Table of contents

Back to the main page

List of equations Christoffel symbols, tensors etc. that might be useful when solving problems Old Exams
1 Problems relevant for Milestone I - Background Cosmology

2. Problems relevant for Milestone II - Thermodynamics/statistical mechanics and the thermal history of our Universe

3. Problems relevant for Milestone III - Cosmological perturbation theory

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

5. Problems relevant for Milestone IV - From perturbations to statistical observables

Old Exams

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


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.

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).

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 $\gamma$ ($w=1/3$), 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 in the different eras

It's often hard to solve equations exactly in cosmology (which is why this course is numerical), however we can still learn a lot from simple estimates where we just use the scaling of different quantities in different regimes. Analytical estimates are also good to testing that whatever you solve for numerically is done correctly. This problem is to get some experience in that.

Density parameters:
We have three eras in our Universe: the radiation dominated, matter dominated and dark energy dominated eras. What is the equation of state $w$ in each of these eras (for the dominating component)? How can we roughly approximate $\Omega_i(a)$ in the different eras?

Scaling of $\rho$:
How does $\rho(a)$ evolve for the different components (baryons, dark matter, dark energy, radiation)? For the general scaling when we have a constant $w$ see a previous problem.

Scaling of $H$:
In the radiation dominated era we can approximate $H^2 \approx \frac{\Omega_{R0}}{a^4}$ or just simply $H \propto \frac{1}{a^2}$. How can you estimate $H$ in 1) the matter dominated era and 2) the dark energy dominated era? What is $\frac{a}{H}\frac{dH}{da}$ approximately equal to in these eras? Do the same for the conformal Hubble factor $\mathcal{H} = aH$ and estimate $\frac{a}{\mathcal{H}}\frac{d\mathcal{H}}{da}$ in the different regimes (this will be useful for testing your result in the numerical project).

Scaling of $t$:
In the radiation era $H^2 \propto 1/a^4 \to \dot{a} \propto 1/a$ so taking $a \propto t^n$ we see that $n=1/2$ and $a\propto t^{1/2}$. What is the scaling of $a(t)$ in the different eras?

Scaling of $\eta$:
The conformal time is given by $\frac{d\eta}{da} = \frac{1}{a^2H}$. Show that $\eta \approx \frac{1}{aH} = \frac{1}{\mathcal{H}} \propto a$ in the radiation dominated era.

Change of coordinates:
Often equations are more easy to solve when working with $a$ or $\eta$ (defined via $dt=ad\eta$) or $x=\log a$. Use the chain rule to express $\frac{d}{da}$ in terms of $\frac{d}{dt}$. Also express $\frac{d}{dx}$ and $\frac{d}{d\eta}$ in terms of $\frac{d}{dt}$. The conformal Hubble factor is defined in the same way as the normal Hubble factor (which has a $t$ derivative of $a$ instead): $$\mathcal{H} = \frac{1}{a}\frac{da}{d\eta}$$ Show that $\mathcal{H} = aH$. Show that expressing the continuity equation $$\frac{d\rho}{dt} + 3H(1+w)\rho = 0$$ in terms of $\eta$ gives the same equation just with $\mathcal{H}$ in place of $H$. What is the continuity equation in terms of $x$? Use this to verify that the solution is $\rho \propto e^{-3(1+w)x} = 1/a^{3(1+w)}$.

A problem using the above:
Later in this course we will meet the growth equation for how fast matter perturbations grows: $$\frac{d^2\delta}{dt^2} + 2H\frac{d\delta}{dt} = 4\pi G\rho_M(a) \delta$$ Express this equation in terms of $x$ and show that $$\frac{d^2\delta}{dx^2} + \left(2 + \frac{\frac{dH}{dx}}{H}\right)\frac{d\delta}{dx} = \frac{3}{2}\Omega_M(a) \delta$$ What is $\frac{\frac{dH}{dx}}{H}$ in the radiation dominated era? How can we approximate $\Omega_M(a)$ in the same era? Use this to derive an approximative solution for $\delta$ and show that it can be expressed as $\delta \approx A + B\log(a)$ for some constants $A,B$. What is the equivalent approximations you can make in the matter dominated era? Use this to show that the solution in the matter era is $\delta \propto a$. This tells us that matter perturbations grows very slowly in the radiation era and we have to wait till the Univers becomes matter dominated before density perturbations really start to grow. This will be important when we try to understand the matter power-spectrum later in this course.

Einstein's static Universe

Show that in a Universe with curvature, a cosmological constant and normal matter its mathematically possible to have a static Universe. What is the relation between $\rho_k \equiv -\frac{3k}{8\pi G a^2}$, $\rho_M$ and $\rho_\Lambda$ in such a Universe? Hint: use the two Freidman equations and enforce $\dot{a} = \ddot{a} = 0$. You can take $a=1$.

If we add a tiny perturbation $\rho_M \to \rho_M + \delta\rho_M$ where $\rho_M$ is the density you found above and $\delta\rho_M$ is a constant what happens to the resulting Universe regardless of how small $\delta\rho_M$ is? If we introduce spatial perturbations, what do you think would happen to regions where we have some over- or under-densitity? Use this to explain what the problem with this "static" Universe model is.

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 (here also including the physical constants we usually put to unity) $$\rho_\gamma = \frac{g_\gamma}{(2\pi)^3 \hbar^3}\int_{\mathbb{R}^3} \frac{E}{e^{\frac{E}{k_bT}} - 1} d^3{\bf p}$$ $$\rho_\nu = \frac{g_\nu}{(2\pi)^3\hbar^3}\int_{\mathbb{R}^3} \frac{E}{e^{\frac{E}{k_bT}} + 1} d^3{\bf p}$$ 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 $3.046$).

Note that the combination $\Omega_{\gamma 0} h^2$ and $\Omega_{\nu 0} h^2$ only depends on physical constants and the (very well) measured value of the temperature so this is known to high precision so in effect this combination is not really a free parameter. The degeneracy between $\Omega$'s and $h$ in the Hubble equation means its often convenient to work with this combination of parameters called "physical density parameters".

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 $\eta = 2.75\cdot 10^{-8} \Omega_{b 0} h^2$, i.e. there are over a billion photons per baryon! 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 and the equation of state for matter

We are going to derive the ideal gas-law $P = nT$ for a non-relativistic gas using the Boltzmann formalism and thereby showing that the equation of state for such a gas is $w\approx 0$.

The energy density, number density and pressure follows from the distribution function as $$\rho = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} E(p)f(p) d^3{\bf p}$$ $$n = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} f(p) d^3{\bf p}$$ $$P = \frac{g}{(2\pi)^3}\int_{\mathbb{R}^3} \frac{p^2}{3E}f(p) d^3{\bf p}$$ 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). Compare the integral for $\rho$ (using the non-relativistic energy relation) to that for $P$ and $n$ and show that $$\rho \approx mn + \frac{3}{2}P = mn + \frac{3}{2}nT \approx mn$$ and note that this gives us that the equation of state for the gas $$w = \frac{P}{\rho} = \frac{T}{m} \ll 1$$ Assuming the baryon temperature to be the same as that of photons and taking the mass to be the hydrogen mass we get $w \approx 10^{-15}$ today. This is the justification for $w = 0$ that we presented without proof in the lectures.

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.

Constraining the mass of the neutrinos

The aim here is to set an upper limit on the sum of the neutrino masses. In the early Universe the neutrinos are effectively massless. Show that the number density of a single flavour of neutrinos is given by $n_\nu = \frac{3\zeta(3)}{2pi^2}T_\nu^3$ (you can look up the integrals). Use this and the result for the neutrino temperature in a previous exercise to compute how many neutrinos there are per ${\rm cm}^3$ today (you should get a number that is between $10$ and $1000$). Neutrinos have mass and eventually they become (or will become) non-relativistic. There are $3$ flavours of neutrinos with masses $m_1,m_2,m_3$. Assume they are all non-relativistic today and compute the total energy density in neutrinos $\rho_\nu$ in terms of the sum of neutrino masses $\sum m_i = m_1+m_2+m_3$. What is the numerical value of the massive neutrino density parameter today, $\Omega_{m\nu 0} \equiv \frac{\rho_{\nu 0}}{\rho_{c0}}$, as function of $\frac{\sum m_i}{100 h^2 eV}$. Observations constrain the total energy density that is in form of non-relativistic matter is $\lesssim 0.3$. Use this and your previous results to set an upper limit on the sum of the neutrino masses. Finally be sure to check all the assumptions (e.g. if they happen to not be non-relativistic today what can you say about the mass?).

Boltzmann equation for dark matter production

If you want a (much more extensive) problem (both theoretical and numerical) on dark matter freeze-out than what we do here then take a look at the project given in the course AST3220.

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 ($N_X = a^3 n_X$) $$\frac{dN_X}{dt} = -a^{-3}\left\lt\sigma v\right\gt(N_X^2 - (N_X^{\rm Eq})^2)$$ 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 $T \leq M_X$ ($x \geq 1$). If you cannot see why, write down the equilibrium distribution for $T \leq M_X$. 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.

Exam problem - Recombination with Helium in the Saha approximation

In our treatment of recombination in the lectures we assumed there only was hydrogen in our Universe. However there is also a sizable amounts of helium in our Universe. Helium comes in the forms ${\rm He^0}$, ${\rm He^+}$ and ${\rm He^{++}}$ (i.e. neutral, singly ionized an doubly ionized helium) with corresponding masses $m_{\rm He^0}$, $m_{\rm He^+}$ and $m_{\rm He^{++}}$. The mass ratio of helium to hydrogen that is generated in big bang nucleosynthesis is given by $$Y_p \equiv \frac{4n_{\rm He}}{n_B}$$ where $n_{\rm He} = n_{\rm He}^0 + n_{\rm He}^{+} + n_{\rm He}^{++}$. Likewise hydrogen comes in the two forms $H^0$ and $H^+$ and $n_{\rm H} = n_{\rm H^0} + n_{\rm H^+}$ with corresponding masses $m_{\rm H^0}$ and $m_{\rm H^+}$ and the total baryon number density is given by $n_B = 4n_{\rm He} + n_{\rm H}$ (which basically is the number density of protons/neutrons). In our Universe $Y_p \approx \frac{1}{4}$.

The interactions between helium, hydrogen and electrons relevant for recombination is the following: $${\rm He}^{++} + e^{-} \rightleftharpoons {\rm He}^{+} + \gamma$$ $${\rm He}^{+} + e^{-} \rightleftharpoons {\rm He}^{0} + \gamma$$ $${\rm H}^{+} + e^{-} \rightleftharpoons {\rm H}^{0} + \gamma$$

You are in this problem going to derive a closed system of equations for recombination including both hydrogen and helium in the Saha approximation. You can use without proof that the equilibrium distribution $n_i^{\rm Eq} = g_i\left(\frac{m_i T}{2\pi}\right)^{3/2}e^{-\frac{m_i - \mu_i}{T}}$ for a non-relativistic particle in the low temperature limit and that $g_e = 2$, $\frac{g_{\rm He^{++}}}{g_{\rm He^{+}}} = 2$, $\frac{g_{\rm He^{+}}}{g_{\rm He^{0}}} = 1$ and $\frac{g_{\rm H^{+}}}{g_{\rm H^{0}}} = \frac{1}{2}$.

  • a) Write down the Saha equations for the first two interactions above. You are free to assume that the interactions are in chemical equilibrium such that $\mu_1 + \mu_2 = \mu_3 + \mu_4$ for a $1+2 \rightleftharpoons 3+4$ process. Simplify the equations you find where possible and explain the approximations you are making. The Saha equation for hydrogen, the last ineraction above, was derived in the lectures and the result we found was $$\frac{n_{\rm H^{+}} n_{\rm e^-}}{n_{\rm H^{0}}} = \frac{g_{\rm H^{+}}g_e}{g_{\rm H^{0}}}\left(\frac{m_eT}{2\pi}\right)^{3/2}e^{-\frac{\epsilon_0}{T}} = \left(\frac{m_eT}{2\pi}\right)^{3/2}e^{-\frac{\epsilon_0}{T}}$$ where $\epsilon_0$ is the ionization energy of hydrogen.
  • b) Introduce the ionization fractions $x_{\rm He^{++}} = \frac{n_{\rm He^{++}}}{n_{\rm He}}$, $x_{\rm He^+} = \frac{n_{\rm He^+}}{n_{\rm He}}$ and $x_{\rm H^+} = \frac{n_{\rm H^+}}{n_{\rm H}}$ and show that the system above can be written as a closed system for the three $x_i$'s (i.e. the equations should only depend on $T$, $n_{\rm e^-}$, the $x_i$'s and physical constants). For hydrogen the result we derived in the lectures reads $$\frac{x_{\rm H^{+}} n_{\rm e^-}}{1-x_{\rm H^{+}}} = \left(\frac{m_eT}{2\pi}\right)^{3/2}e^{-\frac{\epsilon_0}{T}}$$
  • c) The final part is to obtain an expression for the electron number density. What is the number density of free electrons in terms of $n_{\rm He}^{+},n_{\rm He}^{++},n_{\rm H}^{+}$ and what is the assumption you are making to get this? Use this to express $n_{\rm e^-}$ in terms of $n_B$, i.e. $n_{\rm e^-} = Fn_B$ for some $F$ depending on the $x_i$'s. Use this to express the system above in terms of $n_B$ and $F$ and write it on such a way that the left hand side depends only depends on the $x_i$'s (and $F$) and the right hand side consists only of known quantities. As a check that you got the correct result show that when there is no helium in the Universe (i.e. $Y_p = 0$ and in this case $F = x_{\rm H^+}$) then one of the equations above reduces to the result we derived in the lectures $$\frac{X_e^2}{1-X_e} = \frac{1}{n_B}\left(\frac{m_eT}{2\pi}\right)^{3/2}e^{-\frac{\epsilon_0}{T}}$$ where $X_e = \frac{n_{\rm e^-}}{n_{\rm H}}$.
  • d) What we are mainly interested in for recombination is the free electron fraction $X_e \equiv \frac{n_{\rm e^-}}{n_H}$ as this determines the optical depth. In the figure below we show a sketch for the evolution of $X_e$ as function of temperature. Explain from the equations you have derived (and give a physical reason for it) this evolution; what do the three regimes seen in the figure correspond to? If you did not manage the last problem try to explain the evolution on purely physical grounds.

    Figure: Evolution the free electron fraction as function of temperature. This is just a sketch and the numbers does not correspond to when the transitions happen in our Universe.

    Derive a formula for the value of $X_e$ in terms of $Y_p$ at the three plateaus you see (the three dashed lines). For this discussion you can use that $m_{\rm He^{++}} + m_{\rm e} - m_{\rm He^{+}} \approx 55$ eV, $m_{\rm He^{+}} + m_{\rm e} - m_{\rm He^{0}} \approx 25$ eV and $\epsilon_0 = m_{\rm H^{+}} + m_{\rm e} - m_{\rm H^{0}} \approx 13$ eV. Hint: You are not meant to try to solve the equations exactly, but rather find approximate solutions for the $x_i$'s in the different regimes either from the equations or from physical reasoning. Recall that if $T\ll T_*$ then $e^{-T_*/T} \approx 0$.

The Boltzmann equation in a perturbed Universe

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 $\frac{1}{1+2\Phi} \approx 1 - 2\Phi$).

Working in perturbation theory

When we do calculations in perturbation theory we can ignore terms that are small. If we have some function $f(\epsilon)$ where $\epsilon$ is a perturbation (a small quantity) then in first order perturbation theory we simply approximate this with its first order Taylor expansion, i.e. we take $$f(\epsilon) \approx f(0) + f'(0)\epsilon$$

Show that $(1+\epsilon)^n \approx 1 + n\epsilon$. Expand $\frac{1}{\sqrt{1+2\Phi}}$, $\frac{1}{1-2\Psi}$ and $(1+2\Phi)^3$ to first order in perturbation theory.

What is $\Phi^2 \exp((1+\Phi)^{1000} + \exp(\exp(\Phi))))$ to first order (easy)? What is it to third order? (Hint: the prefactor is second order so we only need the exponential to first order. Recursively replace the functions with its Taylor expansion until you get to the result. If you see $4$ and $1001$ you did it correctly).

Consider the following expression $$\frac{d\rho}{dt} + 3H\rho +\nabla \cdot (\rho \vec{v}) = 0$$ When there are no perturbations $\rho = \overline{\rho}(t)$, $\vec{v} = \overline{\vec{v}} = 0$ where $$\frac{d\overline{\rho}}{dt} + 3H\overline{\rho} = 0$$ Perturb $\rho$ and $\vec{v}$, i.e. take $$\rho = \overline{\rho} + \delta\rho$$ $$\vec{v} = \overline{\vec{v}} + \delta \vec{v}$$ where we also define the overdensity $\delta$ via $\delta\rho = \overline{\rho}\delta$. Show that to first order in perturbation theory we get $$\frac{d\delta}{dt} + \nabla\cdot \delta\vec{v} = 0$$ This is the perturbed continuity equation that tells us how density perturbations evolve (when we ignore metric perturbations, we will do it properly in the lectures by deriving it from scratch).

The inverse of the perturbed metric

Finding the inverse of the metric when its diagonal is easy, but how can we do it in general? 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 $\delta g_{\mu\nu}$ and the background metric to first order in perturbation theory (i.e. second order terms in $\delta g$ can be put to zero) by using the defining relation above. Apply this to the metric in the Newtonian gauge $g_{00} = -(1+2\Psi)$, $g_{ij} = a^2(1+2\Phi)\delta_{ij}$ and $g_{0i} = 0$. Figure out the inverse in the way we did it in the lectures 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.

Christoffel symbols for the perturbed metric

In this problem we will derive the Christoffel symbols $$\Gamma^{\mu}_{\alpha\beta} = \frac{1}{2}g^{\mu\delta}(g_{\alpha\delta,\beta} + g_{\delta\beta,\alpha} - g_{\alpha\beta,\delta})$$ for the Newtonian guage metric where $g_{00} = -(1+2\Psi)$, $g_{ij} = a^2(1+2\Phi)\delta_{ij}$ and $g_{0i} = 0$. In the lectures we showed that in linear perturbation theory $$\Gamma^0_{ij} = (H - 2H\Psi + \dot{\Phi})g_{ij} = a^2(H + 2H\Phi - 2H\Psi + \dot{\Phi})\delta_{ij}$$ and used without proof that $$\Gamma^0_{00} = \dot{\Psi}$$ $$\Gamma^0_{0i} = \Gamma^0_{i0} = \Psi_{,i} = \frac{\partial \Psi}{\partial x^i}$$ Compute these symbols. If you want more work compute the remaining symbols that we will need later on: $$\Gamma^i_{00} = \frac{1}{a^2}\Psi_{,i}$$ $$\Gamma^i_{0j} = \delta_{ij}(H + \dot{\Phi})$$ $$\Gamma^i_{jk} = \delta_{ij}\Phi_{,k} + \delta_{ik}\Phi_{,j} - \delta_{jk}\Phi_{,i}$$

Trajectories of photons in a perturbed Universe

In the lectures we used the definitions $g_{\mu\nu}P^\mu P^\nu \equiv -m^2$, $g_{ij}P^i P^j \equiv p^2$ and defined the direction $\hat{p}^i$ via $P^i = C\hat{p}^i$ for some constant $C$ such that $\hat{p}^i$ becomes a unit vector $\hat{p}^i\hat{p}^j\delta_{ij} \equiv 1$. We then showed that in linear perturbation theory $$P^0 = E(1-\Psi)$$ $$P^i = \frac{p\hat{p}^i}{a}(1-\Phi)$$ where $E = \sqrt{p^2 + m^2}$. Redo this derivation and use then evaluate the two geodesic equations $$\frac{dP^\mu}{d\lambda} + \Gamma^\mu_{\alpha\beta}P^\alpha P^\beta = 0$$ for $\mu = 0$ and $\mu = i$ (see the previous problem for the Christoffel symbols) to get an equation for $\frac{dE}{d\lambda}$ and $\frac{d\hat{p}^i}{d\lambda}$. For a smooth Universe ($\Phi=\Psi=0$) solve the equations for photons ($m=0$). For this its useful to use that $\frac{d}{d\lambda} = \frac{dt}{d\lambda}\frac{d}{dt} = P^0 \frac{d}{dt}$. Check that the answers agrees with the expectations you have for how photons evolve in a smoooth Universe.

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

The Boltzmann equation for dark matter reads $$\frac{df}{d\lambda} = \frac{\partial f}{\partial t}P^0 + \frac{\partial f}P^i + \frac{\partial f}{\partial E} \frac{dE}{d\lambda} + \frac{\partial f}{\partial \hat{p}^i} \frac{d\hat{p}^i}{d\lambda} = 0$$ or by dividing by $P^0$ $$\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x^i}\frac{P^i}{P^0} + \frac{\partial f}{\partial E} \frac{1}{P^0}\frac{dE}{d\lambda} + \frac{\partial f}{\partial \hat{p}^i} \frac{1}{P^0}\frac{d\hat{p}^i}{d\lambda} = 0$$ where $P^\mu = \frac{dx^\mu}{d\lambda}$ is the 4-momentum of the CDM particles. Argue why we can drop the last term to first order in perturbation theory.

The momentum 4-vector for dark matter particles $P^\mu$ we have derived in general in previous problems how it evolves. From $g_{\mu\nu}P^\mu P^\nu = -m^2$ and $p^2 \equiv g_{ij}P^iP^j$ we showed that in the Newtonian gauge we have $$P^0 = E(1-\Psi)$$ $$P^i = \frac{p\hat{p}^i}{a}(1-\Phi)$$ where $E = \sqrt{p^2 + m^2}$ (which is $\simeq m + \frac{p^2}{2m}$ in the non-relativistic limit). We have also shown that $$\frac{1}{P^0}\frac{dE}{d\lambda} = -\frac{p^2}{E} \left[H + \frac{\partial \Phi}{\partial t} + \frac{E}{p}\frac{\partial \Psi}{\partial x^i}\frac{\hat{p}^i}{a} \right]$$ With all of this we get $$ \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x^i}\frac{\hat{p}^ip}{aE} - \frac{\partial f}{\partial E}\left[\frac{p^2}{E}(H + \frac{\partial \Phi}{\partial t}) + \frac{\hat{p}^ip}{a}\frac{\partial \Psi}{\partial x^i}\right] = 0$$ which is the starting point of this problem.

Take the zeroth moment of the Boltzmann equation above, i.e. multiply it by $\frac{d^3p}{(2\pi)^3}$ and integrate it over all momenta using the definitions $$n \equiv \int \frac{d^3 p}{(2\pi)^3} f$$ $$n v^i \equiv \int \frac{d^3 p}{(2\pi)^3} f \frac{p\hat{p}^i}{E}$$ and show that $$\frac{\partial n}{\partial t} + 3Hn + 3\frac{\partial \Phi}{\partial t}n + \frac{\partial}{\partial x^i}(n v^i) = 0$$ What is this equation called and what does it equation represent physically? Explain what the different terms in the equation above represents. Hint: 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).

Solve the equation above 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 $$\frac{\partial \delta}{\partial t} + \frac{1}{a}\frac{\partial v^i}{\partial x^i} + 3\frac{\partial \Phi}{\partial t} = 0$$

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 the Boltzmann equation by $\frac{d^3p}{(2\pi)^3}\frac{p\hat{p}^j}{E}$ and integrate over all momenta. 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 the lecture notes or Dodelson page 105 for more details if you have problems with some of the integrals.

What is this equation called and what does it represent physically?

The dark matter equations in Fourier space

The velocity field in 3D can be decomposed into two parts: a curl-free (irrotational) part and a divergence free part. The latter (the rotational component) counts as a vector perturbation and since we only work with scalar perturbations our velocity field(s) are assumed to be irrotational. A vector field is irrotational if $\nabla\times \vec{F} = 0$. Show that in Fourier space this implies that the fourier transform of $\vec{F}$, $\hat{\vec{F}}$, can be written $$\hat{\vec{F}} = i\frac{\vec{k}}{k}\hat{F}$$ for some scalar $\hat{F}$ (the magnitude of $\hat{\vec{F}}$ - the presence of the "$i$" here is just a convention). Hint: either use a Helmholtz decomposition or take the Fourier transform of $\nabla\times \vec{F} = 0$ and ask yourself what $\vec{a} \times \vec{b} = 0$ says about the two vectors.

Take the Fourier transform of the continuity and Euler equations assuming that the velocity field is irrotational $$\frac{\partial \delta}{\partial t} + \frac{1}{a}\frac{\partial v^i}{\partial x^i} + 3\frac{\partial \Phi}{\partial t} = 0$$ $$\frac{\partial v^i}{\partial t} + Hv^i + \frac{1}{a}\frac{\partial \Psi}{\partial x^i} = 0$$ to obtain two scalar equations for $\hat{\delta}$ and $\hat{v}$ (defined via $\hat{\vec{v}} = i\frac{\vec{k}}{k} \hat{v}$). Express the equations in terms of derivatives with respect to $x=\log a$ instead of $t$ and compare to the equations given in Milestone III.

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 Thompson (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?

Legendre multipole math

When we expanded in mutipoles we had to compute some integrals involving Legendre polynomials. Evaluate the following ones (you can compare your result to Eq. 55-56 here) $$I^{(1)}_{\ell,\ell'} = \int_{-1}^1 \mu^1 P_\ell(\mu) P_{\ell'}(\mu)dx$$ $$I^{(2)}_{\ell,\ell'} = \int_{-1}^1 \mu^2 P_\ell(\mu) P_{\ell'}(\mu)dx$$ and show that $I^{(1)}_{\ell,\ell'}$ can only be non-zero if $\ell'=\ell \pm 1$ and that $I^{(2)}_{\ell,\ell'}$ can only be non-zero if $\ell' = \ell \pm 0,2$. Thy to convince yourself that this pattern holds in general: if the integrand contains a $\mu^{2n}$ term then the resulting integral can only be non-zero if $\ell' = \ell \pm 2k$ for $k=0,1,2,\ldots,n$ and if it contains $\mu^{2n+1}$ then it can only be non-zero for $\ell' = \ell \pm 2k+1$ for $k=0,1,\ldots,n$ (this can be formally proved using induction).

Hint: use the relation $$(\ell + 1)P_{\ell+1}(\mu) = (2\ell+1)\mu P_{\ell}(\mu) - \ell P_{\ell-1}(\mu)$$ to eliminate the $\mu$'s and get integrals of just two Legendre polynomials for which you can directly use the orthogonality relation $$I^{(0)}_{\ell,\ell'} = \int_{-1}^1 P_\ell(\mu) P_{\ell'}(\mu)dx = \frac{2\delta_{\ell \ell'}}{2\ell +1}$$

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 $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu = -(1+2\Psi)dt^2 + a^2(1+2\Phi)(dx^2+dy^2+dz^2)$$ We have computed these in previous problems: $$\Gamma^0_{00} = \Psi_{,0}$$ $$\Gamma^0_{ij} = a^2\delta_{ij}[H + 2H(\Phi-\Psi) + \Phi_{,0}]$$ $$\Gamma^i_{00} = \frac{1}{a^2}\Psi_{,}^{i},\,\, \Gamma^0_{0i} = \Psi_{,i},\,\,\Gamma^i_{j0} = \delta_{ij}(H+\Phi_{,0})$$ $$\Gamma^i_{jk} = [\delta_{ij}\Phi_{,k} + \delta_{ik}\Phi_{,j} - \delta_{jk}\Phi_{,i}]$$ Use this to compute the $R_{00}$ component of the Ricci tensor $$R_{00} = \Gamma^\mu_{00,\mu} - \Gamma^\mu_{0\mu,0} + \Gamma^\mu_{\nu\mu}\Gamma^\nu_{00} - \Gamma^\mu_{\nu 0}\Gamma^\nu_{0\mu}$$ The other components are $$R_{0i} = 0$$ $$R_{ij} = \delta_{ij}[(2a^2H^2 + a \ddot{a})(1+2\Phi - 2\Psi) + a^2(6\dot{\Phi} - \dot{\Psi}) + a^2\ddot{\Phi} - \nabla^2\Phi] - (\Phi + \Psi)_{,ij}$$ Evaluate the Ricci scalar and use this to get the $G_{00}$ component and check that if you put the perturbations to zero then you recover the background result $\overline{G}_{00} = 3\frac{\dot{a}^2}{a^2}$.

Finally, what is the $G_{00}$ component in Fourier space?

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. Note that since for our metric $P_1 = \sqrt{g_{11}}\vec{p}_1$ and so on we have $\frac{dP_1dP_2dP_3}{\sqrt{-g}} = \frac{\sqrt{g_{11}g_{22}g_{33}}}{\sqrt{-g}} d^3p = \frac{d^3p}{\sqrt{-g_{00}}}$.

The $00$ component for photons is given by the usual expression (using $P_0 = \sqrt{-g_{00}}E$) $$ T^0_0 = g_\gamma \int \frac{d^3 p}{(2\pi)^3} E(p) f_i(t,\vec{x},\vec{p}) $$ 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 the Einstein equation. 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) was computed in the previous exercise: $$\delta \hat{G}^0_0 = -6H\Phi_{,0} + 6\Psi H^2 - 2\frac{k^2\Phi}{a^2}$$ 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 massless 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.

Momentum Flux - Velocity in terms of the multipoles for photons

In this problem we will evaluate the momentum flux $T^j_0$ in perturbation theory and use this to show that the multipole $-3\Theta_1$ is the same quantity as the velocity $v$ is for baryons/CDM (just as we in the lectures showed that $4\Theta_0$ is the density contrast $\delta$).

Background material:
The energy momentum tensor is in general given by $$T^\mu_\nu = \frac{g}{(2\pi)^3} \int \frac{dP_1dP_2dP_3}{\sqrt{-\det g}} \frac{P^\mu P_\nu}{P^0} f$$ For scalar perturbations in the Newtonian gauge we have $$\frac{dP_1dP_2dP_3}{\sqrt{-\det g} P^0} = \frac{d^3p}{E}$$ so $$T^\mu_\nu = \frac{g}{(2\pi)^3} \int d^3p \frac{P^\mu P_\nu}{E} f$$ where $$P^i = \frac{\vec{p}^i}{a}(1-\Phi) = \frac{p\hat{p}^i}{a}(1-\Phi)$$ $$P_i = a\vec{p}^i(1+\Phi) = ap\hat{p}^i(1+\Phi)$$ where $\hat{p}^i\hat{p}_i = 1$.

What quantities are the integrals $\frac{g}{(2\pi)^3} \int f d^3p$ and $\frac{g}{(2\pi)^3} \int E f d^3p$ describing?

Momentum flux for baryons/CDM:
The $T^j_0$ component represents momentum flux. What is the momentum for a non-relativistic particle? For baryons and CDM show that $$T^j_0 = -\overline{\rho}\frac{\vec{v}^j}{a}$$ Its useful to remember that the mean of some quantity $Q(\vec{p})$ is defined as $$\left\lt Q \right\gt \equiv \frac{1}{n}\frac{g}{(2\pi)^3} \int d^3p Q(\vec{p}) f$$ Evaluate $\frac{ik_j}{k}T^j_0$ using that the velocity is curl-free so $\vec{v} = \frac{i\vec{k}}{k} v$ where $v$ is the velocity we derived equations for in the lectures.

Momentum flux for photons:
Now lets move on to photons. What is $E$ for a photon? Show that we can write (use spherical coordinates in the integral for $\rho$) $$\overline{\rho} = \frac{g}{2\pi^2}\int p^3 \overline{f}\, dp$$ This result will be useful below. The expansion of the distribution function is given by $$f = \overline{f} - p\Theta \frac{\partial \overline{f}}{\partial p}$$ where $\Theta$ does not depend on $p$ and only on the direction through $\mu = \frac{\hat{p}\cdot \vec{k}}{k}$. In the lectures we showed that for photons $$T^0_0 =-\overline{\rho}(1+4\Theta_0)$$ $$\left(\frac{k_ik^j}{k^2} - \frac{1}{3}\delta_i^j\right)T^i_j = -\frac{8}{3}\overline{\rho} \Theta_2$$ Show that in linear perturbation theory $$T^j_0= -\frac{g}{2\pi^3} \int d^3p \frac{p^2\hat{p}^j}{a}\Theta \frac{\partial \overline{f}}{\partial p}$$ and evaluate $\frac{ik_j}{k}T^j_0$. For this its useful to use spherical coordianates, use integration by parts for the $p$ integral and compare to a previous result, remember that $P_1(\mu) = \mu$ and the definition of the multipoles $$\Theta_\ell = \frac{i^\ell}{2}\int_{-1}^1 \Theta P_\ell(\mu)\, d\mu$$

Photon velocity in terms of the multipoles:
The general expression for the momentum flux for a general component is $$\frac{ik_j}{k}T^j_0 = (\overline{\rho} + \overline{P})\frac{v}{a}$$ where $v$ is the velocity. Compare this to your results for photons and show that $v_\gamma = -3\Theta_1$.

Exam problem - Vector perturbations

In the course you have studied both scalar and tensor perturbations, but now we will take a closer look at vector perturbations and how they evolve in an expanding universe.

A general 3D vector field, $v_i$, can always be decomposed into two parts, the part with no curl, and the part with no divergence: $$ v_i = v_i^\parallel + v_i^\perp,$$ where $\nabla \times \mathbf{v}^\parallel = 0$ and $\nabla \cdot \mathbf{v}^\perp = 0.$ The first part, $\mathbf{v}^\parallel = \nabla \phi$, can always be written as the gradient of some scalar field, $\phi$ and is therefore considered the scalar part of $v_i$ (In the project, when you used the dark matter and baryon velocity, $v$ and $v_b$, they were assumed to only consist of this scalar part). The latter part of the velocity, $v^\perp_i$, is what is called the vector component of $v_i$ and it is what we will consider in this problem.

We will only work with perturbations to first order in this problem, so you can safely neglect all higher order terms.

  • a) Consider a general perturbed metric $$ g_{\mu\nu} = \bar{g}_{\mu\nu} + h_{\mu\nu}, $$ where $h_{\mu\nu}$ is a small perturbation and a $\,\bar{ }\,$ denotes an unperturbed quantity.

    The Christoffel symbols are given by: $$\Gamma^\mu_{\nu\lambda} = \frac{1}{2} g^{\mu\rho} \left[\partial_\lambda g_{\rho\nu} + \partial_\nu g_{\rho\lambda} -\partial_\rho g_{\lambda\nu}\right].$$

    Let us define the perturbed Christoffel symbols $\delta\Gamma^\mu_{\nu\lambda}$ by the following equation $$ \Gamma^\mu_{\nu\lambda} = \bar{\Gamma}^\mu_{\nu\lambda} + \delta\Gamma^\mu_{\nu\lambda}.$$ Show that to first order in the perturbation $h_{\mu\nu}$ $$ \delta\Gamma^\mu_{\nu\lambda} = \frac{1}{2} \bar{g}^{\mu\rho} \left[\partial_\lambda h_{\rho\nu} + \partial_\nu h_{\rho\lambda} -\partial_\rho h_{\lambda\nu} - 2 h_{\rho\sigma} \bar\Gamma^\sigma_{\nu\lambda}\right].$$ Hint: $h^{\mu\nu} \equiv g^{\mu\nu} - \bar{g}^{\mu\nu} = -\bar{g}^{\mu\rho} h_{\rho\sigma} \bar{g}^{\sigma\nu} \ne \bar{g}^{\mu\rho} h_{\rho\sigma} \bar{g}^{\sigma\nu}$.
  • b) Consider a vector perturbation to a flat Fridmann-Robertson-Walker spacetime $$ \begin{eqnarray*} \bar{g}_{00} &=& -1,\\ \bar{g}_{i0} &=& 0,\\ \bar{g}_{ij} &=& a^2(t) \delta_{ij},\\ h_{00} &=& 0, \\ h_{i0} = h_{0i} &=& a(t) G_i(t, \bf{x}),\\ h_{ij} &=& 0, \end{eqnarray*} $$ where $\delta^{ij} \partial_j G_i = 0$ (it has zero divergence).

    In the lectures you have already calculated the background evolution of this spacetime, and you can use these results here if you need them: $$ \begin{eqnarray*} \bar{\Gamma}^0_{ij} &=& \delta_{ij} a^2 H, \\ \bar{\Gamma}^i_{0j} = \bar{\Gamma}^i_{j0} &=& \delta_{ij} H, \end{eqnarray*} $$ the other unperturbed Christoffel symbols are zero.

    Show that (compute four of the five symbols below for maximum marks) $$ \begin{eqnarray*} \delta\Gamma^i_{00} &=& \frac{\partial_0 h_{i0}}{a^2} \\ \delta\Gamma^0_{0i} = \delta\Gamma^0_{i0} &=& H h_{0i}\\ \delta\Gamma^i_{j0} = \delta\Gamma^i_{0j} &=& \frac{\partial_j h_{i0} - \partial_i h_{0j}}{2a^2}\\ \delta\Gamma^0_{ij} &=& -\frac{\partial_j h_{0i} + \partial_i h_{0j}}{2}\\ \delta\Gamma^i_{jk} &=& -\delta_{jk} h_{i0} H. \end{eqnarray*} $$ All the other perturbed Christoffel symbols are zero.
  • c) The perturbed Ricci tensor is then given by: $$ \begin{eqnarray*} \delta R_{\mu\nu} = \partial_\alpha \delta \Gamma^\alpha_{\mu\nu} - \partial_\nu \delta \Gamma^\alpha_{\mu\alpha} + \delta \Gamma^\beta_{\mu\nu}\bar{\Gamma}^\alpha_{\alpha\beta} + \delta\Gamma^\alpha_{\alpha\beta}\bar{\Gamma}^\beta_{\mu\nu} - \delta\Gamma^\beta_{\mu\alpha}\bar{\Gamma}^\alpha_{\nu\beta} - \delta\Gamma^\alpha_{\nu\beta}\bar{\Gamma}^\beta_{\mu\alpha}. \end{eqnarray*} $$ Show that $$ \delta R_{i0} = h_{0i} \left( \frac{\ddot a}{a} + 2 H^2\right) - \frac{\nabla^2h_{i0}}{2a^2}. $$ Hint: Remember that $\delta^{ij} \partial_j G_i = 0$.
  • d) Consider a vector perturbation to the energy momentum tensor $$ T_{\mu\nu} = \bar p g_{\mu\nu} + (\bar \rho + \bar p) u_\mu u_\nu,$$ where $u_\mu = (-1, u_i)$, $\delta^{ij} \partial_j u_i = 0$ and $u_i$ is small.

    You can always rewrite Einstein equation as $$ R_{\mu\nu} = 8\pi G S_{\mu\nu},$$ where $$ S_{\mu\nu} \equiv T_{\mu\nu} - \frac{1}{2} g_{\mu\nu} T^\lambda {}_\lambda. $$ In this case, it is a very good idea to write the equation on this form, because it allows us to extract the $i0$'th component of the Einstein equation without having to calculate the Ricci scalar (which would take a considerable effort)!

    Calculate $\delta S_{i0}$.

    Show that the corresponding component of the Einstein equation becomes: $$\frac{1}{2} \nabla^2G_i = 8 \pi G (\bar \rho + \bar p) a u_i,$$ where $\nabla^2 \equiv \delta^{ij} \partial_i \partial_j $. Hint: You can use the Friedmann equations for a flat spacetime to write $\bar \rho$ and $\bar p$ in terms of $a, \dot{a}, \ddot{a}$.
  • e) The momentum conservation equation $\nabla^\mu T_{\mu \nu} = 0$ for the vector perturbation takes the following form (you do not need to derive this): $$ \partial_0 \left[(\bar \rho + \bar p)u_i\right] + \frac{3\dot a}{a}\left[(\bar \rho + \bar p)u_i\right] = 0.$$ Let us define the vector density $V_i \equiv \left[(\bar \rho + \bar p)u_i\right] $. How do $V_i$ and $G_i$ evolve as functions of $a$ as the universe expands? How is this evolution different for modes inside and outside the horizon?

    What does this mean for the impact of vector perturbations in cosmology? Contrast this with the case for scalar and tensor fluctuations.

  • Exam problem - Cold Dark Matter Perturbations

    In the lectures (and in the book) we derived evolution equation for the cold dark matter (CDM) overdensities, $\delta(\vec{x}, t),$ and velocity, $v^{i}(\vec{x}, t),$ using the Boltzmann equation for CDM. Today, however, you are going to use the fact that CDM can be treated as a perfect fluid to derive the same evolution equations from the conservation equation $$ \nabla_{\mu} T_{\nu}^{\mu}=0 $$ The energy momentum tensor of the CDM fluid is given by $$ T_{\nu}^{\mu}=\rho U^{\mu} U_{\nu} $$ where $\rho(\vec{x}, t)$ is the energy density and $U^{\mu}(\vec{x}, t)$ is the four-velocity of the CDM fluid. It is often useful to separate the energy momentum tensor into a zero'th order part and a perturbed part: $$ T_{\nu}^{\mu}(\vec{x}, t)=\overline{T}_{\nu}^{\mu}(t)+\delta T_{\nu}^{\mu}(\vec{x}, t) $$ We will be working in the Newtonian gauge, given by: $$ d s^{2}=-[1+2 \Psi(\vec{x}, t)] d t^{2}+a^{2}(t)[1+2 \Phi(\vec{x}, t)] \delta_{i j} d x^{i} d x^{j} $$

    • a) The four-momentum is given (to first order) by $$ P^{\mu}=\left((1-\Psi) E, \frac{(1-\Phi)}{a} p^{i}\right) $$ Using this, show that for CDM, the four velocity, $U^{\mu} \equiv d x^{\mu} / d \tau=P^{\mu} / m,$ is given by $$ U^{\mu}=\left((1-\Psi), \frac{(1-\Phi)}{a} v^{i}\right) $$ and that $$ U_{\mu}=\left(-(1+\Psi), a(1+\Phi) v^{i}\right) $$
    • b) The four-velocity of the CDM fluid is given by Eqs 15 and $16,$ only that $v^{i}$ is then the velocity of the fluid (i.e. the velocity of the particles), which we assume to be small, and not the velocity of any individual particle. Defining the CDM overdensity $$ \delta(\vec{x}, t) \equiv \frac{\rho(\vec{x}, t)-\overline{\rho}(t)}{\overline{\rho}(t)} $$ show that, to first order, the perturbed part of the CDM energy momentum tensor is given by $$ \delta T_{\nu}^{\mu}=\overline{\rho}\left(\begin{array}{cccc} -\delta & a v^{1} & a v^{2} & a v^{3} \\ -v^{1} / a & 0 & 0 & 0 \\ -v^{2} / a & 0 & 0 & 0 \\ -v^{3} / a & 0 & 0 & 0 \end{array}\right) $$

    • c) Use the time component of the conservation equation $$ \nabla_{\mu} T_{0}^{\mu}=0 $$ to derive the evolution equations for the CDM density. You can use the following result (without deriving it): $$ \nabla_{\mu} \overline{T}_{0}^{\mu}=-\frac{d \overline{\rho}}{d t}-3\overline{\rho}\left(H+\frac{\partial \Phi}{\partial t}\right) $$ Remember also that the covariant derivative of $\delta T^{\mu}{ }_{0}$ is given by: $$ \nabla_{\mu} \delta T_{0}^{\mu}=\frac{\partial \delta T^{\mu}_0}{\partial x^{\mu}}+\Gamma_{\alpha \mu}^{\mu} \delta T_{0}^{\alpha}-\Gamma_{\mu 0}^{\alpha} \delta T_{\alpha}^{\mu} $$ Separate the zero'th order equation from the first order one, and show that you get the following $$ \begin{aligned} \frac{d \overline{\rho}}{d t}+3 H \overline{\rho} &=0 \\ \frac{\partial \delta}{\partial t}+\frac{1}{a} \frac{\partial v^{i}}{\partial x^{i}}+3 \frac{\partial \Phi}{\partial t} &=0 . \end{aligned} $$ Hint 1: Remember that you can drop any terms second order in the small quantities.
    • d) Using the spatial components of the conservation equation, we can also derive an equation for the CDM velocity (you do not have to do this!). Give a physical/intuitive explanation of the different terms in each of the three equations: $$ \begin{aligned} \frac{d \overline{\rho}}{d t}+3 H \overline{\rho} &=0 \\ \frac{\partial \delta}{\partial t}+\frac{1}{a} \frac{\partial v^{i}}{\partial x^{i}}+3 \frac{\partial \Phi}{\partial t} &=0 \\ \frac{\partial v^{i}}{\partial t}+H v^{i}+\frac{1}{a} \frac{\partial \Psi}{\partial x^{i}} &=0 \end{aligned} $$
    • e) How would the CMB power spetrum change (roughly) if you changed the ratio of CDM to baryons (i.e. replacing some of the CDM by baryons or vise versa, keeping everything else fixed)? Please draw a sketch and explain your reasoning.

    Initial conditions and inflation

    Estimating how much inflation we need

    In this problem we assume inflation starts at some scale-factor $a_{\rm ini}$, ends at $a_{\rm end}$ and from then the Universe evolves as normal until today $a_{\rm today}=1$. The size of the Hubble radius at any time is $d_H = \frac{1}{\mathcal{H}}$.

    Express the size of the Hubble sphere today relative to that at the end of inflation, $d_H^{\rm today} / d_H^{\rm end}$, in terms of $a_{\rm end}$. You can assume for simplicity that the Universe is always radiation dominated from the end of inflation till today. Assuming that inflation happened roughly at the GUT scale $k_bT(a_{\rm end}) \sim 10^{15}$ GeV (where $k_b \sim 10^{-13}$ GeV/K) estimate $a_{\rm end}$. For the temperature today you can use $T(a_{\rm today}) \approx 1$ K (we only care about order of magnitude estimates).

    During inflation $H$ is approximately constant. Express the size of the Hubble radius at the end of inflation to that at the start of inflation, $d_H^{\rm end} / d_H^{\rm ini}$, in terms of $a_{\rm ini}$ and $a_{\rm end}$.

    For inflation to do its job and solve the Horizon problem we must require that the whole observable universe today fits in the comoving Hubble radius at the beginning of inflation, i.e. we must have $$d_H^{\rm today} \lt d_H^{\rm ini}$$ During inflation the Universe expanded in size by a factor $\frac{a_{\rm end}}{a_{\rm ini}}$. From the condition above and the previous results estimate the smallest factor the Universe expanded by. We often express this in terms of $N = \log(a_{\rm end} / a_{\rm ini})$ - the so-called number of e-folds of inflation. In other words the Universe expands by a factor $e^N$ during infation. Estimate how many e-fold we need to solve the Horizon problem.

    Notice the symmetry in the result (i.e. in the relation between $a_{\rm end}/a_{\rm ini}$ to $a_{\rm today}/a_{\rm end}$ that you found in the case where the inequality above is a strict equality), explain this based on how $d_H$ evolves during inflation and in radiation domination (see figure 2.4 of Baumann for an intuitive picture).

    The flatness problem

    We saw that curvature enters the Friedmann equation as it was a normal matter component with density parameter $$\Omega_k(a) = \frac{\Omega_{k0}}{a^2H^2/H_0^2}$$ 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 a non-zero curvature today, i.e. $\Omega_{k0} \not=0$, then estimate how close to flat did it had to have been at the end of inflation ($a_{\rm end} \sim 10^{-28}$).

    Why is this considered a problem? Turned around: if $\Omega_k(a_{\rm end})$ has a "reasonable" value (i.e. not super close to $0$), what would the curvature of the Universe be today? Does this conclusion change if the Universe was matter domainated instead?

    Compute how $\Omega_k(a)$ evolve during inflation ($H = $ const) and explain how inflation solves the flatness problem (you can use that the scalefactor increases by atleast a factor $10^{28}$ from the start to the end of inflation as showed in the previous 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.

    Quantum fluctuation during 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 (gauge invariant) curvature perturbations is in the Newtonian gauge given by $$\mathcal{R} = \Phi + \frac{\mathcal{H}(\Phi' - \mathcal{H}\Psi)}{4\pi G a^2 (\overline{\rho} + \overline{P})}$$ where $\overline{\rho},\overline{P}$ is the total energy density and pressure in the Universe at any time. On super-horizon scales ($k\ll \mathcal{H}$) we have $\Phi'\approx 0$ and $\Psi' \approx 0$. In this problem assume that we are always on super-horizon scales. Use this to show that $\mathcal{R}' \approx 0$ (you should consider both a radiation dominated and a matter dominated Universe) so that $\mathcal{R}$ is also frozen outside the horizon. This is the key property that allows us to relate predictions from inflation to the initial conditions when we solve numerically - the perturbations on a given scale generated during inflation is frozen in when the mode leaves the horizon and stays frozen until the mode later enters the horizon.

    When solving the Einstein-Boltzmann system 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. Explain why this is possible, i.e. what property of the equation system is key to this? What is $\mathcal{R}$ in terms of $\Phi,\Psi$ on super-horizon scales. If there is no anisotropic stress then $\Phi + \Psi = 0$, use this to compute the initial conditions for $\Phi$ and $\Psi$. If we include neutrinos we no longer have $\Phi + \Psi = 0$ so we must do a more thorough analysis which is the aim of the rest of this problem.

    The equations for the neutrino multipoles are $$ \begin{align} \mathcal{N}^\prime_0 &= -\frac{k}{\mathcal{H}} \mathcal{N}_1 - \Phi^\prime, \\ \mathcal{N}^\prime_1 &= \frac{k}{3\mathcal{H}} \mathcal{N}_0 - \frac{2k}{3\mathcal{H}}\mathcal{N}_2 + \frac{k}{3\mathcal{H}}\Psi \\ \mathcal{N}^\prime_2 &= \frac{2 k}{5\mathcal{H}} \mathcal{N}_{1} -\frac{3k}{5\mathcal{H}}\mathcal{N}_{3} \end{align} $$ On super-horizon scales the higher order multipoles are small $\mathcal{N}_0 \gg \mathcal{N}_1 \gg \mathcal{N}_2$. Use this to show that $$\mathcal{N}_2'' + \mathcal{N}_2' \approx \frac{2}{15}\left(\frac{k}{\mathcal{H}}\right)^2(\mathcal{N}_0 + \Psi)$$ The Einstein equation we derived in the lectures says $$\Phi+\Psi = -\frac{12H_0^2}{k^2a^2}(\Omega_{\gamma 0} \Theta_2 + \Omega_{\nu 0} \mathcal{N}_2)$$ What can you say about the photon multipole $\Theta_2$ in the early Universe? Use this equation and the equation you derived for $\mathcal{N}_2$ above to show that $$\frac{5}{12}(\Phi + \Psi) \approx f_\nu(\mathcal{N}_0 + \Psi)$$ where $f_\nu = \frac{\Omega_{\nu 0}}{\Omega_{\gamma 0} + \Omega_{\nu 0}}$. Recall the initial condition we derived for $\mathcal{N}_0 = \frac{-\Psi}{2}$ and show that $$\Phi = -\Psi\left(1 + \frac{2f_\nu}{5}\right)$$ We want to set the initial such that $\mathcal{R} = 1$. Use this to show that this is that case if $$\Psi = \frac{1}{\frac{3}{2} + \frac{2f_\nu}{5}}$$

    Evolution of $\Phi$ on super-horizon scales

    Recall that the curvature perturbation $$\mathcal{R} = \Phi + \frac{\mathcal{H}(\Phi' - \mathcal{H}\Psi)}{4\pi G a^2 (\overline{\rho} + \overline{P})}$$ is frozen outside the horizon. The same is also true for $\Phi$ and $\Psi$ inside the matter and radiation era. However as we transition from the radiation era to the matter era $\Phi$ evolves. Your job is to estimate how much. Use the expression for $\mathcal{R}$ to relate the value, $\Phi_{\rm matter-era}$, of the gravitational potential in the matter era to, $\Phi_{\rm radiation-era}$, the gravitational potential in the radiation era and show that $\Phi_{\rm matter-era} = \frac{9}{10}\Phi_{\rm radiation-era}$. You can assume that $\Phi+\Psi = 0$. This shows that going from a radiation dominated Universe to a matter dominated Universe the gravitational potential on super-horizon scales drops by roughly $10\%$. Repeat the calculation with the result for $\Phi+\Psi$ derived in the previous exercise and show that we get $\frac{9}{10}\cdot \frac{1 + \frac{4f_\nu}{15}}{1 + \frac{6f_\nu}{25}}$ and conclude that the suppression is in $[0.9,0.94]$ no matter what (physical) value $f_\nu$ takes.

    Spherical harmonics identity

    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 $$\int d\Omega_{\hat{p}} \mathcal{P}_\ell(\hat{k}\cdot\hat{p}) Y_{\ell'm}^*(\hat{p}) = \delta_{\ell\ell'} \frac{4\pi}{2\ell+1} Y_{\ell m}^*(\hat{k}) $$

    Growth of matter perturbations on subhorizon scales

    We are going to take the perturbations equations we have derived and figure out how fast the matter density perturbations grow on subhorizon scales $k\gg \mathcal{H}$. The equation for both the dark matter perturbations and the baryon perturbations (when Thompson scattering can be ignored) takes the form $$\delta' = \frac{k}{\mathcal{H}}v - 3\Phi'$$ $$v' = -v - \frac{k}{\mathcal{H}}\Psi$$ Here and below a prime denotes a derivative wrt $x = \log(a)$. Use the equations above to eliminate $v$ and derive the following ODE for $\delta$ in terms of $\Psi,\Phi$: $$\delta'' + \left(1+ \frac{\mathcal{H}'}{\mathcal{H}}\right)\delta' = -\frac{k^2}{\mathcal{H}^2}\Psi - \left(3\Phi'' + 3\Phi'(1+ \frac{\mathcal{H}'}{\mathcal{H}})\right)$$ Argue why at sub-horizon scales we can ignore the $\Phi$ terms such that $$\delta'' + \left(1+ \frac{\mathcal{H}'}{\mathcal{H}}\right)\delta' \simeq -\frac{k^2}{\mathcal{H}^2}\Psi$$ Argue why the Poisson equation on sub-horizons scales can be written $$\frac{k^2}{\mathcal{H}^2}\Phi \simeq \frac{3}{2} (\Omega_b(a)\delta_b + \Omega_{\rm CDM}(a)\delta_{\rm CDM})$$ What is the relation between $\Phi$ and $\Psi$ in the regime we are considering? Use this to show that the total matter density contrast $$\delta_M \equiv \frac{\Omega_{b}(a) \delta_b + \Omega_{\rm CDM}(a) \delta_{\rm CDM}}{\Omega_M(a)}$$ where $\Omega_M(a) = \Omega_{b}(a) + \Omega_{\rm CDM}(a)$ satisfies $$\delta_M'' + \left(1+ \frac{\mathcal{H}'}{\mathcal{H}}\right)\delta_M' \simeq \frac{3}{2}\Omega_M(a)\delta_M$$ Solve this equation in the matter era (how can we approxmate $\mathcal{H}(a)$ and $\Omega_M(a)$ in this era?) by taking an ansatz $\delta_M \propto a^n = e^{nx}$ and extract an equation for $n$. Show that we must have $$n = 1\,\,\,\text{or}\,\,\, n = -\frac{3}{2}$$ Why do we have two solutions, i.e. what does each of these two solutions represent?

    Solve the equation above in the radiation era (how can we approximate $\Omega_M(a)$ there?) and show that we have $\delta_M \simeq A + Bx = A + B\log(a)$ for some constants $A,B$.

    Exam problem - Line of sight integration for neutrinos

    In the lectures we derived the line of sight integral for the photon multipoles at the present time. We started with the Boltzmann equation for the photon distribution $\Theta = \Theta(\eta,k,\mu)$ (here ignoring the quadrupole correction $\Pi$) $$\frac{d\Theta}{d\eta} + ik\mu(\Theta + \Psi) = -\frac{d\Phi}{d\eta} - \frac{d\tau}{d\eta}\left(\Theta_0 - \Theta + i\mu v_b\right)$$ and showed that this implies the line of sight integral $$\Theta_\ell(k,\eta_0) = \int_0^{\eta_0}\left[g(\Theta_0+\Psi) + \left(\frac{d\Psi}{d\eta}-\frac{d\Phi}{d\eta}\right) e^{-\tau} - \frac{1}{k}\frac{d}{d\eta}(gv_b)\right]j_\ell(k\eta_0 - k\eta)d\eta$$ We are going to do this same exercise for neutrinos. The Boltzmann equation for the neutrino temperature perturbations $\mathcal{N} = \mathcal{N}(\eta,k,\mu)$ is given by $$\frac{d\mathcal{N}}{d\eta} + ik\mu(\mathcal{N} + \Psi) = -\frac{d\Phi}{d\eta} $$ In this problem you can use that the multipole expansion is given by $$\mathcal{N} = \sum_\ell \frac{2\ell+1}{i^\ell}\mathcal{N}_\ell P_\ell(\mu)$$ where $P_\ell(\mu)$ is the Legendre polynomials and the multipoles are given by $$\mathcal{N}_\ell = i^\ell \int_{-1}^1 \frac{d\mu}{2}\mathcal{N} P_\ell(\mu)$$ The spherical Bessel functions $j_\ell$ are defined by $$j_\ell(x) = i^\ell \int_{-1}^1 P_\ell(\mu)\frac{d\mu}{2} e^{-i\mu x}$$

    • a) Integrate the equation for $\frac{d\mathcal{N}}{d\eta}$ to obtain an expression for $\mathcal{N}$ at the present time. State any approximations/assumptions you are making.
    • b) Expand this result in multipoles $\mathcal{N}_\ell$ and show that for all $\ell > 0$ the result can be written on the form $$\mathcal{N}_\ell(\eta=\eta_0,k) = \int_0^{\eta_0}S_\nu j_\ell(k\eta_0 - k\eta)d\eta$$ for some source-function $S_\nu = S_\nu(k,\eta)$. State any approximations/assumptions you are making. If needed you can assume the initial neutrino distribution only has a monopole. Hint: to simplify you can use the relation $f(\eta_*) = \int f(\eta)\delta(\eta - \eta_*)d\eta$ if needed.
    • c) Compare your result to the known result for photons (given above) and discuss similarities and differences. Could you have obtained this result from that without doing any calculations? Give a physical interpretation of $\mathcal{N}_\ell$ just as we did for the photon multipoles in the lectures.

    Exam problem - Scattering processes involving baryons

    When you derived the Boltzmann equation for baryons, you found one equation for the baryon density and one equation for the baryon velocity. Give a short physical/intuitive explanation for what each of these two equations tell us.

    Consider all these scattering processes involving baryons:

    • $p^+ + \gamma \leftrightharpoons p^+ + \gamma$
    • $e^- + \gamma \leftrightharpoons e^- + \gamma$
    • $e^- + e^- \leftrightharpoons e^- + e^-$
    • $e^- + e^+ \leftrightharpoons \gamma + \gamma$
    • $e^- + p^+ \leftrightharpoons e^- + p^+$
    • $e^- + \nu_e \leftrightharpoons e^- + \nu_e$
    • $n \rightarrow e^- + p^+ + \bar{\nu}_e$

    where $p^+$ denotes a proton, $e^-$ an electron, $e^+$ a positron, $n$ a neutron, $\nu_e$ an electron neutrino and $\gamma$ a photon.

    For each of these scattering processes give a short physical/intuitive explanation for why that process would contribute or not contribute (or be negligible) to the collision term for the equation for baryon density and/or the equation for baryon velocity.

    Note that process number 4 can go in both directions, consider the process in each direction separately. Hints: For Thompson scattering $\sigma_T \propto 1/m^2$. Neutrons have a decay time of $\sim$ 15min when they are free, but don't decay (apart from radioactive decay) when they are tied up in atoms. Neutrinos only interact via weak interactions.

    Exam problem - Warm up questions

    Milestone I:

    • Why is the temperature of the cosmological neutrinos smaller than the temperature of the photons today?
    • What is the equation of state for a perfect fluid? Mathematically what does it mean that the Universe is accelerating and what condition must the equation of state of a fluid responsible for an accelerated expansion of the Universe satisfy?

    Milestone II:

    • What is the visibility function and what is the physical interpretation of this? Why it is practically zero at very early times?
    • At what temperature do we naively expect recombination to happen at? The actual temperature is a factor $\sim 40$ smaller, why is this?

    Milestone III:

    • What are the main reason(s) we solve the CMB equations in Fourier space instead of doing it in real space?
    • In the Boltzmann equation for photons we encountered the term $$\frac{\partial f}{\partial \hat{p}}\frac{d \hat{p}}{dt}$$ Explain why we can ignore this term when deriving an equation for the photon perturbation $\Theta$. If this term could not be ignored then how would you compute $\frac{d \hat{p}}{dt}$? (Don't do the calculation)

    Milestone IV:

    • A new physical effect is found that causes the theoretical CMB quadrupole ($C_\ell$ for $\ell = 2$) to increase by $1\%$ while keeping everything else the same. Would this be possible to detect in a modern CMB experiment, why/why not?
    • What are the main advantages of line of sight integration over the traditional Boltzmann hierarchy approach?

    Exam problem - The CMB and matter power-spectrum

    This problem is about understanding features in the CMB and matter power-spectrum. You might need to do some very small calculations in a few questions, but its mainly about the physical understanding. Keep the answers brief.

    • a) The line of sight integration expression for $\Theta_\ell$ (ignoring quadrupolar corrections) is given by $$\Theta_\ell(k,\eta_0) = \int_0^{\eta_0}\left[g(\Theta_0+\Psi) + \left(\frac{d\Psi}{d\eta}-\frac{d\Phi}{d\eta}\right) e^{-\tau} - \frac{1}{k}\frac{d}{d\eta}(gv_b)\right]j_\ell(k\eta_0 - k\eta)d\eta$$ Explain what the different terms above represent physically. Use what you know about the visibility function to write down an approximate expression for the integral of the first term. How do we explain this result physically, i.e. what does this tell us about how the anisotropies we observe today are formed?
    • b) You are given an theoretical spectrum showing the Sachs-Wolfe plateau ($\ell \lt 30$) of the CMB power spectrum. $C_\ell $ is seen to increase as we go from large to small $\ell$. What are the cosmological parameters and the related physical effects that can cause such a signal?
    • c) Explain briefly the effects of changing the optical depth of reionizaiton on the CMB (including polarization) and matter power-spectrum. Give an example of a cosmological parameter the optical depth is degenerate with in the CMB power-spectrum.
    • d) The evolution equation for tensor perturbation $h$ are given by $$\frac{d^2h}{d\eta^2} + 2\frac{1}{a}\frac{da}{d\eta}\frac{dh}{d\eta} + k^2h = 0$$ What kind of equation is this? The tensor perturbation $h$ acts as a source for tensor perturbations in the photon distribution which adds to the scalar perturbation signal we computed in the lectures for the CMB power-spectrum. It only sources these only for scales where the amplitude $h$ is sizable in the period around recombination. On the basis of how $h$ evolves explain why we would not see any effects of tensor perturbations in the temperature power-spectrum for large $\ell$.
    • e) The matter power-spectrum today $$P(k) = \Delta^2_M(a=1,k)P_{\rm priomordial}(k)$$ where $P_{\rm priomordial}(k) = \frac{2\pi^2}{k^3}A_s (k/k_{\rm pivot})^{n_s-1}$ is the primordial power-spectrum and $\Delta_M(a,k) \equiv \frac{\Phi(a,k) k^2}{4\pi G\overline{\rho}(a) a^2}$ is the matter density contrast has a peak around $k\sim 0.01 h/{\rm Mpc}$. Explain the reason for why we get this peak based on how $\Delta_M$ grows in different regimes and give an expression for the wave-number $k_{\rm peak}$ it correspond to? Use this to discuss what cosmological parameter combination the position of the peak depends on (you can assume that the only relevant forms of energy in the Universe is matter and radiation and you don't need to derive how $\Delta_M$ evolves in different regimes - its enough to explain how it evolves).

    Which parameters have changed?

    a) Below is the CMB power-spectrum for two cosmologies. In the second cosmology I have changed two parameters (significantly; you would be able to tell the difference by eye if I changed each of them) among $\Omega_{M0},\Omega_{b0},h,A_s,n_s,\tau_{\rm reion}$. What two parameters do you think I have changed and why? Bonus question: can you give a combination of those two parameters that should be the same in both cosmologies?

    b) Below I show the matter power-spectrum for two cosmologies which both have a fixed value of $\Omega_{M0}$, but where one cosmology has a higher baryons density. Which curve correspond to more baryons and why?

    c) Below I show the matter power-spectrum for two models. One with $\Omega_{M0} = 0.3,\,\Omega_{\Lambda 0} = 0.7$ and one with $\Omega_{M0} = 1.0,\,\Omega_{\Lambda 0} = 0.0$ (so no dark energy). The baryon density is the same in the two models. What is the key feature we can use to figure out which is which here? How does this feature depend on the cosmological parameters? Based on this explain which curve is which cosmology.