Thermodynamics in the expanding universepersonal.psu.edu/duj13/ASTRO545/notes/ch4-ExpandingThermodanamics.pdfThermodynamics in the expanding universe We have shown the expansion of

Chapter 4

Thermodynamics in the expandinguniverse

We have shown the expansion of the Universe reduces the momentum of indivisual particle proportionalto the scale factor p(t)∝ 1/a(t). That means, at earlier times in the Universe, each particle moves withhigher and higher momentum. On top of that, the number density of particles increases as the cubic ofthe scale factor a3(t) to make the early universe a hot and dense state. In this chpater, we will studythe thermal history of the Universe: what happens at those hot, dense states in the early Universe?

As the Universe is an expanding and everything in the Universe changes in time, the usual notion ofequilibrium thermodynanics cannot be directly applied here. Luckly, however, for many cases, we donot have to deal with the full kinetic theory that describes the time evolution non-equilibrium state.That is, for the cases where particles interact so frequently that the interaction time scale ti = (nσv)−1

is much smaller than the expansion time scale tH = H−1:

ti tH, (4.1)

these particles are in a local equilibrium state. If this is the case, we can describes the thermodynamicalstates of particles by maximizing their entropy (this is exactly what we mean by the local equilibriumstate).

Eventurally, however, the interaction time scale typically had dropped more quickly than the Hubbletime scale, and the interaction went out of equilibrium. Then, we need to invoke the kinetic theory toset up the Boltzmann equation describing the out-of-equilibrium process. That will be the subject of thenext chapter.

4.1 Brief thermal history

We start our discussion from brief thermal history of the Universe. Before beginning, however, we haveto clarify one thing: what do we mean by the temperature of the Universe? As discovered by Pensiazand Wilson in 1965, and determined much more precisely in the early 1990s by NASA’s COBE (CosmicBackground Explorer) FIRAS (Far Infrared Absolute Spectrometer), the Unierse is filled with a gas ofmicrowave photons with a blackbody spectrum:

Iν =8πν3

exp(ν/T0)− 1, (4.2)

1

2 CHAPTER 4. THERMODYNAMICS IN THE EXPANDING UNIVERSE

with T0 = 2.726K. This gas of photons is called Cosmic Microwave Background (CMB). It is, however,important to clarify that the CMB photons are not in thermal equilibrium, although the energy distribu-tion has the form of that for the massless particle in thermal equilibrium. Thermal equilibrium impliesthe frequent energy-momentum exchange via collision between particles. But, the photon gas that weobserve in the microwave only rarely interact with electrons (recall the optical depth of τ ' 0.06 fromz = 1089 to now), and its mean-free-path is comparable to the size of the observable universe!

The extremely rare interaction, on the other hand, preserves the form of the spectral energy distribution.The only change is the frequency of photons due to cosmological redshift: a CMB photon with frequencyν0 now had ν = (1 + z)ν0 at redshift z. That is, to keep the form of the spectral energy distributioninvariant, one has to increases the temperature with the same factor T (z) = (1 + z)T0. As we havestressed before, this does NOT mean that the CMB photons are in thermal equilibrium, but it meansthat the form of the spectral energy distribution is the same as a massless boson gas in the thermalequilibrium with temperature T (z).

The fact that the spectral energy distribution of the CMB is extremely close to the Planck curve suggeststhat the CMB photons indeed were in thermal equilibrium at early history of the Universe. This happensat redshift higher than the CMB recombination time z ' 1100, before which the Universe is opaquefor the CMB photons and the CMB photons were in thermal equilibritum. The temperature of theUniverse was about 0.2 eV and the Universe was about 400,000 years after the Big-Bang. Before therecombination, the temperature of the Universe is the thermal temperature of photon.

Following is the list of important events that happens prior to the CMB recombination:

• T ∼ 3 eV, t ∼ 104∼5 yrs: matter-radiation equality; before this time, the energy density of theUniverse is dominated by radiation.

• T ∼ keV, t ∼ 105 sec: photons fall out of chemical equilibrium; at earlier times, free-free anddouble compton scattering can change the photon number rapidly compared with the expansionrate; no CMB photons are created or destroyed after this time.

• T ∼ MeV, t ∼ 100∼2 sec: Big-bang nucleosynthesis (BBN) happens; this is the time when neutronscombine with protons to form Deuterium, Tritium, Helium, Lithium, Beryllium nuclei. This isprobably the earliest epoch that we test our theoretical prediction with direct observation. Thestandard calculation of BBN matches impressively well with observed light element abundance.

• T ∼ 150 MeV, t ∼ 10−5 sec: The quantum-chromodynamics (QCD) phase transition. This is whenquarks and gluons first become bound into neutrons and protons. Before this time, u-quarks, d-quarks and gluons are in the plasma state called QGP (quark-gluon plasma).

• T ∼ 100GeV, t ∼ 10−11 sec: Electro-weak symmetry breaking. Around this energy, Higgs mech-anism takes place to give mass to weak bosons: W± and Z . Before this time, they were massless.

• T ∼ 1012 GeV, t ∼ 10−30 sec: The Peccei-Quinn phase transition, if the Peccei-Quinn mechanismis the correct explanation for the strong-CP problem. This is highly speculative.

• T ∼ 1016 GeV, t ∼ 10−38 sec: The GUT phase transition, before which the stron and electroweakinteractions are indistinguishable. This is highly speculative.

• T ∼ 1019 GeV, t ∼ 10−43 sec: The Planck scale. Fundamental strings? look quantum gravity?quantum birth of the Universe? This is all highly speculative. We do not yet know the physics atthis scale, a quantum theory of gravity.

4.2. MAXIMUM ENTROPY STATES 3

4.2 Maximum entropy states

When the scattering is frequent enough, the local equilibrium state in the expanding Universe maximizesthe entropy:

S = ln Γ , (4.3)

where Γ is the number of possible microscopic state for a given macrostate. In this section, we calculatethe maximum entropy states for bosons (with integer spin, no exclusion) and fermions (with half-integerspin, exclusion).

4.2.1 Boson-Einstein distribution function

Let us consider an ideal gas of N -bosons with total energy E in a box of volume V . We denote thenumber of bosons in energy state between ε and ε+δε as∆Nε and the number of possible microscopicstates that a particle can occupy in the one particle phase space (degeneracy) as ∆gε.

Then, we calculate the total number of possible microscopic states that the ∆Nε particles can have as

∆Gε =

∆Nε +∆gε − 1∆Nε

=(∆Nε +∆gε − 1)!(∆Nε)!(∆gε − 1)!

(4.4)

The total number of microstates for the whole system is then

Γ (∆Nε) =∏

ε

∆Gε, (4.5)

from which we calculate the entropy as

S = ln Γ =∑

ε

ln [(∆Nε +∆gε − 1)!]− ln [(∆Nε)!]− ln [(∆gε − 1)!]

=∑

ε

ln [(∆Nε +∆gε)!]− ln(∆Nε +∆gε)− ln [(∆Nε)!]− ln [(∆gε)!] + ln∆gε. (4.6)

Using Stirling’s formula ln N !' N ln N − N , we approximate the entropy as

S '∑

ε

(∆Nε +∆gε) ln (∆Nε +∆gε)− ln(∆Nε +∆gε)− (∆Nε) ln (∆Nε)− (∆gε) ln (∆gε) + ln∆gε.

(4.7)

Now we define the mean occupation number nε as

∆Nε = nε∆gε, (4.8)

to have

S '∑

ε

∆gε(1+ nε) ln [∆gε(1+ nε)]

− ln [∆gε(1+ nε)]− (nε∆gε) ln (nε∆gε)− (∆gε) ln (∆gε) + ln∆gε

'∑

ε

∆gε [(1+ nε) ln (1+ nε)− nε ln nε] , (4.9)


where we approximate ∆gε 1 in the last line. We want to maximize Eq. (4.9) with the constraintthat the total energy and number of particles in the system are

E =∑

ε

∆Nεε =∑

ε

εnε∆gε, N =∑

ε

∆Nε =∑

ε

nε∆gε. (4.10)

This can be done by using Lagrangian multiplier method with Lagrangian

L =S +λ1

∑

ε

εnε∆gε − E

+λ2

∑

ε

nε∆gε − N

=∑

ε

∆gε [(1+ nε) ln (1+ nε)− nε ln nε +λ1εnε +λ2nε]−λ1E −λ2N . (4.11)

Usign Euler-Lagrange equation, we have

∂ L∂ nε

=∑

ε

∆gε [ln (1+ nε)− 1− ln nε + 1+λ1ε +λ2]

=∑

ε

∆gε [ln (1+ nε)− ln nε +λ1ε +λ2] = 0, (4.12)

or

nε =

e−λ1ε−λ2 − 1−1

. (4.13)

To match with the usual definition in thermodynamics1, we identify

λ1 =−

∂ S∂ E

N≡ −

1T

(4.17)

λ2 =−

∂ S∂ N

E≡µ

T, (4.18)

then the occupation density in the local equilibrium becomes

nε =1

e(ε−µ)/T − 1. (4.19)

Here, T and µ are, respectively, temperature and the chemical potential, and this function is calledBose-Einstein distribution function.

1This comes from the thermodynamic identity:

dS =1T

dE +PT

dV −µ

TdN , (4.14)

ordE = T dS − PdV +µdN . (4.15)

The interpretation of chemical potential follows from Eq. (4.15) as

µ=

∂ E∂ N

S,V(4.16)

the energy required to change the number of particle when fixing entropy and volume. The minus sign in the definitionEq. (4.14) comes about to define the poticle flow from higher to lower chemical potential.

4.2. MAXIMUM ENTROPY STATES 5

4.2.2 Fermi-Dirac distribution function

Due to the Pauli exclusion principle, no two Fermi particles occupy the same quantum states. Therefore,the maximum number of particles occupying∆gε internal states is∆gε, and the total number of possiblemicroscopic states with ∆Nε particles and ∆gε degenerate states is

∆Gε =

∆gε∆Nε

=(∆gε)!

(∆Nε)!(∆gε −∆Nε)!. (4.20)

We repeat the same calculation as before. The entropy is

S = ln Γ =∑

ε

ln [(∆gε)!]− ln [(∆Nε)!]− ln [(∆gε −∆Nε)!]

'∑

ε

∆gε ln∆gε −∆Nε ln∆Nε − (∆gε −∆Nε) ln (∆gε −∆Nε)

=∑

ε

∆gε ln∆gε − nε∆gε ln nε − nε∆gε ln∆gε − (1− nε)∆gε ln (1− nε)− (1− nε)∆gε ln∆gε

=−∑

ε

[nε ln nε + (1− nε) ln (1− nε)]∆gε, (4.21)

then we maximize the Lagrangian:

L = S = −∑

ε

[nε ln nε + (1− nε) ln (1− nε)−λ1εnε −λ2nε]∆gε −λ1E −λ2N . (4.22)

The Euler-Lagrangian equation is

∂ L∂ nε

= −∑

ε

ln nε − ln (1− nε)−λ1ε −λ2∆gε = 0, (4.23)

solving that reads

nε = eλ1ε+λ2

1+ eλ1ε+λ2−1=

e−λ1ε−λ2 + 1−1

, (4.24)

and using the same identification, we find the occupation number of

nε =1

e(ε−µ)/T + 1. (4.25)

This is called Fermi-Dirac distribution function.

4.2.3 A word for chemical potential

Let us consider the following chemical reaction:

ν1X1 + ν2X2 + · · · ν′1Y1 + ν′2Y2 · · · , (4.26)

in the chemical equilibrium. One can also re-define the variables ν′i = −ν j and Yi = X j to rewrite abovechemical reaction as

K∑

i

νiX i = 0. (4.27)


The conservation of number density indicates ∀i ∈ (1, · · · , K),

δNX i

νi= const.≡ δN . (4.28)

To get the equilibrium condition let us assume that the reaction happens at constant temperature T andvolume V . In that case, we have three variables T , V , N , and the Helmholtz free energy A= E − TS isa good function to use:

dA= dE − T dS − SdT = −SdT − PdV +µdN (4.29)

from Eq. (4.15). In the equilibrium state, δA= 0, which implies

0= δA=K∑

i=1

µidNX i=

K∑

i=1

µiνiδN . (4.30)

Since δN is an arbitrary constant, we find a condition for chemical equilibrium:

K∑

i=1

µiνi = 0. (4.31)

For example, from the pair creation-annihilation process of particle X and anti-particle X :

X + X 2γ, (4.32)

and µγ = 0, we have µX = −µX in the chemical equilibrium.

4.3 Phase space in an expanding Universe

The maximum entropy states that we calculated in the previous section is the same as the equilibriumstate phase space distribution function in the usual statistical mechanics. When applying this to theexpanding Universe, however, one have to carefully define the phase space first. As a phase space is asix dimensional space of space coordinate x and the momentum coordinate p, constructing the phasespace requires to fix the time coordinate. Also, in order to enjoy the compressless property in the phasesapce thanks to Liouville theorem, momentum must be chosen to be the conjugate momentum of thespace coordinate. Of course, the coordinate can be naturally chosen in the FRW Universe that we aredealing with here, and we find the conjugate momentum from the classical action (that we choose timecoordinate as a variable):

S =12

∫

dλ

gµνd xµ

dλd xν

dλ

=12

∫

d t

d tdλ

gµνd xµ

d td xν

d t(4.33)

with constraint

gµνd xµ

d td xν

d t

d tdλ

2

= −m2. (4.34)

The conjugate momentum is

pi =∂ L∂ x i

=

d tdλ

giνd xν

d t= giν

d xν

dλ. (4.35)

4.3. PHASE SPACE IN AN EXPANDING UNIVERSE 7

That is, the conjugate momentum is simply a three spatial component of the four momentum Pµ =d xµ/dλ. Corresponding Hamiltonian is

H = pi xi − L, (4.36)

and the position x i and conjugate momentum pi satisfy Hamilton’s equation of motion:

x i =∂ H∂ pi

, pi = −∂ H∂ x i

. (4.37)

Therefore, by using Liouville’s theorem2, we can define a single-particle phase space with x i and pi .Note that the phase space volume element d3 xd3p = d x1d x2d x3dp1dp2dp3 is invariant under thecoordinate transformation (t,x)→ ( t, x) = ( t(t,x), x(t,x)) 3.

We define the distribution function as the number of particles found in the volume element d3 xd3p:

dN = f (x,p, t)d3pd3 x . (4.43)

Liouville’s theorem says in the absence of the particle sink or source, the phase space density f (x,p, t)stays the same along the path of particles in the phase space. In addition, the phase space volume isinvariant under the coordinate choice, which makes f (x,p, t) a scalar.

4.3.1 Cosmological dimming, revisited

Applying this to cosmic radiation reads cosmological dimming, or the brightness theorem. The specificsurface brightness Iν of a beam of radiation is flow of energy δu per time interval δt, normally through

2Recap on Liouville’s theorem: without source or sink of particles, the phase space density ρ(x,p, t) must satisfy thecontinuity equation in 6− d, denoted by q= (x,p):

∂ ρ

∂ t+∇i

q(ρqi) =∂ ρ

∂ t+∂ (ρ x i)∂ x i

+∂ (ρ pi)∂ pi

= 0 (4.38)

Then, convective time derivative along the trajectory is given by

dρd t=∂ ρ

∂ t+ x i ∂ ρ

∂ x i+ pi ∂ ρ

∂ pi

= −∂ (ρ x i)∂ x i

−∂ (ρ pi)∂ pi

+ x i ∂ ρ

∂ x i+ pi ∂ ρ

∂ pi

= −ρ

∂ x i

∂ x i+∂ pi

∂ pi

= −ρ

∂

∂ x i

∂ H∂ pi

+∂

∂ pi

−∂ H∂ qi

= 0. (4.39)

Here, we use the Hamilton’s equation in the last line. The Liouville’s theorem says that the phase space density is conservedalong the trajectories if there is no source or sink of particles.

3The phase space volume calculated at the constant- t-hypersurface is

d3 x d3 p =∂ ( x1, x2, x3, p1, p2, p3)∂ (x1, x2, x3, p1, p2, p3)

d3 xd3p, (4.40)

where ∂ (· · · )/∂ (· · · ) is the Jacobian of the coordinate transform

x i = x i( t, x j), pi =

∂ xµ

∂ x i

xpµ. (4.41)

Caution: here, I am emphasizing agagin that we calculate the phase space volume calculated at constant- t-hypersurface! TheJacobian is,

J = det

∂ x i

∂ x j

t=const.

det

∂ p j

∂ pk

t=const.

= det

∂ x i

∂ x j

t=const.

det

∂ x j

∂ x k

t=const.

= 1. (4.42)


the area δA, into the solid angle δΩ, and unit bandwidth δν:

Iν =δu

δtδAδΩδν. (4.44)

From Eq. (4.43), the photon number flux density per unit interval of p and solid angle is p2 f . The energyof photon is, E = p, which makes Iν = p3 f . Let pe the photon’s momentum observed at the galaxy thatemits the photon. When it get observed, the photon momentum is redshifted to po = pe/(1+ z), andbecause f is unchanged, the observed specific surface brightness is

Iνo=

Iνe

(1+ z)3, (4.45)

then the total surface brightness is

Io =

∫

Iνodνo =

∫

Iνe

(1+ z)3dνe

(1+ z)=

Ie

(1+ z)4. (4.46)

This is the cosmological dimming factor that we have calculated in the previous section! Note that itonly depends on the redshift.

4.4 Phase space density (distribution function) in the FRW universe

In the FRW universe, the phase space distribution function must be independent on the spatial coordi-nate x (spatial homogeneity), and on the direction of the momentum p (isotropy): f (x,p, t) = f (p, t).For the local equilibrium state, the distribution function should have the form that we find in the previ-ous section, and

f (p) =1

e(ε−µ)/T ± 1, (4.47)

where (+) sign correspond to the fermions, and (−) sign corresponds to the bosons.

The number of particles in the phase space volume element d3 xd3p is now written as

dN(x,p, t) = g f (x,p, t)d3 xd3p(2π)3

= g f (p, t)d3 xd3p(2π)3

. (4.48)

Here, g is the internal degeneracy factor, and we divide the phase space volume by the uncertainty cellof one dimensional size of ∆p∆x ≡ (2π). The degeneracy ∆gε that we introduced in the previoussection is then

∆gε =g

(2π)3

∫

Vd3 x

∫

E=εd3p =

gV(2π)3

∫

E=εd3p (4.49)

4.4.1 Energy density, number density, pressure

From a generic distribution function f (x,p, t), we calculate the number density, energy density andpressure as following.

• number densityThe number density of particle is simply from Eq. (4.48)

n=δNδV= g

∫

d3p(2π)3

f (x,p, t). (4.50)

4.4. PHASE SPACE DENSITY (DISTRIBUTION FUNCTION) IN THE FRW UNIVERSE 9

• energy densityFrom the relativistic expression of the energy ε =

p

p2 +m2, for a particle of mass m and mo-mentum, p2 = a2 gi j p

i p j , we find the energy density

ρ =∑

εδNδV= g

∫

d3p(2π)3

f (x,p, t)Æ

m2 + p2. (4.51)

• pressureTo get the expression for the pressure, let us consider a small area element ∆σ normal to adirection n. Particles within the spherical shell of radius R= |v|t and |v|(t +∆t) now will hit theelement between time t and t+∆t, if the velocity vector falls within the solid angle∆σ/R2(n · v).Total number of such particles within the solid angle ∆Ω is then given by:

∆Nσ =

∆σ/R2(n · v)4π

× n(v)R2|v|∆t∆Ω= n(v) (n · v)∆σ∆t∆Ω

4π. (4.52)

Here, the curly braket is the fraction of particles targetting the area element ∆σn assuming thestatistical isotropy. If the particles are reflected elastically, each transfers the momentum 2(p · n)to the target. Therefore, the contribution of the particle with velocity v to the pressure is

∆p =∑

∆Ω

2 (p · n)∆Nσ∆σ∆t

=∑

∆Ω

2 (p · n) (n · v) n(v)∆Ω

4π= 2

n(v)ε

∫

(p · n)2dΩ4π

, (4.53)

where we use p= εv. Without loss of generality, we can set n ‖ z to do the integral:

∆p = 2n(v)|p|2

ε

∫ 1

0

µ2 dµ2

∫ 2π

0

dϕ2π=

13

n(v)|p|2

ε. (4.54)

Here, we integrated over the hemisphere (µ = 0...1), where particles are facing tothe surfacetoward −n direction. Using the phase space density function, we find the pressure as

P = g

∫

e3p(2π)3

f (x,p, t)|p|2

3p

m2 + p2. (4.55)

Using the distribution function that we found in the previous section, the number density ni , energydensity ρi and pressure Pi for a given particle i are given by

ni = gi

∫

d3p(2π)3

1e(ε−µi)/Ti ± 1

, (4.56)

ρi = gi

∫

d3p(2π)3

ε

e(ε−µi)/Ti ± 1, (4.57)

Pi = gi

∫

d3p(2π)3

p2

3ε1

e(ε−µ)/T ± 1, (4.58)

where − sign is for Bosons and + sign is for Fermions, and ε =q

m2i + p2 is the energy of a particle.


4.4.2 Entropy density

Let us come back to the entropy and consider it as a function of number density, energy denstity andpressure. In terms of the phase space distribution function fε, the entropy for the boson is (Eq. (4.9))

S =∑

ε

∆gε [(1+ fε) ln(1+ fε)− fε ln fε] =gV(2π)3

∫

d3p [(1+ fε) ln(1+ fε)− fε ln fε] (4.59)

and for the fermion is (Eq. (4.21))

S = −∑

ε

∆gε [(1− fε) ln(1− fε) + fε ln fε] = −gV(2π)3

∫

d3p [(1− fε) ln(1− fε) + fε ln fε] (4.60)

Note that I simply replaced nε to fε.

Let us substitute the equilibrium distibution function

fε =1

e(ε−µ)/T ∓ 1, (4.61)

to the entropy equations above. First, we note the identity

ln (1± fε) =ε −µ

T+ ln fε, (4.62)

from which we calculate the derivative

d ln(1± fε)dε

=1T+

d ln fεdε

=1T

1− fεe(ε−µ)/T= ∓

fεT

. (4.63)

Then the expression for the entropy becomes

S =±gV2π2

∫ ∞

0

p2dp [(1± fε) ln (1± fε)∓ fε ln fε]

=±gV2π2

∫ ∞

0

p2dp [ln (1± fε)± fε (ln (1± fε)− ln fε)]

=±gV2π2

∫ ∞

0

p2dph

ln (1± fε)± fεε −µ

T

i

(4.64)

and using Eq. (4.63), the first term in the integral can be rewritten by means of integration by part as∫ ∞

0

p2dp ln (1± fε) =13

p3 ln (1± fε)

p=∞

p=0−

13

∫ ∞

0

dpp3 d ln (1± fε)dp

=±1

3T

∫ ∞

0

dpp3 dεdp

fε = ±1

3T

∫ ∞

0

dpp3 pε

fε. (4.65)

Combining all, we have the expression for the entropy

S =gV2π2

1T

∫ ∞

0

p2dp fε

p2

3ε+ ε −µ

= V

ρ + P −µnT

. (4.66)

Note that the equation above is consistnet with the standard definition of the temperature, presure andchemical potential. That is, the temperature, pressure, chemical potential are the quantities equalized

4.4. PHASE SPACE DENSITY (DISTRIBUTION FUNCTION) IN THE FRW UNIVERSE 11

between two systems when, respectively, energy (E = ρV ), volume, and the particle number (N = nV )reach the equilibrium (maximum entropy) state:

1T=

∂ S∂ E

V,N,

PT=

∂ S∂ V

E,N,µ

T= −

∂ S∂ N

E,V. (4.67)

As we have discussed earlier, in an expanding Universe, the equilibrium state exist only locally, and wedefine the entropy density as

s =SV=ρ + P −µn

T. (4.68)

One important aspect of the entropy density in the expanding Universe is the time evolution of thecomoving entropy density (sa3):

d(sa3)d t

= −µ

Td(na3)

d t, (4.69)

which is the outcome of the energy-momentum conservation ρ+3H(ρ+P) = 0, and following identities:

∂ P∂ T

µ

= s,∂ P∂ µ

T= n. (4.70)

You will show this in the homework.

Eq. (4.69) implies that the comoving entropy sa3 of a particle species is separately conserved whencomoving particle number density is conserved, d(na3)/d t = 0, or chemical potential is smaller than thetemperature (µ T , non-degenerate). This is the case for all the particle species that we are interestedin—although we do not know precise lepton number (thus, chemical potential of the nentrinos), thereis a good reason to believe that it is of the same order as the baryon number in the standard model ofparticle physics (namely, B − L is conserved, and takes a value of order unity).

When it comes to the entropy of the whole Universe, the comoving entropy of the homogeneousand isotropic universe must be conserved as there is no external souces of entropy. Again, youwill explicitly show this in the homework by combining the entropies of all particle species. The totalentropy density of universe may be written as

stot =ρtot + Ptot

T, (4.71)

and d(a3stot)/d t = 0. Therefore, the FRW universe is, in some sense, such a boring system. Every partof the universe is evolving exactly the same way (homogeneity!), without building up complexity. Thismay be similar to the stage that Boltzmann referred as ‘thermal death of the Universe’, where every partof the Universe has reached the maximum entropy state.

The story completely changes with tiny small initial fluatuations, which might be originated from quan-tum fluctuations during inflation. However small it is, the gravitational instability will enhance thestructures and evolves into the complex large-scale structures, galaxies, stars, planetary system, andeventually the life. We will talk more about it later in the class.

————————————————————————————————————————————–By the way, there is an alternative—perhaps easier—way of reaching at the same conclusion. But, Idon’t follow the path because I don’t find it easy to accept the idea of pdV work done in the expandingUniverse. For your information, I summarize you how to do that here.Starting from the second law of thermodynamics, T dS = dU + PdV and replacing U = ρV , we find


T dS = (ρ + P)dV + V dρ = d[(ρ + P)V ]− V dp. Then, use the identity (only true when µ = 0. see,Eq. (4.70)) dP/dT = (ρ + P)/T , we have

dS =d[(ρ + P)V ]

T− (ρ + P)V

dTT2= d

(ρ + P)VT

+ const

. (4.72)

From the first law of thermodynamics (energy conservation), d[(ρ + P)V ] = V dp, then dS = 0. Thatconserves the entropy per comoving volume as S ∝ sa3. This is, for example, how Kolb & Turner’stextbook derives the conservation of comoving entorpy.————————————————————————————————————————————–

4.5 Thermodynamical quantities

In the previous sections, we have calculated equilibrium phase space density in the FRW universe. Oncethe phase space density is given, we can calculate the thermodynamical quantities from Eqs. (4.56)–(4.58). But, these integrals cannot be done analytically except for some extreme cases and must bedone numerically. In this section, we study the evolution of thermodynamical quantities as temperatureof the Universe drops.

4.5.1 Ultra-relativistic particles

When particles are ultra-relativistic (T m), we ignore the mass of the particles, and Eqs. (4.56)–(4.57)reduce to

ni =gi

2π2

∫ ∞

0

dpp2

e(p−µi)/Ti ± 1(4.73)

ρi =gi

2π2

∫ ∞

0

dpp3

e(p−µi)/Ti ± 1, (4.74)

and Pi = ρi/3, which is the equation of state for the relativisitic particles.

Let us first consider chemical potential of particles in the Universe. As we have discussed before, thechemical potential of the antiparticles (X ) is given by µX = −µX . Using this, we calculate the particlenumber excess over antiparticles as

ni − ni =gi

2π2

∫ ∞

0

dpp2

1e(p−µi)/Ti ± 1

−1

e(p+µi)/Ti ± 1

. (4.75)

For fermions, the integral reads exact answer:

n(F)i − n(F)i =gi T

3i

6

µi

Ti

+1π

µi

Ti

3

. (4.76)

For bosons, we need to make an approximation, and we find

n(B)i − n(B)i 'gi T

3i

3

µi

Ti

, (4.77)

4.5. THERMODYNAMICAL QUANTITIES 13

for the non-degenerate Bose gas (µ T). Therefore, the small baryon-to-photon ratio indicates thatthe chemical potential for baryons (mostly carried by protons and neutrons) is also small, of the sameorder as the baryon-to-photon ratio: µi/T ' (n− n)/s ' η10. Then, electric neutrality of the Universedictates the chemical potential of the electron must be the same as that of proton (they share the sametemperature in equilibrium state). You will estimate the chemical potential in the early Universe in thehomework. Find out how small it is!

When chemical potential is much smaller compared to temperature, µ T , the number density andenergy density become

ni =gi

2π2

∫ ∞

0

dpp2

ep/Ti ± 1=

gi

2π2T3

∫ ∞

0

d xx2

ex ± 1, (4.78)

ρi =gi

2π2

∫ ∞

0

dpp3

ep/Ti ± 1=

gi

2π2T4

∫ ∞

0

d xx3

ex ± 1, (4.79)

where the integrals are, for bosons∫ ∞

0

d xx2

ex − 1=2ζ(3) (4.80)

∫ ∞

0

d xx3

ex − 1=π4

15, (4.81)

with ζ(x) being Riemann zeta function ζ(3) ' 1.20206. The integrals for fermions can be written interms of the integrals above because

∫ ∞

0

d xxn

ex − 1−∫ ∞

0

d xxn

ex + 1=

∫ ∞

0

d x2xn

e2x − 1=

12n

∫ ∞

0

d ttn

et − 1(4.82)

reads∫ ∞

0

d xxn

ex + 1=

1−12n

∫ ∞

0

d xxn

ex − 1. (4.83)

At temperature T (when µ T), the number density, energy density, and entropy density for bosonsare

n(B)i =ζ(3)π2

gi T3i , ρ(B)i =

π2

30gi T

4i , s(B)i =

ρ(B)i + P(B)i

Ti=

2π2

45gi T

3i , (4.84)

and for fermions are

n(F)i =3ζ(3)4π2

gi T3i , ρ(F)i =

78π2

30gi T

4i , s(F)i =

78

2π2

45gi T

3i . (4.85)

Note the number density and entropy density both have T3 dependence, which keeps the entropy perbaryon and fermion constant:

s(F)i

n(F)i

=76

s(B)i

n(B)i

=76

2π4

45ζ(3)' 4.202,

s(B)i

n(B)i

' 3.602 (4.86)

average energy per particles drops proportional to the reciproccal of temperature:

ρ(F)i

n(F)i

=76

ρ(B)i

n(B)i

=76

π4

30ζ(3)Ti ' 3.151Ti ,

ρ(B)i

n(B)i

' 2.701Ti , (4.87)


which is consistent with the cosmological redshift.

For low-z universe, T ® 10 keV, when the photon (boson with gi = 2 for two polarization states)and massless neutrinos (fermion with gi = 2 for two helicity states: νL , νR) are the only relativisticparticles, both photon temperature and neutrino temperature scales as Ti ∝ a−1, and the expressionsabove are consistent with the result from particle number conservation n∝ a−3 and energy-momentumconservation ρR∝ a−4. For high-z universe, T ¦ 1 MeV, this simple temperature scaling does not workanymore because of the events such as QCD phase transition, EW phase transition, etc, which changerelativistic particle content. We will come back to this point in the next section.

4.5.2 Non-relativistic particles

For non-relativisitic particles, m T , the exponential factors in the denominator of the distributiondominates over ±1 and fermions and bosons follows essentially the same distribution function:

f (ε)' e−(ε−µi)/Ti ' e−(m−µi)/Ti e−p2/(2mTi), (4.88)

where we use ε ' m + p2/(2m) + · · · . We calculate equilibrium number density with the distibutionfunction as4

ni =gi

2π2e−(m−µi)/Ti

∫ ∞

0

dpp2e−p2/(2mTi) =gi


14

Æ

π(2mTi)3

=gi

mTi

2π

3/2

e−(m−µi)/Ti , (4.93)

which makes the particle abundance over antiparticle as

ni − ni =2gi

mTi

2π

3/2

e−m/Ti sinh

µi

Ti

. (4.94)

That is, the antiparticle number density is suppressed compared to the particle number density by afactor of

ni = nie−2µi/Ti . (4.95)

4Gaussian integral can be most conveniently done from∫ ∞

−∞d xe−x2

2

=

∫ ∞

−∞d xe−x2

∫ ∞

−∞d ye−y2

=

∫

d2re−r2= 2π

∫ ∞

0

drre−r2= π

∫ ∞

0

d(r2)e−r2= π. (4.89)

Equivalently,∫ ∞

−∞d xe−ax2+bx = e

b24a

∫ ∞

−∞d xe−a(x− b

2a )2

= eb24a

s

π

a. (4.90)

From here, you have two choices to calculate the integral of form∫ ∞

−∞d x x2e−ax2

=−d

da

∫ ∞

−∞d xe−ax2+bx

b=0

=12

s

π

a3

=∂ 2

∂ b2

∫ ∞

−∞d xe−ax2+bx

b=0

=s

π

a

dd b

b2a

eb24a

b=0=

12

s

π

a3. (4.91)

Of course, they agree. Choose one that you like. By the way, if you master this way of doing the Gaussian integral, then youare ready to solve quantum field theory! I use the former to get

∫ ∞

−∞d x x4e−ax2 =

34

s

π

a5. (4.92)

4.5. THERMODYNAMICAL QUANTITIES 15

Energy density is

ρi =gi


∫ ∞

0

dpp2Æ

m2 + p2e−p2/(2mTi)

'mn+gi


12m

∫ ∞

0

dpp4e−p2/(2mTi) = n

m+32

Ti

, (4.96)

and pressure is

Pi =gi


∫ ∞

0

dpp2 p2

3p

m2 + p2e−p2/(2mTi)

'gi


13m

∫ ∞

0

dpp4e−p2/(2mTi) = nTi , (4.97)

that gives the entropy density as

si =ρi + Pi +µin

Ti=

m−µi

Ti+

52

ni . (4.98)

Comparing equilibrium number density of non-relativistic particles, Eq. (4.93), with that of relativisticparticles in Eqs. (4.84)–(4.85), the particle number density of non-relativistic particles (m T) issuppressed by a factor of

n(NR)i

n(B)i

∼m

T

3/2e−m/T , (4.99)

for particles with µ/T 1, and the entropy density is also suppressed by the same exponential factor. Inthe homework, you will calculate contribution of proton entropy to the total entropy at T 'MeV (whenprotons are non-relativistic).

4.5.3 The end of cosmic-antiparticle-background

The exponential factor in the equilibrium density of the matter particles comes about because it isharder and harder to create the particle-antiparticle pairs from the thermal bath when average energyof photon is less than the rest-mass energy of the particle: ⟨E⟩ ∝ T ® m. But, still, because we have somany photons for one baryon, the photons at the exponential tail of the Planck curve can produce somepairs, but not as much as when the particles were relativistic because number of high energy photonsis exponentially suppressed in the Wein tail.

Eventually, when temperature of the Universe drops sufficiently below, T m particle-antiparticle pairproduction essentially stops so that we can completely ignore the antiparticle. Let us estimate whenthis happens. We do it as a function of

β =n− n

s' 10−9 (4.100)

and from Eq. (4.93),nns2'm

T

3e−2m/T . (4.101)


Then, n/s and −n/s are two solutions of the quadratic equation

x2 + β x −m

T

3e−2m/T =

x +β

2

2

−β2

4−m

T

3e−2m/T = 0. (4.102)

Solving the equation, we find

ns=

√

√

√

β

2

2

+m

T

3e−2m/T +

β

2,

ns=

√

√

√

β

2

2

+m

T

3e−2m/T −

β

2. (4.103)

That means, the antiparticle number density is neglegibly small if (β/2)2 (m/T )3 exp[−2m/T]. Tak-ing the log (and now > may be enough to ensure),

ln

β

2

>32

logm

T

−mT

. (4.104)

We iteratively invert it as log(m/T ) changes much slower compare to m/T :

mT> ln

2β

+32

ln

ln

2β

+ · · ·

. (4.105)

For the observed value of β ' 10−9, we find

T ®m

ln(2/β)'

m21

. (4.106)

That is, when the temperature of the Universe drops below 5 % of the mass of the particle, then particle-antiparticle pair production ceases and we can practically ignore the antiparticles. For protons andneutrons, this happens at T ® 50MeV, and for for electrons, this happens at T ® 20 keV. Again, Iemphasize one more time that it is because we have about a billion times more photons than baryonsand electrons.

4.6 Thermal history of the early Universe

In the previous section, we study the energy density, pressure, number density and entropy for twoextreme cases: ultra-relativistic and non-relativistic particles. Let us now turn to the thermal history ofthe entire Universe. The goal of this section is to calculate the time evolution of the temperature andscale factor in the early Universe.

The various events that we have discussed at the beginning of this chapter affect the thermal evolutionof the Universe. As the Universe expand and temeprature drops, two main things happens: [1] whentemeprature of the Universe drops below the mass energy of some particle, the particle undergoesrelativistic-to-nonrelativistic transition, then particle-antiparticle pair production stops at T ® m/20.[2] when temperature of the Universe drops below the symmetry breaking scale, some drastic changehappens to the nature of particles involved in the symmetry: e.g. QCD phase transition, EW phasetransition.

4.6.1 Effective relativistic degrees of freedom

As we have seen in the previous sections, once a particle speices transits from relativistic to nonrela-tivisitc, then the number density, energy density and entropy of the particles are significantly reduced.That is, the thermal evolution of the early Universe is controlled mainly by the relativistic particles.

4.6. THERMAL HISTORY OF THE EARLY UNIVERSE 17

10−2 10−1 100 101 102

mi/T

0.0

0.2

0.4

0.6

0.8

1.0

rela

tivis

ticde

gree

sof

free

dom

peri

nter

nals

tate

gB⋆

gF⋆

gB⋆s

gF⋆s

10−2 10−1 100 101 102 103 104 105 106

Tγ [MeV]

101

102

effe

ctiv

ere

lativ

istic

degr

ees

offr

eedo

m

g⋆g⋆s

Figure 4.1: (Left) Efective relativistic degrees of freedom as a funciton of x i ≡ mi/T . As temperaturedrops, the particle species becomes non-relativistic and the effective degrees of freedom is reduced forT < mi . (Right) Efective relativistic degrees of freedom as a funciton of the photon temperature Tγ inthe early Universe. We include all the standard model particles and assuming that QCD phase transitionhappens sharply at Tγ = 150 MeV.

But, the relativistic-nonrelativistic transition does not happen instantaneously, as we have seen in theprevious section, and one must take into account the smooth transition properly to calculate correctthermal history.

This motivates us to defin the effective relativistic degrees of freedom as the coefficient between tem-perature and the total energy density (g?) and the entropy ednsity (g?s):

ρ(T ) =π2

30g?(T )T

4 (4.107)

s(T ) =2π2

45g?s(T )T

3. (4.108)

n(T ) =ζ(3)π2

g?n(T )T3. (4.109)

Compare these two equations to Eq. (4.84), we find that relativistic bosons contribute

∆g? =∆g?s =∆g?n = gi , (4.110)

and relativistic fermions contribute

∆g? =∆g?s =78

gi , ∆g?n =34

gi . (4.111)

to g? and g?s, repectively, when the particles are in the equilibrium states with photon (T here is thetemperature of photons). If some particles are thermally decoupled from photon bath and evolvesindependently (with temperature T ′, e.g. neutrinos at T ® 1.5 MeV), the contribution to the effectiverelativistic species is revised as

∆g? = gi

T ′

T

4

, ∆g?s = gi

T ′

T

3

, ∆g?n = gi

T ′

T

3

(4.112)


for Bosons, similarly, for Fermions with appropriate factors (3/4 for g?m, 7/8 for g? and g?n).

The effective relativistic d.o.f for massive particles changes drastically when temperature is around themass of the paticle Ti ∼ mi . When ignoring the chemical potential (this is a good approximation as wesee in the previous section), we find that

g?n,i =π2

ζ(3)T3gi

∫ ∞

0

p2dp2π2

1

eq

p2+m2i /Ti ± 1

=gi

2ζ(3)

Ti

T

3∫ ∞

x i

udu

q

u2 − x2i

eu ± 1(4.113)

g?,i =30π2T4

gi

∫ ∞

0

p2dp2π2

q

m2i + p2

eq

p2+m2i /Ti ± 1

=15π4

gi

Ti

T

4∫ ∞

x i

u2du

q

u2 − x2i

eu ± 1(4.114)

g?s,i =45

2π2T3

gi

Ti

∫ ∞

0

p2dp2π2

m2i + 4p2/3

q

m2i + p2

eq

p2+m2i /Ti ± 1

=15π4

gi

Ti

T

3∫ ∞

x i

u2du

q

u2 − x2i

(eu ± 1)

1−x2

i

4u2

.

(4.115)

For the second equalities, we simplify the integral by setting u2 ≡ (m2i + p2)/T2

i , so that udu= pdp/T2i

and define x i ≡ mi/Ti: At x i → 0 limit, we reproduce the formula for the relativistic particles.

Occasionally, we need to evaluate the derivatives of g?’s with respect to the temperature Ti:

d g?n,i

d ln Ti= −

d g?n,i

d ln x i=

gi

2ζ(3)

Ti

T

3∫ ∞

x i

udu1

eu ± 1

x2i

q

u2 − x2i

, (4.116)

and similarly,

d g?,id ln Ti

=15π4

gi

Ti

T

4∫ ∞

x i

u2du1

eu ± 1

x2i

q

u2 − x2i

, (4.117)

d g?s,id ln Ti

=−15π4

gi

Ti

T

3∫ ∞

x i

u2du1

eu ± 1

3x2i (x

2i − 2u2)

4u2q

u2 − x2i

. (4.118)

4.6.2 Thermal history with standard model of particle physics

Now let us consider the effective relativistic degrees of freedom in our Universe with the standard modelparticles. This may be the correct description of the thermal history at temperature T ® TeV —exceptfor some exceptional cases where beyond-standard-model effects show up at lower energy scales (e.g.with massive neutrinos). Let use first review the particle contents in the standard model.

The standard model of particle physics contains three families of fermion particles, which are doubletunder the electro-weak interaction. In the lepton sector, they are (e,νe), (µ,νµ), (τ,ντ), and in thebaryon sector, they are (u, d), (c, s), (t, b) quarks. On top of the electric charge, each quark has one ofthree color charges (r,g,b). These are all spin-1/2 fermions with gi = 2 and all have correpsonding anti-particleis except for neutrinos. In the standard model, neutrinos care about the handedness (helicity,or chirality), so that all neutrinos in standard model are left-handed: so, we write them as νL. Allanti-neutrinos, at the same time, are right-handed (νR)5. When all the standard model fermions are

5This picture may be changed with massive neutrino, which indicates the physics beyond the standard model—neutrinosare massless in the standard model. There, right handed neutrinos (with a large mass so that we cannot observe them in thelab experiments up until now) are possible.


relativistic, therefore, the relativistic degrees of freedome from them is

gF =3(r, g, b)× 2(↑,↓)× 6(u, d, c, s, t, b)× 2(q) + 2(↑,↓)× 3(e,µ,τ)× 2(e, µ, τ) + 3(ν′s)× 2(ν′s)

=90. (4.119)

Bosons in the standard model either mediate three fundamental interactions—electromagnetic interac-tion (γ), weak interaction (W±, Z0), and strong interaction (g, 8 gluons)— or controls the electroweaksymmetry breaking (H, the Higgs boson). The electroweak symmetry breaking, where the Higgs fieldmoves away from the origin to its vacuum-expectation-value (VEV) v ' 246 GeV, happens around thecritical temperature6 TEW ' 196.7 GeV. The W± bosons and Z0 bozon get their mass after this criticaltemperature. Therefore, around T ' 200GeV, all the standard model bosons are massless, and theircontribution to the relativistic degrees of freedom is

gB = 2(γ) + 3× 3(W±, Z0) + 2× 8(g) + 1(H) = 28. (4.120)

Here, I use that the photon and gluons are massless spin-1 particles with g = 2, and W and Z bosonsare massive spin-1 particles with g = 3. The Higgs boson is a scalar (s = 0) particle and therefore g = 1.At around T ' 200GeV, the relativistic degree of freedom is

g? = g?s = 28+ 90×78= 106.75, g?n = 95.5, (4.121)

which makes

ρ(T )'35.12T4 ' 8.149× 1024

T100 GeV

3

g/cm4

s(T )'61.98T3 ' 8.056× 1048

T100 GeV

3

cm−3

n(T )'11.6313T3 ' 1.514× 1048

T100GeV

3

cm−3, (4.122)

when the temperature of the Universe is

T = 1.160× 1015

T100GeV

K. (4.123)

Yes, the whole Universe was in a hot-dense state! Note that if there is no other physics beyond thestandard model, then the estimation given above should work all the way to the Planck scale—althoughfor scales above electroweak phase transition, we should change gW and gZ to 2 as they are masslessfor T > TEW.

After the electroweak phase transition, standard model particles become non-relativistic and drop outof the thermal bath in the descending order of mass: t = 173 GeV, H = 125 GeV, Z0 = 91 GeV,W± = 80 GeV, b = 4 GeV, τ = 1.777 GeV, c = 1 GeV. At around T ∼ 150 MeV QCD phase transitionhappens so that free quarks and gluons form baryons such as protons (p), and neutrons (n) as wellas mesons (π0, π±). Then, pions π ' 139 MeV and µ = 106 MeV go non-relaivistic. We show theevolution of g? and g?s during this period in the right panel of Fig. 4.1.

The Tablea 4.1 summarizes the particle content of the standard model of particle physics.

6The critical temperature of the electroweak symmetry breaking is Tc = mH/g, where mH is the mass of the Higgs bosonand g is the weak coupling constant, GF =

p2/8(g/mW )2 = 1.1663787(6)× 10−5 GeV−2. Now we know this number pretty

well, because we have measured the mass of the Higgs boson from LHC (mH = 125.7± 0.4GeV).


symbol mass spin anti-particle gu 2.3 MeV 1/2 u 2(u, u)× 2(↑,↓)× 3(color)d 4.8 MeV 1/2 d ”c 1.275 GeV 1/2 c ”s 95 MeV 1/2 s ”t 173.21 GeV 1/2 t ”b 4.18 GeV 1/2 b ”e 0.511 MeV 1/2 e+ 2(e, e+)× 2(↑,↓)µ 105.7 MeV 1/2 µ+ ”τ 1.777 GeV 1/2 τ+ ”νe <2 eV 1/2 νe,R 2(νe,L , νe,R)ντ <2 eV 1/2 ντ,R ”νµ <2 eV 1/2 νµ,R ”γ 0 1 2g 0 1 8(gluons)× 2

W+ 80.4 GeV 1 3W− 80.4 GeV 1 3Z 91.2 GeV 1 3H 125.7 GeV 0 1

Table 4.1: Particle content of the standard model of particle physics. Data from particle data group(http://pdg.lbl.gov/index.html).

4.6.3 Neutrino decoupling and e/e+-annihilation

At high temperatures, neutrinos are in thermal equilibrium with cosmic plasma by electron-neutrinoscattering with weak interaction cross section given by

σwk ' G2F T2 (4.124)

where GF ' 1.166× 10−5 GeV−2 is the Fermi constant for weak interaction coupling. When electronsare relativistic, the number density of electrons are

ne∝ T3, (4.125)

that give rises to the collosion rate as

1/tcoll = neσwk ' O (1)G2F T5. (4.126)

Now we compare this to the Hubble time scale

1/tH = H ' O (1)

√

√8πG3

T2. (4.127)

The electron-neutrino scattering freezes out when

tcoll > tH→ G2F T5 < O (1)

√

√8πG3

T2, (4.128)

or

Tν−decoupling < O (1)

8πG

3G4F

1/6

' O (1)0.0012GeV. (4.129)


More accurate calculation show that the numerical coefficient is in fact very close to unity and Tνe'

1.34MeV, Tνµ,ντ ' 1.5 MeV for different neutrino species [?]. Therefore, after T ® 1.5MeV, neutrinosdecouple from the cosmic plasma of photon, electron and positron; then, the neutrinos evolve indepen-dently by conserving their comoving entropy separately from the other particles.

Neutrino decouples from the thermal bath right before the e−/e+ annihilation period happens at aroundT = me ' 0.511 MeV. Before e−/e+ annihilation time, although decoupled, temperature of neutrinosis the same as that of photons during its free-streaming as for both cases T = Tν∝ 1/a (we will showthat in the next section with kinetic theory). But, the pair annihilation of electrons transfer entropy onlyto the photons, and not to neutrinos, and, as a result, the temperature of photon drops a bit slower than1/a leading to the heating of photon relative to neutrinos.

The final temperature difference between photon and neutrinos can be calculated by using the conser-vation of comoving entropy of the cosmic plasma consisting of photon, electron and antielectron. Beforee−/e+ annihilaiton, all three particles are relativistic, and we have

gγ (before)?s = 2+

78× 2× 2=

112

. (4.130)

Here, gγ?s refers to the relativistic degrees of freedom counting only for the species coupled to photon.After the annihilation is complete (T ® me/21= 24.3keV), only photon remains to be relativistic and

gγ (after)?s = 2 (4.131)

The conservation of comoving entropy before and after the e-annilation reads

a(before)3

gγ (before)?s

T (before)3=

a(after)3

gγ (after)?s

T (after)3

, (4.132)

then

T (after) =

gγ (before)?s

gγ (after)?s

1/3a(before)

a(after)T (before) =

gγ (before)?s

gγ (after)?s

1/3

T (after)ν . (4.133)

Plugging in the numbers, we findTTν=

114

1/3

, (4.134)

at T < Tν ' 1.5MeV. Using this, we calculate the neutrino temperature now is

Tν =

411

1/3

TCMB ' 1.946K. (4.135)

You will calculate the details of this process in the homework.

One final remark. Like any other processes in nature, the neutrino decoupling process does not happeninstantaneously. That is, the e + νe e + νe scattering is not completely shut off at T < Tν, andsome, albeit small, energy transfer from electron-positron pair to neutrino sector is allowed to heat theneutrino temperature and lead to the spectral distortion of neutrino. Result of this leackage is oftenparameterized as the effective number of neutrino species, Neff, and a detailed study of this processshows that Neff = 3.035 by using the standard model of particle physics [?]. There are additional∆Neff = 0.011 from the plasma effects (QED correction at finite temperature) that changes the neutrinointeraction rate [?], to make the total Neff = 3.046 [?,?]. You can measure Neff from the abundance ofHelium and Deuterium as well as the temperature anisotropy of CMB. We will come back to this pointlater in the lecture. But, for now, let’s ignore and stick on N = 3 families of neutrino species.


10−1210−1010−810−610−410−2

100102104

age

ofth

eU

nive

rse

(s)

exactapprox.

10−2 10−1 100 101 102 103 104 105

Temperature (MeV)

−0.14−0.12−0.10−0.08−0.06−0.04−0.02

0.00

resi

dual

(app

rox.

/exa

ct-1

)

100 101 102 103 104

age of the Universe (s)

10−2

10−1

100

Tem

pera

ture

(MeV

)

neutrino temperatureTCMBa0/a(t)

photon temperature

Figure 4.2: Left: (Top) Age of the Universe as a function of temperature from exact integration,Eq. (4.143), and constant-g? approximation, Eq. (4.140), and its residual (Bottom). The approximationworks pretty well except for the abrupt change at QCD phase transition. Even so, the residual is only14% at maximum. Right: Time evolution of the photon temperature and neutrino temperature duringe−/e+-annihilation.

4.6.4 Temperature and redshift

We have thus far written all quantities as a function of temperature of the Universe. How is the temper-ature of the Universe related to the scale factor? We use that the comoving entropy density sa3 staysconstant along the evolution. From Eq. (4.108), the comoving density is

g?sT3a3 = const, (4.136)

then we calculate the temperature-redshift relation, through a(1+ z) = a0, as

T (a) =

g?s0g?s

1/3 a0

aTCMB. (4.137)

Here, TCMB = 2.726 K is the temperature of CMB at present, g?s0 is the relativistic degrees of freedomat present:

g?s0 = 2+78× 2× Neff ×

Tν0

TCMB

3

' 3.909, (4.138)

where Neff = 3 is the effective number of neutrino species, and Tν0 ' 1.946 K is the present day neutrinotemperature.

4.6.5 Redshift and time

The observational bound of the spatial curvature of the Universe at present is |Ωk| ® 0.01, that meanswe can completely ignore the curvature fraction at earlier times. Then, the Friedmann equation is

3H2 = 8πGρ = 8πG

π2

30g?(T )T

4

. (4.139)


When g?(T ) is constant, H2∝ t−2∝ T4 yield H = 1/(2t) and we have

34t2= 8πG

π2

30g?(T )

T4 −→ t =

4516π3Gg?(T )

1/2 1T2= 2.420g−1/2

? T−2MeV s. (4.140)

For example, at T = 1MeV, the relativistic degree of freedome is

g?(1MeV) = 2+78(2× 2+ 2× 3) = 10.75, (4.141)

and the Universe is t ' 2.420/p

10.75 = 0.738s old, and at T = 200GeV, the relativistic degree offreedom is g?(200GeV) = 106.75 and the Universe is t = 5.8556× 10−12 s old. Corresponding Hubbleparameter is

H ' 1.360× 10−25pg?T2MeV GeV '

4.831 g−1/2? T−2

MeV s−1'

2.081 g−1/2? T−2

MeV R−1

. (4.142)

That means, the physical size of the Universe at 1MeV is about 0.6 R.

But, as you can see from the right panel of Fig. 4.1, g? is not always a constant, and rather have a broadfeature whenever there is a thermal event. In that case, we can form a differential equation for a(t) byusing Eq. (4.139) along with Eq. (4.137) as:

3H2 = 3

1a

dad t

2

=8πGπ2

30g?(T )

g?sp

g?s(T )

4/3 ap

a

4T4

p , (4.143)

where the subscript p refers to some pivot scale to which we normalize the temperature and scale factor.If we choose current CMB temperature as the pivot scale, then the equation becomes

3H2 =8πGρrad

g?(T )g?0

g?s0g?s(T )

4/3 a0

a

4= 3H2

0Ωrad

g?(T )g?0

g?s0g?s(T )

4/3 a0

a

4, (4.144)

with

g?0 = 2+78× 2× Neff ×

Tν0

TCMB

4/3

' 3.363, (4.145)

being the relativistic degrees of freedom g?0 at present time.

To see the effect of time-varying relativistic degrees of freedom, let us assume that nothing has happenedbefore Tp = 200GeV, and the relativistic degrees of freedom is fixed to g? = g?s = 106.75 for T ¦200GeV. Then, from the analysis above, tp = 5.8556× 10−12 s old, and the scale factor (and redshift)is

ap

a0=

g?s,0g?s,p

1/3 TCMB

Tp= 3.90× 10−16, (4.146)

and zp = 2.56× 1015. We then solve the differential equation by integrating

1y

d yd x=tp

√

√

√8πG3π2

30g?(y)

g?sp

g?s(y)

4/3 1y4

T4p =

√

√

√14

g?(y)g?p

g?sp

g?s(y)

4/3 1y2

'1.08901

q

g?(y) [g?s(y)]−4/3

y2, (4.147)


where we rescale the variables as y = a/ap and x = t/tp, and use Eq. (4.140) to eliminate constants.The solution for the differential equation is

∫ y

1

d y y

√

√

√[g?s(y)]4/3

g?(y)= 1.08901(x − 1). (4.148)

We show the result in Fig. 4.2. It turns out that the constant-g? approximation in Eq. (4.140) workspretty well!

Documents

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