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Exam problems 1 Problems relevant for Milestone I - Background Cosmology

2. Problems relevant for Milestone II - Thermodynamics and the Boltzmann equation

3. Problems relevant for Milestone III - The Boltzmann equation in a perturbed Universe

4. Problems relevant for Milestone III and IV - Initial conditions and inflation


Exam problems


Here is a list of old exam problems:

Home exam 2020, (with solutions)
Exam 2018 postponed, (with solutions)
Exam 2018, (with solutions)
Exam 2016, (with solutions)
Exam 2015, (with solutions)
Exam 2014, (with solutions)
Exam 2013, (with solutions)
Exam 2012, (with solutions)
Exam 2011
Exam 2010


Problems


This document contains some selected problems from the stuff we have gone through in the lectures. Most of the problems are there for you to see if you are able to perform the derivations we have gone through in the lectures (and understand the physics of it) so they might not be that exciting. For more problems see the end of each section in Dodelson or in Baumann's lecture notes. I have added a brief summary of some of the theory needed in this course on this page which might be useful.


Background cosmology



Calculations with General Relativity


This first problem set is to make you confortable with doing GR calculations. This is some work, but you should atleast do parts of this to get familiar with doing such calculations. The results you derive here will also be useful for the rest of the course. The aim is to compute all of the following starting from the metric components $g_{\mu\nu}$ for a flat FRLW metric in Cartesian coordinates:

  • The inverse metric components $g^{\mu\nu}$
  • The Christoffel symbols $\Gamma^\alpha_{\mu\nu}$
  • The Ricci tensor $R_{\mu\nu}$
  • The Ricci scalar $R$
  • The Einstein tensor $G_{\mu\nu}$
  • The covariant derivative $\nabla_\mu T^{\mu\nu}$ for a perfect fluid

With all this information its straight forward to derive all the relevant equations we need in Milestone I (Friedmann equations, continuity equation).


1 Compute the inverse metric for the flat FRLW metric


Start from the flat FRLW metric in Cartesian coordinates $$ds^2 = -dt^2 + a^2(t)(dx^2 + dy^2 + dz^2)$$ i.e. $g_{00} = -1$ and $g_{ij} = a^2(t) \delta_{ij}$ (so $g_{11} = g_{22} = g_{33} = a^2(t)$ and the other terms being zero) and compute the inverse metric $g^{\mu\nu}$ (remember that its a symmetric tensor). Use the definition: $$g_{\mu\nu}g^{\alpha\nu} = \delta^\alpha_\mu$$ or in matrix form ${\bf g} {\bf g^{-1}} = {\bf 1}$ (Hint: what is the inverse of a diagonal matrix?).


2 Derive the Christoffel symbols for the flat FRLW metric


Compute the Christoffel symbols $$\Gamma^\sigma_{\mu\nu} = \frac{1}{2}g^{\delta\sigma}(g_{\mu\delta,\nu} + g_{\delta\nu,\mu} - g_{\mu\nu,\delta})$$ for all the different $\sigma,\mu,\nu$ which are (remember that its symmetric in the lower two indices) $\Gamma^0_{00}$, $\Gamma^0_{i0}$, $\Gamma^0_{ij}$, $\Gamma^i_{00}$, $\Gamma^i_{j0}$, $\Gamma^i_{jk}$ and show that the only non-zero terms are $$\Gamma^0_{ij} = a\frac{da}{dt}\delta_{ij} = a^2 H \delta_{ij}$$ $$\Gamma^i_{j0} = \frac{1}{a}\frac{da}{dt}\delta^i_j= H\delta^i_j$$ where $H = \frac{1}{a}\frac{da}{dt}$.


3 Derive the Ricci tensor for the FRLW metric


Compute the Ricci tensor $$R_{\mu\nu} = \partial_\alpha \Gamma^\alpha_{\mu\nu} - \partial_\mu \Gamma^\alpha_{\alpha\nu} + \Gamma^\alpha_{\alpha\beta}\Gamma^\beta_{\mu\nu} - \Gamma^\beta_{\mu\alpha}\Gamma^\alpha_{\beta\nu}$$ for all the different $\mu,\nu$ which are $R_{00}$, $R_{0i}$, $R_{ij}$ (its a symmetric tensor). Show that $$R_{00} = -3\frac{\ddot{a}}{a}$$ $$R_{i0} = 0$$ $$R_{ij} = (a\ddot{a} + 2\dot{a}^2)\delta_{ij}$$ Use this to compute the Ricci scalar $R = g^{\mu\nu}R_{\mu\nu}$. Show that $$R = g^{00}R_{00} + g^{ij}R_{ij} = 6\left(\frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2}\right)$$


4 Derive the Einstein tensor


Use the results in the previous exercises to evaluate the components Einstein tensor $$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}$$ i.e. $G_{00}$, $G_{0i}$, $G_{ij}$ (its symmetric). Show that $$G_{00} = 3\frac{\dot{a}^2}{a^2}$$ $$G_{i0} = 0$$ $$G_{ij} = \left(-2\ddot{a}a - \dot{a}^2\right)\delta_{ij}$$ Combine this with the energy momentum tensor for a perfect fluid $$T_{\mu\nu} = (\rho + p)u_\mu u_\nu + pg_{\mu\nu}$$ where $u^\mu = (1,0,0,0)$ thus $T_{00} = \rho$ and $T_{jk} = p a^2\delta_{ij}$ and show that all the full set of Einstein equations gives us two indepdent equations (the Friedmann equations): $$H^2 = \frac{8\pi G}{3} \rho$$ $$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho + 3p)$$


5 Derive the continuity equation for a perfect fluid


The Einstein equations imply the conservation equation $\nabla_\mu T^{\mu \nu} = 0$. Show that $$\nabla_\mu T^{\mu \nu} = \partial_\mu T^{\mu \nu} + \Gamma^\mu_{\mu\alpha} T^{\alpha \nu} + \Gamma^\nu_{\mu\alpha} T^{\mu \alpha}$$ For a perfect fluid $$T^{\mu\nu} = U^\mu U^\nu (\rho + p) + p g^{\mu\nu} $$ with $U^\mu = (1,0,0,0)$. Evaluate this for $\nu = 0$ and $\nu = i$ and show that this gives just one new equation $$\frac{d\rho}{dt} + 3H(\rho + p) = 0$$ However show that this equation is a consequence of the Einstein equations derived in the previous exercise (Hint: take the time-derivative of the $H^2$ equation to get $\dot{\rho}$ and use the two Friedmann equations to simplify show that the equation above is satisfied).


Solve the continuity equation for a perfect fluid


The continuity equation following from $\nabla_\mu T^{\mu\nu} = 0$ when $T^{\mu\nu}$ is that of a perfect fluid was derived in a previous exercise and reads $$\frac{d\rho}{dt} + 3H(\rho + p) = 0$$ Solve this equation when the equation of state $w = \frac{p}{\rho}$ is a constant to get an expression for $\rho = \rho(a)$ as function of the scale-factor (and $w$).

Some methods you can use to solve it:

  • Use the method of an integrating factor $\frac{dy}{dx} + A(x)y = 0 \implies \frac{d}{dx}[y e^{\int A(x) dx}] = 0$
  • Change variables to $R = \log \rho$ and $x = \log a$. First show $\frac{d}{dt} = H\frac{d}{dx}$. If you do this then $\frac{dR}{dx}$ can be integrated directly.
  • Take an ansatz $\rho \propto 1/a^n$, insert it into the ODE and solve for $n$.

If $w = -1/3$ then the solution is $\rho \propto \frac{1}{a^2}$ (i.e. $\rho = \frac{C}{a^2}$ for some constant $C$). Find the solution for $w = 0$ (dust/dark matter), $w = 1/3$ (radiation) and $w=-1$ (vacuum energy). You might want to treat $w=-1$ as a special case (what does the continuity equation look like for this value?).


Simplified form for the Friedman equation


The first Friedman equation is given by $$H^2 = \frac{\dot{a}^2}{a^2} = \frac{8\pi G\rho}{3}$$ where $\rho = \sum \rho_i$ is the sum over all the different components in the Universe. Each component evolves according to $\dot{\rho}_i + 3H(\rho_i+p_i) = 0$ with a constant $w_i = p_i/\rho_i$ and in a previous problem we found that $\rho_i = \frac{\rho_{i0}}{a^{3(1+w_i)}}$ where $\rho_{i0}$ is the energy density today ($a=1$). Define the energy density parameter today as $\Omega_{i0} = \frac{\rho_{i0}}{\rho_{c0}}$ where $\rho_{c0} = \frac{3H_0^2}{8\pi G}$ is the critical density today and $H_0$ is the value of the Hubble function $H$ today ($a=1$). Show that we can write the Friedman equation on the form $$H^2 = H_0^2\sum \frac{\Omega_{i0}}{a^{3(1+w_i)}}$$ For a Universe with baryons $b$ ($w=0$), cold dark matter ${\rm CDM}$ ($w=0$), radiation $r$ ($w=1/3$), masseless neutrinos $\nu$ ($w=1/3$) and a cosmological constant $\Lambda$ ($w=-1$) show that this becomes $$H^2 = H_0^2\left( \frac{\Omega_{b0}}{a^3} + \frac{\Omega_{{\rm CDM}0}}{a^3} + \frac{\Omega_{\gamma 0}}{a^4} + \frac{\Omega_{\nu 0}}{a^4} + \Omega_{\Lambda 0}\right)$$ This is the form you will implement in the numerical project. The Hubble constant can be written as $H_0 = 100h km/s/Mpc$ where $h$ is a free parameter. In the literature people often work with a sightly different set of parameters namely $h^2\Omega_i$ instead of $\Omega_i$'s. This combination is often called the "physical density parameters". Why do you think this combination is used? (Hint: consider what happens to $H$ when we, for example, increase the $\Omega$'s by a factor of $4$ and reduce $h$ by a factor of $2$).


Curvature in the Friedman equations


The Friedman equations with curvature is given as $$\frac{k}{a^2} + H^2 = \frac{8\pi G}{3}\rho$$ $$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho + 3p)$$ where $\rho = \sum \rho_i$ and $p = \sum p_i$ is the sum over all the different components in the Universe. Show that curvature enters the two Friedman equations as if it we had a "curvature energy component" with equation of state $w = -\frac{1}{3}$ in a flat Universe. To do this define an approprivate $\rho_k$ and $p_k$ such that the Friedmann equation with curvature reduces to the one without curvature but with an additional energy component.


Extra (long calculation): Friedman equations for a curved Universe


If we don't assume flatness then the general form of the line-element in spherical coordinates ($t,r,\theta,\phi$) is given by $$ds^2 = -dt^2 + a^2\left(\frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right)$$ where $k$ is a constant describing the geometry of the Universe ($k>0$ spherical, $k=0$ flat and $k<0$ hyperbolic). Derive the Einsteins equation for this metric. You will have to compute all the Christoffel symbols from scratch, then $R_{\mu\nu}$ (you only need $R_{00}$ and $R_{ij}$) and finally $R$ and evaluate the $00$ component of Einsteins equation to get the first Friedmann equation. Evaluate the $ij$ component to get the second Friedman equation. Check that the equations reduce to what we derived in the lectures when $k=0$.


Lifetime of the Universe


Assume we have a Universe with a single component with equation of state $w$, i.e. $H^2 = \frac{8\pi G \rho}{3}$ with $\rho \propto a^{-3(1+w)}$. Compute the life-time of the Universe $t = \int dt = \int_{a=0}^{a=1}\frac{da}{aH(a)}$ in this model. Express the result on the form $t = \frac{C}{H_0}$ where $C$ is some constant and $H_0$ is the current value of the Hubble parameter $H_0 = H(a=1)$. Explain the result you get when $w=-1$.


Scalefactor as function of time


Use the Friemann equation to compute $a$ as a function of $t$ for a Universe dominated by a component with equation of state $w$. Its useful to take an ansatz $a(t) \propto t^{n}$ and solve for $n$. Compare this with the known expressions $n = 2/3$ in matter domination ($w=0$) and $n=1/2$ in radiation domination ($w=1/3$). What is $a(t)$ when $w=-1$?


Approximations for $\eta$


In the numerical project you are to compute $\eta$. Here we will derive an analytical approximation that is valid in the early Universe and can be used to test your numerical results. Start with $$\eta = \int_0^a\frac{cda}{a^2H}$$ Assume that relativistic matter is dominating the energy budget so that $H^2 \approx H_0^2 \frac{\Omega_{R 0}}{a^4}$ (here $\Omega_{R 0} = \Omega_{\gamma 0} +\Omega_{\nu 0}$ is the total radiation density parameter). Compute $\eta$ and show that $\eta = \frac{c}{aH}$. This approximation will be valid as long as $\Omega_R(a) \approx 1$ so until $a\approx \mathcal{O}(10^{-3})$. Derive similar approximations for the matter (and also dark energy if you want) era. To do this write $$\eta = \eta(a_*) + \int_{a_*}^a \frac{cda}{a^2 H}$$ for some $a_*$ such that $\Omega_m(a_*) \approx 1$. Approximate $H$ in this era as we did above and solve the integral. This will give you an approximation on the form $\eta = \eta(a_*) + C\left(\frac{1}{(aH)(a)} - \frac{1}{(aH)(a_*)}\right)$ for some constant $C$ that you can derive and which will be valid as long as $\Omega_m(a_*) \approx 1$. Check that your analytical approximation agres with your numerical result (you will have to adjust $\eta(a_*)$ to get the curves to match or match it together with the previous approximation).

Note that these approximations will not be perfect, but it will give you an indication if you are doing it correctly or not.


Thermal history



Temperature of neutrinos


Entropy is a useful concept for working with the early Universe. The second law of thermodynamics reads $$dS = \frac{d(\rho V) + PdV}{T}$$ where $V$ is the volume in question, $P$ the pressure and $T$ the temperature. Use the conservation equation $\dot{\rho} + 3H(\rho+P) = 0$ together with $V\propto a^3$ (the volume we consider expands with the Universe) to show that $S$ is a conserved quantity.

The entropy of a relativistic gas is given by $S = \frac{4\rho V}{3T} \propto T^3$ and the total entropy in the early Universe is given by $$S \propto g_{\rm eff}T^3$$ where $g_{\rm eff} = \sum_{\text{Boson-degreees-of-freedom}} + \frac{7}{8} \sum_{\text{Fermion-degreees-of-freedom}}$ is the number of relativistic degrees of freedom (recall the factor of $7/8$ that comes from what we talked about in the lectures - the difference between Fermi-Dirac and Bose-Einstein statistics). Non-relativistic matter carries negligible entropy and can be ignored.

When neutrinos decouple the temperature of the plasma is $T$ (and this is the temperature of both photons and neutrinos $T_\gamma^{\rm before} = T_\nu = T$). Then electrons and positrons annihalate which increases the temperature to the plasma from to $T_\gamma^{\rm before}$ to $T_\gamma$. Use conservation of entropy to relate $T_\gamma^{\rm before}$ and $T_\gamma$. For this you will need to count the number of relativistic degrees of freedom before (before: e+, e-, and photons) and after (we just have photons) electron-positrons annihalate. All of these have $2$ polarizations (degrees of freedom).

Show that $$T_\nu = T_\gamma^{\rm before} = \left(\frac{4}{11}\right)^{1/3}T_\gamma$$


Evaluating Boltzmann Integrals


In the lectures we enounted integrals like $\int_0^\infty \frac{x^2}{e^{x} \pm 1}dx$ and $\int_0^\infty x^2 e^{-x^2}dx$, but did not bother to solve them. This is your task here.

Show that we can write $$\frac{1}{e^x - 1} = e^{-x}\frac{1}{1-e^{-x}}$$ Recall the geometrical series $\frac{1}{1-r} = 1 + r + r^2 + r^3 + \ldots$. Apply this to the expression above and express the integrand of $\int_0^\infty \frac{x^3}{e^{x} - 1}dx$ as an infinite series. Recall the expression for the $\Gamma$ function $$\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}dx$$ Show that under a change of variables and show that the integral of each of the terms in the infinite series can be put on this form. Evaluate these integrals in terms of the $\Gamma$ function for some value of $s$. You will then be left with an infinite series on the form (the Riemann zeta function) $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$ for some $s$ (if you were to do the same derivation with $\frac{1}{e^x+1}$ instead then the sum will have an addition $(-1)^n$ term and the function is the closely related Dirichlet eta function. The $7/8$ factor difference for the energy density of fermions and bosons comes from the fact that $\eta(4) / \zeta(4) = 7/8$). Use this to give an expression for the integral in terms of these two functions. The values of the $\zeta$ function at even integers have a simple closed form (see e.g. $\zeta(2) = \pi^2/6$ and $\zeta(4) = \pi^4/90$ for possible derivations).

For the second integral lets start by generalizing it to $\int_0^\infty x^2 e^{-\lambda x^2}dx$ where $\lambda=1$ is the value we want. The integral without the $x^2$ term is the well known Gaussian integral $\int_0^\infty e^{-\lambda x^2}dx = \frac{\sqrt{\pi}}{2\sqrt{\lambda}}$. Compute the derivative wrt $\lambda$ of both sides and use this to find the result for the integral we want. What is the result if the integrand is $x^4$?


Photon and neutrino density parameters


We will derive an expression for $\Omega_{\gamma 0}$ and $\Omega_{\nu 0}$ that you will need in milestone I of the numerical project.

The energy density for photons (bosons) and a single neutrino (fermions) is given by $$\rho_\gamma = \frac{g_\gamma}{(2\pi)^3 \hbar^3}\int_{\mathbb{R}^3} \frac{E}{e^{\frac{E}{k_bT}} - 1} d^3{\bf p}$$ $$\rho_\nu = \frac{g_\nu}{(2\pi)^3\hbar^3}\int_{\mathbb{R}^3} \frac{E}{e^{\frac{E}{k_bT}} + 1} d^3{\bf p}$$ where $g_\gamma = g_\nu = 2$ (two internal polarization states) and $E = |{\bf p}|c = pc$. Use spherical coordinates and evaluate these integrals using the results of the previous exercise. Show that the density parameters for photons and massless neutrinos today is given by $$\Omega_{\gamma 0} = 2\cdot \frac{\pi^2}{30} \frac{(k_bT_{\rm CMB 0})^4}{\hbar^3 c^5} \frac{8\pi G}{3H_0^2}$$ where $T_{\rm CMB 0}$ is the temperature of the CMB measured today and $$\Omega_{\nu 0} = N_\nu\cdot \frac{7}{8}\left(\frac{4}{11}\right)^{4/3}\Omega_{\gamma 0}$$ where $N_\nu = 3$ is the number of neutrinos. There is a small correction to this latter formula. Neutrinos have not fully decoupled when electrons and positrons annhialate so their temperature is a tiny bit larger than we estimated. A way to correct for this is to use a slightly larger value for $N_\nu$. A detailed calculation gives $N_\nu = 3.046$. This is called the effective number of relativitic degrees of freedom $N_{\rm eff}$ and is sometimes used as a free parameter and fit to the data (and this can tell us if there are hints of new relativitic species beyond the standard model present in the early Universe if data is not consistent with $3.046$).


Baryon to photon ratio


We are going to estimate how many photons there are per baryon in our Universe. For relativistic bosons the number density is $$n = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} \frac{1}{e^{\frac{p}{T}} - 1} d^3{\bf p} = \frac{g\zeta(3)T^3}{\pi^2}$$ and for non-relativistic particles of mass $m$ we have $\rho \approx m n$. Use this to compute the number-density of photons today ($T_{\rm CMB 0} = 2.725$ K) and the number density of baryons today (assume all baryons are hydrogen atoms with mass $m_H = 939$ MeV and that $\Omega_{b 0} = 0.05$) and use this to estimate the baryon-to-photon ratio $\eta \equiv \frac{n_b}{n_\gamma}$ in our Universe (you will have to restore the constants $\hbar,c,k_b$ to get the number density in $1/{\rm m}^3$ and you can use that the little 'h' in the Hubble constant is $h = 0.7$). We computed it only for today, but how does $n_\gamma,n_b$ and thereby the ratio evolve with time (scalefactor)? As a check you can compare to the exact number $\eta = 2.75\cdot 10^{-8} \Omega_{b 0} h^2$. This number is what models of baryogenesis wants to reproduce and the small ratio has implications for when recombination happen as we will see later in this course.


The ideal gas law


We are going to derive the ideal gas-law $P = nT$ for a non-relativistic gas using the Boltzmann formalism.

The number density and pressure follows from the distribution function as $$n = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} f(p) d^3{\bf p}$$ $$P = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} \frac{p^2}{3E}f(p) d^3{\bf p}$$ where in the low temperature limit both the Bose-Einstein and Fermi-Dirac distribution reduces to the Maxwell one $f(p) \approx e^{-\frac{E(p)-\mu}{T}}$ where $E(p) = \sqrt{p^2 + m^2}$ is the relativistic energy relation and $p = |{\bf p}|$. You can assume the Maxwell distribution in this exercise.

Show that $E \approx m + \frac{p^2}{2m}$ in the non-relativistic limit $p\ll m$. Show that in the non-relativistic limit $T \ll m$ we have $$n \approx g\left(\frac{mT}{2\pi}\right)^{3/2}e^{-\frac{m-\mu}{T}}$$ $$P \approx g\left(\frac{m}{2\pi}\right)^{3/2}\sqrt{T}e^{-\frac{m-\mu}{T}}$$ and therefore $P = nT$ (if we restore $k_b$ and use $n = N/V$ it becomes $PV = Nk_bT$ which is a more familiar form).

To do this derivation remember that the integrand only depends on $p$ (not the direction) so you can use $\int_{\mathbb{R}^3} d^3 {\bf p} = \int_0^\infty 4\pi p^2 dp$ and the non-relativistic approximation for $E(p)$ will also be useful (why can you use this when the integral is over all $p$?). You can also assume $\frac{p^2}{E} \approx \frac{p^2}{m}$ for the pressure. To compute the resulting integral see the previous weeks exercises or simply use the results: $$\int_0^\infty x^2 e^{-x^2}{\rm d}x = \frac{\sqrt{\pi}}{4}$$ $$\int_0^\infty x^4 e^{-x^2}{\rm d}x = \frac{3\sqrt{\pi}}{8}$$


The Boltzmann equation in a smooth Universe



Boltzmann equation basics


We have four particle species $1,2,3,4$ that interact via the process $1+2 \rightleftharpoons 3+4$ and we has number densities $n_i$ and corresponing equilibrium values $n_i^{\rm Eq}$. The Boltzmann equation for species $1$ reads $$\frac{1}{a^3}\frac{d(a^3 n_1)}{dt} = -\alpha n_1 n_2 + \beta n_3 n_4$$ Describe what the different terms in this equation represent physically. Derive an expression for $\beta$ in terms of $\alpha$ and the equilibrium values. Explain why the equation for $n_2$ takes the same form as the one for $n_1$ (and likewise the $n_3$ equation is the same as the $n_4$ equation). In a co-moving volume $V_0$ what does $N_0 = a^3V_0(n_1+n_2+n_3+n_4)$ represent and what is $\frac{dN_0}{dt}$? Use just this to derive an equation for $n_3$, $n_4$.


Boltzmann like equation for a very simple toy model


Here we are just going to use the intuition we have for the Boltzmann equation to write it down for a different process. Let's assume we have two particle species $1$ and $2$ and they can transform into each other $1 \rightleftharpoons 2$ (if you want you can think of $1$ and $2$ as different flavours of neutrinos and this being a very simple toy model for neutrino oscillations). $1$ can turn into $2$ with rate $\alpha$ and $2$ and turn into $1$ with rate $\beta$. Write down a Boltzmann-like equation (on the same kind of form as in the previous exercise) that describes this toy model. There will be two terms: one representing $1$'s turning into $2$' and one representing $2$'s turning into $1$'s. Remember that the total number of particles stays the same. To simplify the equation let $a^3 n_1 = N_1$ and $a^3 n_2 = N_2$ denote the number of particles of each species in some fixed co-moving volume and rephrase it in terms of $N_1$ and $N_2$.

Show that in equilibrium a fraction $\frac{\beta}{\alpha+\beta}$ of the particles will be $1$'s and a fraction $\frac{\alpha}{\alpha+\beta}$ of the particles will be $2$'s.

Solve the Boltzmann equation you put up and show that we always reach the equilibirum within a time of a few $t_*$'s where $t_* = \frac{1}{\alpha+\beta}$. This demonstrates the obvious thing that the higher the rate the quicker equilibrium is reached. However this simple toy model does not have decoupling (and freeze-out) since the particle "interacts" with themselves so in the next exercise we'll do a realistic example with dark matter.


Boltzmann equation for dark matter production


We are going to see what we can learn about the dark matter mass and the cross-section from the observed dark matter abundance for a common dark matter production mechanism. We assume dark matter is in equilibrium with the rest of the standard model in the early Universe with the process $$X + \overline{X} \leftrightharpoons \ell + \overline{\ell}$$ where $\ell$ is some light standard model particle and an overbar denotes the antiparticle. At some point dark matter decouples and freezes out leaving us with some non-zero abundance.

In the radiation dominated regime we have $H^2 \propto \rho \propto g_*T^4$ where $g_* = g_*(T)$ is the number of relativistic degrees of freedom at any time. Expressing $H$ in terms of $T$ is quite useful in the radition dominated Universe and we will use this in this problem. Below you will need $\frac{dT}{dt}$ and for this purpose you can just assume $T\propto a^{-1}$ (this holds except for when $g_*$ changes - i.e. in the short time-frame when a species decouple).

The Boltzmann equation for dark matter thermally produced by annhilation into light standard model particles is given by ($N_X = a^3 n_X$) $$\frac{dN_X}{dt} = -a^{-3}\left<\sigma v\right>(N_X^2 - (N_X^{\rm Eq})^2)$$ Introduce a new time-variable $x = \frac{M_X}{T}$ where $M_X$ is the mass of the dark matter particles and we will here $T$ is the temperature of the primordial plasma. Show that we can write $H = \frac{H(T=M_X)}{x^2}$. Show next that $$\frac{dN_X}{dx} = -\frac{\lambda}{x^2}(N_X^2 - (N_X^{\rm Eq})^2)$$ for some constant $\lambda$. Argue why we can neglect $N_X^{\rm Eq}$ for $T < M_X$ ($x > 1$). If you cannot see why, write down the equilibrium distribution for $T < M_X$. The equation for large $x$ (after decoupling) can therefore be approximated as $$\frac{dN_X}{dx} = -\frac{\lambda}{x^2}N_X^2$$ Solve this equation by integrating it from $x = x_f$ (decoupling) till $x = \infty$ and use derive an expression for the relic dark matter density $N_X^\infty$ assuming $N_X(x_f) \gg N_X^\infty$ which is typically the case.

The dark matter density is $\rho_{\rm DM} = M_X n_X = \frac{M_XN_X}{a^3}$ so the dark matter density parameter today is $$\Omega_X = \frac{M_X N_X^{\infty}}{\rho_{c0}}$$ Derive an expression for this and show that it doesn't depend directly on the mass $M_X$ (only weakly via the freeze-out time $x_f$ and $g_*(T=M_X)$) so the observed dark matter abundance does not tell us that much about the mass of the dark matter particles, but does tell us a lot about the cross-section.


Figure taken from Daniel Baumann's lecture notes.


The Boltzmann equation in a perturbed Universe



The inverse of the perturbed metric


In the problems below we work to first order in perturbation theory so terms like $\Phi^2$ and $\Phi\Psi$ can be ignored as they are second order. A useful relation is $(1+x)^n \approx 1+ nx$ which is valid when $|x|\ll 1$ (so for example $\frac{1}{1+2\Phi} \approx 1 - 2\Phi$). The metric always satisfy $g_{\mu\alpha}g^{\alpha\nu} = \delta^{\nu}_{\alpha}$ ($g\cdot g^{-1} = I$). We have a perturbed metric $g_{\mu\nu} = \overline{g}_{\mu\nu} + \delta g_{\mu\nu}$ where $\overline{g}_{\mu\nu}$ is the background metric (and also satisfy the relation above). Derive a general expression for $\delta g^{\mu\nu}$ in terms of $\delta g_{\mu\nu}$ to first order in perturbation theory (i.e. second order terms in $\delta g$ can be put to zero). Apply this to the metric in the Newtonian gauge $ds^2 = -(1+2\Psi)dt^2 + a^2(1+2\Phi)(dx^2+dy^2+dz^2)$ to compute $\delta g^{\mu\nu}$ and thereby $g^{\mu\nu}$. Figure out the inverse in another way by considering $g_{\mu\alpha}g^{\alpha\nu} = \delta^{\nu}_{\alpha}$ as a matrix equation and use what you know about the inverse of a diagonal matrix. Show that to first order this expression agrees with what you got above.


Trajectories of photons in a perturbed Universe


The left hand side of the Boltzmann equation for photons (we will go through this in the lectures) can be written (chain-rule) $$\frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial p}\frac{d p}{dt} + \frac{\partial f}{\partial \hat{p}^i} \frac{d\hat{p}^i}{dt} + \frac{\partial f}{\partial x^i}\frac{d x^i}{d t}$$ To evaluate this we need to know $\frac{dx^i}{dt}$ and $\frac{dp}{dt}$. We will do this here. Consider the trajectory of a photon, $x^\mu(\lambda)$, where $\lambda$ is parametrizing the path. Since photons move on the light cone we have $g_{\mu\nu}P^\mu P^\nu = 0$ where the 4-momentum of the photon $P^\mu = \frac{dx^\mu}{d\lambda}$, i.e. $P^0 = \frac{dt}{d\lambda}$ and $P^i = \frac{dx^i}{d\lambda}$. Derive an expression for $P^0$ in terms of the metric components and $p$. We define $p^2 \equiv g_{ij}P^i P^j$. Use the chain-rule to derive an expression for $\frac{dx^i}{dt}$ in terms of $P^0$ and $P^i$. We also define the direction of the momentum $\hat{p}^i$ as a unit-norm vector: $\hat{p}^i\hat{p}^j\delta_{ij} = 1$. Use this, plus the definition of $p^2$, to compute what $P^i$ is in terms of $\hat{p}^i$ and the metric components? The geodesic equation for the 4-momentum of the photon is given by $$\frac{dP^\mu}{d\lambda} + \Gamma^\mu_{\alpha\beta} P^\alpha P^\beta = 0$$ Use the geodesic equation for $\mu = 0$ to derive an expression for $\frac{dp}{dt}$. Do this in general (i.e. without evaluating the Christoffel symbols).


Christoffel symbols for perturbed metric


Show that to first order $$\Gamma^0_{00} = \dot{\Psi}$$ $$\Gamma^0_{0i} = \Psi_{,i}$$ $$\Gamma^0_{ij} = a^2 H(1 + 2\Phi - 2\Psi + \frac{\dot{\Phi}}{H})\delta_{ij}$$ Express the last Christoffel symbol in terms of $g_{ij}$ (useful for the next point). Evaluate $\Gamma^\mu_{\alpha\beta} P^\alpha P^\beta$ in the problem above to get an expression for $\frac{dp}{dt}$.


Boltzmann equation for photons


Compute the term $$\frac{\partial f}{\partial p}\frac{d p}{d t}$$ to first order ($f^{(0)} = \frac{1}{e^z-1}$ with $z = \frac{p}{T(t)}$). For this you can use that $$f = f^{(0)} - \frac{\partial f^{(0)}}{\partial z} \frac{p\Theta}{T}$$ to first order and $$\frac{d p}{d t} = -p\left(H + \dot{\Phi} + \frac{\hat{p}^i}{a}\frac{\partial \Psi}{\partial x^i}\right)$$ Compare with the result in Dodelson.


Fourier transform basics


In case you are not familiar with the Fourier transform in this problem we will do some problems related to this. To start with, for simplicity, we will do this in 1D. There the transform is defined via: $$f(x) = \int_{-\infty}^\infty \hat{f}(k) e^{ikx}dk$$ $$\hat{f}(k) = \frac{1}{2\pi}\int_{-\infty}^\infty f(x) e^{-ikx}dx$$ Show that the Fourier transform of $\nabla f$ is $(i\vec{k})$ times the Fourier transform of $f$ by performing the explicit integrals. For simplicity you can do this in 1D (for which $\nabla f = \frac{df}{dx}$). Start with $\frac{d}{dx} f(x)$, multiply by $\frac{e^{-ikx}}{2\pi}$ and integrate over all $x$. Perform integration by parts to move the derivative from $f$ to the exponential (ignore boundary terms - we assume they vanish).

Repeat the calculation for $3D$ for which $$f(\vec{x}) = \int_{-\infty}^\infty \hat{f}(\vec{k}) e^{i\vec{k}\cdot \vec{x}}d^3k$$ $$\hat{f}(\vec{k}) = \frac{1}{(2\pi)^3}\int_{-\infty}^\infty f(\vec{x}) e^{-i\vec{k}\cdot \vec{x}}d^3x$$ You have to do integration by parts for each of the three integrals.

The Fourier transform transforms linear PDEs to (uncoupled) ODEs - one ODE for each wavenumber. Take the Fourier transform (wrt $\vec{x}$) to get the resulting ODEs for the following equations: $$\ddot{u} - c^2\nabla^2 u = 0\,\,\,\text{(wave-equation)}$$ $$\dot{\delta_m} + \nabla\cdot \vec{v} = 0\,\,\,\text{(continuity-eq)}$$

Compute the Fourier transform of the Dirac $\delta$-function (in $1D$ and $3D$). Take the inverse transform to get an integral expression for $\delta(x)$ (in $1D$ and $3D$).

This next problem is not something you will need for this course, but it just demonstrates how the Fourier transform can be useful to solve PDEs. Take the Fourier transform (wrt $x$) of the Poisson equation (again we are in 1D so $\nabla$ is just $\frac{d}{dx}$) $$\nabla^2\Phi(x) = 4\pi G \rho(x)$$ Solve this for $\hat{\Phi}(k)$ in terms of $\hat{\rho}(k)$ and take the inverse transform to get an expression for $\Phi(x)$ as an integral over $k$. Insert the expression for $\hat{\rho}(k)$ in terms of $\rho(x)$ and show that $$\Phi(x) = -2G\int_{-\infty}^{\infty}\left( \int_{-\infty}^{\infty}\frac{e^{ik(x-x')}}{k^2}dk\right)\rho(x')dx'$$ To simplify this further use that $\frac{d^2}{dx^2}|x-x'| = 2\delta(x-x')$ and $\delta(x-x') = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ik(x-x')} dk$. Use this to show that the solution can be written $$\Phi(x) = 2\pi G\int_{-\infty}^{\infty}|x-x'|\rho(x')dx'$$ The equivalent expression in 3D, which can be derived in a similar way, is: $$\Phi(\vec{x}) = -G\int_{-\infty}^{\infty}\frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x'$$ These PDEs can also easily be solved using Greens functions (which is basically what we have found here - just via a different route) so maybe not the best example of how great Fourier transforms are. However numerically Fourier transforms is for sure the best approach for this equation and its solved in four simple steps: 1) take the fourier transform of $\rho$ 2) divide by $k^2$ and 3) Fourier transform back to get $\Phi$ 4) profit!


Boltzmann equation for dark matter I


The Boltzmann equation for dark matter can be written (using the variables $t,E,\hat{p}^i$ and $x^i$) reads $$\frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x^i}\frac{dx^i}{d t} + \frac{\partial f}{\partial E} \frac{dE}{dt} + \frac{\partial f}{\partial \hat{p}^i} \frac{d\hat{p}^i}{dt} = 0$$ Argue why we can drop the last term to first order in perturbation theory. To evaluate this we need to know how dark matter moves, i.e. $\frac{dx^i}{dt}$ and $\frac{dE}{dt}$. This is what we are going to do here.

The momentum 4-vector for dark matter particles $P^\mu = m\frac{dx^\mu}{d\lambda}$ satisfy $$g_{\mu\nu}P^\mu P^\nu = -m^2\tag{1}$$ Just as for photons we define the momentum $p$ and the direction $\hat{p}^i$ via $$p^2 \equiv g_{ij}P^iP^j\tag{2}$$ $$P^i \equiv C\hat{p}^i \,\,\,\text{such that}\,\,\, \hat{p}^i\hat{p}^j\delta_{ij} = 1\tag{3}$$ Use the definitions $(1),(2)$ and $(3)$ above to show that to first order in perturbation theory we have $$P^0 = E(1-\Psi)$$ where $E = \sqrt{p^2 + m^2}$ and $$C = \frac{p}{a}(1-\Phi)$$ Compute $\frac{dx^i}{dt}$ (use the chain-rule) in terms of $p$ and $\hat{p}^i$. Evaluate the $0$-th component of the geodesic equation $$\frac{dP^\mu}{d\lambda} + \Gamma^\mu_{\alpha\beta} P^\alpha P^\beta = 0$$ to get an equation for $\frac{dE}{dt}$. The Christoffel symbols were computed in a previous exercise. Use this to show that $$ \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x^i}\frac{\hat{p}^ip}{aE} - \frac{\partial f}{\partial E}\left[H \frac{p^2}{E} + \frac{p^2}{E}\frac{\partial \Phi}{\partial t} + \frac{\hat{p}^ip}{a}\frac{\partial \Psi}{\partial x^i}\right] = 0$$


Boltzmann equation for dark matter II


Take the zeroth moment of the Boltzmann equation $$ \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x^i}\frac{\hat{p}^ip}{aE} - \frac{\partial f}{\partial E}\left[H \frac{p^2}{E} + \frac{p^2}{E}\frac{\partial \Phi}{\partial t} + \frac{\hat{p}^ip}{a}\frac{\partial \Psi}{\partial x^i}\right] = 0$$ i.e., multiply it by $\frac{d^3p}{(2\pi)^3}$ and integrate it using $$n \equiv \int \frac{d^3 p}{(2\pi)^3} f$$ $$n v^i \equiv \int \frac{d^3 p}{(2\pi)^3} f \frac{p\hat{p}^i}{E}$$ For the integrals that are not straight forward (i.e. following directly from the relations above) you will need to use spherical coordinates (i.e. $d^3p = p^2 dp d\Omega$) and then integration by parts to move the derivative off the distribution function (the relation $\frac{dE}{dp} = \frac{p}{E}$ is useful). Show that we get the continuity equation $$\frac{\partial n}{\partial t} + 3Hn + 3\frac{\partial \Phi}{\partial t}n + \frac{\partial}{\partial x^i}(n v^i) = 0$$ Explain what the different terms in the equation above represents. Solve this for the case of no perturbations (i.e. $\Phi = v^i = 0$, $n = n^{(0)}(t)$). Show that the first order equation for the dark matter density contrast $\delta$ (we set $n = n^{(0)}(1+\delta)$) can be written $$\frac{\partial \delta}{\partial t} + \frac{1}{a}\frac{\partial v^i}{\partial x^i} + 3\frac{\partial \Phi}{\partial t} = 0$$


Boltzmann equation for dark matter III


To close the equation system we also need to know how $v^i$ evolves. We can find an equation for $v^i$ by taking the first moment of the Boltzmann equation. Multiply $$ \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x^i}\frac{\hat{p}^ip}{aE} - \frac{\partial f}{\partial E}\left[H \frac{p^2}{E} + \frac{p^2}{E}\frac{\partial \Phi}{\partial t} + \frac{\hat{p}^ip}{a}\frac{\partial \Psi}{\partial x^i}\right] = 0$$ by $\frac{d^3p}{(2\pi)^3}\frac{p\hat{p}^j}{E}$ and integrate. We assume that the motion is non-relativistic so that $\frac{p}{E}$ (the velocity) is small and we can ignore second order terms $\frac{p^2}{E^2}$ (this assumption is what effectively truncates the Boltzmann hierarchy so that we end up with a closed system - otherwise the equation we get would depend on the next order moment and so on). See Dodelson page 105 for more details if you have problems with some of the integrals.


The dark matter equations in Fourier space


A vector field is irrotational if $\nabla\times \vec{F} = 0$. Take the Fourier transform of this relation and show that $$\hat{\vec{F}} = \frac{\vec{k}}{k} \hat{F}$$ where $\hat{F}$ is the norm of $\hat{\vec{F}}$. Take the Fourier transform of the continuity and Euler equations assuming that the velocity field is irrotational (i.e. $\nabla\times \vec{v} = 0$) $$\frac{\partial \delta}{\partial t} + \frac{1}{a}\frac{\partial v^i}{\partial x^i} + 3\frac{\partial \Phi}{\partial t} = 0$$ $$\frac{\partial v^i}{\partial t} + Hv^i + \frac{1}{a}\frac{\partial \Psi}{\partial x^i} = 0$$ to obtain two scalar equations for $\hat{\delta}$ and $\hat{v}$ (the norm of $\hat{v}^i$).


Boltzmann equation for baryons


Write down (schematically) the Boltzmann equations for cosmological baryons (for simplicity we assume we only have protons and electrons). What are the relevant interactions that takes place?

To get an equation for how the number density of baryons evolve we take the zeroth moment of the Boltzmann equation (just as for dark matter). Explain why this equation is the same as for cold dark matter (i.e. why none of the interactions contribute to this equation).

The equation for the baryon velocity $v_b$, found from taking the first moment, is (here in Fourier space) $$\frac{\partial v_b}{\partial t} + Hv_b + \frac{ik\Psi}{a} = \frac{\partial\tau}{\partial t} \frac{4\rho_\gamma}{3\rho_b}(3i\Theta_1 + v_b)$$ This has a contribution from Compton scattering, but not from Coulomb scattering. What role does Coulomb scattering play in all of this? Why is the collision term small if the baryon density (i.e. the proton mass) is huge?


The perturbed Einstein equations



The perturbed Einstein equations: The left hand side


To evaluate the perturbed Einstein equations we first need the Christoffel symbols for the metric $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu = -(1+2\Psi)dt^2 + a^2(1+2\Phi)(dx^2+dy^2+dz^2)$$ Show that to first order in perturbation theory we have $$\Gamma^0_{00} = \Psi_{,0}$$ and $$\Gamma^0_{ij} = a^2\delta_{ij}[H + 2H(\Phi-\Psi) + \Phi_{,0}]$$ The rest of the symbols are given by $$\Gamma^i_{00} = \frac{1}{a^2}\Psi_{,}^{i},\,\, \Gamma^0_{0i} = \Psi_{,i},\,\,\Gamma^i_{j0} = \delta_{ij}(H+\Phi_{,0})$$ and $$\Gamma^i_{jk} = [\delta_{ij}\Phi_{,k} + \delta_{ik}\Phi_{,j} - \delta_{jk}\Phi_{,i}]$$ Compute the $R_{00}$ component of the Ricci tensor $$R_{00} = \Gamma^\mu_{00,\mu} - \Gamma^\mu_{0\mu,0} + \Gamma^\mu_{\nu\mu}\Gamma^\nu_{00} - \Gamma^\mu_{\nu 0}\Gamma^\nu_{0\mu}$$ Do this calculation in Fourier space (i.e. compute $\hat{R}_{00}$; this just amounts to replacing $A_{,i}\to ik_i \hat{A}$ and for terms without spatial derivatives $B\to \hat{B}$). The rest of the components can be found in Dodelson.


The perturbed Einstein equations: The right hand side


The general expression for the energy momentum tensor from the distribution function of species $i$ is given by $$ T^\mu_\nu = g_i \int \frac{dP_1 dP_2 dP_3}{(2\pi)^3\sqrt{-g}} \frac{P^\mu P_\nu}{P^0} f_i(t,\vec{x},\vec{p}) $$ where $P^\mu = \frac{dx^\mu}{d\lambda}$ is the 4-momentum and $g$ is the determinant of the metric.

The $00$ component for photons is given by the usual expression (using $p^2 = g^{ij}P_iP_j$ etc.) $$ T^0_0 = g_\gamma \int \frac{d^3 p}{(2\pi)^3} E(p) f_i(t,\vec{x},\vec{p}) $$ where $E=p$. Evaluate the $00$ component for photons to first order in perturbation theory for which $f = f^{(0)}(t,p) - p\frac{\partial f^{(0)}(t,p)}{dp}\Theta(t,\vec{x},\hat{p})$. To do this note that $\int d^3 p = \int p^2 dp d\Omega$, that $f^{(0)}$ does not depend on $\hat{p}$ and recall the definition of the monopole $\Theta_0 = \frac{1}{4\pi}\int \Theta d\Omega$ and recall that whenever we have an integral of derivatives of the distribution function then integration by parts is the way to go. You can use that $T^0_0 = -\overline{\rho}_\gamma$ when $\Theta = 0$ (i.e. you don't have to evaluate any integrals explicitly).

For dark matter and baryons we perturbed the number density directly, i.e. $$n = n^{(0)}(1+\delta)$$ The energy density for massive particles is simply $\rho \approx mn$ in the non-relativistic limit where $m$ is the mass of the particles (recall $\rho \sim \int E f$ and for the relevant parts of the integral $E\sim m$ so $\rho \sim m \int f = mn$). Express $T^0_0$ for dark matter and baryons in terms of the background energy density and the perturbation.

Use this to write down $T^0_0$ for the sum of all components we have in our Universe. Split it into a background part and a peturbation part. This component is needed to find an evolution equation for the metric potential $\Phi$ using Einsteins equation (as done in Dodelson). The left hand side of the Einstein equation is $G^0_0 = \overline{G}^0_0 + \delta G^0_0$ where the perturbation (in Fourier space - skipping the hat over the metric potentials - its too much typing) is $$\delta \hat{G}^0_0 = -6H\Phi_{,0} + 6\Psi H^2 - 2\frac{k^2\Phi}{a^2}$$ Use the results above to find an equation for how the metric potientials evolve. Check that this agrees with expectations: if we have a static Universe ($a=1$, $H=0$ and $\Phi_{,0}=0$) then the perturbed metric is nothing but the Newtonian limit of General Relativity (small perturbations about Minkowski space) for which $\Phi$ should just follow the usual Newtonian Poisson equation (but the right hand side will have a negative sign which is just due to how we defined $\Phi,\Psi$).

To find the second equation for the metric potentials we evaluate $P_i^j (\hat{G}^i_j - 8\pi G \hat{T}^i_j) = 0$ where $P_i^j = \frac{k^ik_j}{k^2} - \frac{1}{3}\delta^i_j$ is a projection tensor and a 'hat' means we consider this equation in Fourier space. As shown in Dodelson we have $$P_i^j \hat{G}^i_j = \frac{2k^2(\Phi + \Psi)}{3a^2}$$ Evaluate the Einstein equation above to first order in perturbation theory in terms of Legendre multipoles of $\Theta$ using that for photons $$T^i_j = g_\gamma\int \frac{d^3 p}{(2\pi)^3}\frac{p\hat{p}^ip\hat{p}_j}{E(p)}f$$ You only have to consider photons (and massive neutrions, but that is the same calculation). Use that (in Fourier space $\Theta$ only depends on $\mu = \cos\theta$ not the angle $\phi$ so the last integral is trivial) $\int d^3 p = \int_0^\infty p^2 dp \int_{-1}^1d\mu \int_0^{2\pi}d\phi$ where $\mu \equiv \frac{\vec{k}\cdot \hat{p}}{k}$ and the definition of the Legendre multipoles $\Theta_\ell = \frac{1}{(-i)^\ell}\int_{-1}^1 \frac{d\mu}{2}\Theta P_\ell(\mu) d\mu$. Note that $P_2(\mu) = \frac{3\mu^2-1}{2}$ and recall the orthogonality relation $\int_{-1}^1 P_\ell(\mu)P_{\ell'}(\mu)d\mu = \frac{2\delta_{\ell\ell'}}{2\ell+1}$. The expression for $f$ is the same as given above. From the result for photons explain why dark matter and baryons does not contribute any significant amount here.


Initial conditions and inflation



Estimating how much inflation we need


Assuming that inflation happened roughly at the GUT scale $k_bT \sim 10^{15}$ GeV (where $k_b \sim 10^{-13}$ GeV/K). Use how the temperature of the primordial plasma evolves to estimate the scale-factor $a_E$ after inflation ended.

The Hubble radius is $r = \frac{1}{aH}$. To solve the Horizon problem we require that the whole observable Universe fits into the Hubble sphere before inflation happened, i.e. $$r_I = \frac{1}{a_I H_I} > \frac{1}{H_0} = r_0$$ At the end of inflation $a=a_E$ the Universe is radiation dominated and for simplicity you can assume that the Universe only has radiation (even today). Estimate $\log(a_E / a_I)$, i.e. how many e-folds of inflation do we need to solve the Horizon problem. You can assume that $H_E = H_I$, i.e. that the Hubble factor at the end of inflation is the same as the Hubble factor when inflation started. Notice the symmetry in the result (i.e. in the relation between $a_E/a_I$ to $a_0/a_E$ that you found), explain this based on how $r$ evolves (see figure 2.4 of Baumann for an intuitive picture).


The flatness problem


We saw that curvature enters the Friedmann equation as a term $$\Omega(a) = 1 - \Omega_k(a)$$ where $\Omega(a)$ is the critical density (i.e. $\Omega(a) = 1$ when the Universe is perfectly flat) and $\Omega_k(a) = \frac{\Omega_{k0}}{a^2H^2/H_0^2}$. The Universe we observe is very close to flat, i.e. $|\Omega_{k0}| \ll 1$. For simplicity we assume the Universe only has curvature and radiation. How does $\Omega_k(a)$ evolve with $a$ in a radiation dominated Universe? If the Universe has some curvature so $\Omega_{k0} \not=0$ then how close to flat did it have to be in the very early Universe (say $a = a_E \sim 10^{-28}$)?

Why is this considered a problem? Turned around: if $\Omega_k(a_E)$ has a "reasonable" value (i.e. not super close to $0$), what would the curvature of the Universe be today?

Compute how $\Omega_k(a)$ evolve during inflation ($H = $ const) and explain how inflation solves the flatness problem. Also give an "intuitive" answer, i.e. without having to rely on the mathematics.


Relating the initial perturbations to inflation


Here we will go through how to set the initial conditions for $\Phi,\Psi$ (from which all the other initial conditions are given in terms off). This is to get the same normalization as used when quoting results from CMB experiments.

Inflation sets up curvature perturbation $\mathcal{R}$ with a power-spectrum $$\left<\mathcal{R}_k(\vec{k})\mathcal{R}_k(\vec{k}')^*\right> = (2\pi)^3\delta(\vec{k}-\vec{k}')P(k)$$ where $P(k) = \frac{2\pi^2}{k^3}\mathcal{P}(k)$ and $\mathcal{P}(k) = A_s(k/k_{\rm pivot})^{n_s-1}$ is the usual expression to see for the initial power-spectrum as is standard in the litterature and used by CMB experiments. The curvature perturbations in the Newtonian gauge is given by (see Baumann, but note he uses a different sign convention for $\Phi$ and some assumptions that is equivalent to assuming no neutrinos) $$\mathcal{R} = \Phi + \frac{\mathcal{H}(\Phi' - \mathcal{H}\Psi)}{4\pi G a^2 (\overline{\rho} + \overline{P})}$$ and can be shown to not evolve when outside the horizon. When solving it numerically we want to set the initial conditions such that $\mathcal{R} = 1$ at the initial time. That way, we only have to multiply $|\Theta_\ell|^2$ by $\mathcal{P}(k)$ when computing the $C_\ell$'s. Compute what the initial conditions for $\Phi$ and $\Psi$ should be. If your code does not include neutrinos then you can ignoring them here (however neutrinos does change the result). Here are some pointers:

Show using the Einstein equation for $\Phi$ to find the solution on super-horizon scales. Conclude that $\Phi$ is also frozen when outside the horizon.

The Einstein equation relating $\Phi$ and $\Psi$ is $$\Phi+\Psi = -\frac{12H_0^2}{k^2a^2}(\Omega_{\gamma 0} \Theta_2 + \Omega_{\nu 0} \mathcal{N}_2)$$ See Dodelson Exercise 2 in Chapter 6 for how to compute the neutrino quadrupole initial condition if you are including neutrinos. Use this to relate $\Phi$ and $\Psi$. The photon quadrupole is practically zero early on, why? (See the Boltzmann equation for photons).


Spherical harmonics


Use the orthogonality condition $$\int d\Omega_{\hat{p}} Y_{\ell m} Y^*_{\ell' m'} = \delta_{\ell\ell'}\delta_{mm'}$$ and the expansion of Legendre polynomials in spherical harmonics $$\mathcal{P}_\ell(\hat{a}\cdot \hat{b}) = \frac{4\pi}{2\ell+1}\sum_{m=-\ell}^\ell Y_{\ell m}(\hat{a}) Y^*_{\ell m}(\hat{b})$$ to derive the identity $$\int d\Omega_{\hat{p}} \mathcal{P}_\ell(\hat{k}\cdot\hat{p}) Y_{\ell'm}^*(\hat{p}) = \delta_{\ell\ell'} \frac{4\pi}{2\ell+1} Y_{\ell m}^*(\hat{k}) $$


Line of sight integration


The Boltzmann equation for the photon temperature perturbations (before taking multipoles) can be written $$\frac{d\Theta}{d\eta} + ik\mu \Theta = -\frac{d\Phi}{d\eta} - ik\mu \Psi - \frac{d\tau}{d\eta}\left(\Theta_0 - \Theta + i\mu v_b - \frac{1}{2}\mathcal{P}_2(\mu)\Pi\right)$$ Integrate this equation to obtain an equation for $\Theta$ as an integral over $\Theta_0,\Psi,\Phi,\tau,\Pi$ (but not $\Theta$ itself). Take the multipoles of this expression and show that we get $$\Theta_\ell = \int_0^\infty S(k,\eta) j_\ell(k(\eta_0-\eta))d\eta$$ (TODO: Make this into a proper problem. Add the same for neutrinos and polarization)