6th Week: The Early Universe and Big Bang Nucleosynthesis

The Big Bang Cosmology

Cosmological principle

The cosmological principle, which is the more general version of the Copernican principle, states that on large spatial scales, the Universe is homogeneous and isotropic. This means that there is no special point in the Universe. Homogeneity of the universe means that the universe has the same property at any regions from point to point. Isotropy of the universe means that the universe looks the same from all directions. We know that at small scales the universe is not homogenous and not isotropic otherwise any structures e.g. galaxies, stars, planets and humans would not even exist. However provided that we consider the universe on average on large scales, it looks approximately homogenous and isotropic. The observed cosmological scales are therefore approximately $300$ to $500$ $Mpc$ in which the cosmological principle works.

The expanding Universe

Before 1915, it was believed that the cosmos was static and infinite. But the infinite Universe (Newtonian Universe) was ruled out soon due to the Olbers paradox, that states for such a Universe the dark night should not exist. The Einsteins theory of gravitation suggested that the Universe is no more static. But however in order to get a static Universe solution Einstein added a so called cosmological constant, $\Lambda$ which later he called it his greatest blunder. In 1922 Friedmann solved the Einsteins equations for isotropic and homogeneous universe and found that the Universe is either expanding or collapsing. This was experimentally discovered by Hubble in 1928, that the Universe is expanding and the expansion law is the following

${\vec {v}}=H_{0}{\vec {R}}\ ,$

where $H_{0}$ is the so called Hubble constant, which is today believed to be

$H_{0}=70.1\pm 1.3\quad (km/s)/Mpc\ .$

The Hubble "constant" is not a constant, but is actually a time varying quantity. Defining a scale factor, $a$,

${\vec {R}}=a{\vec {x}}\ ,$

where ${\vec {x}}$ is the comoving coordinate, one can find the folowing relation

$H={\frac {\dot {a}}{a}}\ .$

Friedmann equations

From the large-scale distribution of galaxies and the near-uniformity of the CMB temperature, we have good evidence that the universe is nearly homogeneous and isotropic. Under this assumption, the space-time metric can be written in the form

$ds^{2}=-c^{2}dt^{2}+a^{2}(t)\left[{\frac {\mathrm {d} r^{2}}{1-kr^{2}}}+r^{2}\mathrm {d} \mathbf {\Omega } ^{2}\right]\ ,$

where

$\mathrm {d} \mathbf {\Omega } ^{2}=\mathrm {d} \theta ^{2}+\sin ^{2}\theta \,\mathrm {d} \phi ^{2}\ .$

Figure 1. Curvature and the fate of the Universe

There are two independent Friedmann equations for modeling a homogeneous, isotropic universe. They are:

$H^{2}=\left({\frac {\dot {a}}{a}}\right)^{2}={\frac {8\pi G}{3}}\rho -{\frac {kc^{2}}{a^{2}}}+{\frac {\Lambda c^{2}}{3}}$

which is derived from the 00 component of Einstein's field equations, and

${\dot {H}}+H^{2}={\frac {\ddot {a}}{a}}=-{\frac {4\pi G}{3}}\left(\rho +{\frac {3p}{c^{2}}}\right)+{\frac {\Lambda c^{2}}{3}}$

which is derived from the trace of Einstein's field equations.

Using the first equation, the second equation can be re-expressed as

${\dot {\rho }}=-3H\left(\rho +{\frac {p}{c^{2}}}\right)\ ,$

which eliminates $\Lambda \!$ and expresses the conservation of mass-energy.

If the scale factor $a$ is taken to be 1 at the present time, $k$ describes the spatial curvature when $a=1$ (i.e. today). If $k$ is positive, then the Universe is geometrically viewed as a hyperspherical, or also called as a closed Universe. If $k$ is zero, then the universe is flat. If $k$ is negative, then the universe is hyperbolic in geometry and also called an open Universe. In the closed Universe the expansion stops at some point and then recollapses to a Big Crunch. The flat Universe expands forever. The open Universe expands with acceleration and the end result is known as Big Chill.

Solution of the Friedmann equations

For simplicity, from now on we work in the units of $c=1$. Then the fluid equation is just

${\dot {\rho }}=-3H\left(\rho +p\right)\ .$

The equation of state generally can be written as

$p=w\rho \ ,$

where $w=0$ for cosmological dust, $w=1/3$ for radiation, $w=-1$ for cosmological constant or vacuum energy. The general solution is then

$\rho =\rho _{0}\left({\frac {a}{a_{0}}}\right)^{-3(1+w)}\ .$

And the solution to the Friedmann equation gives

$a=a_{0}\left({\frac {t}{t_{0}}}\right)^{2/[3(1+w)]}\ .$

Evolution of density with time

Putting the las two equations together we obtain

$\rho (t)=\rho _{0}\left({\frac {t}{t_{0}}}\right)^{-2}\ ,$

where the subscript $0$ denotes as the quantity for today. Thus, the density evolves the same regardless of the type of the fluid. Further substitution defines the Hubble constant in terms of time

$H(t)=\left({\frac {2}{3(1+w)}}\right){\frac {1}{t}}\ .$

Relying on the observation of Hubble constant and using the equation of state parameter one can find the approximate age of the Universe. For a dust-filled Universe this time is equal to

$t_{0}={\frac {2}{3H_{0}}}\approx 9.2\times 10^{9}\quad years\ .$

=

Primordial Nucleosynthesis

Early Universe

Freeze -out time The reactions on the free nucleons, the proton and the neutron, are most important in the time of equilibrium and, after freeze out, when the free neutrons decay.

weak reactions eventually end reacting slower than the cooling time of the

At 0.3 Mev,nuclei become favoured over free nucleons Apparently here, the production rate is larger than the photon destruction rate All free neutrons left (number of neutrons after freeze out minus number of neutrons destroyed due to free neutron decay) get bound up to He-4

$T_{freeze}=1.5x10^{4}K(\Omega _{b}h^{2})$

Most of the mass in the universe is not in the form of ordinary baryonic matter (electrons & atomic nuclei).