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Gennaro Miele » 1.Rèsumé of standard cosmology in FRWL

The homogeneous and isotropic Universe

The assumption that our Universe is spatially homogeneous and isotropic implies that it is possible to choose a suitable set of coordinates in which the space- time metric is

$g_{\mu \nu} dx^\mu dx^\nu = - dt^2 + a^2(t) \left( \frac{dr^2}{1-k r^2} + r^2 \left( d \theta^2 + \sin^2 \theta d\phi^2 \right) \right) \label{FRW}$

where t is the “physical time”, while r, θ and Φ are the spatial comoving coordinates which labels the points of the 3-dimensional constant time slice. They are not sensitive to the expansion dynamics and define comoving observers, namely those which have constant values of r, θ and Φ as time flows.

The physical distances are weighted by the scale factor a(t), and are thus increasing with time in an expanding Universe. The parameter k specifies the spatial curvature of the model. By a suitable redefinition of the radial variable r it can be reduced to the canonical values k = +1, 0, −1 corresponding to a constant positive curvature 3-space (a 3- sphere), a flat 3-dimensional plane and a negative curvature 3-dimensional space, respectively. For the flat case using cartesian coordinates the FRW metric becomes

$g_{\mu \nu} dx^\mu dx^\nu = - dt^2 + a^2(t) \left( dx^2+dy^2+dz^2 \right)$

By using the conformal time η defined as

$\eta \equiv \int \frac{dt}{a(t)}$

FRW metric in conformal time

the metric then becomes

$g_{\mu \nu} dx^\mu dx^\nu = a^2(\eta) \left( - d\eta^2 + \frac{dr^2}{1-k r^2} + r^2 \left( d \theta^2 + \sin^2 \theta d\phi^2 \right) \right)$

In terms of the comoving distance χ(t) travelled by the photon in the time interval t-t_i

$\chi(t) = \int_{t_i}^t \frac{dt'}{a(t')}= \eta-\eta_i$

it is possible to rewrite the FRW metric using χ as a coordinate, namely

$g_{\mu \nu} dx^\mu dx^\nu = - dt^2 + a^2(t) \left( d \chi^2 + r^2(\chi) \left( d \theta^2 + \sin^2 \theta d\phi^2 \right) \right)$

where the function r²(χ) is

$r^2(\chi) = \left\{ \begin{tabular}{c} $\sinh^2 \chi$ \,, \,\,\,\,$k=-1$ \\ $\chi^2$ \,, \,\,\,\,\,\,\, $k=0$ \\ $\sin^2 \chi$ \,, \,\,\,\, $k=+1$ \end{tabular} \right.$

FRW metric in conformal time

In general the freely falling motion of a particle in a given spacetime metric is given by the geodesic equation

$\frac{d u^\mu}{d \lambda} +\Gamma_{\nu \rho}^\mu u^\nu u^\rho =0$

where λ is some monotonically increasing variable along the particle path and the Christoffel symbols are related to first derivatives of the metric as.

Mass-shell condition in FRW

$\Gamma_{\nu \rho}^\mu = \frac{g^{\mu \sigma}}{2} \left( \frac{\partial g_{\sigma \rho}}{\partial x^\nu} + \frac{\partial g_{\nu \sigma }}{\partial x^\rho} - \frac{\partial g_{\nu \rho}}{\partial x^\sigma} \right)$

In the FRW metric and for k=0 the only non vanishing components are the following

$\Gamma_{0i}^j = \frac{\dot a}{a} \delta^j_i \,,\,\,\,\, \Gamma_{ij}^0 = \frac{\dot a}{a} g_{ij}$

with the dot denoting the derivative with respect to physical time. For a particle of mass m one can introduce the energy-momentum four vector as P^μ = m u^μ with u^μstanding for the 4-velocity. The components of P^μ satisfy the mass shell condition P² = – m² , and also obey the geodesic equation

$P^0 \frac{dP^\mu}{dt} + \Gamma_{\nu \rho}^\mu P^\nu P^\rho=0$

Mass-shell condition in FRW

The geodesic equation allows to compute the evolution in time of the energy and momentum of particles, and gives one of the main observational consequences of an expanding Universe, the redshift of energy/momentum of photons, and more generally the fact that physical linear momentum of all particles decreases as 1/a as the Universe expands.

Using the mass shell condition for k=0 (the general case can be treated in close analogy using χ as radial variable)

$-E^2 + a^2(t) \sum_i (P^{i})^2 = -m^2$

where E ≡ P⁰ .We can also introduce the physical momentum, which measures the rate of change in physical distances as pⁱ =a(t) Pⁱ

Definition of redshift

so that

$-E^2 + p^2 = -m^2$

By using the connection we can compute how the momentum components Pⁱ changes with the expansion, namely

$E \left( \frac{d P^i}{dt} + 2 \frac{1}{a} \frac{da}{dt} P^i \right) =0$

from which we see that

$P^i \sim a^{-2}$ ; $P_i \sim \mbox{constant}$ ; $p^i \sim a^{-1}$

In the case of photons, this result gives a simple description of the observed redshift of wavelength. If t a photon is emitted by some astrophysical source (galaxy, star, etc.) at t_emwith energy E_emand is then received at t_obs. The energy which is measured is then E_obs= E_ema(t_em)/a(t_obs). In terms of the radiation wavelength

$\lambda_{obs}/\lambda_{em} \equiv 1+z = a(t_{obs})/a(t_{em})$

which defines the redshift z . The redshift of a given object is related to the scale factor at emission time through the relation

$1+z = \frac{a_0}{a}$

where a₀ is the scale factor today, usually set to unity in the case of a flat Universe.

Hubble parameter

For small redshifts we can expand up to the first order the previous expression around present time t_obs

$1+z = 1+ \left .\frac{1}{a} \frac{d a}{dt} \right|_{t_{obs}} (t_{obs}-t)$

Since the time difference t_obs- t = d/c with d the physical distance of the source, we get the famous Hubble’s law

$c z \sim \left .\frac{1}{a} \frac{d a}{dt} \right|_{t_{obs}} d = H_0 d$

The Hubble constant has dimension t^-1 and its order of magnitude is of a few tenth of km s^-1Mpc^-1. It is usually expressed in terms of the adimensional parameter h

$H_0 = 100 \, h \,\mbox{km s}^{-1} \mbox{ Mpc}^{-1}$

The dynamics of the expansion is completely encoded in the function a(t) which appears in the metric, and it is the dynamical variable in the Einstein equations

$G_{\mu \nu} \equiv R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R= 8 \pi G T_{\mu \nu} - \Lambda g_{\mu \nu}$

where R_μν and R are the Ricci tensor and scalar, respectively. Moreover Λ is the Einstein cosmological constant and G the Newton constant. In natural units G defines a mass scale, the Planck mass scale $m_{Pl} = G^{-1/2} = 1.22 \cdot 10^{19} \, \mbox{GeV}$

Energy-stress tensor

Finally, T_μν stands for the energy-stress tensor of all matter species filling the Universe. This tensor is symmetric and is covariantly conserved

$\nabla_\mu T^{\nu \mu} = \frac{\partial T^{\nu \mu}}{\partial x^\mu} + \Gamma^\nu_{\mu \rho}T^{\rho \mu}+ \Gamma^\mu_{\mu \rho}T^{\nu \rho}=0$

In many cases one can approximate a gas of particles as a perfect fluid with no viscosity, thus the energy-stress tensor reads

$T^\mu_\nu = (\rho+P) u^\mu u_\nu + P \delta^\mu_\nu$

If the fluid has vanishing velocity in the comoving frame we get

$T^0_0 = -\rho \,, \,\,\,\, T^i_j=P\, \delta^i_j$

The characteristic of the fluid filling the universe is encoded into its equation of state P=P(ρ). For relativistic particles (usually denoted as “radiation”) P=ρ/3, while for nonrelativistic particles (dust or simply “matter”), P = 0. In these relevant cases the equation of state is a simple linear relationship P = ω ρ. Even the cosmological term can be seen as a further contribution to the energy-stress tensor

$T_{\mu \nu, \Lambda} \equiv -\frac{\Lambda}{8 \pi G} g_{\mu \nu}$

that implies …

Energy-stress tensor for a gas of particles

$\rho_\Lambda=\Lambda/(8 \pi G)=-P_\Lambda$

this means a fluid with ω = -1. The energy-stress tensor gets contribution from all particles filling the universe can be seen,at some extent, as a gas with Bose-Einstein or Fermi-Dirac statistics (this is not true in the inflationary and accelerating stages, which will not be considered for the moment). In this case the energy-stress tensor can be given in terms of the distribution of particles in phase space. In particular we obtain

$n_\mu(t) = g \int \frac{d^3p}{(2 \pi)^3} \frac{P_\mu}{E} f(p,t)$

and

$T^\mu_\nu(t) = g \int \frac{d^3p}{(2 \pi)^3} \frac{P^\mu P_\nu}{E} f(p,t)$

from which we get

$n(t) =g \int \frac{d^3p}{(2 \pi)^3} f(p,t)$

and

$T^0_0 = -\rho = -g\int \frac{d^3p}{(2 \pi)^3} E f(p,t)$ $T^i_j = P \delta^i_j= \delta^i_j\, g \int \frac{d^3p}{(2 \pi)^3} \frac{p^2}{3 E} f(p,t) \$

Friedmann equation

For massless particles E=p hence we get P=ρ/3, whereas for non-relativistic particles E ≈ m and thus ρ ≈ m n and

$P \sim \rho \, \int \frac{d^3p}{(2 \pi)^3} \frac{p^2}{3 m^2} f(p,t) \left( \int \frac{d^3p}{(2 \pi)^3} f(p,t) \right)^{-1} \ll \rho$

The covariant conservation of stress-energy tensor, for an ideal fluid , using ν=0 gives

$\dot \rho + 3 \frac{\dot a}{a} (\rho+P) =0$

which can be solved by ρ = a^-3(1+ω). In particular one gets ρ_R ≈ a^-4 for ω=1/3 (Radiation), ρ_M ≈ a^-3 for ω=0 (Matter), ρ_Λ ≈ const. for ω=-1 (Cosmological constant). The 0-0 component of the Einstein equations gives (Friedmann Equation)

$\frac{\dot a^2}{a^2} + \frac{k}{a^2}= \frac{8 \pi G}{3} \rho + \frac{\Lambda}{3}$

that together with the previous equation allows for determining a=a(t). Note that the off-diagonal components of the equation do not give contribution due to the isotropy of space-time, and that the i – i component does not add further conditions. Differentiating the Friedmann Equation and using the relation for ρ we have

$\frac{\ddot a}{a} = -\frac{4 \pi G}{3} (\rho + 3P) + \frac{\Lambda}{3}$

Critical energy and spatial curvature

For an Universe radiation dominated, i.e. the main contribution to ρ comes from relativistic particles, one gets a(t)≈ √t. If the Universe is dominated by non–relativistic particles, i.e. matter dominated, one gets a(t)≈ t^2/3. A cosmological constant yields an exponentially growing scale factor.

Defining the critical density as

$\rho_{cr} \equiv \frac{3 H^2}{8 \pi G}$

the Friedmann equation can be rewritten in the form

$\Omega(a)-1 = \frac{k}{H^2 a^2}$

where $\Omega(a)=\rho/\rho_{cr}$ . From the previous equation one can see that the spatial curvature is fixed by the value ρ. A critical Universe, namely with Ω=1, is spatially flat. Defining the spatial contribution to the energy density as

$\Omega_k \equiv -k/(H^2 a^2)$

we have the sum rule $\Omega(a)+\Omega_k(a) =1$ . The critical density today (a=1) is

$\rho_{0,cr} = \frac{3 H_0^2}{8 \pi G} = 1.878 \times 10^{-29} h^2 \mbox{g cm}^{-3}$

Deceleration parameter

by using this quantity we can measure the contributions of Radiation, Matter, Curvature and Cosmological Constant in terms of

$\Omega_R=\frac{\rho_{r,0}}{\rho_{cr,0}}= \frac{8 \pi G \rho_{r,0}}{3 H_0^2} \, , \,\,\,\,\Omega_M=\frac{\rho_{m,0}}{\rho_{cr,0}} \, , \,\,\,\, \Omega_k = -\frac{k}{H_0^2} \, , \,\,\,\, \Omega_\Lambda=\frac{\rho_\Lambda}{\rho_{cr,0}}$

Hence the Friedmann equation gives

$H(z) = H_0 \left( \Omega_R (1+z)^{4} + \Omega_M (1+z)^{3} + \Omega_k (1+z)^{2} + \Omega_\Lambda \right)^{1/2}$

For small redshifts the expansion rate can be Taylor expanded. Up to the first order the deceleration parameter

$q = \left. -\frac{\ddot a}{a } \right|_{0} \frac{1}{H_0^2}$

reads

$q= \frac{\Omega}{2} \left( 1+ 3 \frac{P}{\rho} \right)_{0}$

Distances in the Universe

To define a distance in an expanding Universe is more tricky than in a static background. To determine how far is a source from the observation point it is necessary to take into account the cosmological expansion between the emission and observation times. According to the particular operational definition of distance one gets different results (luminosity distance, angular distance, etc.).

Let usconsider the so-called comoving particle horizon χ_p(t), defined as the comoving distance travelled by light between a very early initial time t_i and some time t. Since photons propagate along null geodesics we get in FRW

$\chi_p(t) = \int_0^r \frac{dr'}{\sqrt{1-k r'^2}} = \int_{t_i}^t \frac{dt'}{a(t')} = \int_0^{a(t)} \frac{da'}{a'^{2} H(a')} = \int_z^{\infty} \frac{d z'}{H(z')}$

where we have assumed a vanishing initial scale factor. This quantity represents the largest distance from which an observer can receive information at a given t. Analogously, the light received today from a source with redshift z has travelled the comoving distance

$\chi = \int_{a(t_{em})}^{a_0} \frac{da'}{a'^{2} H(a')} = \int_0^{z} \frac{d z'}{H(z')}$

t_em is the emission time, and can be obtained in terms of z if we know the expansion history of the Universe, namely inverting the relation a(t) = a₀/(1+z(t)).

Last scattering surface

CMB Photons were emitted from a thin surface corresponding to their last interaction with charged particles. Such surface of last scattering surface is placed at redshift z_LS≈ 10³. For a universe MD the comoving distance of this surface is therefore

$\chi_{LS} = \frac{1}{H_0}\int_0^{z_{LS}} d z' \frac{1}{(1+z')^{3/2} } = \frac{2}{H_0} \left(1- \frac{1}{\sqrt{1+z_{LS}}} \right)$

Such value would increase if we had included the contribution of Ω_Λ(for Ω_Λ= 1 one would get χ_LS= z_LS/H₀. In case of a particle with mass m, such as neutrinos, we have

$\chi = \int_{a(t_{em})}^{a(t_{obs})} \frac{dt}{a(t)} \frac{y}{\sqrt{y^2+m^2 a(t)^2}}= \int_0^z \frac{dz'}{H(z')} \frac{y}{\sqrt{y^2+m^2/(1+z)^2}}$

As stated before, the distance we get depends on the operational definition we use. If we look at a point–like source, like galaxies, stars etc, of which we know the absolute luminosity L (denoted as standard candle), the physical distance d of the source can be defined by measuring the energy flux Φ observed at the detector

$\Phi = \frac{L}{4 \pi d^2} \,,\,\,\,\Rightarrow d= \sqrt{L/(4 \pi \Phi)}$

Such relationship allows to define the luminosity distance as

$d_L\equiv\sqrt{L/(4 \pi \Phi)}$

Luminosity Distance I

To compute the expression of d_L we have to take into account that:

1) the signal received have propagated over a physical distance a(t_obs) χ , with χ the comoving distance travelled by photons to reach us. Since we assume an isotropic emission, the sources are homogeneously distributed over a thin spherical shell with surface 4 π a(t_obs)² r²(χ)

2) the energy of photons received is smaller than at the emission, by a factor a(t_em)/a(t_obs) due to redshift

3) the arrival rate is also lower by the same factor. In fact we know that

$\chi = \int_{t_{em}}^{t_{obs}} \frac{dt}{a(t)}$

Moreover a radiation emitted at t_em+δt will be received at the time t_obs+δt’ such that

$\chi = \int_{t_{em}+\delta t}^{t_{obs}+ \delta t'} \frac{dt}{a(t)}$

which implies

$\frac{\delta t}{a(t_{em})} = \frac{\delta t'}{a(t_{obs})}$

Thus the flux Φ results

Luminosity Distance II

$\Phi = \frac{L a^2(t_{em})}{4 \pi a^4(t_{obs}) r^2(\chi)}$

and the luminosity distance results

$d_L = a^2(t_{obs}) r(\chi)/a(t_{em}) = a (t_{obs})r(\chi) (1+z)$

For a spatially flat metric (k=0) and for Ω_R << Ω_M this expression simplifies for small redshift, namely

$d_L = (1+z) \frac{1}{H_0}\int_0^{z} d z' \frac{1}{(\Omega_M(1+z')^3+\Omega_\Lambda)^{1/2} }$

Expanding at small redshifts up to the first order in z, assuming k=0 and Ω_M +Ω_Λ=1, we get again the Hubble law. Up to the second order we have instead

$d_L \sim \frac{z}{H_0} \left(1+ \frac{1}{2}(1-q) z \right)$