Cosmology

Following Baumann’s Cosmology.

Conventions

• We are assuming a (- + + + ) signature for the metric.
• Greek letters for spacetime indices, and Latin letters for spatial indices
• overdots for physical time derivatives and primes for conformal time derivatives

Qualitative Overview

• Meters are a cumbersome unit to work with at universe scales. We use lightyears and parsecs instead
  • 1 ly is roughly 9.5E15 m
  • 1 parsec (pc) is defined as the distance at which the radius of the Earth’s orbit around the Sun subtends an angle of one arcsecond (re: $\frac{1}{3600}$ of a degree)
    • In light years, 1 parsec is around 3.26 lightyears
• For a sense of scale:
  • The nearest star (Proxima Centauri) is 4.2 lightyears away
  • Our Galaxy, the Milky Way, is around 30 kpc across
  • The nearest galaxy (Andromeda) is around 2 million light years away (2 Mpc)
  • The local group (~ 50 nearby galaxies) is around 10 Mpc in diameter
  • The local supercluster (Laniakea) is 500 Mpc
  • The observable universe is around 14 Gpc or 46.5 billion lightyears
    • This is ostensibly larger than the age of the universe (13.8 lightyears) due to expansion
• The consituents of the universe are:
  • Ordinary matter: This is us. It’s 5%
  • Dark matter: This is the vast majority of the matter contents
    • Major experimental evidence for this is increased rotational speeds of hydrogen gas in the outer reaches of galaxies (Vera Rubin)
      • Only can be explained if galaxies embedded in halos of dark matter
    • Another major piece is lensing of the CMB is in tension with measurements of element abundances without non-baryonic dark matter
      • Also, the anisotropies of the CMB wouldn’t manifest as they do without dark matter
  • Dark Energy: This is the biggest component, and is roughly 70% of the universe’s energy content
    • Some negative pressure associated with the universe’s expansion
• Big Bang
  • In the beginning, it was hot and dense, with all particle species in roughly equal abundances
  • At some point, it expanded rapidly, and cooled off in the process
  • We can map out the various phases by converting these temperatures to energies:
    • EW phase transition (100 GeV): the electroweak symmetry is broken, allowing EM and the weak force to become distinct entities and particles gain mass via the Higgs
    • As the temperature drops below the mass of particle species, particle antiparticle annihilation begins while pair production becomes inefficient
      • QCD phase transition (150 MeV): the remaining quarks condense to hadrons
  • At the grand old age of 1, the neutrinos decouple from equillibrium due to the expansion of the universe creating the cosmic neutrino background ($c\nu B$)
  • Big Bang Nucleosynthesis (BBN) occurs around 1 minute in, creating elements lighter than lithium-7
  • 370,000 years later, the universe cooled enough to allow recombination to occur. This allowed the photons to stream through the universe (CMB)
  • There are two periods which we don’t know exactly when they occured
    • dark matter production: lots of theories here (WIMPs, axions etc.)
    • baryogenesis: The symmetry breaking between matter and antimatter (1 part in $10^{10}$) to give the matter to photon ratio observed
• Structure formation
  • The first stars (Population III stars) are boorn 100 millions years after the Big Bang
    • They were massive, and died off rapidly. The UV light emitted heated up the surrounding gas, leading to reionization
    • They could have created the supermassive black holes
    • They created the heavier elements in the universe
  • The first galaxies formed around 1 billion years after the Big Bang
    • The distribution of galaxies is not uniform (on small scales): there are spatial correlations
• Inflation
  • We observe some objects in the universe to be older than the universe’s age
  • Inflation helps solve this: if the distances between objects suddenly increases in a short time frame, you can explain these apparent causality violations
    • This period would need to happen before the Big Bang
  • In 1 billionth of a trillionth of a trillionth of a second, the universe doubled in size about 80 times

The Expanding Universe

GR Concept Refresher

• Spacetime is defined by some metric: $ds^{2} = \Sigma_{\mu,\nu}^{3} g_{\mu\nu}dx^{\mu}dx^{\nu}$
  • $ds^{2}$ is invariant to all observers
• In Minkowski space, this reduces to $ds^{2} = -c^{2}dt^{2} + \delta_{ij} dx^{i} dx^{j}$
• the metric is used to convert between contravariant vectors ($A_{\mu}$) and covariant vectors ($A_{\mu}$)
• You can contract a convariant with a contravariant vector via the metric to produce an invariant scalar

FRW Metric

• On large enough scales, the distribution of mass is isotropic (same in all directions) and homogeneous (same at every point in space)
• The metric then takes the form $ds^{2} = -c^{2} dt^{2} + a^{2}(t) d\vec{l}^{2}$
• homogeneous and isotropic spacetimes must have constant intrinsic curvature
  • This means that we can write $d\vec{l}^{2} = dx^{2}+ k \frac{(x \cdot dx)^{2}}{R_{0}^{2}-kx^{2}}$
    • $R_{0}$ denotes the intrinsic curvature of the space
    • In spherical coordinates, we have $dl^{2} = \frac{dr^{2}}{1-k\frac{r^{2}}{R_{0}^{2}}}+ r^{2} d\Omega^{2}$
  • k can be 0 for flat space, 1 for spherical, and -1 for hyperbolic
• The RW metric becomes $ds^{2} = -c^{2} dt^{2} + a^{2}(t)(\frac{dr^{2}}{1-k\frac{r^{2}}{R_{0}^{2}}}+ r^{2} d\Omega^{2})$
  • The line element has a rescaling symmetry
    • $a \rightarrow \lambda a$, $ r\rightarrow r/ \lambda$ and $R_{0} \rightarrow R_{0}/\lambda$
    • This allows $a(t_{0}) = 1$ (re: the scale factor at the present time) as well as $R_{0} = R(t_{0})$ to be interpreted as the physical curvature today
  • we call r a comoving coordinate, while $r_{phys} = a(t) r$ is the physical coordinate
  • The physical velocity is just the time derivative of $r_{phys}$
    • $v_{phys} = H r_{phys} + v_{pec}$
      • We define $H = \frac{\dot{a}}{a}$ to be the Hubble parameter
      • $v_{pec}$ to be the velocity of a comoving observer (re: someone who follows the Hubble flow)
  • We can make the change of coordinates $d\chi = \frac{dr}{\sqrt{1-\frac{kr^{2}}{R_{0}^{2}}}}$
    • The FRW metric becomes: $ds^{2} = -c^{2} dt^{2} + a^{2}(t) (d\chi^{2}+$

Kinematics

• Define a free particle lagrangian in an arbitrary coordinate system as $\mathbf{L} = \frac{m}{2} g_{ij}(x^{k}) \dot{x}^{i}\dot{x}^{j}$
• Substitute into the Euler-Lagrange equution to get
  • $\frac{d^{2} x^{i}}{dt^{2}} = - \Gamma_{ab}^{i} \frac{dx^{a}}{dt}\frac{dx^{b}}{dt}$
  • We define $\Gamma_{ab}^{i} = \frac{1}{2} g^{ij} (\partial_{a} g_{jb} + \partial_{b} g_{ja} + \partial_{j} g_{ab})$
    • These are the Christoffel symbols
• This is easily extended to 4 dimensions. Simply replace the latin indices for greek ones (re: add time equations in)
  • The derivaion for massless particles require some care though
• If you define the 4-momentum as $P^{\mu} = m \frac{dx^{\mu}}{d\tau}$, the geodesic equation becomes $P^{\alpha}(\partial_{\alpha}P^{\mu} + \Gamma_{\alpha \beta}^{\mu} P^{\beta}) = 0$
  • The bracketed term is the covariant derivative, denoted as $\nabla_{\alpha} P^{\mu}$
• Substituting in the FRW metric to the geodesic equation tells you how the 4-momentum depends on the scale factor
  • Looking at $\mu=0$, we see that:
    • For massless particles, we have $\frac{1}{E} \frac{dE}{dt} = - \frac{\dot{a}}{a}$
      • This implies an energy scalinig of $E = \frac{1}{a}$
    • For massive particles, we have that the momentum scales like $\frac{1}{a}$

Redshift

• Since spacetime is getting stretched out, the wavelength of light gets stretched as well. This increase in observed wavelength is called redshifting
• Since energy of photons scales like $\frac{1}{a}$, the wavelength is proportional to a. Hence $\lambda_{0} = \frac{a(t_{0})}{a(t_{1})} \lambda_{1}$
• redshift is defined as the fractional shift in wavelength:
  • $z = \frac{\lambda_{0}-\lambda_{1}}{\lambda_{1}}$
  • Using $a(t_{0})=1$ for today’s scale factor, we have that $1+z = \frac{1}{a(t_{1})}$
• Conventionally, we labe events in the universe’s history by their redshift:
  • the surface of last scattering happens at z=1100 and the first galaxies formed around z= 10
• For nearby sources (z< 1), we can expand the scale factor around $t_{0}$ and define $H_{0} = \frac{\dot{a_{0}}}{a(t_{0})}$ as Hubble’s constant to get that
  • $a(t_{1}) = 1+ (t_{1}-t_{0})H_{0}$
  • For close objects, $\Delta t$ is simply $frac{d}{c}$, which means that redshift increases linearly with distance
  • Alternatively, the recession speed $v = cz$ becomes $v \approx H_{0}d$
  • The above is called the Hubble-Lemaitre law
• It’s convention to describe Hubble constant measurements as $H_{0} = 100 h \frac{km}{s Mpc}$ since the measurements originally have very large uncertainties
  • h is $0.73 \pm 0.01$ from supernovae measurements
  • while h is $0.674 \pm 0.005$ from CMB measurements
  • There is a statistically significant difference between the two, giving rise to the “Hubble tension”

Distances

• These are hard to measure. The metric distance is unobservable, and the physical distance assumes a fixed time
• To make measurements, we need definitions which take into account the expansion and the finite travel time of light

Luminosity Distance

• We can use “standard candles” (re: objects of known intrinsic brightness) and compare against their observed brightness to calculate distances
• We have the Cepheids, which have a periodic brightness. This period is correlated with the intrinsic brightness of a star
  • More explicitly, the brightness of the star versus the logarithm of the period shows a linear relationship
  • So if you can measure the distance to a Cepheid via parallax, you can extrapolate the distances to the rest of the Cepheids
• The next rung up on the cosmic distance ladder are the Type 1a supernovae
  • These happen when a white drawf accretes too much matter from a companion star and explode
  • They are so bright that they outshine the stars in their local group
  • These explosions occur at a precise moment (re: when the white drawf exceeds the Chandrasekhar limit) and hence have a fixed brightness
• To generalize, suppose that we have a source of known luminosity L (re: energy per unit time)
  • There is some observed flux (re: energy per unit time per unit area) from with L can be inferred
  • Suppose that the source is at redshift z. The comoving distance is $\chi(z) = c\int_{0}^{z} \frac{dz}{H(z)}$
    • The flux in a static space is $F = \frac{L}{4 \pi \chi^{2}}$ since the source is isotropic
    • the expansion complicates this for 3 reasons:
      • the radius of the sphere expands so that the area becomes $4\pi a^{2}(t_0) d_{M}^{2}$
      • The arrival rate of the photons is smaller than the rate at which they are emitted by a factor of $\frac{1}{1+z}$
      • The energy of the photons get red shifted, which contributes another factor of $\frac{1}{1+z}$
    • The end result is that $F = \frac{L}{4\pi d_{M}^{2} (1+z)^{2}} = \frac{L}{4\pi d_{L}^{2}}$ where $d_{L}$ is the luminosity distance
      • $d_{L} = (1+z) d_{M}(z)$ where $d_{M}$ is the metric distance

Angular Diameter Distance

• Standard rulers are objects of know physical size
  • An example is the typical size of the hot and cold spots in the CMB from theory
• Once again, assume a static space. The angular size of an object that is of known length D at some comoving distance $\chi$ away is $\delta \theta = \frac{D}{\chi}$
  • With expansion, the formula becomes $\delta \theta = \frac{D}{a(t_{1}) d_{M}} = \frac{D}{d_{A}}$
    • $d_{A}(z) = \frac{d_{M}}{1+z}$ relates the angular diameter distance to the metric distance

Dynamics

• To calculate the scale factor, we need to solve the Einstein equation: $G_{\mu\nu} = \frac{8\pi G}{c^{4}} T_{\mu\nu}$
  • G is the Einstein tensor (curvature measurement) which T is the energy momentum tensor
• First, let’s look at number density $N^{\mu}$
  • $\mu=0$ is the number density of particles
  • $\mu=i$ component is the flux of the particles in direction $x^{i}$
  • What constraints must $N^{\mu}$ obey to satisfy homogeneity and isotropy?
    • Isotropy implies that $N^{i} = 0$ while homogeneity implies $N^{0} = c n(t)$
  • In a comoving frame, we have $N^{\mu} = n U^{\mu}$
  • Particle number should be conserved, which implies $\partial_{\mu} N^{\mu} = 0$
    • In curved spacetime, we have $\nabla_{\mu} N^{\mu} = 0$
      • $\nabla_{\mu} A^{\nu} = \partial_{\mu}A^{\nu} + \Gamma_{\mu\lambda}^{\nu} A^{\lambda}$
      • $\nabla_{\mu} B_{\nu} = \partial_{\mu}B_{\nu} - \Gamma_{\mu\nu}^{\lambda} B_{\lambda}$
  • So we have $\partial_{\mu} N^{\mu} = -\Lambda_{\mu \lambda}^{\mu} N^{\lambda)$
    • In the rest frame, we have $ \frac{\dot{n}}{n} = -3 \frac{\dot{a}}{a}$
      • This implies that $n(t) \propto a^{-3}$
• In an analagous way to the number density, we can constrain the form of $T_{\mu\nu}$
  • We can break the tensor into 3 components: an energy density (scalar), a vector (momentum density/energy flux. Same values because of symmetry), and a smaller 3x3 stress tensor
    • Isotropy forces the vector components to be 0
    • Homogeneity requires the energy density to be indpendent of position, but dependent on time
    • Isotropy around the origin constrains the mean value of the stress tensor to be proportional to $\delta_{ij}$. Homogeneity requires that the proportionality constant be only a function of time
  • All of that allows us to write down the perfect fluid stress-energy tensor: $T_{\mu\nu} = (\rho+\frac{P}{c^{2}})U_{\mu}U_{\nu}+P g_{\mu\nu}$
  • Local energy and momentum conservation imply $\nabla_{\mu} T^{\mu}{\nu} = 0$
    • Expand out in terms of Christoffel symbols and examine the $\nu=0$ term
    • The $T^{i}_{0}$ term vanishes by isotropy
    • In what remains, $\Gamma_{\mu 0}^{\lambda}$ is 0 unless $\lambda$ and $\mu$ are the same spatial indices, which implies $\Gamma_{i0}^{i} = \frac{3}{c} \frac{\dot{a}}{a}$
    • From the continuity equation, we arrive at: $\dot{\rho} + 3 \frac{\dot{a}}{a} (\rho + \frac{P}{c^{2}}) = 0$
      • Since the time translation is broken in an expanding space, global energy conservation doesn’t hold
• Most cosmological fluids are parametered as $P = w(\rho c^{2})$
  • This implies $\rho \propto a^{-3(1+w)}$
  • Matter is a fluid whose pressure is much smaller than it’s energy density. w=0, which implies $\rho \propto a^{-3}$
    • This could be baryons or dark matter
  • Radiation is anything which obeys $P = \frac{1}{3} \rho c^{2}$. So $w=\frac{1}{3}$ and $\rho \propto a^{-4}$
    • This can be very light particles, photons, neutrinos, gravitons
  • Dark energy has negative pressure, which implies $\rho \propto a^{0}$

Einstein Tensor for FRW metric

• The Einstein tensor is defined as $G_{\mu\nu} = R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}$
  • $R_{\mu\nu}$ is the Ricci tensor and $R = g^{\mu\nu}R_{\mu\nu}$ is the Ricci scalar
• You can write the Ricci tensor in terms of the Christoffel symbols: $\partial_{\lambda} \Gamma_{\mu\nu}^{\lambda} - \partial_{\nu} \Gamma_{\mu\lambda}^{\lambda} + \Gamma_{\lambda\rho}^{\lambda} \Gamma_{\mu\nu}^{\rho} - \Gamma_{\mu\lambda}^{\rho}\Gamma_{\nu\rho}^{\lambda}$
  • $R_{0i} = R_{i0} = 0$ because of isotropy (re: vector component)
  • $R_{00} = -\frac{3}{c^{2}} \frac{\ddot{a}}{a}$
  • $R_{ij} = \frac{1}{c^{2}} (\frac{\ddot{a}}{a} + 2 (\frac{\dot{a}}{a})^{2} + 2 \frac{kc^{2}}{a^{2} R_{0}^{2}}) g_{ij}$
    • The author doesn’t directly calculate this. Instead, he computes $R_{ij}$ around x=0, then argues that since this is a tensorial reslationship, the equation holds everywhere
    • The spatial metric is $\gamma_{ij} = \delta_{ij} + \frac{k x_{i} x_{j}}{R_{0}^{2}-k(x_{k}x^{k})}$
      • They expand to 2nd order since you need derivatives of $\gamma_{ij}$, and $\gamma_{ij}(x=0) = \delta_{ij}$ doesn’t have this information
    • $\gamma_{jk}^{i} = \frac{k}{R_{0}^{2}} x^{i} \delta_{jk}$
• The Ricci scalar is then $R = \frac{6}{c^{2}}(\frac{\ddot{a}}{a} + (\frac{\dot{a}}{a})^{2} +\frac{kc^{2}}{a^{2} R_{0}^{2}})$

Friedmann Equations

• Plugging in the FRW metric into the Einstein equations gives the Friedmann Equations
• The time-time component gives: $(\frac{\dot{a}}{a})^{2} = \frac{8\pi G}{3} \rho - \frac{kc^{2}}{a^{2}R_{0}^{2}}$
  • This is often written in terms of the hubble parameter $H = \frac{\dot{a}}{a}$
  • The first Friedmann equation can also be written as $\frac{H^{2}}{H_{0}^{2}} = \Omega_{r} a^{-4} + \Omega_{m} a^{-3} \Omega_{k} a^{-2} + \Omega_{\Lambda}$
    • Define $\Omega_{i} = \frac{\rho_{i}}{\rho_{crit}}$ where $\rho_{crit} = \frac{3H_{0}^{2}}{8 \pi G}$ os the density of a flat universe evaluated at today
    • $\Omega_{\Lambda} = -\frac{kc^{2}}{(R_{0}H_{0})^{2}}$
    • If you evaluate at $t_{0}$, you get that $1 = \Omega_{r} + \Omega_{m} + \Omega_{\Lambda} + \Omega_{k}$
• The spatical component(s) yield: $\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho+\frac{3P}{c^{2}})$
  • The 2nd Friedmann equation is also called the Raychaudhuri equation
  • It can also be derived by taking the time derivative of the 1st equation and applying continuity for $\dot{\rho}$

Exact Solutions of the Friedmann Equations

• In general, you need numerics to solve the Friedmann equation

Single Component Universes

• Assume a flat universe with just a single component
  • You can parameterize the specific component by its associated w factor
• We have $\frac{d \ln a}{dt} \approx H_{0} \sqrt{\Omega_{i}} a^{-\frac{3}{2}(1+w_{i})}$
• Tranforming to conformal time, we find $a(\eta} \propto \eta^{\frac{2}{1+3 w_{i}}$
  • The special case of $w_{i} = 1/3$ is a universe dominated by spatial curvature
• A pure matter universe is a called a Einstein-de Sitter universe
  • This is a good approximation for long matter-dominated periods in the universe
  • Using a Matter dominated universe implies that the Hubble constant is $\frac{2}{3} \frac{1}{t_{0}}$. This gives a universe age of 9 billion years, which is the famous age problem
• A pure curvature universe is only permitted if k=-1. This is called a Milne universe* A pure curvature universe is only permitted if k=-1. This is called a Milne universe
• If $w=-1$, then we have a de Sitter space, which is a good approximation in the far past and far future
• All of these solutions (baring de Sitter space) have a singularity at t=0. This singularity (which goes like $\rho \propto t^{-2}$ is a generic feature of all Big Bang cosmologies (Hawking and Penrose)
  • This assumes a strong energy condition ($\rho c^{2} + 3P \geq 0 $ which implies that $\ddot{a} \leq 0$ at all times
    • Gravity is attractive, so it slows down the expansion

Our Universe

• There are a handful of parameters that parameterize our universe
• First off is the temperature of the cosmic microwave background (CMB), which is given as $T_{0} = 2.7255 \pm 0.0006 K$
  • The energy density of the photons is given by $\Omega_{\gamma} = 5.4E-5$
  • The energy density of the neutrinos is 68% that of the photons
    • This is currently being constrained by $0.0012 < \Omega_{\nu} < 0.003$, where the lower limit comes from neutrino oscillation experiments and the upper bound comes from cosmology
  • All of these measurements came from COBE
• The anisotropy of the CMB places constrains on the curvature energy density ($|\Omega_{k}| < 0.005$
  • This implies that curvature is a very negligable part of the universe’s energy contents (and more so in the past)
• We have the baryon density, which we can infer from the abundances of light chemical elements produced in the Big Bang. This coems to around $\Omega_{b} \approox 0.05$
• The dark matter density is $\Omega_{c} \approx 0.27$
• Dark energy, responsible for the expansion of the universe, makes up $\Omega_{\Lambda} \approx 0.68$ of the energy content of the universe
• Integrating the Friedmann equation yields: $H_{0} t = =\int_{0}^{a} \frac{da}{\sqrt{\Omega_{r}a^{-2}+\Omega_{m}a^{-1}+\Omega_{\Lambda}a^{2}+\Omega_{k}}}$
  • We can infer the age of the universe (a=1) to be $t_{0} \approx 13.8$ GYrs
  • We can evaluate the time scales at which matter-radiation and matter-dark energy equality occur at:
    • $t_{mr} \approx 50000$ yrs
    • $t_{m\Lambda} \approx 10.2$ Gyrs
      • The coincidence problem: Why did dark energy come to dominate the expansion so close to the present time?

The Hot Big Bang

Natural Units

• We define natural units, which set the speed of light and $\hbar$ to unity
• We also define the reduced Planck mass as $M_{PI} = \sqrt{\frac{\hbar c}{8\pi G}}$
• The Boltzmann constant is also typicall set to unity, equating temperature and mass

Statistical Mechanics Refresher

• The probability of a particle having some momentum and position at time t is described by a distribution function
  • From homogeneity, f can’t depend on x
  • From isotropy, f can only depend on the magnitude of p
• In equillibrium, we have that the gas is in a state of maximum entropy and has the following distribution function
  • $f(p,T} = \frac{1}{e^{\frac{E(p)-\mu}{T}}\pm 1}$ Where the + is for fermions and the - is for bosons
• The density of state is the number of particles assigned to each probability bin
  • For a box normalized system, we have $\frac{g}{h^{3}} = \frac{g}{(2\pi)^{3}}$
  • This gives rise to:
    • The number density: $n(T) = \frac{g}{(2\pi)^{3}} \int d^{3} p f(p,T)$
    • The energy density: $n(T) = \frac{g}{(2\pi)^{3}} \int d^{3} p E(p) f(p,T)$
      • $E(p) = \sqrt{m^{2}+p^{2}}$
    • The pressure: $n(T) = \frac{g}{(2\pi)^{3}} \int d^{3} p f(p,T)\frac{p^{2}}{3E(p)}$
      • Pressure is momentum change per unit time per unit area. The change in momentum is 2|p| along one axis. The volume swept out in unit time is $v_{x} dA = p_{x} \frac{dA}{E}$. We then calculate $\frac{p_{x}^{2}}{E}$ over the particles moving in the correct direction (divide by 2) and along the correct axis (divide by 3)
• For now, we assume that the chemical potential is much smaller than the temperature

The Primordial Plasma

• Using a relativistic energy relationship: $E = \sqrt{p^{2}+m^{2}}$ and setting $\mu=0$, we have:
  • $n = \frac{g}{2\pi^{2}} \int_{0}^{\infty} dp \frac{p^2}{exp(\frac{\sqrt{p^{2}+m^{2}}}{T})\pm 1 }$
  • $\rho = \frac{g}{2\pi^{2}} \int_{0}^{\infty} dp \frac{p^2 \sqrt{p^{2}+m^{2}}{exp(\frac{\sqrt{p^{2}+m^{2}}}{T})\pm 1 }$
• We can define the dimensionless variables $x = \frac{m}{T}$ and $\epsilon = \frac{p}{T}$
  • The integrals become $I_{\pm}(x) = \int_{0}^{\infty} d\epsilon \frac{\epsilon^{2}}{exp(\sqrt{\epsilon^{2} + x^{2}}\pm 1}$ and $J_{\pm}(x) = \int_{0}^{\infty} d\epsilon \frac{\epsilon^{2}\sqrt{\epsilon^{2}+x^{2}}}{exp(\sqrt{\epsilon^{2} + x^{2}}\pm 1}$

Relativistic limit

• The momentum dominates the rest mass, so x approaches 0
• $I_{\pm}(x=0) \int_{0}^{\infty} d\epsilon \frac{\epsilon^{2}}{e^{\epsilon}\pm 1}$
  • Expand the denominator in a geometric series
  • For bosons (-) we get $I_{-}(0) = 2 \zeta(3)$ with $\zeta$ being the Reimann zeta function
  • For fermions, we find that $I_{+}(0) = \frac{3}{4} I_{-}(0)$
    • A cute trick to see this is to note that $\frac{1}{e^{\epsilon}+1} = \frac{1}{e^{\epsilon}-1}-\frac{2}{e^{2\epsilon}-1}$
    • Hence we have $I_{+}(0) = I_{-}(0)- 2 * (\frac{1}{2})^{3}* I_{-}(0) = \frac{3}{4} I_{-}(0)$
• Hence we get that $n = \frac{\zeta(3)}{\pi^{2}} g T^{3}$ and $\rho = \frac{\pi^{2}}{30} gT^{4}$ with a multiplicative factor of 1, $\frac{3}{4}$ and 1, $\frac{7}{8}$ for bosons and fermions respectively
  • This tells use the number density and energy density of relic photons
  • Setting $p = E$ yields $P = \frac{1}{3} \rho$, as expected of a relativistic gas
• The early universe consisted of a collection of different species, which makes the total energy density $\rho = \Sigma_{i} \frac{g_{i}}{2\pi^{2}} T_{i}^{4} J_{\pm}(x_{i})$
  • It’s standard to write the density in terms of the photon temperature: $\rho = \frac{\pi^{2}}{30} g_{*}(T) T^{4}$
  • $g_{*}(T) = \Sigma_{i} g_{i} (\frac{T_{i}}{T})^{4} \frac{J_{\pm}(x_{i})}{J_{0}}$

Calculating $g_{*}$

• If we assume only relativistic species, we have that $g_{*} = \Sigma_{i=b} g_{i} (\frac{T_{i}}{T})^{4} + \frac{7}{8} \Sigma_{f} g_{i} (\frac{T_{i}}{T})^{4}$ where b and f represent summing over bosons and fermions respectively
• How many internal degrees of freedom are there?
  • A massive particle has g = 2s+1, while a massless particle oof any spin has g=2
  • There are 4 force carriers (re: gauge bosons)
    • The photon is massless (g=2)
    • The weak force carriers are all spin 1 gauge bosons, and there are 3 of them (g=9)
    • The Gluons are massless, and there are 8 of them (because 8 generators of SU(3)) (g=16)
  • The leptons are massive spin $\frac{1}{2}$. (g=12, including the antiparticles)
  • The quarks have 6 flavors and each come in 3 colors. (g=72, including antiparticles)
  • The nutrinos, despite being massive spin $\frac{1}{2}$ particles only contribute 1 degree of freedom (only left-helicity neutrinos have ever been observed)

Non-relativistic limit

• we let x» 1 which makes the integral the same for bosons and fermions
• $I(x) = \int_{0}^{\infty} d \epsilon \frac{\epsilon^{2}}{e^{\sqrt{\epsilon^{2}+x^{2}}}}$
  • Taylor expand the square root to 1st order and perform the Gaussian integral
    • Rational is that most of the integral contributions come from $\epsilon « x$
• $I_{\pm}(x) = \sqrt{\frac{\pi}{2}} x^{\frac{3}{2}} e^{-x}$
• $n = g (\frac{mT}{2\pi})^{\frac{3}{2}} e^{-\frac{m}{T}}$
  • The exponential suppresses massive particles at these low temperatures (called Boltzmann suppression)

Entropy Conservation

• Recall the first law of thermodynamics: $TdS = dU+PdV$ (dropped chemical potential since it’s small)
• Recalling that $U = \rho V$, define the entropy density $s = \frac{S}{V}$
• Since s and $\rho$ are intrinsic quantities (re: volume independent), we we re write the first law as $(Ts-\rho-P) dV + V(T\frac{ds}{dT}-\frac{d\rho}{dT}) = 0$
  • Both bracketed terms must vanish seperately since dV and dT can be arbitrary
  • so we have $s = \frac{\rho+P}{T}$ and $\frac{ds}{dT} = \frac{1}{T} \frac{d\rho}{dT}$
    • We can use the continuity equation $\frac{d\rho}{dt} = -3H (\rho+P) = -3H Ts$ to rewrite the above as $\frac{d(sa^{3})}{dt} = 0$
    • This implies that the total entropy is conserved in equillibrium and that the entropy density scales as $s \propto a^{-3}$

Cosmic Neutrino Background

• Since the neutrinos are the most weakly interacting particles in the Standard Model, we expect them to decouple from the thermal plasma first
• Neutrinos couple to the bath via processes like:
  • $\nu_{e} + \bar{\nu_{e}} \leftrightarrow e^{+} + e^{-}$
  • $e^{-} + \bar{\nu_{e}} \leftrightarrow e^{-} + \bar{\nu_{e}}$
• The interaction rate (per particle) is $\Gamma = n \sigma |v|$ where n is the number density of the target particle, $\sigma$ is the cross section and v is the relative velocity
  • v is c is the relativistic limit
• We can get a rough scale of the decoupling temperature:
  • We know that $\sigma \approx G_{F}^{2} T^{2}$ where $G_{F} \approx 1.2E-5 GeV^{-2}$ is the Fermi constant
  • n scales as $T^{3}$
  • The interaction rate becomes $\Gamma \approx G_{F}^{2} T^{5}$
  • The hubble rate scales like $\frac{T^{2}}{M_{Pl}}$
  • These scales are roughly equal to each other when T is around 1 MeV
• After decoupling, the neutrinos preserve the relativistic Fermi-Dirac distribution along all the geodesics
• The number density scales like $n \propto a^{-3} \int d^{3}q \frac{1}{exp(\frac{q}{aT_{\nu}})+1}$
  • q = ap is the a time-independent momentum (needed since $p \propto a^{-1}$)
  • particle number conservation requires $n_{\nu} \propto a^{-3}$. This is only consistent if $T_{\nu} \propto a^{-1}$
    • As long as photon temperature still scales like the neutrino temperature, we will be in equllibrium. Particle annihilation breaks this scaling

Electron-Positron Annihilation

• Shortly after neutrino decoupling, the temperature drops below the electron mass, which means that the electrons and positrons annihilate with the photons
• This causes the photons to be “heated” (re: the photon bath temperature decreases more slowly compared to the neutrinos)
  • We can characterize this by comparing the effective number of degrees of freedom. For $T<m_{e}$, we have g=2, while for $T \geq m_{e}$ we have $2+\frac{7}{8}\otimes 4$
  • electron-positron annihilation causes the $e_{\pm} \gamma$ plasma to evolve quasistatically into a $\gamma$ only plasma due to the timescale of annihilation
  • This implies that the $T_{nu} = (\frac{4}{11})^{\frac{1}{3}} T_{\gamma}$ (Follows from conservation of entropy ($S = g(aT)^{3}$)

Cosmic Microwave Background

• At temperatures of 1 eV, the universe consisted of a plasma of free electrons and nuclei. Photons tightly coupled to electrons via Thomson scattering, while Coulomb scattering coupled electrons to protons
• At temperatures below 0.3 eV, electrons and nuclei formed neutral atoms, which caused the density of free electrons to decrease sharply
  • This caused the mean free path to quickly exceed the Hubble length. Around 0.25 eV, photons decoupled from matter enough to allow the photons to escape (re: the universe became transparent)

Chemical Potential

• Photons have 0 chemical potential, since their number is not conserved
  • Ex: double Compton scattering ($e^{-} + \gamma \rightarrow e^{-}+\gamma + \gamma$)
• The chemical potential of an antiparticle is $\mu_{X} = -\mu_{\bar{X}}$
  • Follows from particle-antiparticle annihilation in equillibrium with a photon bath