Thermal history of the Universe

The expansion of the Universe is now well described by the flat ΛCDM model ($\Omega_k^0=0$). The proportions of each of these components are now estimated at Planck Collaboration et al. (2020):

In this chapter, we will study the thermal history of the Universe and the evolution of its composition. Up to now in this course, non-relativistic matter has been treated as a single entity, slowing down the expansion of the Universe through its gravitational interaction. But to study its evolution with temperature and its interactions with the other components, we need to separate them into two contributions: dark matter $\Omega_{c}^0$ and baryonic matter ^[1] $\Omega_b^0$. In 1933, while studying the Coma cluster, astrophysicist Fred Zwicky showed that the mass deduced from the motion of the seven galaxies that make up the cluster was 400 times greater than the mass deduced from counting luminous objects. This measurement was repeated in 1936 on the Virgo cluster, this time giving a factor of 200. However, these somewhat inaccurate measurements were forgotten until the 1970s, when astronomer Vera Rubin observed that the rotational speed of the stars in the Andromeda Galaxy was much higher than its observed luminous mass would suggest Rubin & Ford, W. Kent, 1970. The observation is quickly repeated for many galaxies: part of the matter making up the galaxy is therefore dark matter, escaping detection and often representing the majority of the total mass of galaxies. The presence of abundant dark matter can even be seen in the amplitude of the temperature anisotropies of the cosmic microwave background (see end of chapter). Today, it is estimated that the proportion of these two forms of cold matter is Planck Collaboration et al. (2020):

1Description of the primordial Universe¶

The Cosmic Microwave Background¶

If the Universe is expanding today, then it was smaller in the past. Cosmic expansion reduces the momentum of particles by a factor $\propto a^{-1}$ and the density of particles by another $a^{-3}$. In the early days, the Universe was therefore hot and dense. There must therefore have been a moment when the Universe was hot enough for atoms to be ionised, and therefore in a plasma state where photons interact with free electrons. Through these frequent interactions, if thermodynamic equilibrium is reached the radiation follows a blackbody spectrum defined by the temperature of the medium (wiki:Planck’s_law). During the transition from the plasma state to the neutral state, at around $3\,000\,\kelvin$ for a hydrogen gas, the Universe suddenly becomes transparent and photons propagate freely. This high-temperature black-body radiation is released at this instant. This so-called fossil radiation has been cooled by the expansion of the Universe. This cosmic microwave background was predicted in 1948 by Ralph Alpher, Robert Herman Alpher & Herman, 1948 and George Gamow Gamow, 1948 around $5,\kelvin$, and discovered fortuitously by Arno Penzias and Robert Wilson in 1964 Penzias & Wilson, 1965Penzias & Wilson, 1965 at a temperature of $3.5\,\kelvin$ (Figure 1).

The spectrum of the cosmic microwave background has been characterised thanks to the COBE satellite, and its temperature has now been established at Mather et al., 1999:

by modelling its data by Planck’s radiation law:

This is the best black-body radiation ever detected (Figure 2). The cosmic microwave background radiation (CMB) is probably the most direct evidence that the Universe was indeed in the form of a hot, dense plasma at equilibrium in the distant past.

Orders of magnitude¶

Let’s go well beyond the redshift of the last scattering surface, and consider the Universe at, say, $z \sim 10\,000$. What can we say?

Temperature¶

For a gas of photons, we know that :

At thermal equilibrium, the energy density of a gas of photons is the integral of Planck’s law:

The equilibrium temperature of the photons $T_\gamma$ evolves as follows:

The photon temperature $T_\gamma$ can therefore be used as a time parameter like $a$ or $z$ if $T$ is isotropic.

Densities¶

We can now calculate the current contribution of CMB photons to the critical density of the universe using the blackbody temperature :

It is therefore a negligible energy density compared to cold matter and dark energy. Of course, other ultra-relativistic particles such as neutrinos contribute to the remaining $\Omega_r^0$. But with 3 massless neutrinos, we would only end up with $\Omega_r^0 \sim 9 \times 10^{-5}$ as we will see at the end of the chapter.

We define equivalence as the moment when relativistic and non-relativistic matter are in equal proportion. Let’s calculate the redshift $z_{\rm eq}$ at the moment when the proportions of matter and radiation are equal:

We deduce that :

So at $z> z_{\mathrm{eq}}$, the contents of the Universe are dominated by relativistic matter.

Photons¶

Let’s now concentrate on the properties of photons. We know that $T_\gamma \propto a^{-1}$, so at $z=10^4$ their temperature is :

The average energy of the photons at $z=10^4$ is :

By dimensional analysis, the density of photons at a redshift z is, apart from the numerical factors we will see later :

Baryons¶

Let’s now estimate the density of baryons (particles with 3 quarks like protons and neutrons) at $z \sim 10\,000$. The baryon density today is $\Omega_b^0 = 0.049$. With a critical density of $\rho_c^0 \sim 6\,m_p / m^{3}$, this gives approximately $n_b^0 \approx 0.3\,\mathrm{baryons / m^{3}}$ today, then at $z \sim 10\,000$ :

The Universe is therefore largely dominated by photons in terms of particle density, and this proportion remains constant throughout the history of the Universe:

Expansion rate¶

The value of the Hubble expansion rate can be deduced from the Friedmann equation:

Let’s take the canonical values $\Omega_m^0 = 0.315$, $\Omega_\Lambda^0 = 0.685$ and $\Omega_r^0 \sim 9 \times 10^{-5}$ for the relativistic matter density (photons and neutrinos). At $z \sim 10\,000$, this gives :

The expansion of the Universe was much faster than it is today!

Mean photon free path¶

Finally, we can look at the mean free path of photons. Photons interact preferentially with electrons by Thomson scattering $e^- + \gamma \rightarrow e^- + \gamma$ and a good approximation of the mean free path of photons is given by :

where $\sigma_T$ is the effective Thomson scattering cross section ($6.6529times 10^{-29}\ \mathrm{m^2}$). For the electron density, let’s assume that since the Universe is neutral, there is one electron for each proton, so $n_e = n_p \approx 0.3\,e^-/\mathrm{m^{3}}$. The typical time between two $\tau_T$ interactions is then :

today if matter is in an ionised state, and at the time :

So we can see that, in the past, the interactions between matter and photons were sufficiently frequent to reach thermal equilibrium in a short time, given the expansion of the Universe. But today, even if all matter were ionised, these photons no longer interact with it. The CMB photons are therefore not in thermal equilibrium with anything else today. The vast majority of CMB photons have never been in contact with particles since they were emitted. However, this absence of interactions has preserved the original shape of the CMB spectrum, which has only been affected by the red shift. A photon detected at frequency ν was originally emitted at frequency $(1+z)\nu$. In other words, the original spectrum was Condon & Matthews, 2018:

i.e. still black-body radiation, but with a temperature $(1+z) T_0$.

Big Bang scenario¶

The Universe at $z \approx 10\,000$ was much hotter and denser. At this temperature, atoms are ionised and we therefore have a plasma. Just as today, the density of photons was significantly greater than that of baryons. Finally, interactions between photons and charged particles were much more frequent (several per Hubble time), so it is quite logical to consider the Universe as a plasma in thermal equilibrium.

On the basis of this description, we can sketch out a scenario for the evolution of primordial plasma by cataloguing the various physical phenomena that can occur as it cools. Here is a non-exhaustive summary.

First of all, at the end of inflation (about $10^{-34}\,$s after the Big Bang), there must have been a phase known as baryogenesis, where all the particles and antiparticles are created with a slight advantage for matter over antimatter leading to $\eta \sim 10^{-9}$. Below a temperature of about $100\,\GeV$ ($t \sim 20\,$ps), the electroweak phase transition takes place, giving mass to the particles and giving rise to the Z-gauge bosons, W^±. Under $150\,\MeV$ ($t\sim 20\,\mathrm{\mu s}$), this is the QCD phase transition: the strong interaction takes over from the thermal effects. The quarks and gluons coagulate to form baryons (three quarks) and mesons (two quarks). Then, $6\,\mathrm{s}$ later, electrons and positrons annihilate as the temperature of the photon bath falls below the mass of the electron $T < m_e=511\,\keV$. During the first three minutes of the Universe ($T > 100\,\keV$), the atomic nuclei of the light elements are formed. After 380,000 years, the electrons bind to the atomic nuclei ($e^- + p \rightarrow \mathrm{H} + \gamma$), this is recombination, and the photons decouple from the matter ($\tau_T \ll 1/H)$. Free to propagate, these photons form the cosmic microwave background and provide a snapshot of the primordial plasma at the end of recombination.

2Statistical thermodynamics at equilibrium¶

We now turn to a more detailed description of what happened in the primordial Universe using statistical physics.

Statistical description¶

Let’s model the contents of the Universe as a gas of weakly interacting particles. We can then use the formalism of statistical physics and describe the gas by the positions and momentum of all its particles, defined on the space $\vec{x}, \vec{p}\}$.

In a gas of particles at thermodynamic equilibrium, the number of particles that can occupy an energy state $(\vec{x}, \vec{p})$ follows a statistical distribution function $f(\vec{x}, \vec{p}, t)$. In cosmology, because of the homogeneity of the Universe, $f$ cannot depend on the position $\vec{x}$. Furthermore, due to isotropy, $f$ can only depend on the norm of the momentum $p \equiv \Vert \vec{p}\Vert$ and not on its direction.

Armed with the distribution functions, we can deduce the macroscopic properties of the gas by evaluating the probability of occupying the states of the system. But what are they? First of all, quantum mechanics dictates that the density of states in phase space is finite. In fact, if we consider a box of size $L$, with periodic conditions and solve Schrodinger’s equation, we obtain that the possible values of the momentum are :

where $\vec{e}_x, \vec{e}_y$ and $\vec{e}_z$ are the unit vectors and $h$ is Planck’s constant. Consequently, in momentum space, there is one state per elementary cube of volume $h^3/L^3$. The density of state in momentum space is therefore $L^3/h^3$ : Secondly, there is only one particle in the quantum box, so only one state of position: in the space of positions, the density of state is $1/L^3$. In total, if the particle has $g$ internal degrees of freedom, the density of state in phase space is :

The density of state is therefore independent of the volume $L^3$. It remains the same for an arbitrarily large system.

The macroscopic properties (number density, energy density, pressure) are deduced from the probability of occupation of the states $f(\vec{x}, \vec{p}, t)$ and the density of state of the phase space. The volume density of momentum particles between $p$ and $p+\dd p$ is for example given by :

The mean particle density of the gas is:

For the mean energy density, since we consider that the particles interact weakly and are not confined, then the energy levels $E(p)$ are those of a free particle $E(p)=\sqrt{p^2c^2 + m^2 c^4}$. To obtain the energy density of the gas, we simply sum the energy levels weighted by their probability of occupation:

We can obtain the gas pressure in the same way:

In the end, the energy-momentum tensor for a set of quantum particles can be written :

Note that this formula is the quantum version in the continuous limit of the formula (5) obtained for a classical perfect gas.

Kinetic equilibrium¶

When particles can exchange energy and momentum frequently through elastic collisions, the gas reaches a state of maximum entropy, called kinetic equilibrium. The distribution functions $f(p,t)$ can be obtained by evaluating the entropy of the gas ($S = k_B \ln \Omega$) and maximising it, for a given total energy and a given total number of particles.

Depending on the fermionic or bosonic nature of the particles in the gas, the combinatorics giving the occupation probabilities of all the Ω microstates is different because of the Pauli exclusion principle. At a fixed total energy and total number of particles, after using the Lagrange multipliers, these constraints impose that the distribution functions at thermodynamic equilibrium are :

with $T$ the temperature of the gas, μ the chemical potential of the species and $g$ its number of internal degrees of freedom (for example the number of spin states). They give the number of particles that can occupy an energy state $E$, depending on whether they are bosons or fermions, at thermodynamic equilibrium.

At high temperatures, we find the Maxwell-Boltzmann distribution:

valid for fermions and bosons.

The Fermi-Dirac and Bose-Einstein distribution functions depend on two parameters: the temperature of the gas, $T$, and the chemical potential of the species μ, which characterises the variation in entropy or energy when the number of particles varies (see box below).

If the gas contains several interacting species, each species $i$ is described by its own distribution function, its own chemical potential $\mu_i$. chemical potential $\mu_i$, and possibly (if decoupled) its own temperature $T_i$. From this we can deduce the number density, energy density and temperature of each species.

If all the species are in kinetic equilibrium and share the same temperature: $T_i = T$, the system has reached thermal equilibrium.

Chemical equilibrium¶

We saw in the box above that if several species interact through a reaction, for example :

and reach chemical equilibrium (i.e. the state of maximum entropy), the chemical potentials satisfy:

plus any conservation equation imposed by a conserved charge (number of particles, electric charge, baryonic charge, etc.).

For photons, there is no conserved charge. Even the number of photons is not conserved. For example, we have Compton double scattering $e^- + \gamma \rightarrow e^- + \gamma + \gamma$ or Bremstrahlung $e^- + p \rightarrow e^- + p + \gamma$. Hence :

Particles and antiparticles have opposite charges, hence, at equilibrium :

The reaction $X + \bar{X} \rightleftharpoons \gamma + \gamma$ can also be used to reach the same conclusion.

To summarise:

• A system composed of different species has reached kinetic equilibrium if it has reached a state of maximum entropy described by a Fermi-Dirac or Bose-Einstein distribution function.
• A system composed of several species interacting via one or more chemical reactions has reached chemical equilibrium if it has reached a state of maximum entropy, where the sum of the chemical potentials of the reactants is equal to the sum of the chemical potentials of the products.
• A system has reached thermodynamic equilibrium if it has reached chemical equilibrium and if all the species share the same temperature $T$, the temperature of the Universe.

Density and pressure of fermions and bosons¶

We now have everything we need to calculate the particle density, energy density and pressure of the constituents of the Universe. Chemical potentials can be neglected at high temperatures ($\mu \ll T$), and the equations (27), (28) and (29) can be rewritten:

with the sign $+$ for fermions and the sign $-$ for bosons.

In the general case, the above integrals must be calculated numerically. However, there are two interesting limits, which help us to understand the physical processes involved: the case where the particles are relativistic ($k_B T \gg m c^2$) and the opposite case of non-relativistic species ($k_B T \ll m c^2$).

Before continuing, let us define: $x \equiv m c^2/k_B T$ and $\xi \equiv pc/k_B T$, we can then rewrite $n$ and ρ above as :

Relativistic limit¶

In the relativistic limit, we have $x\ll 1$ and the integrals $I_\pm(0)$ and $J_\pm(0)$ can be calculated exactly:

where ζ is the Riemann function.

We find :

We can see that the particle and energy densities are identical for relativistic bosons and fermions to within a numerical factor. Furthermore, photons are bosons with $g_\gamma=2$ possible polarisations, so we have re-demonstrated Stefan-Boltzmann’s law.

To calculate the pressure, we have $p^2 / E \sim p$ for relativistic particles, so :

We find the equation of state already introduced earlier.

Non-relativistic limit¶

In the non-relativistic limit, the energy of particles is equal to their rest mass ($m c ^2 \gg k_B T \Leftrightarrow x \gg 1$). The chemical potential μ is not necessarily negligible either. The integrals $I$ and $J$ defined above are the same for fermions and bosons and we find :

When the temperature falls below the rest mass of the particles, the density of the number of particles falls exponentially. The energy density and pressure are (to first order) proportional to $n$ and decrease accordingly. Non-relativistic species therefore behave like a gas without pressure (because $P \ll \epsilon$ i.e. $w_m=0$). It is this description of non-relativistic matter that we have used to calculate the expansion of the Universe in the so-called ‘matter-dominated’ regime.

Effective number of relativistic species¶

Definition¶

We start with a primordial plasma in thermal and chemical equilibrium, containing $i$ species at temperature $T_i$. Before equivalence, the expansion rate is a direct function of the mass density of relativistic matter:

where $rho_r(T)$ is the sum of the densities of each relativistic species present in the primordial fluid:

We saw in the previous section that $\rho_i \propto T_i^4$ as long as the particle remains relativistic, whereas the density falls exponentially when the temperature falls below the mass of the particle. More precisely, we can write :

We define $g_\star(T)$ as the effective number of relativistic degrees of freedom of the plasma at temperature $T$:

When the species $i$ is still in thermal equilibrium with the photons, then $T_i=T$. When the temperature falls below the mass $m_i$ of one of the species, it becomes relativistic and disappears from the sum above. If it decouples from the photons with a temperature $T_i$ different from the photons, while remaining relativistic, then it remains present in $g_*(T)$ with a weight $(T_i/T)^4$.

Evolution of $g_\star(T)$¶

Let’s now study the evolution of $g_\star(T)$, which simply tells the story of the evolution of the relativistic matter of the primordial plasma as it cools with the expansion. Let’s start around $T \approx 300\,$GeV. All particles in the Standard Model are relativistic (see Figure 3). When all particles are relativistic, the total number of degrees of freedom is :

for fermions, and

for the bosons, which gives

To see what happens next, just look at the masses of the particles listed in the Figure 3. The top quark annihilates first because it is the heaviest particle. For $k_BT<m_{\mathrm{top}}$, the plasma at equilibrium can no longer produce top quarks by annihilating pairs of other particles, reducing the number of degrees of freedom to :

Then we have the Higgs boson, followed by the electroweak bosons $W^\pm$ and $Z^0$: this reduces $g_\star$ to 86.25. Then $b$ and $c$ annihilate, and $g_\star$ is reduced to 61.75.

The next event is the QCD phase transition, which occurs at $T \sim 150\ \mathrm{MeV}$. The quarks combine into hadrons (protons, neutrons and mesons). At this temperature, all but the pions are non-relativistic. At this stage, the only remaining relativistic species are photons, neutrinos, electrons and muons and the 3 spin 0 pions (with $g_\pi = 3 \cdot 1 = 3$ internal degrees of freedom). From this we can deduce the number of remaining relativistic degrees of freedom:

Next, the pions and muons annihilate, giving us

The next two significant events are the decoupling of neutrinos around $1\,\MeV$ and the annihilation of electrons and positrons ($m_e = 511\,\keV$).

Entropy¶

Conservation of entropy¶

The entropy of the Universe $S(U,V,N_i)$ is a function of the extensive variables internal energy $U$, volume $V$ and the numbers $N_i$ of particles of a species $i$. The entropy variation of a comobile volume $V$ of plasma at thermodynamic equilibrium obeys the second principle of thermodynamics:

For a volume of the Universe large enough to be considered homogeneous and isotropic, entropy can only increase or remain constant. Moreover, chemical potentials are negligible in primordial plasma ($\mu / (k_B T) \sim \eta$). At thermodynamic equilibrium we have :

where pressure and energy density are ultimately only functions of equilibrium temperature, whether the species are relativistic or not (see previous formulae). Consequently, the variation in entropy is a function of volume and temperature with (Weinberg (1989) p.532):

We identify the partial derivatives:

and thanks to Maxwell’s relations (or Schwartz’s theorem), we have :

With this last relation, we can complete the calculation of the entropy variation:

If we consider a variation with respect to time $t$ :

and finally by remembering the conservation of energy relation $\dot \epsilon = -3H(\epsilon + P)$,

The entropy in a comobile volume is therefore conserved and is written ^[3] :

We define the volume entropy as a function of temperature only:

Entropy of primordial plasma¶

For relativistic species, such as $P=\epsilon/3$, we obtain the volume entropies :

For a collection of species (fermions and bosons) at equilibrium at temperatures $T_i$, we have :

with

Since the entropy $S$ is conserved, then :

Temperature of the Universe¶

Now that we have a conservation relation, we can establish a link between the expansion of the Universe and its temperature:

This relation gives a link between temperature and scale factor at any instant in the history of the Universe. It varies with the redshift in $(1+z)$ but with a proportionality factor $g_{\star S}^{1/3}(T)$ which changes by threshold according to the composition of the Universe (Figure 5).

Expansion of primordial plasma¶

The expansion law obeys Friedmann’s first equation:

and therefore :

So $H \propto T^2$ to within variations of the effective number of degrees of freedom in the primordial plasma. Keep this in mind, as it will be useful for comparing the expansion rate with the various reaction rates between the different species.

Injecting the temperature evolution with the scaling factor (equation (89)), we find that $a(t) \propto t^{1/2}$ in the primordial Universe (equation (4)) with the proportionality factor changing as $g_{\star S}$ varies. But the expansion rate is then simply $H(t) = 1/(2t)$ which gives :

Thus, when the Universe was one second old, the typical energy of relativistic particles was of the order of $0.9\,\MeV$ with $g_*=10.75$.

3History of matter in the early Universe¶

We now have (almost) everything we need to discuss the evolution of matter in primordial plasma. When the temperature is sufficiently high, the primordial plasma contains all the particles of the Standard Model, in relativistic form (plus all the particles that have not yet been discovered, for example the hypothetical particles that would constitute cold dark matter today). All particle species are in thermal equilibrium (kinetic and chemical, same temperature $T$). But as the Universe expands, the temperature decreases at the same rate as the expansion. One after another, the various massive species become non-relativistic, annihilating each other, and their energy densities become sub-dominant compared with the relativistic species.

If the Universe were in perfect thermal equilibrium, and if this equilibrium had persisted until today, the observed abundances of massive particles would be much lower than they are, since each massive species sees its density exponentially suppressed when it becomes non-relativistic. In fact, thermal and chemical equilibria need frequent collision (and/or reaction) rates to be maintained. As the Universe expands, particles become diluted, making it more difficult to maintain reaction rates.

Since $T \propto a^{-1}$ (92), the rate of temperature change is the Hubble rate:

To be able to consider that the system is in thermodynamic equilibrium, there must be sufficient interactions in a time shorter than the time taken for the temperature to change. The rule of thumb is that you need at least several interactions per Hubble time to maintain thermal and chemical equilibrium. Thus, if we note Γ the rate of interaction, thermal and chemical equilibrium is maintained if $\Gamma \gg H$. When the Γ reaction rate falls below $H$, thermodynamic equilibrium is no longer maintained and the particle densities are frozen at their pre-decoupling values. The freezing of interactions is an essential mechanism in explaining the current abundance of particles.

Neutrino decoupling and electron-positron annihilation¶

Neutrino decoupling is our first illustration of the freezing effect. Neutrinos interact only through the weak interaction. Around $\sim 1\,\MeV$, they are still thermalised with the photon bath by interactions such as :

At these energies, the effective cross section of the weak interaction is $\sigma_w \sim G_F^2 T^2$ with $G_F/(\hbar c)^3=1.166 378 7(6)\times 10^{-5}\,\GeV^{-2}$ the Nakamura & others, 2010 Fermi coupling constant. Consequently, the interaction rate $\Gamma = n_e \sigma_w c \propto G_F^2 T^5$ decreases much faster than the Hubble parameter ($\propto T^2$):