# Gennaro Miele » 2.Thermodynamics of the expanding universe

### Boltzmann Equation in an expanding universe

In general, the primordial plasma does not satisfies the equilibrium conditions, hence  we cannot exploit the powerful methods of equilibrium thermodynamics. This means that we cannot assume that a gas of particles be distributed according to Bose-Einstein or Fermi-Dirac functions. If this occurs, one neds to obtain the time evolution of the statistical system in terms of the kinetic theory, namely solving the Boltzmann equation.

The approach to equilibrium for a system made of photons, neutrinos, charged leptons, etc. is driven by interactions processes such as scatterings (crucial to redistribute particle momenta and to achieve the so-called kinetic equilibrium) and interactions where the number of a particular particle specie can change. Since the Universe is expanding the rate of microscopic interactions which tries to thermalize the system must compel with the expansion rate, measured by the Hubble parameter H, which diluting the plasma tries to freeze the species.

If we consider for simplicity the case of two body processes of the form a +b → c+d,  the interaction rate for the species a is then given by

$\Gamma_a \sim \sigma v \, n_b$

where σ stands for the typical cross section, v is the a-b relative velocity and nb the b number density. Hence the equilibrium is maintained if

$\Gamma_a \gg H$

### Decoupling condition

When for a specie  Γa ≤  H we say that it is decoupled. Typically, at high temperature density and mean energy are large enough to ensure that the equilibrium condition be fulfilled. As the expansion proceeds, if the interaction rate decreases faster than H for some value of redshift z=zD

$\Gamma_a (z_D) \sim H(z_D)$

This defines the so-called decoupling redshift. After a specie has decoupled it is called a relic. If the decoupling takes place when the specie is still relativistic  it is called a hot relic, like neutrinos. On the other hand, if it decouples when it is already non-relativistic  it is defined a cold-relics), like for example cold dark matter (CDM).

In general a specie is characterized by a distribution function in the phase-space f(xμ,Pμ). In absence of particle interactions, the value of f cannot change as we follow a point in phase space along a geodesic motion f(xμ(λ),Pμ(λ)), namely

${\cal L}(f) \equiv \frac{d f}{d \lambda} (x^\mu(\lambda), P^\nu(\lambda)) =0$

which defines the Liouville operator as

${\cal L}(f) =\frac{\partial f}{\partial x^\mu} P^\mu - \frac{\partial f}{\partial P^\nu} \Gamma_{\rho \sigma}^\nu P^\rho P^\sigma= 0$

### Liouville operator

In FRW  the universe is homogeneous and isotropic, hence the distribution function depends on time and either energy E or the modulus of linear momentum P (via the mass shell condition). For this reason we get the following equation for f(t,P)

$E \left( \frac{\partial f(t,P)}{\partial t} - 2 H P \frac{\partial f(t,P)}{\partial P} \right) = 0$

Changing the variables, namely using the physical momentum p=P a we have

$E \left( \frac{\partial f(t,p)}{\partial t} - H p \frac{\partial f(t,p)}{\partial p} \right) = 0$

Finally, if we introduce  y ≡ (Σi Pi2)1/2, namely the modulus of the comoving momentum, we get

$\frac{\partial f(t,y)}{\partial t} = 0$

Thus any function depending on the comoving momentum only is a solution of the collisionless Boltzmann equation. Such equation is modified in presence of particle scatterings, annihilations etc. .Using physical momentum  one can write a set of equations, one for each specie,

$\frac{1}{E} {\cal L}(f_j) = \frac{\partial f_j(t,p)}{\partial t} - H p \frac{\partial f_j(t,p)}{\partial p} = {\mathbf C} (f_j(t,p); f_i)$

by fi we denote the distributions of all species interacting with the one we are interested in.

### Conserved quantities

The right hand side of the previous equation, namely C, defines the so-called collisional integral, which depends on fj and fi due to the hypothesis of molecular chaos, namely that particle momenta are uncorrelated with their positions.

Let us consider the case of two-particle into two-particle interactions, a + b ↔ c + d. For the species “a” we have the collisional integral for two body process given by

${\mathbf C}(f_a; f_b, f_c,f_d) = \frac{1}{E_a}\int d \pi(p_b) d \pi(p_c) d \pi(p_d) \,(2 \pi)^4 \delta^{(4)}(p_a+p_b-p_c-p_d)$

$\times \left[ \left| {\cal M}_{cd,ab} \right|^2 f_c(p_c,t) f_d(p_d,t) (1\pm f_a(p_a,t))(1\pm f_b(p_b,t))\right.$

$- \left. \left| {\cal M}_{ab,cd} \right|^2 f_a(p_a,t) f_b(p_b,t) (1\pm f_c(p_c,t))(1\pm f_d(p_d,t)) \right]$

Note that the factors (1 ± f) are due to the Pauli blocking effect for fermions (minus sign), and stimulated emission for bosons (positive sign). If the system is very diluted and the particle chemical potentials are very small, these factors reduce to unity, as in the case of particles obeying classical statistics. It is possible to prove that by denoting with Q a quantity conserved by interactions (like energy or momentum, electric charge, etc.), then

$\int \frac{d^3 p_a}{(2 \pi)^3} {\cal Q}\, {\mathbf C} =0$

### Equilibrium in the expanding universe

If we have two particle species, namely a and b  which can only interact via the scattering process  a + b → a + b both na and nb are conserved. In this case taking a general conserved charge qa (or analogously qb) one still gets that the collisional integral for the a particle vanishes and qa na is covariantly conserved.

Among the possible quantity one can choose  Q=E, in this case  summing over all species which are mutually interacting, we get

$\sum_a \int \frac{d^3 p_a}{(2 \pi)^3} E(p_a)\, {\mathbf C}(p_a;p_i)=0$

that expresses the covariant conservation of the total energy-stress tensor (namely summed over all interacting species).

### Equilibrium in the expanding universe

At this point one may wonder if the equilibrium distributions satisfy the Boltzmann equation for an expanding Universe. Effectively one can prove that also in this case the equilibrium distribution functions,namely the ones which implies C=0 take the standard form

$f_i = \left[ \exp\left(\frac{E_i-\mu_i}{T_i}\right) \pm 1\right]^{-1}$

whenever all species are at the same temperature (kinetic equilibrium), and the chemical potentials satisfy the condition (chemical equilibrium)

$\mu_a+\mu_b = \mu_c + \mu_d$

To yield a vanishing C is a necessary but not sufficient condition for a distribution function to be an equilibrium distribution. In fact one must prove that it is also in the kernel of the Liouville operator. Let us write a generic equilibrium function in the form

$f(p,t) = \left[ \exp\left(\beta E(p)-\xi\right) \pm 1\right]^{-1}$       with ξ=μ/T and β=1/T.

### Liouville operator and ultra-relativistic particles

One can easily prove that the previous expression is in the kernel of Liouville operator if ot satisfies the equation

$E(p) \dot{\beta}-H \frac{p^2}{E(p)} \beta - \dot{\xi}=0$

For massless particles, namely E=p, one can find an exact solution with

$\xi = const. \,\,\,\,\,\,\,\,\,\,\, \beta \propto a$

This means that if some specie decouples when relativistic (hot relic) it will keep a Fermi-Dirac or Bose-Einstein distribution with the decreasing of the temperature also after decoupling. In fact, untill it is in thermal contact with other species it keeps the equilibrium distribution, which will also be a solution of the collisionless Boltzmann equation after decoupling, being a function of the comoving momentum only.

### Liouville operator and non-relativistic particles

For non-relativistic particles (m ≠ 0) the Liouville relation does not admits trivial solutions. In particular we can state that  in FRW there are no exact equilibrium solution to the Boltzmann equation for massive particles . Nevertheless the distribution can approach closely the equilibrium distribution if interaction rates are sufficiently larger than H.

For values of temperature much larger than the particle mass,  T >>m,  one can neglect m and if for a  specie strongly coupled its distribution is a Fermi-Dirac or Bose-Einstein with T ≈ a-1. When T decreases down to  T  ≈ m, the equilibrium distribution is no more a solution of the Boltzmann equation, but one can look for an approximate equilibrium form

$f(p,t) = \left[ \exp\left(\frac{E_i-\mu_i}{T_i}\right) \pm 1\right]^{-1} \left(1+\delta(p,t) \right)$

con |δ(p,t)|<<1. For  large interactions rate Γ >> H, and δ(p,t) ≈ H/Γ <<1 .

As a relevant example, let us consider e- and e+ which interact quite efficiently with γ via Compton scattering and pair processes e+ e↔ γ γ.  When the temperature is much smaller than the electron mass, the Compton cross section is given by the Thomson formula,  namely σT = 8 π e4/(3 me2) ≈ 0.67 10-24 cm2 .  Photons have a distribution function, satisfying the Boltzmann equation, and whose temperature scales as T ≈ a-1=1+z. Hence when T = m, z = me/T0 ≈ 1010 . From this value of redshift and till the equality point, zeq ≈ 103 the universe is radiation dominated.

### Quasi equilibrium distribution

which implies

$H \sim H_0 \left(\frac{1+z}{1+z_{eq}} \right)^2 (1+z_{eq})^{3/2}\,,\,\,\,\,z>z_{eq}$

Neglecting the photon reheating due to e± annihilations, we have that the number of photons reads

$n_\gamma \sim 2 \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{e^{p/T}-1} = \frac{1}{\pi^2} (1+z)^3 \int y^2 dy \frac{1}{e^{y/T_0}-1}$ $= \frac{2 \zeta(3)}{\pi^2} (1+z)^3 T_0^3$

where ζ(3) = 1.202. From this we get

$\Gamma \sim c\, \sigma_T \frac{2 \zeta(3)}{\pi^2} (1+z)^3 T_0^3$

Substituting the numerical values we get

$|\delta(p,t)| (T \sim m_e) \sim \frac{\pi^2}{2 \zeta(3)} \frac{H_0}{c\, \sigma_T \,T_0^2 \,m_e} (1+z_{eq})^{-1/2} \sim 10^{-17}$

Hence e± are distributed with an almost exact equilibrium Fermi-Dirac function. Since also pair processes have a large rate compared with H, chemical equilibrium holds, and reminding that photons have zero chemical potential, we can write

$\xi_{e} + \xi_{e^+} =0$

Hence we can write fot T ≤ me

### Quasi equilibrium distribution

$f_{e^\pm} \sim \exp \left(-\frac{m_e}{T} \mp \xi_{e} -\frac{p^2}{2 m_e T} \right)$

In many relevant cases one can safely assume that kinetic equilibrium holds due to fast scattering processes. In these cases it is convenient to integrate the Boltzmann to get the evolution of the number densities. Let us consider the example of two body – two body processes. In this case we have

$H a \frac{d n_a}{ da} + 3 H n_a = \int d \pi(p_a)d \pi(p_b)d \pi(p_c) d \pi(p_d)\,(2 \pi)^4 \delta^{(4)}(p_a+p_b-p_c-p_d)$

$\times \left| {\cal M}_{ab,cd} \right|^2 \left[ e^{-(E_c+E_d)/T} e^{\mu_c/T} e^{\mu_d/T} - e^{-(E_a+E_b)/T} e^{\mu_a/T} e^{\mu_b/T}\right]$

We have obtained the previous  result considering the particular case where all particles involved are either non–relativistic or, in case they are relativistic, with small (or vanishing as for photons) chemical potential. Fot the species considered the number density is a simple function  of the chemical potential

$n_i \sim e^{\mu_i/T} \int \frac{d^3 p}{ (2 \pi)^3} e^{-E_i/T}$

In this case the energy conservation can be cast in the following form

$a^{-2} \frac{d}{da} \left( n_a a^3 \right) = \frac{\langle \sigma |v| \rangle n_b}{H} n_a \left( \exp \left(\frac{\mu_c+\mu_d-\mu_a-\mu_b}{T}\right)-1 \right)$

### Saha Equation

With

$\langle \sigma |v| \rangle \equiv \int d \pi(p_a)d \pi(p_b)d \pi(p_c) d \pi(p_d)\,(2 \pi)^4 \delta^{(4)}(p_a+p_b-p_c-p_d)$

$\times \left| {\cal M}_{ab,cd} \right|^2 e^{-(E_a+E_b)/T} \left(\int \frac{d^3 p}{(2 \pi)^3} e^{-E_a/T} \int \frac{d^3 p}{(2 \pi)^3} e^{-E_b/T}\right)^{-1}$

which represents the product of the cross section σ times the relative velocity for the process a+b → c+d, averaged over the particle distributions. Note that on the r.h.s. of the Eq. for d/da (na a3) we have the ratio between Γa = <σ|v|>nb and the Hubble parameter.If this ratio is large the speci “a” is in chemical equilibrium. The chemical equilibrium condition can be cast in the following form

$\frac{n_c n_d}{n_a n_b} = \frac{\int d^3 p\, e^{-E_c/T} \,\,\int d^3 p \,e^{-E_d/T}}{\int d^3 p \, e^{-E_a/T} \,\, \int d^3 p \,e^{-E_b/T}}$

which is called the Saha equation.

### Saha Equation

This equation allows to get the order of magnitude of the time (or redshift) at which some important events take place, like the recombination stage. Such event occurs when electrons and protons recombine to form neutral hydrogen, through the two body process

$p + e^- \leftrightarrow \mbox{H} + \gamma$

or to compute the instant at which Deuterium starts forming (the onset of primordial nucleosynthesis) due to proton neutron fusion

$p + n \leftrightarrow\, ^2\mbox{H} + \gamma$

### In the equilibrium conditions

In equilibrium conditions one can easily compute energy and pressure for a given species “i“. Summing over particles and antiparticles we have

$\rho_{i}= g_i \int \frac{d^3p}{(2 \pi)^3} E(p) \left( \frac{1}{e^{E(p)/T_i-\xi_i}\pm 1} + \frac{1}{e^{E(p)/T_i+\xi_i}\pm 1}\right)$

$P_{i}= g_i \int \frac{d^3p}{(2 \pi)^3} \frac{p^2}{3 E(p)} \left( \frac{1}{e^{E(p)/T_i-\xi_i}\pm 1} + \frac{1}{e^{E(p)/T_i+\xi_i}\pm 1}\right)$

gi is the number of internal degrees of freedom as helicity, colour, etc. In the previous relations we have assumed opposite chemical potentials for particles and antiparticles. This is true if these particles are in chemical equilibrium with photons, by direct interactions or via processes like

$i + \bar{i} \leftrightarrow e^-+e^+$  or   $i + \bar{i} \leftrightarrow p+\bar{p}$

For relativistic particles with ξi = 0 one has

$\rho_i = 3 P_i =\left\{ \begin{tabular}{c} \frac{\pi^2}{30} g_i \, T_i^4 \,,\,\,\,\,\,boson \\ \\ \frac{7}{8}\frac{\pi^2}{30} g_i \, T_i^4 \,,\,\,\,\,\,fermion \end{tabular} \right.$

### Particle- Antiparticle asymmetry

If ξi ≠ 0 and we are always in the relativistic case one gets

$\rho_i =3 P_i= \frac{7}{8}\frac{\pi^2}{30} g_i \, T_i^4 \left(1+ \frac{30 \xi_i^2}{7 \pi^2} +\frac{15 \xi_i^4}{7 \pi^4} \right)$

In this  limit, the particle–antiparticle asymmetry of a fermionic specie takes the exact form

$n_i-n_{\bar{i}} = \frac{g_i}{6} T_i^3 \left(\xi_i+ \frac{\xi_i^3}{\pi^2} \right)$

whereas one can only provide an expansion for the number density in terms of powers of ξi , namely up to the first order

$n_{i,\bar{i}} = \frac{3 \zeta(3)}{4 \pi^2} g_i \, T_i^3 \pm \frac{g_i }{12} T_i^3 \xi_i + {\cal O}(\xi_i^2)$

For relativistic bosons, since the chemical potential cannot be greater than the particle mass (otherwise when ξi approaches the value of the particle mass a Bose condensate develops) we have

$n_{i,\bar{i}} \sim \frac{\zeta(3)}{\pi^2} g_i \, T_i^3$

$n_i-n_{\bar{i}} \sim \frac{g_i \,T_i^3}{3} \xi_i$

where the last result only applies in case μi << mi and no Bose condensate forms.

### Non-relativistic equilibrium expressions

For non-relativistic particles  and μ<< mi the distribution is given by the Maxwell-Boltzmann function

$f_{i} \sim g_i \,\exp \left(-\frac{m_i}{T_i} \pm \xi_{i} -\frac{p^2}{2 m_i T} \right)$

so we have

$n_i \sim g_i\,\left( \frac{m_i T_i}{2 \pi} \right)^{3/2} e^{-m_i/T_i \pm\xi_i}$

$\rho_i \sim m_i n_i + \frac{3}{2} T_i n_i$

$P_i \sim T_i n_i \ll \rho_i$

For particle with positive chemical potential the density of their antiparticles is smaller by a factor

$\exp(-2 \xi_i)$

### The entropy density

For a homogeneous system, such as fluids in a FRW Universe, the entropy density, s=S/V is obtained from the gran potential Ω = -P V as

$s= -\frac{1}{V} \left. \frac{\partial \Omega}{\partial T} \right|_{\mu, V} = \left. \frac{\partial P}{\partial T} \right|_{\mu}$

which yields

$s= \int \frac{d^3 p}{(2 \pi)^3} \frac{\exp \left( \frac{E-\mu}{T} \right)}{\left[ \exp \left( \frac{E-\mu}{T} \right) \pm 1 \right]^2 } \frac{E-\mu}{T^2} \frac{p^2}{3 E}$

or analogously

$s= -\int \frac{d^3 p}{(2 \pi)^3} \left( \frac{\partial}{\partial p} \frac{1}{\exp \left( \frac{E-\mu}{T} \right) \pm 1 } \right) \frac{E-\mu}{3 T} p$

Integrating by parts one then obtains

$s = \frac{\rho-\mu \, n+P}{T}$

### Entropy density for relativistic particles

The total entropy density due to several relativistic species can be written in terms of the contribution of one polarization state for the photons, multiplied by an effective number of “entropy relativistic degrees of freedom” gs

$s_R = \frac{4}{3} \rho_R \equiv g_s \frac{2 \pi^2}{45} T^3$

If we have vanishing chemical potentials we get

$g_s = \sum_{i, boson} \, g_i \left( \frac{T_i}{T} \right)^3+ \frac{7}{8}\sum_{j, fermion } \, g_j \left( \frac{T_i}{T} \right)^3$

where we have assumed that different species can have different temperatures Ti with respect to the photon temperature T. The quantity gs is in general a function of T. When the temperature T reaches the mass threshold mi , the i-th species becomes non-relativistic and its contribution becomes negligible. In this situation gs decreases, and all  species which at this stage are still in thermal equilibrium with “i” particles get reheated due to the entropy released by the rapid “i”- anti”i” annihilations. In this phase from the entropy conservation we get

$T \sim g_s^{-1/3} a^{-1}$

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