Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 (Cosmology)

Set $c=1$ and let $H=\dot{a}/a$ denote the proper-time Hubble parameter. The Friedmann equation is the expansion constraint: density drives expansion, while spatial curvature and the cosmological constant contribute separately. The Friedmann acceleration equation shows that the active gravitational density is $\rho +3P$; positive pressure gravitates, while positive vacuum energy accelerates expansion.

Subtracting the constraint from the acceleration equation gives $\dot{H}=-4\pi G(\rho +P)+k/a^2$. Differentiating the constraint gives $2H\dot{H}=(8\pi G/3)\dot{\rho }+2kH/a^2$. Eliminate $\dot{H}$ to obtainThis cosmological continuity equation is local energy conservation. For a comoving volume $V\propto a^3$, it is $d(\rho V)=-PdV$, the adiabatic first law: expansion does work against pressure.

Define the deceleration parameter $q=-a\ddot{a}/{\dot{a}}^2$, the critical density ${\rho }_c=3H^2/(8\pi G)$, and the cosmological density parameters ${\Omega }_m={\rho }_m/{\rho }_c$ and ${\Omega }_{\Lambda }=\Lambda /(3H^2)$. The dust-plus-vacuum acceleration equation then gives $q={\Omega }_m/2-{\Omega }_{\Lambda }$. Dust conservation gives ${\rho }_m\propto a^{-3}$, while ${\rho }_{\Lambda }=\Lambda /(8\pi G)$ is constant. At the acceleration transition, ${\rho }_m=2{\rho }_{\Lambda }$, so

For spatial flatness and $a_0=1$, the constraint today sets ${\Omega }_{\Lambda 0}=1-{\Omega }_{m0}$. Thus $H^2=H_0^2[{\Omega }_{m0}{a}^{-3}+1-{\Omega }_{m0}]$. Put $M={\Omega }_{m0}$, $L=1-M>0$ and $y=a^{3/2}$. Then $\dot{y}=(3H_0/2)\sqrt{M+L{y}^2}$. Integrating from $y(0)=0$ gives $\mathrm{arsinh}(y\sqrt{L/M})=(3H_0/2)\sqrt{L}t$, hence the flat dust-and-vacuum expansionFor small $t$, $a\simeq [(3/2)H_0\sqrt{M}t{]}^{2/3}$, the matter-dominated law. At late times, $a\simeq [M/(4L){]}^{1/3}{e}^{H_0\sqrt{L}t}$, approaching de Sitter spacetime. Negative spatial curvature adds a positive $a^{-2}$ contribution to $H^2$. This is subdominant to dust's $a^{-3}$ at small $a$ and to the constant vacuum term at large $a$, so it does not change either leading limit.

The model describes the transition from matter domination to the accelerated late-time expansion indicated by cosmological distance observations, with nearly flat geometry and vacuum-like dark energy. Its formal $t\to 0$ dust limit is not the actual earliest universe: radiation, neglected here, dominates then and gives $a\propto t^{1/2}$.

For an expanding universe, cosmic inflation requires $\ddot{a}>0$. Including any vacuum contribution in the total density and pressure, the acceleration equation gives $\boxed{\rho +3P<0}$, or $w=P/\rho <-1/3$ for positive density. Equivalently the comoving Hubble radius $(aH{)}^{-1}={\dot{a}}^{-1}$ decreases. Since $|{\Omega }_{\mathrm{total}}-1|=|k|/(aH{)}^2$, prolonged acceleration suppresses curvature and addresses the flatness problem. A sufficiently long inflationary phase also stretches a region that was initially causally connected to encompass the observed region, addressing the horizon problem. The condition on acceleration must last long enough; an arbitrarily short interval is insufficient.

For a homogeneous canonical scalar field $\Phi$ with potential $V(\Phi )$,Thus $V>{\dot{\Phi }}^2$ is sufficient for inflation. A suitably flat positive potential gives slow-roll inflation: Hubble damping in $\ddot{\Phi }+3H\dot{\Phi }+V_{,\Phi }=0$ keeps the kinetic energy small compared with $V$, so $P_{\Phi }\simeq -{\rho }_{\Phi }$.

A physical positive scale factor here requires $t>1/C$. In that domain $\dot{a}=1/t>0$ but $\ddot{a}=-1/t^2<0$. There is no inflationary expansion. The range $0<t<1/C$ has a negative printed scale factor and cannot supply a positive expanding branch of this model.

For $t>0$, $\dot{a}=n{t}^{n-1}>0$ and $\ddot{a}=n(n-1)t^{n-2}$. Consequently inflation occurs exactly when $\boxed{n>1}$. The case $n=1$ is coasting, and $0<n<1$ is decelerating. The Hubble slow-roll parameter is ${ϵ}_H=-\dot{H}/H^2=1/n$, giving the same acceleration criterion ${ϵ}_H<1$.

The Hubble parameter is $H=An{t}^{n-1}>0$, andFor $n=1$ this is positive at every $t>0$, so expansion is exponential and inflationary throughout. For $0<n<1$, acceleration begins only afterThe interval $0<t<t_{\ast }$ is decelerating, while $t=t_{\ast }$ is the transition. Thus the model admits inflation, but not arbitrarily near its specified time origin when $n<1$; choosing parameters can place the transition early. Equivalently ${ϵ}_H=(1-n)/(An{t}^n)$ becomes smaller than one at the same time.

At temperatures above a few MeV, weak interactions such as $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ maintain neutron-proton chemical equilibrium. Negligible lepton chemical potentials give the neutron-to-proton ratio $r_{\mathrm{eq}}=n_n/n_p=\exp [-\Delta /(k_{B}T)]$, where $\Delta =(m_n-m_p)c^2\simeq 1.293\mathrm{MeV}$. At high temperature this ratio is nearly one.

During radiation domination, the interaction time grows as $T^{-5}$, faster than the expansion time $T^{-2}$. Their crossing therefore produces cosmological weak freeze-out. If $\Theta =T/(10^{10}\mathrm{K})$, equating the displayed order-of-magnitude times gives ${\Theta }_f^{-5}\sim g_{\ast }^{-1/2}{\Theta }_f^{-2}$, soUsing $H^{-1}=2t$ instead of $t$ changes only the order-one coefficient. Before electron-positron annihilation, the usual effective relativistic count is $g_{\ast }=2+(7/8)(4+6)=10.75$. The factor $7/8$ weights fermionic relative to bosonic energy density.

After freeze-out, free neutrons decay while protons increase correspondingly. Define the neutron baryon fraction $X_n=n_n/(n_n+n_p)$. In the instantaneous-freeze-out estimate,where ${\tau }_n=t_{1/2}/\ln 2$ is the neutron mean lifetime. Hence the ratio at the onset of nuclear synthesis isThis expression accounts for the protons produced by neutron decay.

Nuclear burning initially waits at the deuterium bottleneck: although deuterium is bound by about $2.2\mathrm{MeV}$, the many photons per baryon can still photodissociate it. Roughly, ${\eta }_b^{-1}{e}^{-B_D/(k_B{T}_N)}\sim 1$ sets the bottleneck temperature, with refinements from reaction rates and phase space. Around $k_B{T}_N\sim 0.1\mathrm{MeV}$, deuterium survives, reactions through deuterium, tritium and helium-3 rapidly build tightly bound helium-4, and nearly all surviving neutrons are captured.

Thus $n_{\mathrm{He}}\simeq n_n/2$ and the primordial helium mass fraction isFor the standard estimate $r_N\simeq 1/7$, neutron and proton constituent fractions are $1/8$ and $7/8$, respectively, and $Y_{\mathrm{He}}\simeq 1/4$. Free neutrons are negligible after capture. The remaining hydrogen mass fraction is about $3/4$, and the helium-to-hydrogen number ratio is about $1/12$.

The printed numerical time coefficients are schematic, not a precision prediction of this quarter fraction: their literal crossing with $g_{\ast }=10.75$ gives $k_B{T}_f\simeq 1.28\mathrm{MeV}$. Keeping physical weak-rate coefficients and the gradual freeze-out instead gives an effective freeze-out around $0.7$–$0.8\mathrm{MeV}$, followed by decay over a few minutes. The abundance formulas above explain the standard quarter fraction and its parameter dependence without treating the rounded time formulas as exact nuclear-rate data.

Reducing the gravitational constant decreases $H$ at fixed temperature. The helium sensitivity to the expansion rate has two effects: weak interconversion persists to a lower freeze-out temperature, leaving fewer neutrons, and the universe takes longer to reach the deuterium-burning temperature, allowing more free-neutron decay after freeze-out. Both lower the helium-4 abundance. In particular $T_f\propto G^{1/6}$ and $t(T)\propto G^{-1/2}$.

At fixed expansion and conversion rates, a smaller neutron half-life reduces ${\tau }_n$ and hence $X_n(t_N)=X_{n,f}{e}^{-(t_N-t_f)/{\tau }_n}$. Fewer neutrons survive, so less helium-4 forms. If the reduced lifetime arises from a stronger weak coupling, its accompanying faster neutron-proton conversion also delays freeze-out and reinforces this decrease.

A thermally populated extra light neutrino species is relativistic during nucleosynthesis; its $0.001\mathrm{eV}$ mass is negligible at MeV temperatures. One neutrino-antineutrino species adds $(7/8)\times 2=7/4$ to $g_{\ast }$. Thus the expansion rate increases by $\sqrt{12.5/10.75}$ and the freeze-out temperature by $(12.5/10.75{)}^{1/6}$. Earlier freeze-out leaves more neutrons, while faster cooling leaves less time for decay. The helium-4 abundance increases. This assumes the new species was populated at the standard neutrino temperature; an unpopulated species would not contribute this extra density.

The equilibrium neutron-to-proton ratio is $e^{-\Delta /(k_{B}T)}$. Increasing the neutron-proton mass difference suppresses neutron abundance at the same freeze-out temperature. Therefore less helium-4 forms, with other inputs held fixed. Any associated shortening of neutron lifetime further reduces survival, rather than reversing this direct effect.

The helium yield is primarily neutron-limited, so small changes in cosmological baryon density have only a weak effect compared with their effect on residual deuterium. Increasing the baryon-to-photon ratio makes deuterium destruction less effective relative to its formation and releases the deuterium bottleneck at a slightly higher temperature. Nuclear reactions also proceed faster. Fewer neutrons decay before capture, so a small increase in baryon density slightly increases helium-4; a decrease slightly lowers it.

At nucleosynthesis temperatures, $50\mathrm{GeV}$ particles are nonrelativistic. Their equilibrium abundance is negligible because of Boltzmann suppression, so they do not behave as an extra relativistic species contributing its full degree count to $g_{\ast }$. A stable thermal weakly interacting massive particle generally freezes out much earlier; a surviving relic population with the usual dark-matter abundance remains negligible compared with radiation during nucleosynthesis. In that standard relic scenario the helium abundance is essentially unchanged.

The mass and weak coupling alone do not specify the relic density. If an unusually abundant stable population contributes appreciably to the expansion density, it raises $H$ at fixed radiation temperature and increases helium by the same freeze-out and survival mechanism as part (g). Decays or energetic annihilation products would require their own specified history. Thus a universal numerical change cannot be inferred from the mass alone.

A faster expansion at fixed temperature makes the weak reaction rate fall below $H$ sooner, at higher temperature, leaving a larger neutron-to-proton ratio. It also shortens the time before the deuterium bottleneck ends, reducing neutron decay. The helium-4 abundance increases, assuming nuclear rates and the temperature history of the particle bath are otherwise unmodified.

For an equilibrium relativistic bath, $P=\rho /3$ and $\rho \propto g_{\ast s}{T}^4$, so $s=(\rho +P)/T\propto g_{\ast s}{T}^3$. Cosmological entropy conservation gives $g_{\ast s}{T}^3{a}^3=\text{constant}$. A decoupled relativistic species instead has $a{T}_{\mathrm{relic}}=\text{constant}$, so annihilation heats the bath relative to that relic.

Use the sharp relativistic degree-counting approximation specified by the temperature interval. Just before muon annihilation, the coupled bath contains photons, electron-positron pairs, muon-antimuon pairs and three neutrino-antineutrino species:After muon annihilation, neutrinos are still coupled and $g_{\ast s}^{\mathrm{after}}=2+(7/8)(4+6)=43/4$. The decoupled H particles are excluded from these bath counts. Equality of bath and relic temperatures at decoupling therefore gives the relic cooling across muon annihilation

By the later electron-mass epoch, neutrinos have decoupled. Electron-positron annihilation heats only the electromagnetic bath, reducing its entropy degree count from $2+(7/8)4=11/2$ to $2$. Photons are heated by $(11/4{)}^{1/3}$ relative to both neutrinos and H particles. After this transition,The neutrino-to-photon ratio is $(4/11{)}^{1/3}$, while the H-particle-to-neutrino ratio remains $(43/57{)}^{1/3}$. The printed $0.59\mathrm{MeV}$ marks the electron-positron annihilation epoch approximately; it is not an exact annihilation threshold.

In Questions 3–4, use primes for conformal-time derivatives and $\mathcal{H}=a^{\prime }/a$; proper-time derivatives will be stated separately. Let ${\xi }^{\mu }=(T,L^i)$. The metric is a covariant tensor, so under the passive coordinate change ${\tilde{x}}^{\mu }=x^{\mu }+{\xi }^{\mu }$,The inverse Jacobian is ${\delta }_{\mu }^{\alpha }-{\partial }_{\mu }{\xi }^{\alpha }$ to first order. At the same new coordinate argument, the homogeneous background satisfies ${\bar{g}}_{\alpha \beta }(x)={\bar{g}}_{\alpha \beta }(\tilde{x})-T{\bar{g}}_{\alpha \beta }^{\prime }$. Shifting the argument of $\delta g$ or multiplying it by a Jacobian perturbation produces only second-order terms, which we discard. Subtracting the unperturbed metric therefore givesThis is the printed Jacobian expression expanded to linear order, equivalently $\delta \tilde{g}=\delta g-{\mathcal{L}}_{\xi }\bar{g}$. It is a gauge transformation in cosmological perturbation theory, not a change of the physical spacetime.

The perturbations obey $\delta g_{00}=2a^2\psi$, $\delta g_{0i}=-a^2{B}_i$, and $\delta g_{ij}=2a^2(\varphi {\delta }_{ij}-E_{ij})$. Apply part (i). The time-time component changes by $-2a^2(T^{\prime }+\mathcal{H}T)$, and the time-space component changes by $a^2(L_i^{\prime }-{\partial }_{i}T)$. HenceFor the spatial components,Taking the trace and the symmetric trace-free part separately giveswhere ${\partial }_{⟨i}{L}_{j⟩}=\frac{1}{2}({\partial }_i{L}_j+{\partial }_j{L}_i)-{\delta }_{ij}{\partial }_k{L}^k/3$. This explicitly accounts for the trace term in $\varphi$; it must not be absorbed into the trace-free shear.

A matter worldline has coordinate velocity $v^i=d{x}^i/d\eta$. On that same worldline,Thus ${\tilde{v}}^i=v^i+L^{i\prime }$. A time-dependent spatial relabelling makes a physically unchanged worldline move relative to the new coordinate grid. This is the frame dependence of the coordinate peculiar velocity, not an added physical acceleration.

For the scalar gauge transformation with trace-free spatial shear, put $L_i={\partial }_{i}L$. Part (ii) and the velocity rule giveDefine $C=\varphi +{\nabla }^{2}E/3$ and $w=B+v$. Then $\tilde{C}=C+\mathcal{H}T$ and $\tilde{w}=w+T$. Therefore $\mathcal{R}=-C+\mathcal{H}w$ is gauge-invariant, because the two time-shift terms cancel.

To identify the hypersurfaces, lower the index on the matter four-velocity. To first order its spatial components are $u_i=-a(B_i+v_i)=-a{\partial }_{i}w$. The matter worldlines are orthogonal to constant-time slices when $u_i=0$, which for the inhomogeneous scalar modes is $w=0$. Choose $T=-w$ to reach this comoving slicing. Then $C_{\mathrm{co}}=C-\mathcal{H}w=-\mathcal{R}$. Applying the supplied intrinsic-curvature formula on those slices givesSpatially homogeneous additions to the scalar potentials do not affect this curvature.

Work in conformal Newtonian gauge, where $\mathcal{R}=-\varphi +\mathcal{H}v$. SetThe momentum equation gives $v=-({\varphi }^{\prime }+\mathcal{H}\varphi )/A$, henceThis assumes nonzero background enthalpy, as required for the matter comoving slicing. Background energy conservation gives ${\bar{\rho }}^{\prime }=-3\mathcal{H}(\bar{\rho }+\bar{P})$ and thus

The density constraint gives $4\pi G{a}^2\delta \rho ={\nabla }^2\varphi -3\mathcal{H}({\varphi }^{\prime }+\mathcal{H}\varphi )$. Insert this into the pressure equation to obtainDifferentiating the expression for $\mathcal{R}$, using the formula for $A^{\prime }/A$ and then $A={\mathcal{H}}^2-{\mathcal{H}}^{\prime }$, groups the terms asThe preceding equation proves the gradient evolution identity for comoving curvature:For a Fourier mode, ${\nabla }^2\varphi =-k^2\varphi$. On super-Hubble scales, $k/\mathcal{H}\ll 1$, its evolution is gradient-suppressed. With the stated regular adiabatic relation between $\mathcal{R}$ and $\varphi$, $\mathcal{R}$ is constant at leading long-wavelength order. At finite wavelength there are corrections; the conclusion is not an exact claim that the Laplacian vanishes.

Inflation produces the primordial comoving curvature perturbation while modes exit the Hubble radius. Its adiabatic super-Hubble conservation allows that initial amplitude to be carried through the end of inflation and subsequent changes of the background equation of state without solving each mode's full evolution at every intervening time. After Hubble re-entry, cosmological transfer functions relate it to the cosmic microwave background temperature and polarization fluctuations and to the matter power spectrum. The conserved curvature perturbation connects inflationary initial conditions to later cosmological observables. This bridge uses the adiabatic conditions stated in part (v); entropy perturbations can supply an additional evolving source.

Use $\hslash =c=1$. For a canonical light scalar field with effective mass $m^2\ll H^2$, define the canonical Fourier mode $u_k=a\delta {\Phi }_k$. In the nearly de Sitter approximation it obeysDeep inside the Hubble radius, $k\gg aH$, the mode behaves like a Minkowski oscillator. Choosing the Bunch-Davies vacuum sets $u_k\to e^{-ik\eta }/\sqrt{2k}$. The normalized de Sitter solution isInflation stretches the physical wavelength until it exits the Hubble radius. For $|k\eta |\ll 1$, the growing scalar solution freezes with $|\delta {\Phi }_k{|}^2=|u_k/a{|}^2\to H^2/(2k^3)$. The dimensionless power spectrum is thereforeHere $H_k$ is evaluated when $k=aH$, since slow evolution near horizon exit fixes the frozen amplitude. Nearly constant $H$ and a light field make this spectrum almost scale-invariant; slow background evolution and small mass corrections produce a small tilt. The stretched vacuum modes supply the primordial stochastic fluctuations.

Vary the quadratic Einstein-Hilbert action within the transverse trace-free tensor sector. Integration by parts giveswith endpoint terms set to zero. Hence the tensor equation isA Fourier mode on super-Hubble scales has $k^2\ll {\mathcal{H}}^2$, so its leading equation is $(a^2{E}_{ij}^{T\prime }{)}^{\prime }=0$. Its solutions areThere is thus a constant tensor mode. During inflation the second mode decays; in exact de Sitter expansion it is proportional to $a^{-3}$ after shifting the constant of integration. Constancy is the leading super-Hubble behaviour, exactly realized for the zero-gradient solution.

Reality of $E_{ij}^T$ and the given polarization relation imply ${\psi }_{(p)}(-\mathbf{k})={\psi }_{(p)}(\mathbf{k}{)}^{\ast }$. Insert the Fourier expansion twice into each quadratic term. The spatial integral givesThis cancels the two factors of $(2\pi {)}^{-3/2}$. Set ${\mathbf{k}}^{\prime }=-\mathbf{k}$ and use $M_{ij}^{(q)}(-\mathbf{k})=M_{ij}^{(q)}(\mathbf{k}{)}^{\ast }$. The unit-normalization condition eliminates cross-polarization terms, givingFor the gradient term, $(i{k}_l)(i{k}_l^{\prime })$ becomes $k^2$ after the delta-function substitution. ThusThe full momentum-space integral and its reality relation are those of real fields; there is no additional independent copy of the negative-wavevector modes.

The tensor normalization with unit-norm polarization tensors identifies the canonically normalized real scalar for each helicity as $\delta {\Phi }_{(p)}=M_{\mathrm{Pl}}{\psi }_{(p)}$. The two polarization actions are independent and have precisely the massless scalar normalization, so the vacuum fluctuations derived in the introductory solution apply to each of them:In the isotropic Bunch-Davies vacuum, the independent helicities have no cross correlation. Translation invariance gives the wavevector delta function, and the dimensionless-spectrum convention givesThis is power per polarization in the printed metric convention $h_{ij}=2E_{ij}^T$ and unit-norm polarization basis. Summing polarizations or quoting a spectrum for $h_{ij}$ changes normalization factors; neither operation belongs in the displayed per-polarization ${\mathcal{P}}_{\psi }$.

Now use proper-time dots and $H=\dot{a}/a$. Differentiate the slow-roll relation $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$, then substitute $\dot{\bar{\Phi }}\simeq -V_{,\Phi }/(3H)$:Thus the Hubble slow-roll parameter ${ϵ}_H$ agrees with ${ϵ}_V$ to this order. Since ${\mathcal{P}}_{\psi }\propto H_k^2$ and $k=aH$ at horizon exit,Consequently the tensor spectral index isThe final equality is the requested first-order slow-roll approximation; corrections beyond that order also affect the horizon-exit spectrum itself.