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: $ds^{2} = -c^{2} dt^{2} + a^{2}(t) (d\chi^{2}+ S_{k}(x)^{2} d\Omega^{2})$ where $S_{k}(\chi)$ is $\chi$ for k=0, $R_{0} \sin (\frac{\chi}{R_{0}})$, and $R_{0} \sinh(\frac{\chi}{R_{0}})$
• The t in this FRW metric is the coordinate time
• Another useful time variable is the conformal time $d\eta = \frac{dt}{a(t)}$

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 label 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 $\rho \propto a^{0}$
    • In detail: EFE is $\G^{\mu\nu} + \Lambda g^{\mu\nu} = 8\pi G T^{\mu\nu}$. If you take $\nabla_{\mu}$, the $\Lambda$ term doesn’t contribute to the constraint
    • Hence, you can define an additional contribution to the energy tensor $T_{de}^{\mu\nu} = -\frac{\Lambda}{8\pi G} g^{\mu\nu}$ without messing up the constraints. Plugging into EFE gives the $P = -\rho$
      • This iis the standard cosmological model of $\Lambda$ CDM

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 $\dot{\rho}$
    • Explicating use $\nabla_{\mu} T_{\mu\nu} = 0$

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
    • Namely assume that $\frac{P}{\rho} = w $
• We have $\frac{d \ln a}{dt} \approx H_{0} \sqrt{\Omega_{i}} a^{-\frac{3}{2}(1+w_{i})}$
• Transforming 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 radiation universe scales like $a(\eta) = 2 H_{0} \sqrt{\Omega_{\gamma}} \eta$
• A pure matter universe is a called a Einstein-de Sitter universe (ie. w=0, since the pressure is much much smaller than the rest mass)
  • 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
  • $a(\eta) = \frac{1}{4} H_{0}^{2} \Omega_{m} \eta^{2}$ for very late times
• 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
  • $a(t) \propto exp(\sqrt{\frac{\Lambda}{3}}t)$
  • If we have $\Lambda<0$ and $k=-1$, it’s called an Anti-de Sitter spacetime
• 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 typically set to unity, equating temperature and mass

Boltzmann Hierarchy

• From Liouville’s Theorem, we know that the phase space density doesn’t change over time: $\frac{d}{dt}(\Delta^{3}x \Delta^{3} p)=0$

• Alternatively, you can think of the number of particles per phase space volume $N(x,p,t)$

  • Without collisions, we have that $\frac{dN}{dt} = 0$ (Immediate consequence of Liouville’s Theorem)
  • With collisions, we have that $\frac{dN}{dt} = C(N)$ where C is the collision operator

• Focus on collisionless for now, and assume a non-relativistic Hamiltonian: $H = \frac{p^{2}}{2m} + \phi$. The above equation is actually a hierarchy of moments

  • Define the number density as $n(t) = \int d^{3} p N(t,\vec{x},\vec{p})$
  • Define the mean velocity as $v(t) = \int d^{3} \frac{\vec{p}}{m} N(t,\vec{x},\vec{p})/ n(t)$
  • Use chain run on N, use NR Hamiltonian , then assume $N(p \rightarrow \infty) \rightarrow 0$. You get continuity equation $\partial_{t} n \vec{\nabla} \cdot (n \vec{V}) = 0$
  • Notice that you need to know $\vec{v}$ in order to close the system, which you can’t know apriori
    • Try taking time derivative of $\vec{v_{i}}$. The hope is that you can potentially construct some relationship which constrains $\vec{v}$
    • Expand out, use the continuity equation to simplify things. integrate by parts at one point. You will eventually get: $\frac{dV_{i}}{dt} = \partial_{t} V_{i} + V_{j} \frac{\partial v^{i}}{\partial r^{j}} = -\nabla^{i} \phi -\frac{1}{\rho} \frac{\partial}{\partial x_{i}} \Sigma_{ij}$
      • You can define the pressure tensor as $\Sigma_{ij}(r,\vec{x}) = m \int d^{3} p (\frac{p^{i}}{m} - v^{i}) (\frac{p^{j}}{m} - v^{j}) N $
      • You can decompose the pressure tensor into a diagnoal, and trace free term: $\Sigma_{ij} = P \delta_{ij} + \Pi_{ij}$
        • For an ideal fluid, via isotropy, the trace-free term vanishes
      • Notice that you need to know $\Sigma_{ij}$ in order to calculate v.

• This problem of the current moment being dependent on the negative divergence of the higher moment is the Boltzmann hierarchy problem

  • One way of closing the system is to truncate at some moment, then define a closure relationship relating the pressure to the density (re: equation of state)

• If you want to generalize to relativistic Hamiltonians, simply note that N is a scalar Lorentz invariant

  • Define the stress energy tensor as $T^{\mu\nu} = \int d^{3}p \frac{p^{\mu} p^{\nu}}{E(p)} N(t, \vec{x}, \vec{p})$
  • Define the number current as $N^{\mu} (t, \vec{x}) = \int d^{3} p \frac{p^{\mu}}{E(p)} N(t,\vec{x},\vec{p})$
  • The collision Boltzmann equation implies that $\partial_{\mu} T^{\mu\nu} = 0$ and $\partial_{\mu} N^{\mu} = 0$

• 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: $\rho(T) = \frac{g}{(2\pi)^{3}} \int d^{3} p E(p) f(p,T)$
      • $E(p) = \sqrt{m^{2}+p^{2}}$
    • 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

Collisions

• Early universe is very collisional due to high densities
• define the occupation number as $f(t,\vec{x},\vec{p})$ as the number of particles per quantum state. $N = \frac{N_{state}}{d^{3}x d^{3}p} \times f = \frac{g}{h^{3}} f$
  • To derive h factor: In principle, would need to count all states of all different particle types. Let’s simplify:
    • Assume all particles are spin 0, so we can use Klein Gordan
    • Place universe in big box with periodic BCs (near center, fluids acts locally. In large L limit, this converges)
    • Solutions place k vectors at discrete lattice sites with spacing $\frac{2\pi}{L_{i}}$ apart. Can use this to simplify $N_{state}$
  • No deriving g-factor, but this accounts for the degeneracy associated with each individual state (re: spin, flavors etc)
• We will only consider 2 particle scattering. More concretely, we will look at reactions of the form $1+2 \leftrightarrow 3+4$
  • The collision operator takes the form $C(N_{s1})(p_{1}) = \Sigma_{s2,s3,s4} C_{s1s2 \rightarrow s3, s4} (N_{s1})(p_{1}) + … $
    • In principle, you can include n to n’ or arbitrary body scattering
    • 2 to 2 is minimum term (1 to 1 is not a collision. 1 to 2 is decay and is not included initially. 2 to 1 is kinematically excluded (think of center of momentum frame + 4 momentum conservation)
  • 2 to 2 scattering looks like: $C_{12, 34} (N_{s}) (p_{1}) = \int d^{3} p_{2} d^{3} p_{3} d^{3} p_{4} \delta_{D}^{4}(p_{3} + p_{4} - p_{1} - p_{2}) \Gamma (p_{i}^{\mu}) \times (-f_{1}(p_{1})f_{2}(p_{2})(1\pm f_{3}(p_{3})) (1\pm f_{4}(p_{4})) + f_{3}(p_{3})f_{4}(p_{4})(1\pm f_{1}(p_{1})) (1\pm f_{2}(p_{2})) )$
    • $\delta^{4}_{D}$ for momentum conservation and energy conservation
    • $\Lambda$ for collision rate, which depends on all momenta (get from QFT)
    • the bracketed term is the forward scattering process minus the back scattering processes
      • forward is -, since removing particle s1. backwards is + because gaining s1 particles
      • Input terms get assigned $f_{i}(p_{i})$ (number of particles of type $s_{i}$ at a given momentum)
      • Outputs get assigned $(1\pm f_{i}(p_{i}))$ ( $\pm$ for bosons and fermions )
        • 1 denotes case where there is no particle of that type in phase space (so the collision can create an output particle)
        • $\pm$ alters occupancy according to spin statistics
• In equillibrium, the occupancies approach their thermal equillibrium value: $f_{eq} = \frac{1}{exp(\frac{E_{s}(p) - \mu_{s}}{T})\pm 1}$
  • Can erive by setting collisions to 0, grouping same f functions, taking log, realize you have a conservation equation that must be linear in the conserved variables (Cauchy’s functional equation solution!).
    • From H theorem, the only possible equillibrium is the state that maximizes the entropy
    • Exact constraints that we have : $\Sigma f_{i} E_{i}= U$ an $\Sigma f_{i} =N$. Maximize entropy subject to these constraints. Compare to ifferential form of entropy from thermo to get constants
  • Light has $\mu$ of 0
    • Two ways to see this. ouble compton scattering has $e \gamma \rightarrow e \gamma \gamma $. Since chemical potential is conserve when overall particle number is , we can see $\mu_{\gamma} $ is 0
    • Look at Bremstralung: $e p \rightarrow p e \gamma$. Same argument
  • In the early universe, chemical The chemical potential of all species is 0 (re: very little energy needed to create/destroy any particle!)

The Primordial Plasma

• Using a relativistic energy relationship: $E = \sqrt{p^{2}+m^{2}}$ and setting $\mu=0$, we have:
  • $n = \frac{g}{2\pi^{2}} \int_{0}^{\infty} dp \frac{p^2}{exp(\frac{\sqrt{p^{2}+m^{2}}}{T})\pm 1 }$
  • $\rho = \frac{g}{2\pi^{2}} \int_{0}^{\infty} dp \frac{\sqrt{p^2 +m^2}}{exp(\frac{\sqrt{p^{2}+m^{2}}}{T})\pm 1 }$
• 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: $\rho = \frac{\pi^{2}}{30} g_{*}(T) T^{4}$
  • $g_{*}(T) = \Sigma_{i} g_{i} (\frac{T_{i}}{T})^{4} \frac{J_{\pm}(x_{i})}{J_{0}}$
    • It is standard to quote g at $T_{\gamma}$

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)
• Only relativistic particle contribute to g*
  • Follows from Boltzmann suppression (see below)
• A particle can drop out of g* if:
  • they all annihilate with their antiparticles (at the T scale equal to their mass)
    • The photons are no longer energetic enough to feed the back reaction
  • Decoupling: neutrinos don’t annihilate, but the scattering rate to electrons gets so weak that they fall out anyways

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 $s \propto a^{-3}$
  • We thus can figure out the temperature scaling via entropy conservation : $(aT)^{3} g(T) =constant$
    • When g is not transitioning, treat it as constant. Hence we get that $T \propto \frac{1}{a}$

Cosmic Neutrino Background

• The neutrinos are the most weakly interacting particles in the Standard Model and they are (ostensibly) lighter than most other particles
  • Hence, they decouple after most confinement and annihilate events, but before electrons positron annihilation to reheat the photons, and also before recombination
    • The matter antimatter asymmetry comes into play here (ie. there are some left over electrons here needed to form hydrogen)
• 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
• 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 $T_{\nu} \propto a^{-1}$
    • The neutrino distribution stays relativistic after decoupling, even if the origin of temperature scaling is different (first was from entropy conservation, the latter is from universe expansion)

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 $T \geq m_{e}$ we have $2+\frac{7}{8}\otimes 4$ For $T<m_{e}$, we have g=2,
  • electron-positron annihilation causes the $e_{\pm} \gamma$ plasma to evolve quasistatically into a $\gamma$ only plasma due to the timescale of annihilation
  • Since the neutrinos follow the temperature of the photons prior to $m_{e}$, this implies that the $T_{nu} = (\frac{4}{11})^{\frac{1}{3}} T_{\gamma}$ (Follows from conservation of entropy ($S = g(aT)^{3}$)

$N_{eff}$

• In reality, the transitions of g from annihilation are not sharp, so there is some overlap between electron annihilation and neutrino decoupling
• The cross section of neutrino scattering is also energy dependent
• Both of these effects combine to change the effective number of flavors $N_{eff}$ from 3 to about 3.046
  • Can be used as a probe for BSM physics

Matter Antimatter Asymmetry

• Existence of non trivial number of electrons reflects poorly understood matter antimatter symmetry
• Baryons face a similar story, in that there is a slight asymmetry in their count
• Baryon conservation (re: baryons get 1, antibaryons get -1, others get 0; initial baryon number equals final baryon number) yields that this slight asymmetry kills off all of the antibaryons
• The number of baryons in the universe gets frozen after annihilation
• Since the number density of the photons and baryons scales like $a^{-3}$, the ratio of baryons to photons remains roughly constant. People call this $\eta$ = $\frac{n_{b}}{n_{\gamma}}$, and its 6E-10 (probe of BSM physics)
• The remaining electrons and baryons (re: protons and neutrons) thermalize with the photons, hence have a temperature of $T_{\gamma}$ until around z of 150

BBN

• At this point, the first nuclei form via Big Bang Nucleosynthesis
• You get a bunch of out of equillibrium reactions to generate the heavier elements
• One important one is $He^{4}$, which by the end of BBN has $Y_{He} = \frac{\rho_{H}}{\rho_{b}} = 0.26$ , where $Y_{He}$ is the abundance of helium

Recombination

Saha Equation

• The formation of hydrogen occurs via the reaction: $e^{-}+p^{+} \leftrightarrow H + \gamma$
• In equillibrium, when $T< m_{i}$ where i could be e, p, or H, the have equillibrium abundances: $n_{i}^{eq} = g_{i}(\frac{m_{i}T}{2\pi})^{\frac{3}{2}} exp(\frac{\mu_{i}-m_{i}}{T})$
• Using $\mu_{e}+\mu_{p} = \mu_{H}$, we can eliminate the chemical potentials: $\frac{n_{H}}{n_{e}{n_{p}}} = \frac{g_{H}}{g_{e}g_{p}} (\frac{m_{H}}{m_{e}{m_{p}}}\frac{2\pi}{T})^{\frac{3}{2}} exp(\frac{m_{p}+m_{e}-m_{H}}{T})$
  • Make the approximation that $m_{H} \approx m_{p}$ in the prefactor
  • In the exponent, define $E_{I} = m_{p}+m_{e}-m_{H}$ as the ionization energy
  • The number of degrees of freedom are 2, 2 and 4 for e,p and H respectively
    • e and p come from being spin 1/2
    • H comes from the singlet and triplet states
  • We believe the universe to overall be electrically neutral (ie. $n_{e} = n_{p})
  • All the above gives: $\frac{n_{H}}{n_{e}^{2}} = (\frac{2\pi}{m_{e}T})^{\frac{3}{2}} e^{\frac{E_{I}}{T}}$
  • Define $X_{e} = \frac{n_{e}}{n_{p}+n_{H}} = \frac{n_{e}}{n_{p}+n_{H}}$ to be the free-electron fraction (or the ionization fraction)
  • Neglecting the small amount of helium, $n_{p} + n_{H}$ is approximately the baryon density, which we define as $n_{b} = \eta n_{\gamma} = \eta \frac{\zeta(3)}{\pi^{2}} T^{3}$
  • We can combine the above 2 into 1 equation: $\frac{1-X_{e}}{X_{e}^{2}} = \frac{n_{H}}{n_{e}^{2}} n_{b}$
  • Subtituting in the number density ratio yields the Saha equation
• The solution to the Saha equation is $X_{e} = \frac{-1+\sqrt{1+4f}}{2f}$ where $f(T,\eta) = \frac{2\zeta(3)}{\pi^{2}} \eta (\frac{2\pi T}{m_{e}})^{\frac{3}{2}} e^{\frac{E_{I}}{T}}$
  • define $T_{rec}$ as the temperature at which $X_{e} = 0.5$
  • The shape of the Saha equation looks like a sigmoid function w.r.t. T (or z via the conversion $T = T_{0}(1+z)$)
• You find out that for $\eta \approx 6E-10$ that at $z \approx 1300$ or so is when $X_{e} = 0.5$
  • Since matter-radiation equality happens are $z\approx 3400$, we know that recombination occurred in the matter-dominated era

Photon Decoupling

• Early on, photons are strongly coupled to free electrons via Thompson scattering: $e^{-} + \gamma \leftrightarrow e^{-} + \gamma$
• At some point , the expansion rate eclipsed the interaction rate, causing the photons to decouple
  • Denote the temperature where these two rates are equal to be $T_{dec}$
• $\Gamma_{\gamma}(T_{dec}) = n_{b} X_{e}(T_{dec}) \sigma_{T} = \frac{2\zeta(3)}{\pi^{2}} \eta \sigma_{T} X_{e}(T_{dec}) T_{dec}^{3}$
• $H(T_{dec}) = H_{0} \sqrt{\Omega_{m}} (\frac{T_{dec}}{T_{0}})^{\frac{3}{2}}$
• Set the rates equal to each other, plug in the Saha equation and cosmological parameters to find that $T_{dec} \approx 0.27 eV$
  • This places decoupling shortly after recombination, but this small change creates a drastic drop in the ionization fraction

Last Scattering

• The precise moment of photon decoupling is called last-scattering
• We define the probability that a photon will scatter within some time interval to be $\Gamma_{\gamma}(t)dt$
• The optical depth is the integrated probability: $\tau(t) = \int_{t}^{t_{0}} \Gamma_{\gamma}(t)dt$
• Taking $t_{0}$ to be the present time, the moment of last scattering is defined to be $\tau(t_{*}) = 1$
  • To a good approximation, $t_{*} \approx t_{dec}$
    • However, the free electron density at the end of recombination greatly influences when $t_{*}$ happens

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

Dark Matter Freeze-Out

• Define the following interaction $X+ \bar{X} \leftrightarrow l + \bar{l}$
  • X is some heavy fermion and l is some light (basically massless) particle
  • X could be dark matter
• We will assume that the light particle remains coupled to the plasma, meaning that they maintain their equillibrium densities ($n_{l} = n_{eq}$)
• We also assume that there is no initial asymmetry between X and $\bar{X}$ densities ($N_{X} == N_{\bar{X}}$)
• The Boltzmann equation becomes: $\frac{1}{a^{3}} \frac{d(n_{X}a^{3})}{dt} = -<\sigma v > (n_{X}^{2} - (n_{X}^{eq})^{2})$
• Define $Y_{X} = \frac{n_{X}}{T^{3}}$
• Define a new time $x = \frac{M_{X}}{T}$. $M_{X}$ denotes the temperature scale associated with the mass of particle X, which is when most of the dynamics happen at
• The Hubble parameter can then be written as $H = H \frac{M_{X}}{x^{2}}$
• Given the above, you get the Riccati equation:
  • $\frac{dY_{X}}{dx} = - \frac{\lambda}{x^{2}} (Y_{X}^{2} - (Y_{X}^{eq})^{2})$
  • $\lambda = \frac{\Gamma(M_{X})}{H(M_{X})}$ is dimensionless and is taken to be constant
• This is not analytically tractable, but numerically, you see that there is a plateauing of Y for large values of $x_{f}$
  • This is “freeze-out”: massive particles initially are in eq. with the plasma, but are eventually become too far apart to interact fast enough to stay in EQ
  • There is a rough freeze out time $x_{f}$ at which this detailed balance gets broken
  • For large enough times, $Y_{X}» Y_{X}^{eq}$. Hence the Boltzmann equation reduces to $\frac{dY_{X}}{dx} \approx - \lambda \frac{Y_{X}^{2}}{x^{2}}$. Integrate from $x_{f}$ to $x = \infty$, and realize that $Y_{X}^{f} » Y_{X}^{eq}$ gives $Y_{X}^{\infty} = \frac{x_{f}}{\lambda}$
    • Hence, the freeze out density is inversely proportional to the interaction rate

WIMP Miracle

• The number density of particles after freeze out scales like $N_{X} \propto a^{-3}$
• $n_{X,0} = n_{X,1} (frac{a_{1}}{a_{0}})^{3} = Y_{X}^{\infty} T_{0}^{3} (\frac{a_{1}T_{1}}{a_{0}T_{0}})^{3}$
  • $a_{1}$ is an arbitrary time that is late enough for the species to have reached their relic abundance, but before particle annihilation has broken the $T \propto a^{-1}$ scaling
• From conservation of entropy $G_{*S}(aT)^{3} = const$ we have $n_{X,0} = Y_{X}^{\infty}T_{0}^{3} \frac{G_{*S}(T_{0})}{G_{*S}(M_{X})}$
• We can calculate the energy density of the particles to be $\Omega_{X} = \frac{\rho_{X,0}}{\rho_{crit,0}}$
  • Plugging in the observed values yields $\Omega_{X} \approx 0.1 \frac{x_{f}}{\sqrt{g_{*}(M_{X})}} \frac{1E-8 GeV^{-2}}{<\sigma v>}$
  • This is insensitive to the mass of the new particles
  • The required cross section to see the observed dark matter density is about 0.1 $\sqrt{G_{F}}$
    • This is the WIMP miracle, since the weak interaction scale gives the right order of magnitude for the dark matter abundance observed today

Baryogenesis

• You can apply the above Boltzmann Equation analysis to baryon annihilation
  • The cross section scales like the inverse mass squared of the pion (nucleon interations effectively mediated by pions)
    • This gives a density of baryons to be $\Omega_{b} 1E-11$, which is much smaller than the observed value
  • We need a theory of baryogenesis to produce the abundance of baryons in the early universe

Sakharov Conditions

• In 1967, Sakharov identified three necessary conditions that any theory of baryogenesis must satisfy
  • Violation of baryon number: Since we start with a equal number of baryons (B=0), we expect that at present time $B\neq 0$)
    • Standard model lagrangian conserves baryon number via guage symmetry. Baryon violation occurs in SM via “triangle anomaly”
  • C and CP violation
    • C is charge conjugation. If C was an exact symmetry, then $i\rightarrow f$ would have equal probability to $\bar{i} \rightarrow \bar{f}$, and thus there would be no net baryon number generated
    • CP (charge conjugation with parity) is more subtle
      • CPT (Charge, parity, time reversal) is a symmetry of any relativistic quantum field theory
      • So CP invariance is the same as time invariance. If you have a process $i(p_{a},s_{a}) \rightarrow f(p_{a},s_{a})$, where p and s are the momenta and spins respectively, then under time reversal, you have $f(-p_{a}, -s_{a}) \rightarrow i(-p_{a}, -s_{a})$. Integrating over all momenta and spin yields a vanishing net baryon number. Hence, CP must be violated
    • Both C and CP are violated in the weak interaction
  • Departure from equillibrium:
    • processes that generate the baryon asymmetry must occur out of thermal equillibrium

Big Bang Nucleosynthesis (BBN)

• Qualitatively:
  • In the beginning, baryonic matter was mostly protons and neutrons, which were coupled to each other via $\beta$ and inverse $\beta$ decay
    • $n+ \nu_{e} \leftrightarrow p^{+} + e^{-}$
    • $n+e^{+} \leftrightarrow p^{+} + \bar{\nu}_{e}$
  • At 1 MeV, the neutrons decoupled from the bath. These free neutrons decay, further reducing their abundances
  • free neutrons also combine with protons to form deuterium (~0.1 MeV):
    • $n + p^{+} \leftrightarrow D + \gamma $
  • Deuterium nuclei combine with free protons to produce helium:
    • $D+p^{+} \leftrightarrow ^{3}He+\gamma$
    • $D+^{3}He \leftrightarrow ^{4}He + p^{+}$
  • All the promordial neutrons get converted to helium
    • heavy element formation difficult, which is why the protons survive and make hydrogen
• In detail:
  • In equillibrium, the abundances go like $\frac{n_{n}}{n_{p}} = exp(-\frac{m_{n}-m_{p}}{kT})$
    • Assuming chemical potentials are negligable. The prefactor of $(\frac{m_{n}}{m_{p}})^{\frac{3}{2}}$ gets dropped b/c of similar masses, but not in the exponential
    • Let $Q = m_{n}-m_{p}$
  • Neutron Freeze out occurs during decoupling at 1 MeV. The neutron fraction is $ X_{n} = \frac{n_{n}}{n_{n}+n_{p}}$
    • Can use equillibrium ratio to get that $X(T)_{eq} = \frac{e^{-\frac{Q}{T}}}{1+e^{-\frac{Q}{T}}}$
    • Use the Boltzmann equation, with 1 being neutrons, 3 being protons, 2 and 4 being leptons (which remain in equillibrium)
      • You find that the neutron fraction obeys: $\frac{X_{n}}{dt} = -\Gamma_{n}(X_{n} -(1-X_{n})exp(-\frac{Q}{T}))$
        • $\Gamma_{n}$ comes from a cross section analysis
        • Make the change of variables to $x = \frac{Q}{T}$
      • Integrate this with the initial condition of equillibrium of neutrons and protons
    • Numerically, this traces out a flipped sigmoid curve. We find that $X_{n}$ approaches 0.15 for large x
    • One wrinkle to this is the neutron life time
      • The lifetime causes the final “equillibrium” to decay
      • If fusion doesn’t happen fast enough, then in principle, the neutrons would vanish. Lucky for us that this isn’t the case
  • Deuterium is the only efficient intermediate to large nuclei (ie. it’s the first stable one to form). This is the so called deuterium bottleneck
    • neutron to neutron is very unstable
    • proton to proton needs to overcome the Coulomb barrier
    • 3 way interactions don’t really happen on these small time scales (100 seconds)
  • After this point, all of the free neutrons get trapped into tritium and helium via the following reactions:
    • $D + p^{+} \leftrightarrow He^{3} +\gamma$
    • $ D + D \leftrightarrow H^{3} + p^{+}$
    • $D+D \leftrightarrow He^{3} + n$
    • $H^{3} +p^{+} \leftrightarrow He^{3} + n$
    • $H^{3} +D \leftrightarrow He^{4} +n$
    • $He^{3} +D \leftrightarrow He^{3} + p^{+}$
  • Evantually, the final equillibrium helium abundance is half that of the neutron abundance
  • Elements with A larger than 7 don’t exist in large quantities from BBN since their lifetimes aren’t long enough and there aren’t enough of them to construct a chain reaction

Inflation

• The universe is largely homogeneous and flat, which would require a very fine tuning of the initial conditions
  • The horizon problem: Why is the universe homogeneous everywhere you look?
    • Why are causally seperated regions of the CMB roughly the same temperature?
  • Flatness problem is: why is the universe so close to being flat currently when it was not initially? A priori, the universe can’t relax to flatness fast enough
  • Why are fluctuations in the CMB correlated over distances larger than the light travel time since the Big Bang (superhorizon correlations)?
• Inflation, namely the rapid stretching of initial quantum fluctuations, gives a generic way to solve all of these problems that is agnostic to the initial conditions

Horizon Problem

• Define the comoving particle horizon as $d_{h} = \int_{t_{i}}^{t} \frac{dt}{a(t)}$
  • This is the maximum possible distance which light (re: nill geodesic) can influence a future event
  • The event horizon is the opposite: how far in the past can light influence an event
  • First, rewrite the particle horizon as $d_{h} = \int_{\ln a_{i}}^{\ln a} (aH)^{-1} d(\ln a)$
    • We call $(aH)^{-1}$ the comoving Hubble radius. For large times, this is large, hence $d_{h} \approx (aH)^{-1}$
• The recombination time is much smaller than the age of the universe today
  • Imagine that you have two photons which start streaming at recombination time which are sufficiently far apart
  • In standard Big Bang cosmology, this implies that there is not enough time for the two events to be in causal contact with each other (their light cones don’t interact)
    • Then why is the CMB homogeneous? This implies that all of the sky was in causal contact with each other
  • How large of a patch do you need for this to be a problem? Depends on your cosmology model.
    • Find $\theta = \frac{2d_{h}(t_{rec})}{d_{A}(t_{rec})}$ where $d_{A}$ is the angular diameter distance between the events
    • The Friedmann equations tell you what H is
• One caveat to the above:
  • We observe that the conformal time blows up at the origin if a=0
  • For a small $\delta t$ after the Big Bang, we are assuming that the conformal time here doesn’t contribute significantly to the overall conformal time
    • Namely, we are assuming that GR doesn’t break down at these small time scales. We know that it does at scales comparable to the Planck time
  • A quantum gravity theory would elucidate the geometry in this area and potentially avoid this blow up
• Inflation provides an answer in lieu of a new quantum gravity theory
  • We have assumed that the Hubble radius monotonically increases as a function of time. Doing this causes points to spread out too fast
  • What if we demand that, in the early universe, that the Hubble radius decreases over time?
    • Namely: $\frac{d}{dt}(aH)^{-1} <0$. Using $H = \frac{\dot{a}}{a}$, this is equivalent to saying that $\ddot{a} >0$ (ie. accelerated expansion)
    • If we do this, we can allow large $d_{H}$ while maintaining a Hubble radius that allows causal contact
  • Making the Hubble radius decrease as a function of time causes the $d_{H}$ integral to be dominated by the early Hubble radius
  • Unlike Big Bang cosmology, inflation allows allows the singularity to occur at a negative conformal time ($\eta = -\infty$), hence providing more spacetime for events to be causally connected

Flatness Problem

• $\Omega_{k}(t) = \frac{\rho_{c}-\rho}{\rho_{c}} = (\frac{a_{0}H}{aH})^{2} \Omega_{k,0}$
  • This is the time dependent curvature parameter, where $\rho_{c} = 3M_{p}^{2}H^{2}$ is the time dependent critical density
  • This denotes how curved the universe is as a function of time
  • Since the Hubble radius $\frac{1}{aH}$ increases as a function of time, this implies that the curvature of the universe was much smaller in the past
  • However, we observe that the universe is extremely flat. This implies that the early universe was extremely flat. Why is this?
• A shrinking Hubble radius in the early universe would explain the apparent extreme flatness: any initial curvature would be quickly flattened out, after which the curvature would very slowly increase

Superhorizon Correlations

• The universe contains density fluctuations which are correlated over acausal distances (superhorizon)
• " These detailed correlations make it more difficult to appeal to a creator without sounding like a young Earth creationist, arguing that the fossil record was planted to deceive us” - David Tong
• Inflation solves this by letting apparently causally connected regions interact, then quickly seperate to their acausal locations

Inflation Duration

• Of course, inflation needs to last long enough so that the slow ramp up matches the observed values
• We need the Hubble radius of the observable universe $(a_{0}H_{0})^{-1}$ to be smaller than the Hubble radius during inflation $(a_i H_i)^{-1}$
• Define $ N = \ln (\frac{a_{e}}{a_i})$ be the number of e-foldings (think how many doublings of space happen, but instead of base 2 it’s base e)
• Assume that the Hubble rate during inflation stayed roughly constant ($H_{i} \approx H_{e}$)
• Inflation goes up until the Hot Big Bang, after which the Hubble radius grows to the current value
  • Assume that the Hubble radius doesn’t change significantly between the end of inflation and the start of the Big Bang (ie. $(a_{e}H_{e})^{-1} \approx (a_{R}H_{R})^{-1}$)
• Assume a radiation dominated universe for simplicity. We see that $\frac{a_{0}H_{0}}{a_{R}H_{R}} = \frac{a_{R}}{a_{0}} \approx \frac{T_{0}}{T_{R}}$
  • $T_{R}$ is the reheating temperature
• Combining all of the above gives bounds on how many e-foldings you need

Structure Formation

• All of the prior stuff was assumed a homogeneous universe. We now sprinkle in inhomogeneities and see how they evolve over time
• We first look at the Newtonian limit to build some intuition as to what happens
  • non-relativistic velocities
  • low gravity (curvature much smaller than the Hubble radius)
  • P « $\rho$

Fluid Dynamics

• We have three equations which govern fluid flow:
  • $\frac{\partial \rho }{\partial t} + \nabla \cdot (\rho \vec{u}) = 0$
    • Follows from mass conservation
  • $\rho \frac{d\vec{u}}{dt} = \rho(\frac{\partial}{\partial t} + \vec{u} \cdot \nabla ) \vec{u} = - \nabla P$
    • Follows from momentum conservation
  • $P = P(\rho,T)$ is an equation of state which closes the system
• Imagine that we have a static fluid (u=0). This implies that $\rho$ and P and constant. What would happen if we peturb from this steady state?
  • let $\rho = \bar{\rho}+\delta p$ and similarly for P
  • Make this substitution and only keep terms up to 1st order to get:
    • $\partial_{t} \delta \rho = - \nabla \cdot (\bar{\rho} \vec{u})$
    • $\bar{\rho}\partial_{t} \vec{u} = -\nabla \delta P$
    • Combining the above yields: $\partial_{t}^{2} \delta \rho - \nabla^{2} \delta P = 0$
  • For now, assume a barytropic fluid: $P = P(\rho)$. We can then write $\delta P = \frac{\partial P}{\partial \rho} \delta \rho = c_{s}^{2} \delta \rho$
    • $c_{s}^{2} = \frac{\partial P}{\partial \rho}$ is the sound speed
  • With the above substitution, the above becomes a wave equation
    • We can write $\delta \rho(x,t)$ as the sum of all the Fourier modes:
      • $\delta \rho(x,t) = \int \frac{d^{3} k } {(2\pi)^{3}} exp(i k\cdot x ) \delta \rho (k,t)$
    • This then gets converted to a ODE of k , with solutions : $\delta(k,t) = C_{k} exp(i\omega_{k}t)+D_{k}exp(-i\omega_{k}t)$ where $\omega_{k} = c_{s}|k|$

Newtonian Fluid Dynamics with Gravity

• Keep the mass continuity the same, but now add a gravitational potential:
  • $\rho \frac{d\vec{u}}{dt} = \rho(\frac{\partial}{\partial t} + \vec{u} \cdot \nabla ) \vec{u} = - \nabla P - \rho \nabla \Phi$
  • Linearize in an analagous fashion to the previous section to see that $(\frac{\partial^{2} }{\partial t^{2}} + c_{s}^{2}k^{2} -4 \pi G \bar{\rho})\delta \rho (k,t) = 0$
  • $\omega_{k} = c_{s}^{2}k^{2} - 4\pi G \bar{\rho}$ is the new frequency of the system. Setting this to 0 defines a critical wavenumber called the Jeans scale: $k_{J} = \frac{\sqrt{4\pi G\bar{\rho}}}{c_{s}}$
    • for k»$k_{J}$, we recover the previous behavior
    • for k « $k_{J}$, we have that gravity dominates. The solutions look like $\delta \rho \propto exp(\pm \frac{t}{\tau})$ where $\tau = \frac{1}{\sqrt{4\pi G \bar{\rho}}}$. These exponential solutions are called Jean instabilities

Adding Expansion

• Now, we imagine that our coordinates change over time:
  • $r(t) = a(t) x$
  • $u(t) = \dot{r} = Hr+v$ where H is the Hubble flow and v is the physical peculiar velocity
    • “peculiar velocity” is just relative velocity w.r.t. some baseline (In this case, the Hubble flow)
  • Originally, the time and space derivatives were independent. Due to expansion, we now get:
    • $\frac{\partial f}{\partial t_{r}} = ((\frac{\partial}{\partial t_{x}}) - H x \cdot \nabla_{x}) f$
      • $t_{i}$ refers to what variable is being held constant
      • Chain rule $f(t,a(t))$ and transform from r to x
• Previously, we implicitly held r constant in the time derivatives for the continuity equation, momentum equation
  • Now, we need to hold x (the comoving coordinate) constant instead
  • The continuity and momentum equations are derived by subbing in the derivative relationship b/w x and r above, and then using the relationship for the velocity field u
  • The Poisson equation picks up a factor of $a^{2}$ on the RHS
• Define the background solution to be when $\vec{v} = 0$. The equations become:
  • $\frac{\partial \bar{\rho}}{\partial t} = -3H \bar{\rho}$
  • $\frac{\partial \vec{\bar{u}}}{\partial t} = -\frac{1}{a}\frac{\nabla \bar{P}}{\bar{\rho}} - \frac{1}{a} \nabla \bar{\Phi}$
  • The solution follows that of a homogeneous non-relativistic fluid
    • $\bar{\rho} \propto a^{-3}$
    • Solutions to Euler equation take the form $\vec{u} = Ha \vec{x}$, $\bar{P} = const$, $\nabla \bar{\Phi} = -\ddot{a} a \vec{x}$
    • This solution holds as long as the scale factor a satisfies $\frac{\ddot{a}}{a} = -\frac{4\pi G}{3} \bar{\rho}$
      • This this the acceleration equation derived from the Friedmann equations
• Define the linear peturbations:
  • $\rho = \bar{\rho}(1+\delta)$
    • $\delta = \frac{\delta \rho} {\bar{\rho}}$
  • $\vec{u} = Ha\vec{x} + \vec{v}$
    • The velocity $\vec{v}$ is the expansion
  • $P = \bar{P} + \delta P$
  • $\Phi = \bar{\Phi} + \delta \Phi$
• Doing the expansion, and keeping terms up to first order, we find:
  • $\dot{\delta} = -\frac{1}{a} \nabla \cdot \vec{v}$
  • $\dot{v} + Hv = -\frac{1}{a \bar{\rho}} \nabla \delta P - \frac{1}{a} \nabla \delta \Phi$
    • Recall that $(\vec{v} \cdot \nabla) \vec{x} = \vec{v}$
  • Combine the above two into a single 2nd order equation
    • Take time derivative of first, plug $\dot{v}$ from first and $\nabla \cdot v$ from second, then the perturbed Possion equation ($\nabla^{2} \delta \Phi = 4\pi G a^{2} \bar{\rho} \delta$, then the peturbed pressure $\delta P = c_{s}^{2} \bar{\rho} \delta$ (remember we are assuming a barotropic fluid!)
    • You get $\ddot{\delta} + 2H\dot{\delta} - (\frac{c_{s}^{2}}{a^{2}}\nabla^{2} + 4\pi G \bar{\rho}(t)) \delta = 0$
      • This describes the growth of density flunctuations in an expanding universe

Growth of Matter Perturbations

• Make the standard Fourier expansion ansatz: $\delta \rho(x,t) = \int \frac{d^{3}k}{(2\pi)^{3}} exp(i\vec{k}\cdot \vec{x}) \delta \rho(\vec{k},t)$
• In Fourier space, we have: $\ddot{\delta}(\vec{k},t) + 2H \dot{\delta}(\vec{k},t) + c_{s}^{2} (\frac{k^{2}}{a^{2}} - k_{J}^{2}) \delta(\vec{k},t) = 0$
  • The physical Jeans wavenumber is $k_{J}(t) = \frac{\sqrt{4\pi G \bar{\rho}(t)}}{c_{s}(t)}$
    • It’s time dependent now!
• On small scales ($\frac{k}{a} » k_{J}$) you get a damped harmonic oscillator, with the hubble expansion acting as friction
• On large scales ($\frac{k}{a} « k_{J}$), you recover the static space case, but the extra friction term prevents the fluctuations from growing exponentially
  • Explicitly: $\ddot{\delta} + 2H \dot{\delta} - 4\pi G \bar{\rho}(t) \delta = 0$ where $\delta$ is a function of k and t
• Hence, Jeans wavenumber sets the scale where you go from suppression at small k to damped growth at large k
• The time dependent nature of the Jean scale is important
  • For baryons, prior to recombination, the baryons are strongly coupled to the photons, hence have a sound speed like the photons ($c_{s} \approx \frac{1}{\sqrt{3}}$ which gives a Jeans scale on the order of a Hubble radius. Hence, there is no growth in baryon fluctuations on subhorizon scales
  • Post recombination, the sound speed drops and flunctuations start to grow
  • For cold dark matter, the sound speed is negligable, hence subhorizon modes can grow earlier

Linear Growth Function

• Since for the large scale case, the ODE involves only time derivatives, we can write an ansatz of the form: $\delta(k,t) = \delta_{+}(k) D_{+}(t) + \delta_{-}(k) D_{-}(t)$
  • Intuition: Need 2 independent functions to span the solution space, each of which only depends on time. Each of these functions can be modulated by some function of k (ie. The fourier modes)
• $D_{+}(t)$ is called the linear growth function and is normalized such that $D_{+}(t_{0}) = 1$
• plugging in this ansatz to the density flunctuation equation we can determine how the flunctuations grow as a function of the scale factor
• If we have multiple components, then the density becomes $\Sigma_{\alpha} \bar{\rho_{\alpha}} \delta_{\alpha}$

Statistical Properties

Correlation Functions

• By definition, the mean value of the density perturbations is zero ($<\delta> = 0$)
• Hence, we need to use the two-point correlation function to get any statistics on the density field
• $\Epsilon(|x-x’|, t) = <\delta(x,t)\delta(x’,t)> = \int D\delta P(\delta) \delta(x,t) \delta (x’,t)$
  • This is a functional integral over field configurations. $P(\delta)$ is the probability of realizing some config $\delta(x,t)$
  • $<…>$ creates an emsemble average of the stochastic process
  • Homogeneity implies that the two-point function should be invariant under translations (ie. only depends on separation between two points)
  • Isotropy implies that only the distance between separation matters
• We want to compute this two-point function, but it is not what we observe
  • The observed universe is a single realization of a random process
  • We assume ergoicity (ie. ensemble averages are equal to spatial averages as the volume approaches infinity)
    • This allows different parts of the universe to be seen as different realization of the underlying random process
  • Somce we have a finite volume, this introduces statistical fluctuations called sample variance
• We can also write the two point function in Fourier space:
  • $<\delta(\vec{k}), \delta(\vec{k’})> = \int d^{3}x d^{3} x’ exp(-i\vec{k}\cdot \vec{x}) exp(-i\vec{k’}\cdot \vec{x’}) <\delta(x)\delta(x’)>$
    • This can be simplified: $\int d^{3}r d^{3}x’ exp(-i\vec{k}\cdot \vec{r}) exp(-i(\vec{k}-\vec{k’})\cdot x’) \Epsilon(r) = (2\pi)^{3} \delta_{D}(\vec{k}-\vec{k’}) P(k)$
      • $P(k) = \int d^{3}r exp(-i\vec{k}\cdot\vec{r}) \Epsilon(r)$
      • The delta function in k implies that each wavevector is independent of each other
      • $P(k)$ is the power spectrum (ie. the 3D Fourier transform of the correlation function $\Epsilon(r)$
  • Due to rotational inveriance, we know the power spectrum only depends on the magnitude of the wavevector
    • Hence, working in spherical coordinates: $P(k) = \frac{4\pi}{k} \int_{0}^{\infty} dr r\sin(kr) \Epsilon(r)$

Gaussian Random Fields

• For a Gaussian random field, the probability distribution is a Gaussian function of $\delta(x)$
• the PDF for N points in space $x_1$ to $x_{N}$ iS $P(\delta_{1},…\delta_{N}) \propto \frac{1}{\sqrt{det(\epsilon_{ij})}} exp(-\delta_{i} \epsilon_{ij}^{-1} \delta_{j})$ where $\delta_{i} = \delta(x_{i})$ and $\epsilon_{ij} = \epsilon(|i-j|)$

Harrison Zeldovich Spectrum

• $P(k,t) = T^{2}(k) \frac{D_{+}^{2}(t)}{D_{+}(t_{i})} P(k,t_{i})$ (will be derived later on)
• The transfer function is defined as $T(k) = \frac{D_{+}(t_{i})}{D_{+}(t)} \frac{\delta(\vec{k},t)}{\delta(\vec{k},t)}$
  • $t_{i}$ is some initial time which is taken to be after inflation
  • This is necessary since the dependency inside and outside the Hubble radius is different
• Harrison Zeldovich and Peebles argued that $P(k,t_{i}) = Ak^{n}$
  • We expect the gravitational potential to be scale invariant during inflation
    • If there was some characteristic scale, then you would observe clumps of energy during inflation
    • In addition, as inflation happens, larger and larger scales enter the Hubble radius. We need the magnitude of each new scale to be the same as the previous one, hence scale invariance
  • From Fourier transforming the Poisson equation, we see that $P_{\Phi} \propto k^{-4} P_{\delta} \propto k^{n-4}$
  • To have the gravitational potential to be scale free, we need the correlation function/variance to not depend on the wavenumber
  • We know that $\sigma_{\Phi}^{2} = \Epsilon_{\Phi}(0) = \int \frac{dk}{k} \frac{k^{3}}{2\pi^{2}} P_{\Phi} $
    • Hence, to keep scale invariance, we need n=1