The Early Universe

This is a standard practice when dealing with plasma. Via $$ E = k_B T $$ where $k_B$ is the Boltzmann constant, energy is directly linked to temperature. This is because for massless particles (radiations), when $T$ is large, the average energy per particle $E$ is indeed $k_B T$.The probability that a massless particle has energy $E_n = nh\nu$ is given by the Boltzmann factor $$ P(n) = \frac{\exp(-E_n/k_B T)}{\sum_{n=0}^\infty \exp(-E_n/k_B T)}, $$ the mean energy of a particle with frequency $\nu$ is then $$ \overline{E_\nu} = \frac{h\nu }{e^{h\nu/k_B T} -1}. $$ For large $T$, $e^{h\nu/k_B T} -1 = \tfrac{h\nu}{k_B T}$, and so $$ \overline{E} = k_B T. $$

Even if one is not dealing with massless particles, since for relativistic particle gas $\overline{E} = n k_B T$ where $n$ is the number of degrees of freedom, and $\overline{E} = \frac{n}{2} k_B T$ for non-relativistic particle gas, the conversion is legitimate. By this we mean, the word 'temperature' really can be intuited to have the usual meaning of temperature.

Numerically $1\text{ GeV} = 1.1605\times 10^{13}\text{ K}$.

Consider the evolution of the effective temperature for free particles. This is the case when the particles decouple from other particles, e.g. after recombination. The effective temperature of massless bosons and fermions decreases in time according to $$ T_{\text{eff}}(t) \propto \frac{1}{a(t)}. $$ Even for massive particles, as long as $T_{\text{eff}} \gg m$ wehere $m$ is the particle mass, the above relation still holds.

While for nonrelativistic particles, $$ T_{\text{eff}}(t) \propto \frac{1}{a^2(t)}. $$

See Gorbunov & Rubakov for details.

To solve the Einsein equation $$ G_{ab} = R_{ab} -\frac{1}{2}\mathcal{R} g_{ab} = 8\pi T_{ab} $$ one needs, in addition to a metric, the matter content of the universe that is described in terms of its stress-energy tensor $T_{ab}$. This we will take to be the general perfect fluid, $$ T_{ab} = \rho u_a u_b + P(g_{ab}+ u_a u_b). $$

The Einstein equation has only two independent equations, since by isotropy $G^{ab}u_b$ cannot have a spatial component. The independent components are $$ G_{ab}u^a u^b = 8\pi T_{ab}u^a u^b = 8\pi \rho, $$ and $$ G_{ab} s^a s^b = 8\pi T_{ab}s^a s^b = 8\pi P. $$ where $s^a$ is any unit vector tangent to the leaves, leading to

\begin{equation} 3\frac{\ddot{a}}{a} = -4\pi (\rho + 3P);\quad 3\frac{\dot{a}^2}{a^2}=8\pi\rho - \frac{3k}{a^2}. \label{eq} \end{equation}

where the second one is the Friedmann equation. The first equation can also be substituted with the covariant conservation of the stress-energy $$ \nabla_a T^{ab} = 0 $$ which yields $$ \dot{\rho} + 3\frac{\dot{a}}{a}(\rho + P) = 0. $$

Supplemented with the equation of state of matter $$ P=P(\rho), $$ the Friedmann equation and the conservation of the stress-energy completely determine dynamics of the cosmological expansion. Summing up,

\begin{gather} \label{eq:cosmology} 3\frac{\dot{a}^2}{a^2}=8\pi\rho - \frac{3k}{a^2}\\\ \dot{\rho} + 3\frac{\dot{a}}{a}(\rho + P) = 0.\\\ P=P(\rho) \end{gather}

are the equations of standard cosmology.

An observations can be made. Provided that $\rho>0$ and $P\geq 0$, or more precisely $\rho+3P > 0$, the universe cannot be static (from the first equation above in $(\ref{eq})$, $\ddot{a}<0$). Either $\dot{a}>0$ or $\dot{a}<0$ with the possible exception of an instant of time when expansion changes over to contraction.

Between two isotropic observers in a homogeneous surface, at proper time $t$, if the distance is $D$, the rate of change of $D$ is $$ \frac{d D}{dt} = \frac{D}{a}\frac{da}{dt} = \frac{\dot{a}}{a} D = DH(t). $$ This is Hubble's law, and $H(t) = \frac{\dot{a}}{a}$. Here $\frac{dD}{da} = \frac{D}{a}$ since $D$ is linear in $a$. $H_0$ would denote the present observed value of $H(t)$.

Since the universe is expanding, $\dot{a}>0$, $\ddot{a}<0$, the universe must have been expanding at a faster rate before. This means that, at a time less than $H_0^{-1}$ ago, the universe was in a singular state. The Hubble constant is related to the redshift, $$ z(t) = \frac{a_0}{a(t)} - 1. $$ Expanding $a(t)$ around $a_0 = a(t_0)$, to the linear order in $(t_0 -t)$, one has $$ z(t) = H_0\cdot(t_0-t). $$ and upp to corrections of second order the distance $r=t_0 -t$, so $$ z=H_0 r. $$

Realistic models should, of course, account for nonvanishing spatial curvatures, and combine $\rho_M$ (matter), $\rho_{rad}$ (radiation), $\rho_\Lambda$ (dark energy) and the effective "density" of spatial curvature $\rho_{k}$ given by $\frac{8\pi}{3}G\rho_{k} = - \frac{k}{a^2}$. It should also be noted that $\rho_i$ changes through time: relativistic particles becoming non-relativistic by cooling, particle creation, etc. In particular, the Universe should have been radiation dominated in its early age, at least in the Hot Big Bang picture, but now it is dark energy dominated.

In the present one can find the numerical value of $\rho_c = \frac{3}{8\pi G} H_0^2$, the critical density, for spatially flat Universe. Should $\rho_c$ equal $\rho_{M,0} + \rho_{rad,0} + \rho_{\Lambda,0}$, then the Universe is actually spatially flat. The current bound on $|\rho_{k,0}/\rho_c|$ is less than $0.01$, so the Universe is at least approximately flat nowadays. Specifying $\Omega_i := \rho_i / \rho_c$ obtained from observations, one gets the $\Lambda$CDM model.

The present values of $\Omega_i$ are roughly $\Omega_{rad} \lesssim 10^{-4}$, $|\Omega_{curv}| < 0.01$, $\Omega_\Lambda = 0.685$, $\Omega_M = \Omega_B + \Omega_{DM} = 0.05 + 0.265$. $\Omega_{DM}$ is the contribution of dark matter and $\Omega_B$ the baryonic matters. The present universe is dark energy and dark matter dominated. We see that relativistic matter and curvature can all be neglected for the present time in reasonable calculations.

We don't know what dark energy is, and in the framework of $\Lambda$CDM one assumes that $\rho_\Lambda$ is independent of time, since $\Lambda$ is really the cosmological constant that is usually thought of as vacuum energy density. The Friedmann equation in the $\Lambda$CDM model is then

\begin{equation} \label{eq:LCDM} \big(\frac{\dot{a}}{a}\big)^2 = \frac{8\pi}{3}G\rho_c\big[\Omega_M (\frac{a_0}{a})^3 + \Omega_{rad}(\frac{a_0}{a})^4 + \Omega_\Lambda + \Omega_{curv}(\frac{a_0}{a})^2\big]. \end{equation}

The age of the Universe is really a problem of solving the Friedmann equation under reasonable assumptions about the composition of the energy density. Since relativistic matter is relevant at early times only and spatial curvature is neglectable, the form of the Friedmann equation one employs when approximating the age of the Universe is $$ \big( \frac{\dot{a}}{a} \big)^2 = H^2_0 \big[ \Omega_M (\frac{a_0}{a})^3 + \Omega_\Lambda\big] $$ under the constraint $\Omega_M + \Omega_\Lambda = 1$ with positive $\Omega_\Lambda$. The solution is given by $$ a(t) = a_0 \big(\frac{\Omega_M}{\Omega_\Lambda})^{1/3}\big) \Big[\sinh\big(\frac{3}{2}\sqrt{\Omega_\Lambda}H_0 t\big)\Big]^{2/3}. $$ Then for $a(t_0) = a_0$, $$ t_0 = \frac{2}{3\sqrt{\Omega_\Lambda}}\frac{1}{H_0}\operatorname{Arsh} \sqrt{\frac{\Omega_\Lambda}{\Omega_M}}, $$ and for a redshift $z(t)$ (recall $z(t) = \frac{a_0}{a(t)} -1$) $$ t(z) = \frac{2}{3\sqrt{\Omega_\Lambda}}\frac{1}{H_0}\operatorname{Arsh} \sqrt{\frac{\Omega_\Lambda}{\Omega_M(1+z)^3}}. $$

The early Universe should be filled with field excitations ("primordial plasma") that interact with each other. If the interaction rate, denoted $\Gamma$, is much larger thant the rate of expansion $H$, i.e. $\Gamma \gg H$, then typical time scaale of interaction is much smaller than the typical expansion time, and the particles can be seen as in thermal equilibrium with each other; they are in local thermal equilibrium.

The interaction rate $\Gamma \equiv n\sigma v$, where $n$ is the number density of particles, $\sigma$ their interaction cross section, and $v$ the average veolicity of the particle (for high temperature, i.e. relativistic particles, $v\sim 1$). For high temperature, dimensional analysis gives $n\sim T^3$, since particle masses can be ignored for ultra-relativistic particles. $\sigma \sim \frac{\alpha^2}{T^2}$, also by dimensional analysis, where $\alpha \equiv g^2_A /2\pi$ for a gauge boson $A$ with coupling $g_A$. Thus for ultra-relativistic particles, $\Gamma \sim \alpha^2 T$.

By the Friedman equation, $H\sim \sqrt{\rho}$, taking dimensions into account $H\sim\sqrt{\rho}/M_{pl}$. $\rho\sim T^4$, and hence $H\sim T^2/M_{pl}$. Thus $$ \frac{\Gamma}{H} \sim \frac{10^{16} \text{GeV}}{T}, $$ where $\alpha\sim 0.01$ is used. When $T$ ranges from $100\text{ GeV}$ to $10^{16}\text{ GeV}$, i.e. for all particles in the Standard Model when they are relativistic the local equilibrium condition is satisfied.

The temperature decreases and relativistic particles become nonrelativistic nevertheless. However, relativistic particles still dominate the density and pressure, and nonrelativistic particles can be ignored from the primordial plasma. The distribution for particles in thermal equilibrium $$ f(p) = \frac{1}{e^{(E(p)-\mu)/T}\pm 1} $$ where $p$ is the momentum, implies that when $T\ll m$, the non-relativistic distribution is supressed by $e^{-m/T}$, thus relativistic particles dominate the density and pressure of the primordial plasma. It is sufficient to consider relativistic particles for the approximation of total energy density. Let $g_\ast = g_\ast(T)$ denote the effective number of relativistic degrees of freedom at temperature $T$, then $$ \rho_r = \sum_i \int d^3 p f_i(p) E_i(p) = \frac{\pi^2}{30}g_\ast(T) T^4. $$ In high temperature, all the species of partiles are relativistic, and $$ g_\ast = g_b + \frac{7}{8} g_f = 106.75 $$ the factor $7/8$ comes from integration.

The presence of massive particle species implies that the equilibrium didn't persist, since otherwize massive particles would have been exponentially suppressed and the universe would be mostly photons (at least for non-baryonic matters, like neutrinos, the suppression would be present). The massive particles that are still abundant today deviated from equilibrium, decoupled from the primordial plasma, and freezed out, leaving relic particles.

The suppression of number and mass density of particles when $T \ll m$ can be interpreted as particle anti-particle annihilation; at low temperatures the thermal particle energies aren't sufficient for pair production. Around $80\%$ of the particle anti-particle annihilation takes place in the interval $T = m \to m/6$.

The heaviest particles annihilates first, reducing $g_\ast$.

1. The top quarks, with mass aroung $160\text{ GeV}$, annihilates, reducing $g_\ast$ to $106.75-\frac{7}{8} \times 12 = 96.25$.
2. Then the Higgs boson and the gauge boson of weak interaction annihilates, roughly at the same time. At $T\sim 10\text{ GeV}$ we have $g_\ast = 96.25 - (1 + 3\cdot 3) = 86.25$.
3. Next annihilate the bottom quarks, leaving $g_\ast = 86.25 - \frac{7}{8}\cdot 12 = 75.75$.
4. Then the charm quarks and the tau leptons, $g_\ast = 75.75 - \frac{7}{8}\times(12+4) = 61.75$.
5. Next should have been the strange quarks, if there is no QCD phase transition (or crossover) at around $150\text{ MeV}$, during which the quarks combine into baryons and mesons. These hadrons are non-relativistic below the temperature of the QCD phase transition, except the pions. Now left are pions, electrons, muons, neutrinos and photons, leaving $g_\ast = 2 + 3 + \frac{7}{8}\times (4+4+6) = 17.25$.

Then should annihilate electrons and positrons, but before that happens, neutrino decouples from the thermal bath.

After the Electroweak Symmetry Breaking which occured at around $160\text{ GeV}$, the gauge bosons of weak interactions acquired mass, and the cross section of weak interactions became $\sigma \sim G^2_F T^2$ where the Fermi constant $G_F \sim \alpha/ M_W^2$. Now $$ \Gamma/H \sim \frac{\alpha^2 M_{pl} T^3}{M^4_W} \sim (\frac{T}{1\text{ MeV}})^3, $$ meaning that at around $1\text{ MeV}$, the particles that interact with the primordial plasma only through the weak interaction decouple and freeze out. These are the neutrinos. They are coupled to the primordial plasma via weak interaction processes such as

\begin{gather*} \nu_e + \bar{\nu}_e \leftrightarrow e^+ + e^-,\\ e^- + \bar{\nu}_e \leftrightarrow e^- + \bar{\nu}_e. \end{gather*}

The equilibrium state of the interacting medium corresponds to the minimum of the Landau potential (Grand potential). When the temperature $T$ is high ($\sim 1$ GeV), like so in the early Universe, chemical potentials are small, and one can use Helmholtz free energy $F$ instead.

At temperature $T$, let the expectation value of a scalar field $\phi$ be $\langle \phi \rangle_T$. In a system where the average value of $\phi$ is $\langle \phi\rangle_T$ everywhere and in thermal equilibrium otherwise, the free energy $F$ of the system would be $F=\Omega V_{eff}(T,\phi)$ where $\Omega$ is the spatial volume. The effective potential $V_{eff}(T,\phi)$ is the free energy density of the medium at temperature $T$ with the average scalar field $\phi$. In equilibrium, $F$ is at minimum, and hence $\langle \phi \rangle_T$ is the absolute minimum of the effective potential $V_{eff}(T,\phi)$, i.e. the vacuum expectation value (VEV).

At 0 temperature, $F$ is the same as the energy of the system, and $V_{eff}$ coincides with the scalar potential $V(\phi)$ in the Lagrangian of the field theory which $\phi$ takes part in. In the Weinberg-Salam model, the absolute minimum of $V(\phi)$, $v = \langle \phi \rangle_{T=0} = \langle \phi \rangle$, is nonzero. The computation of $v$ is given in the next section. At finite temperature $T$, $V_{eff}(T,\phi)\neq V(\phi)$, and hence the absoluate minimum, $\langle \phi \rangle_{T}$ can take other values.

The Weinberg-Salam model is an $SU(2)_L \times U(1)_Y$ gauge theory with tree $SU(2)_L$ gauge bosons $W^i_mu$, $(i=1,2,3)$ and one $U(1)_Y$ gauge boson $B_\mu$. The kinetic Lagrangian density is $$ \mathcal{L}_{K} = -\tfrac{1}{4} W^i_{\mu\nu} W^{\mu\nu i} -\tfrac{1}{4} B_{\mu\nu}B^{\mu\nu}, $$ where $$ W^i_{\mu\nu} = \partial_{[\nu}W^i_{\mu]} + g\epsilon^{ijk}W^j_\mu W^k_\nu;\quad B_{\mu\nu}=\partial_{[\nu}B_{\mu]}. $$

The most general renoramlizable $SU(2)_L$ invariant potential is $$ V(\Phi) = \mu^2 |\Phi^\dagger \Phi|+\lambda|\Phi^\dagger\Phi|^2, $$ with the complex scalar $SU(2)$ doublet $\Phi$ coupled to the gauge fields. If $\mu^2<0$, then the state of minimum energy is not at $\Phi=0$, leading to a vacuum expectation value of the scalar field. The direction of the minimum is not determined since the potential $V$ is ignorant of the direction, thus one can take the VEV to be $$ \langle\Phi\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\\ v\end{pmatrix}, $$ leading to the $U(1)_Y$ charge $Y_\Phi =1$, and the electromagnetic charge $Q=\frac{\tau_3+Y}{2}$ ($\tau_3$ is the third Pauli matrix). Hence $Q\langle \Phi\rangle =0$, and the symmetry is broken from $SU(2)_L\times U(1)_Y$ to $U(1)_{EM}$.

The scalar doublet has the Lagrangian density of $$ \mathcal{L}_s = (D^\mu \Phi)^\dagger (D_\mu \Phi) - V(\Phi) $$ for covariant derivative $$ D_\mu = \partial_\mu + i \frac{g}{2}\tau\cdot W_\mu + i\frac{g'}{2}B_\mu Y $$ which gives physical gauge fields,

\begin{gather*} W^\pm_\mu = \tfrac{1}{\sqrt{2}}(W^1_\mu\mp i W^2_\mu)\\\ Z^\mu = \frac{-g' B_\mu + g W^3_\mu}{\sqrt{g^2+g'^2}}\\\ A^\mu = \frac{gB_\mu + g' W^3_\mu}{\sqrt{g^2+g'^2}} \end{gather*}

with masses $M^2_W = \tfrac{1}{4}g^2 v^2$, $M^2_Z = \frac{1}{4}(g^2 + g'^2)v^2$ and $M_A = 0$.

The parameter $v$ is found from the charged current of $\mu$-decay, $$ \mu \to e\bar{\nu}_e \nu_\mu $$ and the interaction strength for muon decay is measured to be $G_F = 1.16639\times 10^{-5}\text{ GeV}^{-2}$. The momentum carried by the $W$ boson, of order $m_\mu$, can be neglected in comparison with $M_W$, hence $$ \frac{G_F}{\sqrt{2}}=\frac{g^2}{8 M^2_W} = \frac{1}{2v^2} $$ giving $ v = 246\text{ GeV}$, and the vacuum Higgs VEV $174\text{ GeV}$. It was measured that the Higgs mass $\mu = 125\text{ GeV}$.

The finite temperature potential, with Higgs field $\phi$, has the form $$ V(\phi,T) = D(T^2 - T^2_0)\phi^2 - ET\phi^3 + \tfrac{\lambda_T}{4}\phi^4 $$ where $D,E,T^2_0,\lambda_T$ are some factors depending on particle masses, coupling constants, the Higgs VEV in zero temperature, and the temperature. Observe that there is only one minimum of the potential $V(\phi,T)$ sits at $\phi=0$ for large $T$. The critical temperature $T_c$ let $V(\phi,T)$ obtain two minima. Particles obtain masses via the electroweak symmetry breaking mechanism described in the last section. Calculations show that $T_c \approx 178\text{ GeV}$.

A point that needs to be stressed is that some claims that perturbation theory is actually not applicable. What is given here is only a partial and inaccurate but illlustrative picture. The contemporary value for the critical temperature is $T_c \approx 159.5$ GeV (so the finite-temperature perturbation theory actually gives a result that deviates farther w.r.t. the 'actual' value than 0-temperature theory), see The Standard Model cross-over on the lattice.