LOG#108. Basic Cosmology (III).

The current Universe has evolved since its early phase of thermal equilibrium until the present state. The departure from thermal equilibrium in the early Universe made a fossil record we can observe at current time!

There are some easy rules for thermal equilibrium. The easiest one, is that coming from the “interaction rate” $\Gamma_{int}$. It can be expressed in the following way:

$\Gamma_{int}>\mbox{Expansion rate H}$

and then, at a given temperature T, we get

$\Gamma_{int}(T)=n(T)\langle \sigma \vert v \vert \sigma \rangle^T$

and where $H\approx \dfrac{T^2}{M_p}$

Remark: If $\Gamma_{int}=aT^n$ $\forall n>2$, then

$N_{int}=\int_t^\infty T_{int}(t')dt'=\dfrac{\Gamma (H)\vert_t}{n-2}<1$

and it implies that a particle interacts less than once after the time $\Gamma =H$.

Moreover, we can understand roughly the so-called decoupling era:

1st. Any interaction mediated by a massless gauge boson provides

$\sigma\sim \dfrac{\alpha^2_X}{s}$ with $s\sim E^2$ and $\alpha_X=\dfrac{g^2_X}{4\pi}$

and this implies that

$\Gamma \sim n\langle \sigma v\rangle\sim T^3\dfrac{\alpha_X^2}{T^2}=\alpha^2_XT$

and

$\dfrac{\Gamma}{H}\sim \alpha_X^2\dfrac{M_p}{T}$

so the equilibrium temperature is found whenever $T\leq \alpha_X^2M_p$!

2nd. Interactions mediated by any massive gauge boson provides

$\sigma\sim G_X^2s$ with $G_X\sim \dfrac{\alpha_X}{m_X^2}$

and this implies that

$\Gamma \sim T^3G_X^2T^2=G_X^2T^5$

and

$\dfrac{\Gamma}{H}\sim G_X^2M_pT^3$

and then

$\left( G_X^2M_p\right)^{-1/3}\leq T\leq m_X\longrightarrow \mbox{Equilibrium temperature (E.T.)}$

Moreover,

$T\leq \left( G_X^2M_p\right)^{-1/3}\sim \left(\dfrac{m_X}{100\mbox{GeV}}\right)^{4/3}\mbox{MeV}\longrightarrow \mbox{Freeze out}$

As a consequence, we can realize that the out-of-equilibrium phenomena in the early and current Universe are very important processes! In particular:

1) They provide the formation of light elements during the Big Bang Nucleosynthesis (BBN), also known as primordial nucleosynthesis, i.e., the formation of the first light elements after the Big Bang (circa 300000 years after the Universe “birth”).

2) They provide the path of recombination of electrons and protons into hydrogen atoms.

3) They imply the  (likely) production of dark matter (or equivalently the presence of some kind of “modified gravity” or/and modified newtonian dynamics).

Boltzmann’s equation for annihilation of particles in equilibrium

There is a beautiful equation that condenses the previous physical process of equilibrium at a given temperature and the particle production it yields. Conceptually speaking, we have

$\begin{pmatrix}\mbox{Boltzmann}\\ \mbox{Equation}\end{pmatrix}:$

$\begin{pmatrix}\mbox{Rate of change}\\ \mbox{in the abundance}\end{pmatrix}=\begin{pmatrix}\mbox{Rate of}\\ \mbox{particle production}\end{pmatrix}-\begin{pmatrix}\mbox{Rate of}\\ \mbox{particle erasing/annihilation}\end{pmatrix}$

Consider a process like

$\mbox{particle type 1}+\mbox{particle type 2}\leftrightarrow \mbox{particle type 3}+\mbox{particle type 4}$

and where the particle 1 is the one we are interested in. Then, we deduce that

$\underbrace{\dfrac{1}{a^3}\dfrac{d(n_1a^3)}{dt}}_\text{change in comoving volume}=\underbrace{\int\dfrac{d^3p_1}{(2\pi)^32E_1}\int\dfrac{d^3p_2}{(2\pi)^32E_2}\int\dfrac{d^3p_3}{(2\pi)^32E_3}\int\dfrac{d^3p_4}{(2\pi)^32E_4}}_\text{phase space invariant}\times A$

where A is certain complicated facter involving “delta functions” of the energies and momenta of the particles 1,2,3,4 and an additional term depending on the statistics of the particle. Explicitly, it takes the form

$A=\left[(2\pi)^4\delta^3(p_1+p_2-p_3-p_4)\delta (E_1+E_2-E_3-E_4)\vert M\vert^2\right]\times S$

with $S=\left[ f_3f_4(1\pm f_1)(1\pm f_2)-f_1f_2(1\pm f_3)(1\pm f_4)\right]$

and where

$f_i=\dfrac{1}{e^{(E_i-\mu_i (t))/T}\pm 1}$

is the Fermi-Dirac (FD, -)/Bose-Einstein (BE,+) distribution. In fact, the above FD/BE factors provide the so-called Pauli blocking/”Bose-Einstein” enhancement effects for the particle production in the processes $3+4\rightarrow 1+2$ and $1+2\rightarrow 3+4$. Indeed, particle physics enter into the game here (see above formulae again) and we assume

$M(1+2\rightarrow 3+4)=M(3+4\rightarrow 1+2)$

Do you recognize the principle of detailed balance in this equation?

We can simplify the assumptions a little bit:

1st. The kinetic equilibrium is taken to be a rapid elastic scattering and we input the FD/BE statistics without loss of generality.

2nd. The annihilation in thermal equilibrium will be calculated from the sum of the chemical potential in any balanced equation.

3rd. Low temperature approximation. Suppose that

$T<< (E-\mu)$

then we obtain the Maxwell-Boltzmann approximation to the FD/BE statistics

$f\approx e^{-(E-\mu)/T}$ and $1+f\approx 1$, so, since $E_1+E_2=E_3+E_4$, we get

$f_3f_4(1\pm f_1)(1\pm f_2)-f_1f_2(1\pm f_3)(1\pm f_4)\approx e^{-(E_1+E_2)/T}\left( e^{\frac{(\mu_3+\mu_4)}{T}}-e^{\frac{(\mu_1+\mu_2)}{T}}\right)-\star$

What is $\star$? After a change of variable

$\mu_i (t)\longrightarrow n_i(t)=g_ie^{\mu_i/T}\int \dfrac{d^3p}{(2\pi)^3}e^{-E_i/T}$

$\star$ is the “equilibrium number density” deducen from the expression

$n_i^{0}\equiv g_i\int {d^3p}{(2\pi)^3}e^{-E_i/T}=\begin{cases}g_i\left(\dfrac{m_iT}{2\pi}\right)^{3/2}e^{-m_i/T},\;\; \mbox{if}\;\; m_i>>T\\ g_i\dfrac{T^3}{\pi^2},\;\;\mbox{if}\;\; m_i<

and it yields

$n_i=e^{\mu_i/T}n_\gamma^{0}$

and then we finally get that

$\star$ equals $e^{-(E_1+E_2)/T}\left[\dfrac{n_3n_4}{n_3^0n_4^0}-\dfrac{n_1n_2}{n_1^0n_2^0}\right]$

Now, we can define the thermally averaged cross section

$\langle \sigma v\rangle\equiv \dfrac{1}{n_1^0n_2^0}\int \dfrac{d^3p_1}{(2\pi)^32E_1}\cdots \dfrac{d^3p_4}{(2\pi)^32E_4}e^{-(E_1+E_2)/T}(2\pi)^4\delta^3 (p_1+p_2-p_3-p_4)\times$

$\times \delta (E_1+E_2-E_3-E_4)\vert M\vert^2$

The Boltzmann equation becomes with these conventions

$\dfrac{1}{a^3}\dfrac{d(n_1a^3)}{dt}=n_1^0n_2^0\langle \sigma v\rangle \left(\dfrac{n_3n_4}{n_3^0n_4^0}-\dfrac{n_1n_2}{n_1^0n_2^0}\right)$

Remark (I): LHS is similar to $\dfrac{n_1}{t}\sim n_1H$ and the RHS is similar to $n_1n_1\langle \sigma v\rangle$

Remark (II): If the reaction rate is $n_1\langle \sigma v\rangle >> H$, then it provides the chemical equilibrium condition well known in the nuclear statistical equilibrium as the Saha equation, i.e.,

$\dfrac{n_3n_4}{n_3^0n_4^0}-\dfrac{n_1n_2}{n_1^0n_2^0}\approx 0$