# Table of contents

- Friedman equations for a flat Universe
- Energy conservation
- Simplified form for the Friedman equation
- Curvature in the Friedman equations
- Extra (long calculation): Friedman equations for a curved Universe
- Lifetime of the Universe
- Scalefactor as function of time
- Approximations for $\eta$

- Boltzmann equation basics
- Boltzmann like equation for a very simple toy model
- Boltzmann equation for dark matter production

- The inverse of the perturbed metric
- Trajectories of photons in a perturbed Universe
- Christoffel symbols for perturbed metric
- Boltzmann equation for photons
- Fourier transform basics
- Boltzmann equation for dark matter I
- Boltzmann equation for dark matter II
- Boltzmann equation for dark matter III
- The dark matter equations in Fourier space
- Boltzmann equation for baryons

- The perturbed Einstein equations: The left hand side
- The perturbed Einstein equations: The right hand side

- Estimating how much inflation we need
- The flatness problem
- Relating the initial perturbations to inflation

# 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

### Friedman equations for a flat Universe

In the lectures we derived the Einsteins equation for the flat FRLW metric in Cartesian $(t,x,y,z)$ coordinates $ds^2 = -dt^2 + a^2(dx^2+dy^2+dz^2)$, but we did not compute $$R_{ij} = \Gamma^\alpha_{ij,\alpha} - \Gamma^\alpha_{i\alpha,j} + \Gamma^\alpha_{\beta\alpha}\Gamma^\beta_{ij} - \Gamma^\alpha_{\beta i} \Gamma^\beta_{j\alpha}$$ and we only looked at the $00$ component of the Einstein equations. We found that the non-zero Christoffel symbols are $\Gamma^{0}_{ij} = a\dot{a}\delta_{ij}$ and $\Gamma^{i}_{j0} = \frac{\dot{a}}{a}\delta^i_j$. We found that $R_{00} = -3\frac{\ddot{a}}{a}$. Derive $R_{ij}$ and evaluate the remaining Einstein equations, i.e. the $0i$ component and the $ij$ components. Show that this gives us 1 more independent equation and that this can be written (by using the first Friedmann equation $\frac{\dot{a}^2}{a^2} = \frac{8\pi G\rho}{3}$) as $\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho + 3p)$.

### Energy conservation

Evaluate the conservation equation $\nabla_\mu T^{\mu\nu} = 0$ (this represents $4$ equations, one for each of $\nu = 0,1,2,3$, but only the $0$-equation is non-trivial). For this you will need the expression for the covariant derivative of a tensor and the expression for the Christoffel symbols. See the GR lecture notes for this. Also recall the EM tensor for a perfect fluid $T^{\mu\nu} = (\rho+p)u^\mu u^\nu + g^{\mu\nu}p$ where $u^\mu = (1,0,0,0)$ is the 4-velocity of the fluid.

### 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 $w_i = p_i/\rho_i$. Solve the differential equation and show that we can write $\rho_i = \frac{\rho_{i0}}{a^{3(1+w_i)}}$ 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 mater $m$ ($w=0$), radiation $r$ ($w=1/3$), masseless neutrinos $\nu$ ($w=1/3$) and a cosmological constant $\Lambda$ ($w=-1$) show that this becomes $$H^2 = H_0^2\left( \frac{\Omega_{b0}}{a^3} + \frac{\Omega_{m0}}{a^3} + \frac{\Omega_{r0}}{a^4} + \frac{\Omega_{\nu 0}}{a^4} + \Omega_{\Lambda 0}\right)$$ This is the form you will implement in the numerical project.

### 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 $\rho_k = -\frac{3k}{8\pi Ga^2}$ and find a corresponding expression for $p_k$ (and thereby $w_k$) and show that this gives the same equations as the Friedman equations without curvature, but with this new "energy" component $k$.

### 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 $X$ with equation of state $w$, i.e. $H^2 = \frac{8\pi G \rho_X}{3}$ with $\rho_X \propto a^{-3(1+w)}$. 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 $H_0 = H(a=1)$.

### 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 $H^2 \approx H_0^2 \frac{\Omega_r}{a^4}$. 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 $$T_\nu = T_\gamma^{\rm before} = \left(\frac{4}{11}\right)^{1/3}T_\gamma$$

### Some math...

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).

From the exact expression for this integral and the expression for $\rho$ we showed in the lecture you can show that the density parameter today is given by $$\Omega_r = 2\cdot \frac{\pi^2}{30} \frac{(k_bT)^4}{\hbar^3 c^5} \frac{8\pi G}{3H_0^2}$$ where the $2$ is $g$, $T$ is the temperature of the CMB measured today (and if you combine this with the result for the neutrino temperature derived above and you will get the expression for $\Omega_\nu$ that I just wrote up in the lectures).

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 $x^4$?

### Thermodynamical calculations

Here are some exericises to see if you are able to do some basic thermodynamical calculation.

The number density follows from the distribution function as $$n = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} f(p) d^3{\bf p}$$ 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 $p = |{\bf p}|$.

Derive the relation we have used again and again: in the non-relativistic limit $T \ll m$ we have $$n \approx g\left(\frac{mT}{2\pi}\right)^{3/2}e^{-\frac{m-\mu}{T}}$$ To do this derivation remember that the integrand only depends on $p$ (not the direction) so you can use $\int_{\mathbb{R}^3} d^3 {\bf p} = \int_0^\infty 4\pi p^2 dp$. The non-relativistic approximation $E(p) \approx m + \frac{p^2}{2m}$ is also quite useful here. To compute the resulting integral see the previous weeks exercises or simply use the result $\int_0^\infty x^2 e^{-x^2}{\rm d}x = \frac{\sqrt{\pi}}{4}$

### Derive the ideal gas law

The pressure and number density is given by $$P = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} \frac{p^2}{3E}f(p) d^3{\bf p}$$ $$n = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} f(p) d^3{\bf p}$$ Show that in the non-relativistic limit $P \propto nT$ (see the previous problem for the calculation of $n$).

## 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 lets do a realistic example with dark matter.

### Boltzmann equation for dark matter production

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 $\lambda$. Write down the equilibrium distribution for $T < M_X$ ($x > 1$) in terms of $x$ and show that we can approximate it for large $x$ (after freeze-out) as $$\frac{dN_X}{dx} = -\frac{\lambda}{x^2}N_X^2$$ Solve this equation by integrating it from $x = x_f$ (freeze-out) 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 $M_X$ - but it depends on the cross-section (plus the freeze-out time $x_f$ and $g_*(T=M_X)$).

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 $\hat{v}^i$). (You can also get ridd of the '$i$' by redefining $i\hat{v}\to \hat{v}$).

### 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 $$\Phi+\Psi = -\frac{12H_0^2}{k^2a^2}(\Omega_R \Theta_2 + \Omega_\nu \mathcal{N}_2)$$ 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)