LOG#112. Basic Cosmology (VII).

In my final post of this basic Cosmology thread, I am going to discuss scalar perturbations and to review some of the recent results about cosmological parameters by WMAP and PLANCK.

One of the predictions of the Standard Cosmological Model (LCDM or ΛCDM) is that we should expect some inhomogeneities and anisotropies in the Cosmic Microwave Background (CMB). Indeed, the “perturbations” of densities (matter, radiation,…) are closely related to these anisotropies and the easiest way to trigger them is by using “scalar perturbations”, i.e., perturbations induced by the scalar field in the early Universe! In fact, inflation should be understood at quantum level as a “quantum fluctuation” that triggers scalar perturbations of the so called “inflaton field”.

The scalar field perturbations are “shifts” or “fluctuations” in the scalar field, namely

$\phi (\mathbf{x},t)=\phi^0 (t)+\delta \phi (\mathbf{x},t)$

Now, we have to work out a little bit the equation for $\delta \phi$, if we take into account that our Universe is seemingly a “smoothly” expanding Universe. To first order, we can neglect the “back reaction” on the metric field due to the scalar field $\delta \phi$. Therefore, the equation we get is really simple:

$\square \phi-V'(\phi)=0$

or equivalently

$\ddot{\phi}+2aH\dot{\phi}-\nabla^2\phi+V'(\phi)=0$

where

$V'(\phi^0+\delta\phi)=V'(\phi^0)+V''(\phi^0)\delta\phi\approx V'(\phi^0)$

since the term $V''(\phi^0)\delta\phi\propto \varepsilon$ can be neglected during inflation. Furthermore, we can write

$\delta\ddot{\phi}+2aH\delta\dot{\phi}-\nabla^2\delta\phi\approx 0$

so

$\boxed{\delta\ddot{\phi}_k+2aH\delta\dot{\phi}_k+k^2\delta\phi_k=0}$

for any Fourier mode “k” in the Fourier decomposition of the scalar field $\phi$. If we define $\delta\bar{\phi}=a\delta\phi$, then we can rewrite the above equation as follows

$\boxed{\delta \ddot{\bar{\phi}}+\left(k^2-\dfrac{\ddot{a}}{a}\right)\delta\bar{\phi}=0}$

This last equation describes a harmonic oscillator equation with time-varying frequency! We can quantize $\delta\phi$ using the formalism of creation and annihilation operators ($a_k,a^+_k$, respectively):

$\delta\bar{\phi}(k,\eta)=U(k,\eta)a_k+U^*(k,\eta)a^+_k$

and where U satisfies the equation of a harmonic oscillator as well

$\ddot{U}+\left(k^2-\dfrac{\ddot{a}}{a}\right)U=0$

The variance of perturbations in $\delta\bar{\phi}$ are defined with the equations

$\langle \delta^\dagger (k,\eta)\delta\phi(k',\eta)\rangle=\dfrac{1}{a}\vert U(k,\eta)\vert^2(2\pi)^3\delta^3(k-k')=P_{\delta\phi}(2\pi)^3\delta^3(k-k')$

and where we have defined the power spectrum in terms of correlation function as the quantity

$P_{\delta\phi}(k)=\dfrac{1}{a^2}\vert U(k,\eta)\vert^2$

We require a solution for the equation above for $U(k,\eta)$. During inflation, $\ddot{a}/a$ provides

$\dot{a}=a^2H\approx -\dfrac{a}{\eta}$

and

$\dfrac{\ddot{a}}{a}=-\dfrac{1}{a}\dfrac{d}{dt}\left(\dfrac{a}{\eta}\right)=-\dfrac{1}{a}\dfrac{\dot{a}}{\eta}+\dfrac{1}{a}\dfrac{a}{\eta^2}=\dfrac{2}{\eta^2}$

Then, for the equation above for U becomes

$\ddot{U}+\left(k^2-\dfrac{2}{\eta^2}\right)=0$

And now, a subtle issue…Initial conditions! Any initial condition (with proper normalization) long before the horizon exit implies that $-\eta>>1$. It implies that the $k^2$ term dominates and we can substitute the last equation by this one

$\ddot{U}+k^2U=0$

and this equation has pretty simple solutions with proper normalization. Essentially, they are “plane waves” modulated with the Fourier mode k, i.e.,

$U\sim \dfrac{e^{-ik\eta}}{\sqrt{2k}}$

A more precise proper solution is

$U=\dfrac{e^{-ik\eta}}{\sqrt{2k}}\left[1-\dfrac{i}{k\eta}\right]$

Please, note that if $k\vert\eta\vert>>1$, we recover the previous equation. After many e-folds, we get $k\vert\eta\vert<<1$ and

$U\longrightarrow \dfrac{e^{-ik\eta}}{\sqrt{2k}}\dfrac{-i}{k\eta}$

Thus, the power spectrum is deduced to be

$P_{\delta\phi}(k)=\dfrac{1}{a^2}\dfrac{1}{2k^3\eta^2}=\dfrac{H^2}{2k^3}$

For tensor perturbations (I have not discussed them here for simplicity), this result can be compared with

$P_h(k)=\dfrac{8\pi G_NG^2}{k^3}$

So, at least, scalar perturbations have the same order of magnitude and the same power of the Fourier modes! Indeed, the fluctuations $\delta \phi$ are transfered to scalar perturbations or metric pertubations, since

$P_\phi=\dfrac{4}{9}\left(\dfrac{aH^2}{\dot{\phi}}\right)P_{\delta\phi}\vert_{aH=k}$

After the horizon crossing, we obtain

$P_\phi=\dfrac{2}{9k^3}\left(\dfrac{aH^2}{\dot{\phi}}\right)^2\vert_{aH=k}=\dfrac{8\pi G_N}{9k^3}\dfrac{H^2}{\varepsilon}\vert_{aH=k}=\dfrac{128\pi^2G_N}{9k^3}\left(\dfrac{H^2V^2}{V'^2}\right)\vert_{aH=k}$

Now, we can introduce the important concept of spectral index. For scalar fields, the perturbations are defined in terms of this spectral index as follows

$\boxed{P_\phi (k)=\dfrac{8\pi}{9k^3}\dfrac{H^2}{\varepsilon M_p^2}\vert_{aH=k}=\dfrac{50\pi^2}{9k^3}\left(\dfrac{k}{H_0}\right)^{n-1}\delta^2_H\left(\dfrac{\Omega_m}{D_1(a=1)}\right)^2}$

For metric field perturbations, we write

$\boxed{P_h(k)=\dfrac{8\pi}{k^3}\dfrac{H^2}{M_p^2}\vert_{aH=k}=A_Tk^{n_T-3}}$

and where the spectral index $n$ and its tensor analogue $n_T$ are defined through the simple expressions

$\boxed{n=1-4\varepsilon-2\delta}$

$\boxed{n_T=-2\varepsilon}$

Final conclusions, WMAP and PLANCK

The history of the Universe up to the period of BBN is (more or less) well understood in the framework of the Standard Cosmological Model (LCDM). It is well established and tested, BUT, please, do not forget we have to choose “the proper initial conditions” in order to obtain the right (measured/observed) results. In summary, up to the BBN we get

1. A homogenous and isotropic Universe, with small density perturbations. It is equivalent to a thermal bath with temperature above 1MeV, i.e., $T\geq 1MeV$. Indeed, inflation seems to be the simplest and best solution to many problems of modern cosmology (e.g., the horizon problem or the flatness problem). However, the LCDM and the inflation itself can not answer what the inflaton field is and why the anisotropies are so tiny as we observed today. Moreover, the cosmological constant problem remains unanswered today (even when we have just discovered the Higgs field, and that some people are speculating that the Higgs field could be matched with the inflaton field, this idea has some technical problems).

2. The energy-matter content of the Universe…There is a wide agreement about the current composition of the Universe, based on the “concordance model” by several experiments, now confirmed by the WMAP/PLANCK probes. Generally speaking:

Baryons are negligible since $\eta_b\sim 5\cdot 10^{-10}$. In fact, only about a 5% is “matter” we do know through the Standard Model of particle physics. Most of that 5% is “light” (a.k.a., photons) and the remaining 1% are the elements we do know from the Periodic Table (and mainly hydrogen and helium, of course).

(Cold) Dark Matter is about $\Omega_{DM}\sim 0\mbox{.2}-0\mbox{.}3$. There is no known particle that can be assigned to this dark matter stuff. It does not emit light, it is (likely) non-relativistic and uncharged under the gauge symmetry group of the SM, i.e., under $G_{SM}=U(1)_Y\times SU(2)_L\times SU(3)_c$. Of course, you could avoid the need for this stuff invoking a modified newtonian dynamics (MOND) and/or modified gravity (MOG). However, to my knowledge, any MOND/MOG theory generally ends by requiring some kind of stuff that is “like cold DM”. So, Occam’s razor here punches and says that cold DM is (likely) the most promising solution to the flat rotation curve of galaxies and this mysterious energy density. Remember: Dark Matter can not be any known particle belonging to the Standard Model we know today (circa 2013).

Dark Energy (DE), a.k.a. vacuum energy or the cosmological constant. Reintroduced in 1998 to explain the SNae results and the high redshifts we measured from them, its existence is now fully established as well. Perhaps the dark energy name is “unfortunated” since it is, indeed, a negative pressure terms in the Einstein’s field equations. It is about the 70 or 80% of the current energy-matter budget of the Universe, i.e., it has an energy density about  $\Omega_{DE}=\Omega_\Lambda\sim 0\mbox{.}7-0\mbox{.}8$.

Let me draw again a picture I have showed before in this blog:

The main results for the cosmological parameters can be found here (in the case of the WMAP9 mission): http://arxiv.org/pdf/1212.5226v3.pdf

The main results for the cosmological parameter can be found here (in the case of the Planck mission): http://arxiv.org/pdf/1303.5076v1.pdf

We can compare the cosmological parameters coming from the WMAP9 mission

with those of the Planck mission

We observe that the different experiments are providing convergent values of the main cosmological parameters, so we know we are in the right track!

Cosmology is a fascinating subject. I have only sketched some elementary ideas here. I will be discussing more advanced topics in the near future, but you have to wait for it! :).

See you in my next blog post!

LOG#111. Basic Cosmology (VI).

The topic today: problems in the Standard Cosmological Model (LCDM), inflation and scalar fields!

STANDARD COSMOLOGICAL MODEL: ISSUES

Despite the success of the Standard Cosmological model (or LCDM), today it is widely accepted that it is not complete, even if its main features and observables are known, there are some questions we can not understand in that framework.

Firstly, we have the horizon problem. For any comoving horizon, we obtain

$\eta=\int_0^t\dfrac{dt'}{a(t')}\propto \begin{cases}a^{1/2}\;\;\mbox{for a MD Universe}\\ a\;\;\mbox{for a RD Universe}\end{cases}$

Note that $a\propto$ $t^{2/3}$ for a MD Universe and to $t^{1/2}$ for a RD Universe.

According to the CMB observations, today our Universe is very close to be isotropic, since the deviations are tiny

$\dfrac{\delta T}{T}\approx 10^{-5}$

The issue, of course, is how can it be true?Recall that the largest scales observed today have entered theo horizon just recently, long after decoupling. Moreover, microscopic causal physics can not make it! So, where the above tiny anisotropy comes from? E.g., for a distant galaxy, we get

$\left(\dfrac{\delta T}{T}\right)_{\lambda galaxy}$

and this scale was outside the horizon in the past! Please, note that the entropy within a horizon volume is about

$S_H=s\dfrac{4\pi}{3}d_H^3=\begin{cases}0\mbox{.05}g_\star^{-1/2}\left(\dfrac{M_p}{T}\right)^3,\;\; RD\\ 3\cdot 10^{87}\left(\Omega_0h^2\right)^{-3/2}(1+z)^{-3/2}, \;\; MD\end{cases}$

and so

$S_H(t=t_0)=10^{88}$

$S_H(t=t_{rec})=10^{83}=10^5V_H(today)$

$S_H(t=t_{BBN})=10^{63}$

Another problem is the flatness problem. The physical radius of curvature is given by the term

$R_k=a(t)\vert k\vert^{-1/2}=\dfrac{H^{-1}}{\vert \Omega-1\vert^{1/2}}$

The total energy density as a function of the scale factor is given by

$\Omega-1=\dfrac{k}{H^2a^2}\propto \dfrac{1}{\rho a^2}=\begin{cases}a\;\ MD\\ a^2\;\; RD\end{cases}$

When the primordial nucleosynthesis happened, about $t=t_{BBN}\sim 1s$, it gave $\vert \Omega-1\vert\leq 10^{-16}$ and it gives $R_k\geq 10^8H^{-1}$. By the other hand, at the Planck time, i.e., if $t=t_p\approx 10^{-43}s$, it gives the value $\vert \Omega-1\vert\leq 10^{-60}$, and then $R_k\geq 10^{30}H^{-1}$. This large mismatch means that the Big Bang Universe requires VERY SPECIAL initial conditions, otherwise the Universe would not be (apparently) flat as we observe at current time. Note that if $\Omega\sim 1$ and $R_k\sim H^{-1}$ at Planck time, then there are two options:

1) k>0: the Universe recollapse withim few $10^{-43}s$.

2) k<0: the Universe reaches 3K at $t=10^{-11}s$.

Therefore, the natural time scale for cosmology is the Planck time (at least in the very early Universe and before) $10^{-43}s$. The current age of the Universe is about $10^{60}t_p$.

An important, yet unsolved and terrific, problem is the cosmological constant problem. The vacuum energy we observe today (in the form of dark energy) is given by

$T_{\mu\nu}=\Lambda g_{\mu\nu}$

The equation of state for the vacuum energy is known to be $p=\omega \rho_\Lambda=-\rho_\Lambda$, i.e., $\omega_\Lambda=-1$. It yields

$a(t)\propto e^{Ht}$

i.e., the so called de Sitter space (dS) or de Sitter Universe. It is a maximally symmetric spacetime. In a dS Universe, we obtain

$H=\left(\dfrac{1}{3}k^2\Lambda\right)^{1/2}=const.$

Observations provide (via dark energy) that $\rho_{vac}=\rho_\Lambda\sim\rho_c\sim (3\cdot 10^{-3}eV)^4$

Theoretical (and “natural”) Quantum Field Theory (QFT) calculations give

$\rho_{vac}=\rho_\Lambda\sim \Lambda^4_{cut-off}\sim M_P^4\sim 10^{120}\rho_c$

or so( even the mismatch can be 122 or 123 orders of magnitude!!!!). This problem is far beyond our current knowledge of QFT. It (likely) requires new physics or to rethink QFT and/or the observed value of the vacuum energy density. It is a hint that our understanding of the Universe is not complete.

INFLATION

One of the most “simple” and elegant solutions to the flatness problem and the horizon problem is the inflationary theory. What is inflation? Let me explain it here better. If the (early) Universe experienced an stage of “very fast” expansion, we can solve the horizon problem! However, there is a problem (you can call Houston if you want to…). If we do want a quick expansion in the early Universe, we need “negative pressure” to realize that scenario. Negative pressure can be obtained in an scalar field theory (there are some alternatives with a repulsive vector field and/or higher order tensors like a 3-form antisymmetric field, but the most simple solution is given by scalar fields).

The solution to the horizon problem in the inflationary theory proceeds as follows. Firstly, for the comoving horizon, we get

$\eta=\int_0^tdt'/a(t')=\int_0^a\dfrac{da'}{a'}\dfrac{1}{a'H(a')}$

where the disance over which particles can travel in the course of one expansion time is the comoving Hubble radius $H^{-1}/a$, and it is encoded into the fraction

$\dfrac{1}{a'H(a')}$

We have to make a distinction of the comoving horizon $\eta$ and the comoving Hubble radius $1/aH$. If we observe two particles whose comoving distance is r, then

1st. If $r>\eta$, we can never have a communication between those particles.

2nd. If $r>1/a_H$, then we can never communicate NOW using those two particles.

It shows that it is possible to have $\eta>> 1/aH\vert_{t=t_0}$. A particle with $1/aH ca not communicate today BUT they could be in causal contact early on. We only need that $1/aH\vert_{t=t_0}>>1/aH\vert_{t=now}$. That is, $\eta$ get contributions mostly from early epochs. Indeed, both in RD (Radiation Dominated) and MD(Matter Dominated) Universes, $1/a_H$ increase with time, so the latter epoch contributions dominate over cosmological time scales. Then, a solution for the horizon problem is that in the early Universe, in this “inflationary” (very fast) phase, for at least a brief amount of time (how much is not clear) the comoving Hubble radius DECREASED.

How can the scale factor evolve in order to solve the horizon problem?We do know that $(aH)^{-1}$ must decrease, so $aH$ must increase! Therefore,

$\dfrac{d}{dt}(aH)=\dfrac{d^2a}{dt^2}>0$

i.e., we get an (positively) accelerating expansion or “inflation”.

How can we understand quantitatively inflation? Firstly, suppose that the energy scale of inflation is about $\sim 10^{15}GeV$. It is about the Grand Unified Theory (GUT) energy scale, and close to the Quantum Gravity scale (the Planck scale) about $\sim 10^{19}GeV$. Obviously, it only matters at very high energies, very short distances or very tiny time scales after the Big Bang. Then,

$(aH)^{-1}\vert_{T=10^{15}GeV}=10^{-28}(aH)^{-1}\vert_{T=T_0}$

For a RD Universe, $H\propto a^{-2}$ and thus

$\dfrac{a_0H_0}{a_eH_e}=\dfrac{a_0}{a_e}\dfrac{a_e^2}{a_0^2}=\dfrac{a_e}{a_0}\approx \dfrac{T_0}{10^{15}GeV}$

During the inflation, the comoving Hubble radius had to decrease by, at least, 28 orders of magnitude. The most common way to build such inflationary model is to build a model where $H\approx const.$ and

$\dfrac{\dot{a}}{a}=H=constant$

It gives $a(t)=e^{H(t-t_e)}$ and $t, where $t_e$ is the time when inflation ends. In fact, we obtain that

$(aH)^{-1}\propto e^{Ht}$ and thus $10^{28}$ can be guessed from the so-called “e-folds”, or exponential factors, in a very simple way: note that $10^{28}\approx e^{64}$!!!!!!!!! Therefore, more than 64 “e-folds” (or the 64th power of the number “e”) provide the necessary 10-fold we were searching for.

Remark: The comoving horizon is very similar to an effective time parameter!

$\eta_{total}=\int_0^{t_e}\dfrac{dt'}{a(t')}+\int_{t_e}^t\dfrac{dt'}{a(t')}=\eta_{prin}+\eta=(\mbox{very large number})+\mbox{(new time parameter)}$

NEGATIVE PRESSURE

In order to allow inflation, we require that the “weak energy condition” be violated. That is, we ask that (for inflation to be possible)

$\dfrac{\ddot{a}}{a}=-\dfrac{4\pi G_N}{3}(\rho+3p)>0$

It implies that $\rho+3p<0$ or that $p<-\dfrac{1}{3}\rho<0$

Obviously, there is no particle/field in the Standard Model (at least until the discovery of the Higgs field in 2012) able to do that. So, if inflation happened, it is an unknown field/particle responsible for it! The simplest implementation of inflation uses, as we said before, some scalar field $\phi$. Let us define a scalar field $\phi (x^\mu)$. Its lagrangian is generally given by

$\mathcal{L}=-\dfrac{1}{2}\partial_\mu\phi\partial^\mu\phi -V(\phi)$

The energy-momentum tensor for this scalar field can be easily obtained by the well-known prescription

$T_{\mu\ nu}=\partial_\mu\phi\partial_\nu\phi-g_{\mu\nu}\mathcal{L}$

Thus, we get

$\rho=-T^{0}_{\;\;\;0}=\dfrac{1}{2}\dot{\phi}^2+V(\phi)$

$p=T^i_{\;\;\; i}=\dfrac{1}{2}\dot{\phi}^2-V(\phi)$

Negative pressure is obtained whenever we have $V(\phi)>\dfrac{1}{2}\dot{\phi}^2$

Therefore:

1st. A scalar field with negative pressure is trapped into a “false vacuum”.

2nd. A scalar field “slow-rolling” toward its true vacuum provides a simple model for inflation.

The evolution of the scalar field in the expanding Universe is given by a simple equation

$\ddot{\phi}+3H\dot{\phi}+V'(\phi)=0$

Models where the scalar field “slow-rolls” provide $H\approx constant$ (slow varying in time as a scalar function!). The time variable $t\longrightarrow \eta$ implies that

$\eta\equiv \int_{a_e}^a\dfrac{da}{Ha^2}\approx \dfrac{1}{H}\int_{a_e}^a\dfrac{da}{a^2}\approx -\dfrac{1}{aH}$

and it becomes negative during the inflationary phase! The slow-roll parameters

$\varepsilon= \dfrac{dH^{-1}}{dt}=-\dfrac{\dot{H}}{aH}$

where the dot means derivative with respect to $\eta$. Moreover, we also get

$\delta=\bar{\eta}=\dfrac{1}{H}\dfrac{d^2\phi/dt^2}{d\phi/dt}=-\dfrac{1}{aH\dot{\phi}}\left(aH\dot{\phi}-\ddot{\phi}\right)=-\dfrac{1}{aH\dot{\phi}}\left(3aH\dot{\phi}+a^2V'\right)$

Remark: $\varepsilon<<1$ for inflation and $\varepsilon=2$ for RD Universes imply that $\varepsilon<1$ is a definition of inflationary phase!

Quintessence and phantom energy

For any scalar field, and the pressure and energy density given above, we can calculate the $\omega$ quantity:

$\omega=\dfrac{p_\omega}{\rho_\omega}=\dfrac{\dfrac{1}{2}\dot{\phi}^2-V(\phi)}{\dfrac{1}{2}\dot{\phi}^2+V(\phi)}$

In the case that the kinetic term is negligible, we obtain the dark energy fluid value $\omega_\Lambda=-1$. However, this equation is much more general. If we have a slow-rolling scalar field over cosmological times (a very slowly time dependent scalar field) it could mimic the cosmological constant behaviour ( we are ignoring some technical problems here). If the kinetic term is NOT negligible, the $\omega$ value can differ from the standard value of $-1$ (dark energy/cosmological constant/vacuum energy). However, current observation support a pure cosmological constant term. Moreover, this general scalar field is commonly referred as “quintessence” if $-1<\omega_\phi<1/3$ and as “phantom energy” if $\omega_\phi<-1$. Both, dark energy, quintessence or phantom energy models can affect to the future of the Universe. Instead of a Big Freeze (the thermal death of the Universe), the scalar dominated Universe can even destroy atoms/matter/galaxies at some point in the future (at least on very general grounds) and terminate the Universe in a Big (or Little) Rip. Thus, let me review the possible destinies of the Universe according to modern Cosmology:

1) Big Crunch (or recollapse of the Universe). The Universe recollapses until the initial singularity after some time, if it is massive enough. Current observations don’t favour this case.

2) Big Freeze (or thermal death of the Universe). The Universe expands forever cooling itself until it reaches a temperature close to the absolute zero. It was believed that it was the only possible option with the given curvature until the discovery of dark energy in 1998.

3) Big Rip (or Little Rip, depending on the nature of the scalar field and the concrete model). Vacuum energy expands the Universe with an increasing rate until it “rips” even fundamental particles/atoms/matter and galaxies apart from  eath other. It is a new possibility due to the existence of scalar fields and/or dark energy, a mysterious energy that makes the Universe expanss with an increasing rate overseding the gravitational pull of galaxies and clusters!

A more exotic option is that the Universe suffers “oscillations” of positively accelerated expansion and negatively accelerated expansion (oscillatory/cyclic ethernal Universes)…But that is another story…

See you in my final basic cosmological post!

LOG#110. Basic Cosmology (V).

Recombination

When the Universe cooled up to $T\sim eV$, the neutrinos decoupled from the primordial plasma (soup). Protons, electrons and photons remained tighly coupled by 2 main types of scattering processes:

1) Compton scattering: $e+\gamma \leftrightarrow e+\gamma$

2) Coulomb scattering: $e^-+p\leftrightarrow H+\gamma$

Then, there were little hydrogen (H) and though $B_H>T$ due to small baryon fraction $\eta_b$.

The evolution of the free electron fraction provided the ratio

$X_e\equiv =\dfrac{n_e}{n_e+n_H}=\dfrac{n_p}{n_p+n_H}$

where $n_p+n_H\approx n_b$ and the second equality is due to the neutrality of our universe, i.e., to the fact that $n_e=n_p$ (by charge conservation). If $e^-+p\longrightarrow H+\gamma$ remains in the thermal equilibrium, then

$\dfrac{n_en_p}{n_H}=\dfrac{n_e^0n_p^0}{n_H^0}\longrightarrow \dfrac{X_e^2}{1-X_e}=\dfrac{1}{n_e+n_H}\left[\left(\dfrac{m_eT}{2\pi}\right)^{3/2}e^{-\left[m_e+m_p-m_H\right]/T}\right]$

where we have

$\dfrac{1}{n_e+n_H}\left(\dfrac{m_eT}{2\pi}\right)^{3/2}=n_p+n_H=n_b-4n(He)\approx n_p+n_H=n_b=\eta_b\eta_\gamma$

It gives $\eta_b\eta_\gamma\sim 10^{-9}T^3\approx 10^{15}$

and the last equality is due to the fact we take $T\sim E_0$. It means that $X_e\approx 1$ at $T\sim E_0$. As we have $X_e\longrightarrow 0$, we are out of the thermal equilibrium.

From the Boltzmann equation, we also get

$a^{-3}\dfrac{d(n_ea^3)}{dt}=n_e^0n_p^0\langle \sigma v\rangle \left( \dfrac{n_Hn_\gamma}{n_H^0n_\gamma^0}-\dfrac{n_e^2}{n_e^0n_p^0}\right)$

or equivalently

$a^{-3}\dfrac{d(n_ea^3)}{dt}=n_b\langle \sigma v\rangle \left(\dfrac{n_H}{n_b}\dfrac{n_e^0n_p^0}{n_H^0}-\dfrac{n_e^2}{n_b}\right)$

i.e.

$a^{-3}\dfrac{d(n_ea^3)}{dt}=n_b\langle \sigma v\rangle \left( (1-X_e)\left(\dfrac{m_eT}{2\pi}\right)^{3/2}e^{-E_0/T}-X_e^2n_b\right)$

Using that $n_e=n_bX_e$ and $\dfrac{d}{dt}(n_ba^3)=0$, we obtain

$\dfrac{dX_e}{dt}=\left[(1-X_e)\beta -X_e^2n_b\alpha^{(2)}\right]$

with

$\beta\equiv \langle \sigma v\rangle \left(\dfrac{m_eT}{2\pi}\right)^{3/2}e^{-E_0/T}$, the ionization rate, and

$\alpha^{(2)}\equiv \langle \sigma v\rangle$ the so-called recombination rate. It is taken the recombination to the n=2 state of the neutral hydrogen. Note that the ground state recombination is NOT relevant here since it produces an ionizing photon, which ionizes a neutral atom, and thus the net effect is zero. In fact, the above equations provide

$\alpha^{(2)}=9\mbox{.}78\dfrac{\alpha^2}{m_e^2}\left(\dfrac{E_0}{T}\right)^{1/2}\ln \left(\dfrac{E_0}{T}\right)$

The numerical integration produces the following qualitative figure

The decoupling of photons from the primordial plasma is explained as

$\mbox{Compton scattering rate}\sim\mbox{Expansion rate}$

Mathematicaly speaking, this fact implies that

$n_e\sigma_T=X_en_b\sigma_T$

where $\sigma_T$ is the Thomson cross section. For the processes we are interesting in, it gives

$\sigma_T=0\mbox{.}665\cdot 10^{-24}cm^2$

and then

$n_e\sigma_T=7\mbox{.}477\cdot 10^{-30}cm^{-1}X_e\Omega_bh^2a^{-3}$

Thus, we deduce that

$\dfrac{n_e\sigma_T}{H}=0\mbox{.}0692a^{-3}X_e\Omega_bh\dfrac{H_0}{H}$

$\dfrac{n_e\sigma_T}{H}=113X_e\left(\dfrac{\Omega_bh^2}{0\mbox{.}02}\right)\left(\dfrac{0\mbox{.}15}{\Omega_mh^2}\right)^{1/2}\left(\dfrac{1+z}{1000}\right)^{3/2}\left[1+\dfrac{1+z}{3600}\dfrac{0\mbox{.15}}{\Omega_mh^2}\right]^{-1/2}$

and where $X_e\leq 10^{-2}$ implies that the decoupling of photons occurs during the time of recombination! In fact, the decoupling of photons at time of recombination is what we observe when we look at the Cosmic Microwave Background (CMB). Fascinating, isn’t it?

Dark Matter (DM)

Today, we have strong evidences and hints that non-baryonic dark matter (DM) exists (otherwise, we should modify newtonian dynamics and or the gravitational law at large scales, but it seems that even if we do that, we require this dark matter stuff).

In fact, from cosmological observations (and some astrotronomical and astrophysical measurements) we get the value of the DM energy density

$\Omega_{DM}\sim 0\mbox{.}2-0\mbox{.}3$

The most plausible candidate for DM are the Weakly Interacting Massive Particles (WIMPs, for short). Generic WIMP scenarios provide annihilations

$X_{DM}+\bar{X}_{DM}\leftrightarrow l+\bar{l}$

where $X_{DM}$ is some “heavy” DM particle and the (ultra)weak interaction above produces light particles in form of leptons and antileptons, tighly couple to the cosmic plasma. The Boltzmann equation gives

$a^{-3}\dfrac{d(n_Xa^3)}{dt}=\langle \sigma_X v\rangle \left( n_X^{(0)2}-n_X^2\right)$

Define the yield (or ratio) $Y_X=\dfrac{n_X}{T^3}$. It is produced since generally we have

$Y=\dfrac{n_X}{s}$

and since $sa^3=constant$, then $s\propto T^3$. Thus,

$\dfrac{dY}{dt}=T^3\langle \sigma v\rangle \left( Y_{EQ}^2-Y^2\right)$

and

$Y_{EQ}=\dfrac{n_X^0}{T^3}$

Now, we can introduce a new time variable, say

$x=\dfrac{m}{T}$

Then, we calculate

$\dfrac{dx}{dt}=-\dfrac{m}{T^2}\dfrac{dT}{dt}=-\dfrac{m}{T^2}\left(-\dfrac{\dot{a}}{a}T\right)=xH$

For a radiation dominated (RD) Universe, $\rho\propto T^4$ implies that $H\propto T^2$ and $H(x)=-\dfrac{H(m)}{x^2}$

In this case, we obtain

$\dfrac{dY}{dx}=\dfrac{\lambda}{x^2}\left(Y^2-Y_{EQ}^2\right)$

with $\lambda=\dfrac{m^3\langle \sigma v\rangle}{H(m)}$

The final freeze out abundance is got in the limit $Y_\infty=Y(x\longrightarrow \infty)$. Typically, $\lambda >>1$, and when $Y_{EQ}\sim 1$ and $Y\approx Y_{EQ}$, for $x>>1$, and there, the yield drops exponentially

$\dfrac{dY}{dx}\approx \dfrac{\lambda Y^2}{x^2}$

or

$\dfrac{dY}{Y^2}\approx \dfrac{\lambda dx}{x^2}$

Integrating this equation,

$\displaystyle{\int_{Y_f}^{Y_\infty}\dfrac{dY}{Y^2}=\int_{x_f}^\infty \dfrac{\lambda}{dx}{x^2}}$

and then

$\dfrac{1}{Y_\infty}-\dfrac{1}{Y_f}=\dfrac{\lambda}{x_f}$

Generally, $Y_f>>Y_\infty$ and the freeze out temperature for WIMPs is got with the aid of the following equation

$Y_\infty=\dfrac{x_f}{\lambda}$

Indeed, $n\langle \sigma v\rangle= H\longrightarrow x_f\sim 10$

A qualitative numerical solution of the “WIMP” miracle (and its freeze out) is given by the following sketch

The present abundance of heavy particle relics gives

$\rho_X=mY_\infty T_0^3\left(\dfrac{a_1T_1}{a_0T_0}\right)^3\approx \dfrac{mY_\infty T_0^3}{30}$

and where the effect of entrpy dumping after the freeze-out is encoded into the factor

$\left(\dfrac{a_1T_1}{a_0T_0}\right)^3$ with $\left(\dfrac{g_\star (0)}{g_\star (f)}\right)^3\approx \dfrac{1}{30}$

Moreover, the DM energy density can also be estimated:

$\Omega_X=\Omega_{DM}=\dfrac{x_f}{\lambda}\dfrac{mT_0^3}{30\rho_c}=\dfrac{H (m) x_fT_0^3}{30m^2\langle \sigma v\rangle\rho_c}$

so

$\Omega_X=\left[\dfrac{4\pi^3Gg_\star (m)}{45}\right]^{1/2}\dfrac{x_fT_0^3}{30\langle \sigma v\rangle \rho_c}=0\mbox{.}3h^{-2}\left(\dfrac{x_f}{10}\right)\left(\dfrac{g_\star (m)}{100}\right)^{1/2}\dfrac{10^{-39}cm^2}{\langle \sigma v\rangle}$

The main (current favourite) candidate for WIMP particles are the so called lightest supersymmetric particles (LSP). However, there are other possible elections. For instance, Majorana neutrinos (or other sterile neutrino species), Z prime bosons, and other exotic particles. We observe that here there is a deep connection between particle physics, astrophysics and cosmology when we talk about the energy density and its total composition, from a fundamental viewpoint.

Remark: there are also WISP particles (Weakly Interacting Slim Particles), like (superlight) axions and other exotics that could contribute to the DM energy density and/or the “dark energy”/vacum energy that we observe today. There are many experiments searching for these particles in laboratories, colliders, DM detection experiments and astrophysical/cosmological observations (cosmic rays and other HEP phenomena are also investigated to that goal).

See you in a next cosmological post!

LOG#109. Basic Cosmology (IV).

Today, a new post in this fascinating Cosmology thread…

The Big Bang Nucleosynthesis in a nutshell

When the Universe was young, about $T\sim 1MeV$, the following events happen

1st. Some particles remained in thermodynamical equilibrium with the primordial plasma (photons, positrons and electrons, i.e., $\gamma$, $e^+$, $e^-$).

2nd. Some relativistic particles decoupled, the neutrinos! And the neutrinos are a very important particle there (the $\nu$ “power” is even more mysterious since the discovery of the neutrino flavor oscillations).

3rd. Some non relativistic particles, the baryons, experienced a strong and subtle asymmetry. Even if we do not understand the physics behind this initial baryon asymmetry, it is important to explain the current Universe. The initial baryon asymmetry is estimated to be

$\dfrac{n_b-n_{\bar{b}}}{s}\sim 10^{10}$

At $T\sim 1MeV$, the baryon number was very large compared with the number of antibaryons. The reason or explanation of this fact is not well understood, but there are some interesting ideas for this asymmetric baryogenesis coming from leptogenesis. I will not discuss this fascinating topic today. Moreover, the fraction or ratio between the baryon density and the photon density is known to be the quantity

$\eta_b\equiv \dfrac{n_b}{n_\gamma}\approx 5\mbox{.}5\cdot 10^{-10}\left(\dfrac{\Omega_b h^2}{0\mbox{.}020}\right)$

In that moment, the question is: how the baryons end up? The answer is pretty simple. There are two main ideas:

1) The thermal equilibrium is kept thorugh out the whole phase in the early Universe. It means that the nuclear state will be the one with the lowest energy per baryon. That is, the most stable nucleus is iron (Fe), but, in fact,  most of the baryons end up in hydrogen (H) or Helium (He).

2) Simple nucleosynthesis is based on 2 elementary ideas and simplifications. Firstly, no element heavier than helium $^4He$ are produced at high ratios. Therefore, this means that protons(p), neutrons(n), the deuterium(D), the tritium (T), Helium-3 ($^3He$) and Helium-4 ($^4He$) are the main subproducts of Big Bang Nucleosynthesis in this standard scenario. Sedondly, until the Universe froze below $T=0\mbox{.}1MeV$, no light nuclei could form and there were only a “primordial soup” made of protons and neutrons. The neutron to proton ratio $n/p$ IS an input for the synthesis of D, T, $^3He$, and $^4He$. Indeed, the 0.1 MeV temperature is relatively low considering the typical nuclear binding energy, about some MeV’s. We could have some quantities of “heavy” elements, but this effect is small, at the level of $\eta_b\approx 10^{-9}$, i.e., the effect of large number of photons compared to the number of baryons is comparable to the number of baryons and heavy elements beyond helium isotopes. We can loot at this effec in the deuterium (D) production:

$n+p\longrightarrow D+\gamma$

where the reaction is taken in the thermal equilibrium and $n_\gamma=n_\gamma^0$. In this equilibrium, we get

$\dfrac{n_p}{n_nn_p}=\dfrac{n_D^0}{n_n^0n_p^0}=\dfrac{3}{4}\left(\dfrac{2\pi m_D}{m_nm_pT}\right)^{3/2}e^{(m_n+m_p-m_D)/T}=\dfrac{3}{4}\left(\dfrac{4\pi}{m_pT}\right)^{3/2}e^{B_p/T}$

and where $m_D\approx 2m_p$, $m_n\approx m_p$ and $B_p$ is the binding energy of deuterium (D). Indeed, we get $n_p\sim n_n\sim n_b=\eta_bn_\gamma$ and so $n_\gamma\sim T^3$. Moreover we also get

$\dfrac{n_D}{n_b}\sim \eta_b \left(\dfrac{T}{m_p}\right)^{3/2}e^{B_p/T}$

and the exponential “compensates” $\eta_b$ and $T$, since the smal factor $\eta_b$ must be chosen to be smaller than the binding energy to temperature ratio $B_p/T$.

By the other hand, the nuetron abundance can be also estimated. From a simple proton-neutron conversion, we obtain

$p+e^- \leftrightarrow n+\nu_e$

due to the weak interaction! The proton/neutron equilibrium ratio for temperatures greater than 1MeV becomes

$\dfrac{n_p^0}{n_n^0}=\dfrac{e^{-m_p/T}\int dpp^2e^{-p^2/2m_pT}}{e^{-m_nT}\int dpp^2e^{-p^2/2m_nT}}=e^{Q/T}$

where $Q=m_n-m_p=1\mbox{.}293MeV$. In fact, the exponential will not be maintained below $T\approx 1 MeV$ and we define the neutron fraction as follows. Firstly

$X_n=\dfrac{n_n}{n_p+n_n}$

In equilibrium, this becomes

$X_n (eq)=\dfrac{1}{1+n_p^0/n_n^0}$

Boltzmann equation for the process $n+\nu_e\leftrightarrow p+e^-$ can be easily derived

$a^{-3}\dfrac{d( n_na^3)}{dt}=n_n^0n_\nu^0\langle \sigma v\rangle \left( \dfrac{n_pn_e}{n_p^0n_e^0}-\dfrac{n_nn_\nu}{n_n^0n_\nu^0}\right)=n_\nu^0\langle \sigma v\rangle \left(\dfrac{n_pn_n^0}{n_p^0}-n_n\right)$

and where

$e^{-Q/T}=\dfrac{n_n^0}{n_p^0}$

Therefore, we obtain

$n_n=\left(n_n+n_p\right)X_n$

and from the LHS, we calculate

$a^{-3}\dfrac{d}{dt}\left[a^3(n_n+n_p)X_n\right]=a^{-3}X_n\dfrac{d}{dt}\left(a^3(n_n+n_p)\right)+\dfrac{dX_n}{dt}\left(n_n+n_p\right)$

By the other hand, from the RHS

$n_\nu^0\langle \sigma v\rangle \left[(n_n+n_p)(1-X_n)e^{-Q/T}-(n_n+n_p)X_n\right]$

Thus, $\Gamma_{np}\longrightarrow \lambda_{np}$ implies that

$\dfrac{dX_n}{dt}=\lambda_{np}\left[(1-X_n)e^{-Q/T}-X_n\right]$

If we change the variable $t\longrightarrow x=Q/T$, then we write

$\dfrac{dX_n}{dt}=\dfrac{dX_n}{dx}\dfrac{dx}{dt}=-\dfrac{Q\dot{T}}{T^2}=-x\dfrac{\dot{T}}{T}=+x\dfrac{\dot{a}}{a}=xH$

where

$H=\sqrt{\dfrac{\rho_R}{3M_p^2}}=\sqrt{\dfrac{\pi^2g_\star}{90}}\dfrac{T^2}{M_p}=\sqrt{\dfrac{\pi^2g_\star}{90}}\dfrac{Q^2}{M_p}x^{-2}=H(x=1)$

and

$\dfrac{dX_n}{dt}=\dfrac{\lambda_{np}x}{H(x=1)}\left[e^{-x}-X_n(1+e^{-x})\right]$

with

$\lambda_{np}(x)=\dfrac{255}{\tau_nx^5}\left(12+6x+x^2\right)$

and $\tau_n$ is the neutron lifetime, i.e. $\tau_n\approx 886\mbox{.7}s$

The numerical integration of these equations provides the following qualitative sketch for $X_n$:

At $T$ below 0.1MeV, the neutron decays $n\longrightarrow p+e^-+\nu_e$ via weak interaction. It yields

$X_ne^{-t/\tau_n}$ and \$latex $X_n(T_{BBN})=0\mbox{.}15\times 0\mbox{.74}=0\mbox{.}11$

such as the deuterium production is done through the processes

$n+p\longrightarrow D+\gamma$

and it started at about $T\sim 0\mbox{.}07MeV$ and

$t=132s\left(\dfrac{0\mbox{.}1MeV}{T}\right)^2$

The light element abundances

A good approximation is to consider that light element production happens instantaneously at $T=T_{BBN}$. Of course, the issue is…How could we determine that temperature? If we measure the abundance of deuterium abundance today, and the baryon abundance today (i.e., if we know their current densities), we can use the cosmological equations to deduce the ratio

$\dfrac{n_D}{n_b}\approx \eta_b\left(\dfrac{T}{M_p}\right)^{3/2}e^{B_D/T}\sim 1$

Then, we obtain from these equations and the measured densities that $T_{BBN}\approx 0\mbox{.}07MeV\sim 0\mbox{.}1MeV$

Moreover, since $B(He)>B_D$, it implies that helium-4 ($^4He$) production is favoured by BBN! It means that all neutron are processed inside helium-4 or hydrogen. In fact, the helium-4 abundance is known to be

$X_4=\dfrac{4n(^4He)}{n_b}=2X_n(T_{BBN})\approx 0\mbox{.}22$

We can compare this with an exact solution for the “yield” $Y_p=0\mbox{.}2262+0\mbox{.}0135\ln (\eta_b (10^{-10}))$

The observed helium-4 abundance is in good agreement with the theoretical expectations from the Standard Cosmological Model! What an awesome hit! We can also compare this with the primordial helium abundances from cosmological observations

$0\mbox{.}22\sim 0\mbox{.}25$

Thus, we have learned that the deuterium abundance IS a powerful probe of the baryon density!!!!

Remark: Nowadays, there is a problem with the lithium-7 abundances in stars. The origin of the discrepancy is not known, as far as I know. Then, the primordial lithium abundance is a controversial topic in modern Cosmology, so we understand BBN only as an overall picture, and some details need to be improved in the next years.

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.,

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

LOG#107. Basic Cosmology (II).

Evolution of the Universe: the scale factor

The Universe expands, and its expansion rate is given by the Hubble parameter (not constant in general!)

$\boxed{H(t)\equiv \dfrac{\dot{a}(t)}{a(t)}}$

Remark  (I): The Hubble “parameter” is “constant” at the present value (or a given time/cosmological age), i.e., $H_0=H(t_0)$.

Remark (II): The Hubble time defines a Hubble length about $L_H=H^{-1}$, and it defines the time scale of the Universe and its expasion “rate”.

The critical density of matter is a vital quantity as well:

$\boxed{\rho_c=\dfrac{3H^2}{\kappa^2}\vert_{t_0}}$

We can also define the density parameters

$\Omega_i=\dfrac{\rho_i}{\rho_c}\vert_{t_0}$

This quantity represents the amount of substance for certain particle species. The total composition of the Universe is the total density, or equivalently, the sum over all particle species of the density parameters, that is:

$\boxed{\displaystyle{\Omega=\sum_i\Omega_i=\dfrac{\displaystyle{\sum_i\rho_i}}{\rho_c}}}$

There is a nice correspondence between the sign of the curvature $k$ and that of $\Omega-1$. Using the Friedmann’s equation

$\displaystyle{\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}=\dfrac{\kappa^2}{3}\sum_i\rho_i}$

then we have

$\dfrac{k}{H^2a^2}=\dfrac{\displaystyle{\sum_i\rho_i}}{\rho_c}-1=\Omega-1$

Thus, we observe that

1st. $\Omega>1$ if and only if (iff) $k=+1$, i.e., iff the Universe is spatially closed (spherical/elliptical geometry).

2nd. $\Omega=1$ if and only if (iff) $k=0$, i.e., iff the Universe is spatially “flat” (euclidean geometry).

3rd. $\Omega<1$ if and only if (iff) $k=-1$, i.e., iff the Universe is spatially “open” (hyperbolic geometry).

In the early Universe, the curvature term is negligible (as far as we know). The reason is as follows:

$k/a^2\propto a^{-2}<<\dfrac{\kappa\rho}{3}\propto a^{-3}(MD),a^{-4}(RD)$ as $a$ goes to zero. MD means matter dominated Universe, and RD means radiation dominated Universe. Then, the Friedmann’s equation at the early time is given by

$\boxed{H^2=\dfrac{\kappa^2}{3}\rho}$

Furthermore, the evolution of the curvature term

$\Omega_k\equiv \Omega-1$

is given by

$\Omega-1=\dfrac{k}{H^2a^2}\propto \dfrac{1}{\rho a^2}\propto a(MD),a^2(RD)$

and thus

$\vert \Omega-1\vert=\begin{cases}(1+z)^{-1}, \mbox{if MD}\\ 10^4(1+z)^{-2}, \mbox{if RD}\end{cases}$

The spatial curvature will be given by

$\boxed{R_{(3)}=\dfrac{6k}{a^2}=6H^2(\Omega-1)}$

and the curvature radius will be

$\boxed{R=a\vert k\vert ^{-1/2}=H^{-1}\vert \Omega-1\vert ^{-1/2}}$

We have arrived at the interesting result that in the early Universe, it was nearly “critical”. The Universe close to the critical density is very flat!

By the other hand, supposing that $a_0=1$, we can integrate the Friedmann’s equation easily:

$\boxed{\displaystyle{\left(\dfrac{\dot{a}}{a}\right)^2+\dfrac{k}{a^2}=\dfrac{\kappa^2}{3}\sum_i\rho_i=\dfrac{\kappa^2}{3}\sum_i\rho_i(0)a^{-3(1+\omega_i)}}}$

Then, we obtain

$\dot{a}^2=H_0^2\left[-\Omega_k+\sum_i\Omega_ia^{-1-3\omega_i}\right]$

We can make an analogy of this equation to certain simple equation from “newtonian Mechanics”:

$\dfrac{\dot{a}^2}{2}+V(a)=0$

Therefore, if we identify terms, we get that the density parameters work as “potential”, with

$\displaystyle{V(a)=\dfrac{1}{2}H_0^2\left[\Omega_k-\sum_i\Omega_ia^{-1-3\omega_i}\right]}$

and the total energy is equal to zero (a “machian” behaviour indeed!). In addition to this equation, we also get

$\boxed{\displaystyle{H_0t=\int_0^a\left[-\Omega_k+\sum_i\Omega_i\chi^{-1-3\omega_i}\right]^{-1/2}d\chi}}$

The age of the Universe can be easily calculated (symbolically and algebraically):

$\boxed{t_0=H_0^{-1}f(\Omega_i)}$

with

$f(\Omega_i)=\int_0^1\left[-\Omega_k+\sum_i\Omega_i\chi^{-1-3\omega_i}\right]^{-1/2}d\chi$

This equation can be evaluated for some general and special cases. If we write $p=\omega \rho$ for a single component, then

$a\propto t^{2/3(1+\omega)}$ if $\omega\neq -1$

Moreover, 3 common cases arise:

1) Matter dominated Universe (MD): $a\propto t^{2/3}$

2) Radiation dominated Universe (RD): $a\propto t^{1/2}$

3) Vacuum dominated Universe (VD): $e^{H_0t}$ ($w=-1$ for the cosmological constant, vacuum energy or dark energy).

THE MATTER CONTENT OF THE UNIVERSE

We can find out how much energy is contributed by the different compoents of the Universe, i.e., by the different density parameters.

Case 1. Photons.

The CMB temperature gives us “photons” with $T_\gamma=2\mbox{.}725\pm 0\mbox{.}002K$

The associated energy density is given by the Planck law of the blackbody, that is

$\rho_\gamma=\dfrac{\pi^2}{15}T^4$ and $\mu/T<9\cdot 10^{-5}$

or equivalently

$\Omega_\gamma=\Omega_r=\dfrac{2\mbox{.}47\cdot 10^{-5}}{h^2a^4}$

Case 2. Baryons.

There are four established ways of measuring the baryon density:

i) Baryons in galaxies: $\Omega_b\sim 0\mbox{.}02$

ii) Baryons through the spectra fo distant quasars: $\Omega_b h^{1\mbox{.}5}\approx 0\mbox{.}02$

iii) CMB anisotropies: $\Omega_bh^2=0\mbox{.}024\pm ^{0\mbox{.}004}_{0\mbox{.}003}$

iv) Big Bag Nucleosynthesis: $\Omega_bh^2=0\mbox{.}0205\pm 0\mbox{.}0018$

Note that these results are “globally” compatible!

Case 3. (Dark) Matter/Dust.

The mass-to-light ratio from galactic rotation curves are “flat” after some cut-off is passed. It also works for clusters and other bigger structures. This M/L ratio provides a value about $\Omega_m=0\mbox{.}3$. Moreover, the galaxy power spectrum is sensitive to $\Omega_m h$. It also gives $\Omega_m\sim 0\mbox{.}2$. By the other hand, the cosmic velocity field of galaxies allows us to derive $\Omega_m\approx 0\mbox{.}3$ as well. Finally, the CMB anisotropies give us the puzzling values:

$\Omega_m\sim 0\mbox{.}25$

$\Omega_b\sim 0\mbox{.}05$

We are forced to accept that either our cosmological and gravitational theory is a bluff or it is flawed or the main component of “matter” is not of baryonic nature, it does not radiate electromagnetic radiation AND that the Standard Model of Particle Physics has no particle candidate (matter field) to fit into that non-baryonic dark matter. However, it could be partially formed by neutrinos, but we already know that it can NOT be fully formed by neutrinos (hot dark matter). What is dark matter? We don’t know. Some candidates from beyond standard model physics: axion, new (likely massive or sterile) neutrinos, supersymmetric particles (the lightest supersymmetric particle LSP is known to be stable: the gravitino, the zino, the neutralino,…), ELKO particles, continuous spin particles, unparticles, preons, new massive gauge bosons, or something even stranger than all this and we have not thought yet! Of course, you could modify gravity at large scales to erase the need of dark matter, but it seems it is not easy at all to guess a working Modified Gravitational theory or Modified Newtonian(Einsteinian) dynmanics that avoids the need for dark matter. MOND’s, MOG’s or similar ideas are an interesting idea, but it is not thought to be the “optimal” solution at current time. Maybe gravitons and quantum gravity could be in the air of the dark issues? We don’t know…

Case 4. Neutrinos.

They are NOT observed, but we understand them their physics, at least in the Standard Model and the electroweak sector. We also know they suffer “oscillations”/flavor oscillations (as kaons). The (cosmic) neutrino temperature can be determined and related to the CMB temperature. The idea is simple: the neutrino decoupling in the early Universe implied an electron-positron annihilation! And thus, the (density) entropy dump to the photons, but not to neutrinos. It causes a difference between the neutrino and photon temperature “today”. Please, note than we are talking about “relic” neutrinos and photons from the Big Bang! The (density) entropy before annihilation was:

$s(a_1)=\dfrac{2\pi^2}{45}T_1^3\left[2+\dfrac{7}{8}(2\cdot 2+3\cdot 2)\right]=\dfrac{43}{90}\pi^2 T_1^3$

After the annihilation, we get

$s(a_2)=\dfrac{2\pi^2}{45}\left[2T_\gamma^3+\dfrac{7}{8}(3\cdot 2)T_\nu^3\right]$

Therefore, equating

$s(a_1)a_1^3=s(a_2)a_2^3$ and $a_1T_1=a_2T_\nu (a_2)$

$\dfrac{43}{90}\pi^2(a_1T_1)^3=\dfrac{2\pi^2}{45}\left[2\left(\dfrac{T_\gamma}{T_\nu}\right)^3+\dfrac{42}{8}\right](a_2T_\nu (a_2))^3$

$\dfrac{43}{2}\pi^2(a_1T_1)^3=2\pi^2\left[2\left(\dfrac{T_\gamma}{T_\nu}\right)^3+\dfrac{42}{8}\right](a_2T_\nu (a_2))^3$

and then

$\boxed{\left(\dfrac{T_\nu}{T_\gamma}\right)=\left(\dfrac{4}{11}\right)^{1/3}}$

or equivalently

$\boxed{T_\nu=\sqrt[3]{\dfrac{4}{11}}T_\gamma\approx 1\mbox{.}9K}$

In fact, the neutrino energy density can be given in two different ways, depending if it is “massless” or “massive”. For massless neutrinos (or equivalently “relativistic” massless matter particles):

I) Massless neutrinos: $\Omega_\nu=\dfrac{1\mbox{.}68\cdot 10^{-5}}{h^2}$

2) Massive neutrinos: $\Omega_\nu= \dfrac{m_\nu}{94h^2 \; eV}$

Case 5. The dark energy/Cosmological constant/Vacuum energy.

The budget of the Universe provides (from cosmological and astrophysical measurements) the shocking result

$\Omega\approx 1$ with $\Omega_M\approx 0\mbox{.}3$

Then, there is some missin smooth, unclustered energy-matter “form”/”species”. It is the “dark energy”/vacuum energy/cosmological cosntant! It can be understood as a “special” pressure term in the Einstein’s equations, but one with NEGATIVE pressure! Evidence for this observation comes from luminosity-distance-redshift measurements from SNae, clusters, and the CMB spectrum! The cosmological constant/vacuum energy/dark energy dominates the Universe today, since, it seems, we live in a (positively!) accelerated Universe!!!!! What can dark energy be? It can not be a “normal” matter field. Like the Dark Matter field, we believe that (excepting perhaps the scalar Higgs field/s) the SM has no candidate to explain the Dark Energy. What field could dark matter be? Perhaps an scalar field or something totally new and “unknown” yet.

In short, we are INTO a DARKLY, darkly, UNIVERSE! Darkness is NOT coming, darkness has arrived and, if nothing changes, it will turn our local Universe even darker and darker!

See you in the next cosmological post!

LOG#106. Basic Cosmology (I).

The next thread is devoted to Cosmology. I will intend to be clear and simple about equations and principles of current Cosmology with a General Relativity background.

First of all…I will review the basic concepts of natural units I am going to use here. We will be using the following natural units:

$\hbar=c=k_B=1$

We will take the Planck mass to be given by

$M_P=\sqrt{8\pi G_N}\approx 1\mbox{.}2\cdot 10^{19}GeV$

The solar mass is $M_\odot=2\cdot 10^{30}kg$ and the parsec is given by the value

$1pc=3\mbox{.}26lyr=3\mbox{.}1\cdot 10^{16}m$

Well, current Cosmology is based on General Relativity. Even if I have not reviewed this theory with detail in this blog, the nice thing is that most of Cosmology can be learned with only a very little knowledge of this fenomenal theory. The most important ideas are: metric field, geodesics, Einstein equations and no much more…

In fact, newtonian gravity is a good approximation in some particular cases! And we do know that even in this pre-relativistic theory

$\mbox{Gravitational force}=\mbox{Matter/Mass density}$

via the Poisson’s equation

$\nabla^2\phi =4\pi G_N\rho$

This idea, due to the equivalence principle, is generalized a little bit in the general relativistic framework

$\mbox{Spacetime geometry}=\mbox{Matter content/Energy-momentum}$

The spacetime geometry is determined by the metric tensor $g_{\mu\nu}(x)$. The matter content is given by the stress-energy-momentum tensor $T_{\mu\nu}$. As we know one of these two elements, we can know, via Eisntein’s field equations the another. That is, given a metric tensor, we can tell how energy-momentum “moves” in space-time. Given the energy-momentum tensor, we can know what is the metric tensor in spacetime and we can guess how the spacetime bends… This is the origin of the famous motto: “Spacetime says matter how to move, energy-momentum says spacetime how to curve”! Remember that we have “deduced” the Einstein’s field equations in the previous post. Without a cosmological constant term, we get

$G_{\mu\nu}=\kappa^2T_{\mu\nu}=8\pi G_NT_{\mu\nu}$

Given a spacetime metric $g_{\mu\nu}$, we can calculate the (affine/Levi-Civita) connection

$\Gamma^\sigma_{\;\;\mu\nu}=\dfrac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu}\right)$

The Riemann tensor that measures the spacetime curvature is provided by the equation

$R^\rho_{\;\; \sigma\mu\nu}=\partial_\mu \Gamma^\rho_{\;\;\mu\sigma}-\partial_\mu \Gamma^\rho_{\;\; \mu \sigma}+\Gamma^\rho_{\;\;\mu\lambda}\Gamma^\lambda_{\;\;\nu\sigma}-\Gamma^\rho_{\;\;\nu\lambda}\Gamma^\lambda_{\;\;\mu\sigma}$

The Ricci tensor is defined to be the following “trace” of the Riemann tensor

$R_{\mu\nu}=R^\lambda_{\;\;\mu\lambda \nu}$

The Einstein tensor is related to the above tensors in the well-known manner

$G_{\mu\nu}=R_{\mu\nu}-\dfrac{1}{2}Rg_{\mu\nu}$

The Einstein’s equations can be derived from the Einstein-Hilbert action we learned in the previous post, using the action principle and the integral

$\boxed{S_{EH}=\int d^4x \sqrt{-g} \left(\kappa^{-2}R+\mathcal{L}_M\right)}$

The geodesic equation is the path of a freely falling particle. It gives a “condensation” of the Einstein’s equivalence principle too and it is also a generalization of Newton’s law of “no force”. That is, the geodesic equation is the feynmanity

$\boxed{\dfrac{d^2 x^\mu}{d\tau^2}+\Gamma^\mu _{\rho\sigma}\dfrac{dx^\rho}{d\tau}\dfrac{dx^\sigma}{d\tau}=0}$

Finally, an important concept in General Relativity is that of isometry. The symmetry of the “spacetime manifold” is provided by a Killing vector that preserves transformations (isometries) of that manifold. Mathematically speaking, the Killing vector fields satisfy certain equation called the Killing equation

$\boxed{\xi_{\mu ; \nu}+\xi_{\nu ; \mu}=0}$

Maximally symmetric spaces have $n(n+1)/2$ Killing vectors in n-dimensional (nD) spacetime. There are 3 main classes or types of 2D maximally symmetric that can be generalized to higher dimensions:

1. The euclidean plane $E^2$.

2. The pseudo-sphere $H^2$. This is a certain “hyperbolic” space.

3. The spehre $S^2$. This is a certain “elliptic” space.

The Friedmann-Robertson-Walker Cosmology

Current cosmological models are based in General Relativity AND  a simplification of the possible metrics due to the so-called Copernican (or cosmological) principle: the Universe is pretty much the same “everywhere” you are in the whole Universe! Remarkbly, the old “perfect” Copernican (cosmological) principle that states that the Universe is the same “everywhere” and “every time” is wrong. Phenomenologically, we have found that the Universe has evolved and it evolves, so the Universe was “different” when it was “young”. Therefore, the perfect cosmological principle is flawed. In fact, this experimental fact allows us to neglect some old theories like the “stationary state” and many other “crazy theories”.

What are the observational facts to keep the Copernican principle? It seems that:

1st. The distribution of matter (mainly galaxies, clusters,…) and radiation (the cosmic microwave background/CMB) in the observable Universe is homogenous and isotropic.

2nd. The Universe is NOT static. From Hubble’s pioneer works/observations, we do know that galaxies are receeding from us!

Therefore, these observations imply that our “local” Hubble volume during the Hubble time is similar to some spacetime with homogenous and isotropic spatial sections, i.e., it is a spacetime manifold $M=\mathbb{R}\times \Sigma$. Here, $\mathbb{R}$ denotes the time “slice” and $\Sigma$ represents a 3D maximally symmetric space.

The geometry of a locally isotropic and homogeneous Universe is represented by the so-called Friedmann-Robertson-Walker metric

$\boxed{ds^2_{FRW}=-dt^2+a^2(t)\left[\dfrac{dr^2}{1-kr^2}+r^2\left(d\theta^2+\sin\theta^2d\phi^2\right)\right]}$

Here, $a(t)$ is the called the scale factor.  The parameter $k$ determines the geometry type (plane, hyperbolic or elliptical/spherical):

1) If $k=0$, then the Universe is “flat”. The manifold is $E^3$.

2) If $k=-1$, then the Universe is “open”/hyperbolic. The manifold would be $H^3$.

3) If $k=+1$, then the Universe is “closed”/spherical or elliptical. The manifold is then $S^3$.

Remark: The ansatz of local homogeneity and istoropy only implies that the spatial metric is locally one of the above three spaces, i.e., $E^3,H^3,S^3$. It could be possible that these 3 spaces had different global (likely topological) properties beyond these two properties.

Kinematical features of a FRW Universe

The first property we are interested in Cosmology/Astrophysics is “distance”. Measuring distance in a expanding Universe like a FRW metric is “tricky”. There are several notions of “useful” distances. They can be measured by different methods/approaches and they provide something called sometimes “the cosmologidal distance ladder”:

1st. Comoving distance. It is a measure in which the distance is “taken” by a fixed coordinate.

2nd. Physical distance. It is essentially the comoving distance times the scale factor.

3rd. Luminosity distance. It uses the light emitted by some object to calculate its distance (provided the speed of light is taken constant, i.e., special relativity holds and we have a constant speed of light)

4th. Angular diameter distance. Another measure of distance using the notion of parallax and the “extension” of the physical object we measure somehow.

There is an important (complementary) idea in FRW Cosmology: the particle horizon. Consider a light-like particle with $ds^2=0$. Then,

$dt=a(t)\dfrac{1}{\sqrt{1-kr^2}}$

or

$\dfrac{dr}{\sqrt{1-kr^2}}=\dfrac{dt}{a(t)}$

The total comoving distance that light have traveled since a time $t=0$ is equal to

$\boxed{\eta=\int_0^{r_H}\dfrac{dr}{\sqrt{1-kr^2}}=\int_0^t\dfrac{dt'}{a(t')}}$

It shows that NO information could have propagated further and thus, there is a “comoving horizon” with every light-like particle! Here, this time is generally used as a “conformal time” as a convenient tiem variable for the particle. The physical distance to the particle horizon can be calculated

$\boxed{d_H(t)=\int_0^{r_H}\sqrt{g_{rr}}dr=a(t)\int_0^t\dfrac{dt'}{a(t')}=a(t)\eta}$

There are some important kinematical equations to be known

A) For the geodesic equation, the free falling particle, we have

$\Gamma^0_{ij}=\dfrac{\dot{a}}{a}\overline{g}_{ij}$

$\Gamma^i_{0j}=\Gamma^i_{j0}=\dfrac{\dot{a}}{a}\delta_{ij}$

$\Gamma^i_{jk}=\overline{\Gamma}^i_{jk}$

for the FRW metric and, moreover, the energy-momentum vector $P^\mu=(E,\mathbf{p})$ is defined by the usual invariant equation

$P^\mu=\dfrac{dx^\mu}{d\lambda}$

This definition defines, in fact, the proper “time” $\lambda$ implicitely, since

$\dfrac{d}{d\lambda}=\dfrac{dx^0}{d\lambda}\dfrac{d}{dx^0}=E\dfrac{d}{dt}$

and the 0th component of the geodesic equation becomes

$E\dfrac{dE}{dt}=-\Gamma^0_{ij}p^ip^j=-\delta_{ij}a\dot{a}p^ip^j$

$g_{\mu\nu}p^\mu p^\nu=-E^2+a^2\delta_{ij}p^ip^j=-m^2$

$EdE=a^2\vert \mathbf{p}\vert d\vert \mathbf{p}\vert$

$a^2 p\dfrac{dp}{dt}=-a\dot{a} p^2$

$\dfrac{1}{\vert \mathbf{p}\vert }\dfrac{d\vert \mathbf{p}\vert}{dt}+\dfrac{\dot{a}}{a}=0$

Therefore we have deduced that $\vert \mathbf{p}\vert \propto a^{-1}$. This is, in fact, the socalled “redshift”.  The cosmological  redshift parameter is more generally defined through the equation

$\boxed{\dfrac{a(t_0)}{a(t)}=1+z=\dfrac{\lambda_0}{\lambda}}$

B) The Hubble’s law.

The luminosity distance measures the flux of light from a distant object of known luminosity (if it is not expanding). The flux and luminosity distance are bound into a single equation

$\boxed{F=\dfrac{L}{4\pi d^2_L}}$

If we use the comoving distance between a distant emitter and us, we get

$\chi (a)=\int_t^{t_0}\dfrac{dt'}{a(t')}=\int_a^1\dfrac{da'}{a'^2 H(a')}$

for a expanding Universe! That is, we have used the fact that luminosity itself goes through a comoving spherical shell of radius $\chi (a)$. Moreover, it shows that

$F=\dfrac{L (\chi)}{4 \pi \chi (a)^2 a_0^2}=\dfrac{L}{4\pi (\chi (a)/a)^2}$

The luminosity distance in the expanding shell is

$d_L=\dfrac{\chi (a)}{a}=\left(\dfrac{L}{4\pi F}\right)^{1/2}$

and this is what we MEASURE in Astrophysics/Cosmology. Knowing $a(t)$, we can express the luminosity distance in terms of the redshift. Taylor expansion provides something like this:

$H_0d_L=z+\dfrac{1}{2}(1-q_0)z^2+\ldots$

where higher order terms are sometimes referred as “statefinder parameters/variables”. In particular, we have

$\boxed{H_0=\dfrac{\dot{a}_0}{a_0}}$

and

$\boxed{q_0=-\dfrac{a_0\ddot{a}_0}{\dot{a}_0^2}}$

C) Angular diameter distance.

If we know that some object has a known length $l$, and it gives some angular “aperture” or separation $\theta$, the angular diameter distance is given by

$\boxed{d_A=\dfrac{l}{\theta}}$

The comoving size is defined as $l/a$, and the coming distance is again $\chi (a)$. For “flat” space, we obtain that

$\theta=\dfrac{l/a}{\chi (a)}$

that is

$d_A=a\chi (a)=\dfrac{\chi}{1+z}$

In the case of “curved” spaces, we get

$d_A=\dfrac{a}{H_0\sqrt{\vert \omega_k\vert}}\cdot\begin{cases}\sinh \left( \sqrt{\Omega_k}H_0\chi\right),\Omega_k>0\\ \sin \left( \sqrt{-\Omega_k}H_0\chi\right),\Omega_k<0\end{cases}$

FRW dynamics

Gravity in General Relativity, a misnomer for the (locally) relativistic theory of gravitation, is described by a metric field, i.e., by a second range tensor (covariant tensor if we are purist with the nature of components). The metric field is related to the matter-energy-momentum content through the Einstein’s equations

$G_{\mu\nu}=-\kappa^2 T_{\mu\ nu}$

The left-handed side can be calculated for a FRW Universe as follows

$R_{00}=-3\dfrac{\ddot{a}}{a}$

$R_{ij}=(a\ddot{a}+2\dot{a}^2+2k)\overline{g}_{ij}$

$R=6\left(\dfrac{\ddot{a}}{a}+\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}\right)$

The right-handed side is the energy-momentum of the Universe. In order to be fully consistent with the symmetries of the metric, the energy-momentum tensor MUST be diagonal and $T_{11}=T_{22}=T_{33}=T$. In fact, this type of tensor describes a perfect fluid with

$T_{\mu\nu}=(\rho+p)U_\mu U_\nu+pg_{\mu\nu}$

Here, $\rho, p$ are functions of $t$ (cosmological time) only. They are “state variables” somehow. Moreover, we have

$U_\mu =(1,0,0,0)$

for the fluid at rest in the comoving frame. The Friedmann equations are indeed the EFE for a FRW metric Universe

$3\left(\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}\right)=\kappa^2\rho$ for the 00th compoent as “constraint equation.

$2\dfrac{\ddot{a}}{a}+\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}=-\kappa^2p$ for the iith components.

Moreover, we also have

$G_{\mu\nu}^{;\nu}=T_{\mu\nu}^{;\nu}=0$

and this conservation law implies that

$\dot{\rho}+3\dfrac{\dot{a}}{a}(\rho+p)=0$

Therefore, we have got two independent equations for three unknowns $(a, \rho, p)$. We need an additional equation. In fact, the equation of state for $p=p(\rho)$ provides such an additional equation. It gives the “dynamics of matter”!

In summary, the basic equations for Cosmology in a FRW metric, via EFE, are the Friedmann’s equations (they are secretly the EFE for the FRW metric) supplemented with the energy-momentum conservations law and the equation of state for the pressure $p=p(\rho)$:

1) $\boxed{\dfrac{\dot{a}^2}{a^2}+\dfrac{k^2}{a^2}=\dfrac{\kappa^2}{3}\rho}$

2) $\boxed{\dot{\rho}+3\dfrac{\dot{a}}{a}(\rho+p)=0}$

3) $\boxed{p=p(\rho)}$

There are many kinds of “matter-energy” content of our interest in Cosmology. Some of them can be described by a simple equation of state:

$\boxed{p=\omega \rho}$

Energy-momentum conservation implies that $\rho\propto a^{-3(\omega +1)}$. 3 special cases are used often:

1st. Radiation (relativistic “matter”). $\omega=1/3$ and thus, $p=1/3\rho$ and $\rho\propto a^{-4}$

2nd. Dust (non-relativistic matter). $\omega=0$. Then, $p=0$ and $\rho\propto a^{-3}$

3rd. Vacuum energy (cosmological constant). $\omega=-1$. Then, $p=-\rho$ and $\rho=\mbox{constant}$

Remark (I): Particle physics enters Cosmology here! Matter dynamics or matter fields ARE the matter content of the Universe.

Remark (II): Existence of a Big Bang (and a spacetime singularity). Using the Friedmann’s equation

$\dfrac{\ddot{a}}{a}=-\dfrac{\kappa^2}{6}(\rho+3p)$

if we have that $(\rho+3p)>0$, the so-called weak energy condition, then $a=0$ should have been reached at some finite time in the past! That is the “Big Bang” and EFE are “singular” there. There is no scape in the framework of GR. Thus, we need a quantum theory of gravity to solve this problem OR give up the FRW metric at the very early Universe by some other type of metric or structure.

Particles and the chemical equilibrium of the early Universe

Today, we have DIRECT evidence for the existence of a “thermal” equilibrium in the early Universe: the cosmic microwave background (CMB). The CMB is an isotropic, accurate and non-homogeneous (over certain scales) blackbody spectrum about $T\approx 3K$!

Then, we know that the early Universe was filled with a hot dieal gas in thermal equilibrium (a temperature $T_e$ can be defined there) such as the energy density and pressure can be written in terms of this temperature. This temperature generates a distribution $f(\mathbf{x},\mathbf{p})$. The number of phase space elements in $d^3xd^3p$ is

$d^3xd^3p=\dfrac{d^3\mathbf{x}d^3\mathbf{p}}{(2\pi\hbar)^3}$

and where the RHS is due to the uncertainty principle. Using homogeneity, we get that, indeed, $f(x,p)=f(p)$, and where we can write the volume $d^3x=dV$. The energy density and the pressure are given by (natural units are used)

$\rho_i=g_i\int \dfrac{d^3p}{(2\pi)^3}f_i(p)E(p)$

$p_i=g_i\int \dfrac{d^3p}{(2\pi)^3}f_i (p)\dfrac{p^2}{3E(p)}$

When we are in the thermal equilibrium at temperature T, we have the Bose-Einstein/Fermi-Dirac distribution

$f(p)=\dfrac{1}{e^{(E-\mu)/T}\pm 1}$

and where the $+$ is for the Fermi-Dirac distribution (particles) and the $-$ is for the Bose-Einstein distribution (particles). The number density, the energy density and the pressure are the following integrals

$\boxed{\mbox{Number density}=n=\dfrac{N}{V}=\dfrac{g}{2\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{1/2}}{e^{(E-\mu)/T}\pm 1}dE}$

$\boxed{\mbox{Density energy}=\rho=\dfrac{E}{V}=\dfrac{g}{2\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{1/2}E^2}{e^{(E-\mu)/T}\pm 1}dE}$

$\boxed{\mbox{Pressure}=p=\dfrac{g}{6\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{3/2}}{e^{(E-\mu)/T}\pm 1}dE}$

And now, we find some special cases of matter-energy for the above variables:

1st. Relativistic, non-degenerate matter (e.g. the known neutrino species). It means that $T>>m$ and $T>>\mu$. Thus,

$n=\left(\dfrac{3}{4}\right)\dfrac{\zeta (3)}{\pi^2}gT^3$

$\rho=\left(\dfrac{7}{8}\right)\dfrac{\pi^2}{30}gT^4$

$p=\dfrac{1}{3}\rho$

2nd. Non-relativistic matter with $m>>T$ only. Then,

$n=g\left(\dfrac{mT}{2\pi}\right)^{3/2}e^{-(m-\mu)/T}$

$\rho= mn+\dfrac{3}{2}p$, and $p=nT<<\rho$

The total energy density is a very important quantity. In the thermal equilibrium, the energy density of non-relativistic species is exponentially smaller (suppressed) than that of the relativistic particles! In fact,

$\rho_R=\dfrac{\pi^2}{30}g_\star T^4$ for radiation with $p_R=\dfrac{1}{3}\rho_R$

and the effective degrees of freedom are

$\displaystyle{\boxed{g_\star=\sum_{bosons}g_b+\dfrac{7}{8}\sum_{fermions}g_f}}$

Remark: The factor $7/8$ in the DOF and the variables above is due to the relation between the Bose-Einstein and the Fermi-Dirac integral in d=3 space dimensions. In general d, the factor would be

$(1-\dfrac{1}{2^d})=\dfrac{2^d-1}{2^d}$

Entropy conservation and the early Universe

The entropy in a comoving volume IS a conserved quantity IN THE THERMAL EQUILIBRIUM. Therefore, we have that

$\dfrac{\partial p_i}{\partial T}=g_i\int \dfrac{d^3p}{(2\pi)^3}\dfrac{df}{dT}\dfrac{p^2}{3E(p)}=g_i\int \dfrac{4\pi pE dE}{(2\pi)^3}\dfrac{df}{dE}\left(-\dfrac{E}{T}\right)\dfrac{p^2}{3E}$

and then

$\dfrac{\partial p_i}{\partial T}=\dfrac{g_i}{2\pi^2}\int \left(-\dfrac{d}{dE}\left(f\dfrac{p^3E}{3T}\right)+f\dfrac{d}{dE}\left(\dfrac{p^3E}{3T}\right)\right)dE$

or

$\dfrac{\partial p_i}{\partial T}=\dfrac{1}{T}(\rho_i+p_i)$

Now, since

$\dfrac{\partial \rho}{\partial t}+3\dfrac{\dot{a}}{a}(\rho+p)=0$

then

$\dfrac{\partial}{\partial t}\left(a^3(\rho+p)\right)-a^3\dfrac{\partial p}{\partial t}=0$

$\dfrac{1}{a^3}\dfrac{\partial (a^3(\rho +p))}{\partial t}-\dfrac{\partial \rho}{\partial t}=0$

if we multiply by $T$ and use the chain rule for $\rho$, we obtain

$\dfrac{1}{a^3}\dfrac{\partial}{\partial t}\left(\dfrac{a^3(\rho+p)}{T}\right)=0$

but it means that $a^3s=\mbox{constant}$, where $s$ is the entropy density defined by

$\boxed{s\equiv \dfrac{\rho+p}{T}}$

Well, the fact is that we know that the entropy or more precisely the entropy density is the early Universe is dominated by relativistic particles ( this is “common knowledge” in the Stantard Cosmological Model, also called $\Lambda CDM$). Thus,

$\boxed{s=\dfrac{2\pi^2}{45}g_\star T^3}$

It implies the evolution of temperature with the redshift in the following way:

$T\propto g_\star^{-1/3}a^{-1}$

Indeed, since we have that $n\propto a^{-3}$, $s\propto a^{-3}$, the yield variable

$Y_i\equiv \dfrac{n_i}{s}$

is a convenient quantity that represents the “abundance” of decoupled particles.

See you in my next cosmological post!