# Introduction to Thermal History of Universe - Essay - Physics

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To this point, we have considered the smooth universe that is homogeneous and isotropic. Now we introduce lumpiness into the model in the form of non-homogeneous matter and energy. To do so, we must understand the matter...
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Introduction to the Thermal History of the Universe

September 22, 2006

1 Particle Physics

1.1 Motivation

To this point, we have considered the smooth universe that is homogeneous and isotropic. Now we introduce lumpiness into the model in the form of non-homogeneous matter and energy. To do so, we must understand the matter constituents of the universe. To relate these matter constituents to big bang nucleosynthesis, we must also investigate the thermal history of the universe.

1.2 Elementary Particles

Information about known particles that will be relevant to our study of the history of the universe is given in tables 1 to 5. Dark matter, which we will discuss more, likely cannot be composed of any of these types of known particles, and instead must consist of some type of substance that has not yet been observed.

1.3 Fundamental Interactions

We define a dimensionless coupling strength to parameterize each force. A comparison of the fundamental forces of nature as we currently under-

stand them is shown in 6. As shown in the table, gravity is weak compared to the other forces of nature.

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Lepton Charge Mass Mean lifetime e -1 .510998918± .000000044 MeV > 46× 1026 yr νe 0 < 2 eV oscillates µ -1 106 MeV 2.2 µsec (→ e νeνµ) νµ 0 < 190 MeV τ -1 1.78 GeV 2.9× 10−13 s ντ 0 < 18.2 MeV

Table 1: Leptons which are fermions that are fundamental particles not con- structed from quarks. Like all fermions, leptons have spin 1

2 . Leptons come in

three generations, each consisting of two members. Indirect measurements can place upper limits on neutrino masses, but these masses have not yet been observed directly. Neutrino oscillation experiments suggest a nonzero neutrino mass, although these experiments are only sensitive to mass differ- ences.

Quarks charge mass u 2

3 1.5− 3MeV

d −1 3

3− 7 MeV s −1

3 95± 25 MeV

t 2 3

1, 25± 0.009 GeV b −1

3 4.20± 0.007 GeV

t 2 3

174.1± 3.3 GeV

Table 2: Quarks, the components of Fermions that are not leptons. Particles composed of quarks are called hadrons. Mass measurements are of bound mass, since quarks have not been observed in isolation.

Note. • There is no quantum theory for the weak force alone. However, the weak force can be unified with the electromagnetic force, so the lack of an independent theory is not a serious drawback. The vector- axial (V-A) theory, proposed by Feynman and Gellman in 1958, is only correct to 0th order.

• The strong force can be studied in the context of an abstract vector space, in which the three basis vectors are the three different quark color charges and the generators are the gluons.

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Gauge Charge Mass γ < 5 · 10−30e < 6× 10−17 GeV

W± ±1 80.403± 0.0029 GeV Z0 0 91.1876± 0.0021

gluons 0 (confined)

Table 3: Spin 1 bosons. All bosons have integer spins. The first four of these bosons are the carriers of the unified electroweak theory, and the gluon carries the strong force.

Boson Spin Charge Mass Higgs boson 0 0 ? Graviton 2 0 0

Table 4: Exotic bosons. Neither have yet been observed.

Particles Quark content Charge Mass (MeV) Lifetime Proton (p) uud 2 938.277079± 0.000080 ≥ 1032 yr

Neutron (n) udd 0 939.565360± 0.000081 885.7± 0.8 sec

Table 5: Baryons, hadrons which are bound states of three quarks. Proton decay has been predicted, but has not yet been observed. The quoted lifetime of the proton is thus a lower bound. The lifetime of the neutron is unusually long because it decays via a three-body process which is highly constrained in phase space. Many heavier baryons exist, but they will not be relevant in the formation of the universe.

Gravitation Weak force EM Strong

Coupling GM2proton

~c ≈ 5.9× 10 −39 GFM2proton

(~c)3 ≈ 10 −5 α = e

2

~c = 1 137

1− 10 Classical Newton, Einstein, SR, GR none Maxwell (1864) none Quantum ? V-A (1958) QED (1949) QCD (1970’s)

Table 6: Comparison of fundamental forces. Note that gravity is weak com- pared to other forces. GF that appears in the coupling constant of the weak force is the Fermi constant.

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• The first success at unifying the fundamental forces of nature came in 1970 with the creation of the electroweak theory, which involves the SU(2)×U(1) group. The electroweak and the strong force can be com- bined to form so-called Grand Unified Theories. The first such theory was based on the SU(5) group, although this theory has since been proven false because it predicts an impossibly short proton lifetime.

• We believe the symmetries of the Grand Unified Theory to hold in the early universe until some symmetry breaking mechanism, such as the Higgs boson, splits the forces into the separate phenomena observed today.

1.4 Unit conversion tricks

1.4.1 Particle physics

In the normalized units of particle physics, ~ and c are both taken to be one. This quantity then represents the conversion factor between all units,

~c ≈ 200 fm MeV

Note. 1 Fermi is = 10−13 cm.

Thus taking ~c = 1, 1MeV ≈ 1 200

1 fm

, so that energy is proportional to inverse length.

Taking c = 1 implies that energy is also proportional to mass, via the factor c2. In these units, 1 GeV is roughly equal to the mass of a proton. The mass equivalent of a GeV is 1.67 · 10−24 g.

1.4.2 Solid state physics

In solid state physics, the Boltzmann constant is set equal to 1 so that energy and temperature are expressed in equivalent units. comes up often. A useful conversion between energy and temperature is,

kBT ≈ 1

40 eV at room temperature

For example, the CMB temperature is about 3 K, which is 1 100

of normal room temperature ≈ 300 K,

kBTCMB ∼ 1

40

1

100 eV ∼ 2.5× 10−4eV

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1.4.3 Astrophysics

The gravitational constant G can be related to the Schwarzschild radius of an object, which is defined as the distance from the object at which the escape speed becomes equal to the speed of light. For an object of mass M ,

Rsch = 2GM

c2

Note. • The classical expression for escape velocity from Newton’s law of gravitation surprisingly produces the correct numerical factor in the Schwarzschild radius.

• Rsch is a function of the mass of the body under consideration. A commonly used reference when employing the Schwarzschild radius is

a black hole with mass equal to the solar mass (although the Schwarzschild radius for such black hole is often confusingly called the Schwarzschild radius of the sun). For such a black hole, Rsch is about 3 km, thus providing a conversion factor between solar masses and length.

1.5 Planck Mass

1.5.1 Definition

The Planck mass represents the scale at which gravity and quantum mechan- ics should become unified. To find where these regimes become comparable, we compare the relevant length scales.

For a particle of massM , the corresponding length scale is the Schwarzschild radius,

R = 2GM

c2

The quantum mechanical length associated with an object of mass M is the de Broglie wavelength,

λ = hc

Mc2 =

2π~c Mc2

Note. • We showed above that ~c has units of length·energy, and so we can see immediately that this expression for a length scale has the appropriate units.

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• It is interesting that the length scales associated with the two regimes have opposite dependence on M .

Setting these two length scales equal defines the Planck mass,

2GMpl

�c2 =

2π~c Mpl�c

2

⇒Mpl ≈ √

~c G

= 1.22× 1019 GeV c2

1.5.2 The ugliest number in physics

We expect the cosmological constant to result from some version of gravity that incorporates quantum mechanics. Thus the scale of the cosmological constant should be related to the Planck mass.

The matter density of the cosmological constant is,

ΩΛ = ρΛ ρcrit

≈ 0.7 ≈ 1

Thus the energy density in Λ is similar to ρcrit, which is an observable quan- tity,

ρΛ ∼ ρcrit ∼ 1.88h2 × 1029 g cm−3

∼ 1.054h2 × 10−5 GeV cm−3

Using the conversion mentioned above, energy is proportional to length−1. Hence ρΛ should be proportional to energy

4. To compare ρΛ to the Planck mass in similar units, compare ρΛ to M

4 pl. We will see that the result is not

pretty. The conversion between length and energy is,

1 GeV = 1000 MeV = 1000 · 1 200

fm−1

= 5 ( 10−13 cm

)−1 = 5× 1013 cm−1

⇒ 1 GeV3 = 125× 1039 cm−3

∼ 1041 cm−3

Using this conversion factor with ρΛ,

ρΛ ∼ 10−5

1041 GeV4 = 10−46 GeV4

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Taking the fourth power of the Planck mass,

M4pl ∼ 104·19 GeV4 = 1076 GeV4

The dimensionless ratio of the cosmological constant matter density to the Planck mass is thus,

ρΛ M4pl

∼ 10−122

It is extremely difficult to explain a number that is so small yet nonzero using theory. The discrepancy between the observed and predicted behavior of the cosmological constant is a great unresolved problem in physics.

2 Thermodynamics

2.1 Motivation

We now do a brief review of thermodynamics. The new aspect of the subject that we will encounter in this case is that the temperature of the universe will be continually dropping due to expansion. Understanding the thermal evolution of the universe is vital to predictions about the early universe, in particular comparison of the present composition of the universe to that predicted theoretically by big bang nucleosynthesis.

2.2 Distribution function

The occupation number for a quantum state of energy  in a system that is thermal equilibrium is,

f = 1

e −µ kT ± 1

Note. • The negative sign in the denominator is taken when considering bosons, and the positive sign is taken when considering fermions.

• We will usually consider particles in the thermal radiation background, for which µ = 0. µ is defined to be d

dN , the change in energy per particle

introduced or removed from the system. The number of particles in the radiation background is constant, and so µ = 0 for these particles.

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The distribution function can be interpreted as a phase space distribution of quantum states over an area defined by three spatial coordinates and three momentum coordinates,

phase space volume = 1

h3 d3xd3p

Integrating the distribution function over phase space thus gives the number of particles contained in that volume of phase space,

dN = f(~x, ~p, t) d3xd3p

h3 · g

where g is the degeneracy of the quantum state.

2.3 Density

2.3.1 Number density

The number density of particles is given by an integral over a region of phase space,

n = 1

V

∫ f · d

3xd3p

h3 · g

Note. This is a very general formula based on the definition of the distribution function.

For fermions and bosons in thermal equilibrium, we can use the explicit form of f quoted above. Because this distribution function has no spatial dependence, the integral over space can be performed to obtain a factor of V that cancels with the factor of V −1 in the definition of n. Also, since the distribution function depends only on the magnitude of the momentum and not the direction, the momentum integral can be performed over a sphere of radius p, with the angular integrals producing a factor of 4π,

n = g

h3

∫ fd3p =

g

h3

∫ 4πp2dp

e (p) kT ± 1

Note. The most general expression for the relativistic energy of a particle is  =

√ m2c4 + p2c2.

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2.3.2 Energy density

Using the distribution function, it is also possible to calculate the energy density u, which is equivalent to the mass density via ρc2. The energy den- sity is obtained by averaging over the distribution function, weighing each quantum state according to energy,

u = g

h3

∫ ∞ 0

(p) 4πp2dp

e (p) kT ± 1

2.3.3 Entropy density

We will also need to track the entropy density of the universe,

s ≡ S V

= 1

V

1

T (U + PV − µN)

As above, for the cases of interest to cosmology, we can set µ = 0. Thus the expression for s becomes,

s = u+ P

T =

g

h3

∫ ∞ 0

4πp2dp

e (p) kT ± 1

( (P )

T + p2c2

3T

) where we have used,

P = n

3 · 〈pv〉 = n

3

〈 p2c2



〉 It is possible to evaluate these integral expressions for n, u, and s at any

point in the history of the universe for any type of particle. However, there are some limiting cases that are of particular interest.

2.4 Ultra-relativistic limit

2.4.1 Derivation

The ultra-relativistic regime is that in which kT  mc2, meaning that the kinetic energy due to thermal agitation is much greater than the rest mass of the particle. In this limit, particles are effectively massless. Most particles are not relativistic today, but were at some earlier time in the history of the universe.

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The key simplification of the relativistic regime is that  ≈ pc, i.e. that the rest mass of the particle can be neglected in comparison to the momentum. With this assumption, we can calculate the number density,

n = g

h3

∫ ∞ 0

4πp2dp

e pc kT ± 1

Introduce the dimensionless variable pc kT

= y,

n = 4πg

(2π)3

( kT

~c

)3 ∫ ∞ 0

y2dy

ey ± 1

The integral results in a constant factor, and so n ∝ T 3, which is an important conclusion.

The energy density is computed with a similar integral,

u = 4πg

(2π)3 (kT )4

(~c)3

∫ ∞ 0

y3dy

ey ± 1

Note. The temperature T that appears in these integrals can be thought of as a characteristic temperature of the particle species under consideration. Because we have assumed thermal equilibrium, all particles of a given species should be at the same temperature.

2.4.2 Ultra-relativistic bosons

Photons are an example of bosons that are always treated relativistically. The integral in the expressions for n and u takes the negative sign in the denominator for bosons, and the integral can be computed analytically in terms of the Riemann Zeta function. The general form is,

ζ(n) = 1

Γ(n)

∫ ∞ 0

yn−1dy

ey − 1 The integral in the number density corresponds to n = 3 and that in the energy density corresponds to n = 4 in this formula. Recall that the Γ function is, Γ(n) = (n− 1)!. The ζ function is numerical, but takes a simple form for even n,

n = 3 Γ(3)ζ(3) = 2 · 1.202

n = 4 Γ(4)ζ(4) = 6 · π 2

90 =

π2

15

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Energy density Substituting into the expressions above for n and u,

nUR boson = g · ζ(3)

π2

( kT

~c

)3 uUR boson = g ·

π2

30 · (kT )

4

(~c)3

Note. This expression for the energy density is the the Boltzmann distribu- tion that applies to all blackbody radiation.

Photon flux We can also compute the photon flux, which is the radiation per unit area. The numerical factor of 1

4 results from the details of the full

derivation,

Flux = 1

4 uc =

π2

60

k4

~3c2 T 4 = σT 4

where we have set g = 2 for two possible polarization states of the photon. This result is the Steffan-Boltzmann law, which states that flux ∝ T 4. The constant of proportionality is the Steffan-Boltzmann constant,

σ = π2

60

k4

~3c2 = 5 · 67× 10−8 W m−2K−4

Entropy density We can also compute the entropy density of UR bosons, but first we need an expression for P for such particles. The desired relation comes from the equation of state discussed previously. For UR particles,

P = 1

3 ρ =

1

3 u

where we have taken c = 1. The entropy density can thus be expressed in terms of quantities calcu-

lated above,

s = u+ P

T =

4

3

u

T ∝ T 3 ∝ n

Radiation temperature We derived previously how ρ depends on a using the first law of thermodynamics,

ρ ∝ a−3(1+w)

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where a is the scale factor and w is from the equation of state. For radiation, p ∝ a−4.

Since we found an expression for the energy density of UR bosons above, we can find a relationship between the temperature and the scale factor,

u ∝ T 4 and ρ ∼ u ∝ a−4

T ∝ a−1

Hence the temperature of radiation is inversely proportional to the scale factor, as we would expect from energy loss due to redshifting.

Note. At early times in the universe, many particle species were relativistic. We will talk about all such particles as a single species, for which g will not necessarily be 2. g will change as a function of time as particles fall out of thermal equilibrium. The scaling law obtained here, T ∝ a−1, is only true if g is fixed. Over the history of the universe, this scaling law is generally correct, but the coefficient will change over time.

In the next lecture, we will consider the thermodynamics of fermions.

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