.e. 100.
Their contribution to the present energy density is
grav
T
4
0:018
.
Cosmology:5.Thermodynamics of the Early Universe,Winter Semester 2009/10 46
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