Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 310 (Cosmology)

Use units $c=1$ throughout. Set $\chi =|\mathbf{x}|$, so the spatial Euclidean metric is $d{\mathbf{x}}^2=d{\chi }^2+{\chi }^{2}d{\Omega }^2$. On either hemisphere of the embedded three-sphere, the constraint gives $u^2=R^2-{\chi }^2$ and henceIntroduce the dimensionless comoving coordinate $r=\chi /R$. ThenThis is the positively curved Friedmann-Lemaître-Robertson-Walker metric, with $a(t)$ the physical curvature radius. An equally valid convention retains $r=\chi$, takes $a=b$, and sets $K=R^{-2}$; only the coordinate normalization differs. The chart with $r\le 1$ describes a hemisphere, not the entire closed slice. Writing $r=\sin \psi$ gives $d{\ell }^2=R^2(d{\psi }^2+{\sin }^2\psi d{\Omega }^2)$, with $0\le \psi \le \pi$, displaying the full spherical spatial slices of a closed FLRW universe.

For separately conserved cosmological fluids, the prerequisite is absence of energy or momentum exchange between components. Each matter action then gives its own local conservation equation ${\nabla }_{\mu }{T}_i^{\mu \nu }=0$. In the homogeneous perfect fluid in general relativity, its time component isThis is the cosmological perfect-fluid continuity equation. Equivalently, a physical volume $V\propto a^3$ following the fluid satisfies $d({\rho }_{i}V)=-P_{i}dV$.

The Einstein field equations instead have the single shared metric on its left and ${\sum }_i{T}_i^{\mu \nu }$ on its right. The Friedmann equation and Friedmann acceleration equation therefore involve total density and total pressure; the same geometry cannot be sourced separately by each component's density alone.

The PDF's separate-conservation statement needs this noninteraction qualification. In an interacting system,where $Q_i$ is the energy-transfer rate into component $i$. Total conservation remains true, but the source-free equation does not hold individually during processes such as annihilation or particle decay.

For a constant equation-of-state parameter $w=P/\rho$, the cosmological perfect-fluid continuity equation integrates to $\rho \propto a^{-3(1+w)}$. Comparing the given powers yieldsThis constant-equation-of-state density scaling describes a cosmic string network and a cosmological domain wall network. An intuitive check is that the energy in a fixed comoving cell grows in proportion to the total string length, $a$, or total wall area, $a^2$, while the physical cell volume grows as $a^3$. Thus the densities scale as $a^{-2}$ and $a^{-1}$ respectively. These effective equations of state use the noninteracting, statistically isotropic network assumptions of the problem.

Choose a reference epoch with $H_0=\dot{a}(t_0)/a_0\ne 0$ and let $x=a/a_0$. Define the critical density and each cosmological density parameter byFor a closed universe ${\Omega }_{K,0}<0$. Take the dark-energy component to be a cosmological constant, as indicated by its $\Lambda$ label. The constant-equation-of-state density scaling givesSubstitution into the Friedmann equation givesAt the reference epoch, ${\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{s,0}+{\Omega }_{w,0}+{\Omega }_{\Lambda ,0}+{\Omega }_{K,0}=1$. The Friedmann acceleration equation isStrings have ${\rho }_s+3P_s=0$ and make no direct acceleration contribution, although they affect the Friedmann constraint. The normalized density parameters are undefined at a static epoch, where $H_0=0$; the static solution must therefore be described directly with dimensional densities.

For a single positive-density component, staticity requires $\rho +3P=0$ from the Friedmann acceleration equation. Among the listed components, only the cosmic string network has $w=-1/3$. Put $a(t)=a_E>0$ and use the Friedmann equation to obtainAll other components are absent. This is the static closed cosmic-string universe.

There is a normalization condition hidden in this expression. String conservation fixes $C_s={\rho }_s{a}^2$, and a static solution exists only whenOnce this condition is imposed, every positive radius $a_E$ is a static solution, with its density determined as above. One cannot independently prescribe an arbitrary string normalization and then choose a radius to compensate: both the string density term and curvature term scale as $a^{-2}$.

First perturb the radius consistently with the conserved string normalization. If $a_E\mapsto a_E(1+ϵ)$ while $C_s={\rho }_s{a}^2$ stays at its static value, the density becomesThe Friedmann equation still has an exact cancellation between the string and curvature terms, and the Friedmann acceleration equation still vanishes. ThusThe homogeneous stability of a static cosmic-string universe is neutral in this restricted sense: there is no restoring force and no exponentially growing displacement. Holding the density fixed while changing the radius would instead change $C_s$ and would not be this pure displacement perturbation.

The distinction between the evolution equation and the constraint is essential. For strings alone, the exact equations reduce toLinearizing the acceleration equation gives $\delta a=A+Bt$. At the exactly tuned static $C_s$, however, the full Friedmann equation requires ${\dot{a}}^2=0$, so $B=0$ for an admissible exact perturbation.

A small nonzero velocity $v$ can be admissible if its initial density also changes:Arbitrarily small $v$ and the corresponding second-order density change therefore produce a secular departure. There is no Lyapunov stability against all nearby constrained initial data for the static closed cosmic-string universe. Neutral radius shifts at fixed normalization and drift when the normalization changes are different perturbation classes.

Increase the density at the initial radius by a small fraction $ϵ>0$, maintaining the string equation of state. Then $C_s=C_E(1+ϵ)$ and the Friedmann equation fixesBecause $\ddot{a}=0$, the exact expanding and contracting branches areThe expanding branch grows indefinitely; the contracting branch reaches zero scale factor after $a_E/\sqrt{Kϵ}$. Thus a positive density-normalization perturbation destroys staticity. The departure speed is proportional to $\sqrt{ϵ}$, reflecting the fact that the Friedmann constraint is quadratic in the velocity at the static point. This is another aspect of homogeneous stability of a static cosmic-string universe.

If the initial density is decreased by $ϵ<0$ at the same radius, the string-only Friedmann equation would requirewhich has no real solution. It is not an allowed initial perturbation of a closed string-only universe at fixed curvature normalization. In particular, it should not be interpreted as an oscillating or automatically recollapsing solution.

Likewise, increasing the density while keeping both $a$ and $\dot{a}=0$ fixed violates the constraint; the admissible velocity is the one obtained above. The homogeneous stability of a static cosmic-string universe must always be assessed using both the acceleration equation and the Friedmann constraint. Allowing extra components or changing the spatial curvature would define a different perturbation problem.

Interpret $P(k)$ in the variance integral as the unsmoothed cosmological density power spectrum of $\delta$. A linear smoothing window gives $P_R(k)=P(k)|\tilde{W}(kR){|}^2$ for the smoothed field; the PDF's description of $P(k)$ as already belonging to ${\delta }_R$ would otherwise count the window twice.

For a scale-free matter power spectrum $P(k)=A{k}^{n_{\mathrm{eff}}}$, substitute $q=kR$ to obtainWhenever the dimensionless integral is finite and nonzero, this proves the scale-free smoothed density varianceThe printed condition $n_{\mathrm{eff}}<3$ is insufficient by itself. For a normalized window with $\tilde{W}(0)=1$, infrared convergence requires $n_{\mathrm{eff}}>-3$. Ultraviolet convergence depends on the filter: a spherical top-hat window function requires $n_{\mathrm{eff}}<1$, a Gaussian filter suppresses every ultraviolet power, and a sharp-k smoothing filter has finite Fourier support. In particular, the monotone relation $R\to \infty \Rightarrow S={\sigma }_R^2\to 0$ requires $n_{\mathrm{eff}}>-3$ in the scale-free model.

At a single variance $S_{\ast }$, the Gaussian random field has ${\delta }_{S_{\ast }}\sim N(0,S_{\ast })$. Thus its endpoint tail isHere $\mathrm{erfc}$ is the complementary error function.

For the crossing probability, the key assumption is independent, symmetric increments as the smoothing scale changes. A sharp-k smoothing filter, $\tilde{W}(kR)=1_{k<1/R}$, supplies this property: decreasing $R$ adds independent Fourier shells of the underlying Gaussian field. Parameterizing their accumulated variance by $S$ gives a Brownian motion with covariance $\min (S,S^{\prime })$. An arbitrary real-space window gives correlated increments and does not justify this argument.

Reflect a trajectory after its first hit of ${\delta }_c$. The Strong Markov property and symmetric future increments preserve its probability, and reflection maps an endpoint below the barrier to one above it. Every endpoint above the barrier has already crossed; the reflected paths give an equally probable set that crossed but ended below. The Brownian reflection principle therefore yields the excursion-set description of halo formation:Its derivative is the Brownian first-passage time densityThe factor of two resolves trajectories that crossed a larger-scale barrier but finish below it on the chosen smaller scale.

Define the halo peak height $\nu ={\delta }_c/{\sigma }_R$. The cumulative probability $\mathcal{P}(>M)$ is a dimensionless mass fraction. With $V_M=M/\bar{\rho }$, its conversion to halo number density givesConsequently the Press-Schechter halo mass function isIt is positive when ${\sigma }_R$ decreases with mass. A chosen mass-radius convention $M\propto R^3$ gives $d\log {\sigma }_R/d\log M=-(n_{\mathrm{eff}}+3)/6$. For a sharp Fourier filter that mass assignment needs a convention, since its real-space kernel is not a localized top-hat volume.

For warm dark matter, the cutoff introduces a new length and invalidates the pure power-law variance scaling near that length. With the same sharp-k filter and $n_{\mathrm{eff}}>-3$, the variance is exactlyThe Brownian walk runs only up to $S_{\max }$ and then stops. The cutoff does not destroy independence of the Fourier shells that are still present. Thus the first-crossing density and derivative mass-function formula remain valid on the invertible part of $S(M)$, using the actual cutoff variance. For $R<k_{\mathrm{WDM}}^{-1}$, $d{\sigma }_R/dM=0$ and this idealized sharp-filter model has no new crossings or new low-mass haloes. Its total collapsed fraction isThis is halo first crossing with a finite variance cutoff: some trajectories never cross, so one must not force the collapsed fraction to one. The uncut power-law abundance cannot be extrapolated to small masses. For other windows, a variance plateau still occurs but the Markov/reflection derivation is not exact; the same mass function is then a modelling approximation, rather than a result established by this calculation.

Let $V$ be the physical volume of a region with fixed comoving coordinates, so $V\propto a^3$, and write $U=\rho V$. The first law of thermodynamics is $dU=TdS-PdV+\mu dN$. Use extensive quantities: under replication, $U(\lambda S,\lambda V,\lambda N)=\lambda U(S,V,N)$. The Euler theorem for homogeneous functions givesSetting $\mu =0$ yieldsThis is the entropy form of the Gibbs-Duhem equation; it fixes the normalization by extensivity, rather than adding an arbitrary volume-independent entropy constant.

For equilibrium expansion of the same region, the first law and the cosmological perfect-fluid continuity equation implyTherefore $\boxed{\dot{S}=0}$, establishing cosmological entropy conservation for adiabatic equilibrium evolution. Here “comoving volume” means a patch moving with the cosmological fluid; its physical volume $V$ changes. The corresponding coordinate volume is constant. The local TeX's $\dot{V}/P$ in the continuity-equation hint is a transcription error: the PDF correctly has $\dot{V}/V$.

For a relativistic equilibrium species with temperature $T_i$, the entropy density is $s_i=({\rho }_i+P_i)/T_i=4{\rho }_i/(3T_i)$. In units $k_B=\hslash =c=1$, bosons have ${\rho }_i={\pi }^2{g}_i{T}_i^4/30$, while fermions have $7/8$ times this expression. Relative to a reference bath temperature $T$, define the effective number of relativistic degrees of freedom for entropy byThe sums here concern relativistic species, or their effective entropy contributions. Since a physical comoving patch has volume proportional to $a^3$, cosmological entropy conservation givesfor an isolated adiabatic bath. After thermal decoupling in cosmology, one must apply this law separately to sectors that no longer exchange entropy.

Before electron-positron annihilation in cosmology, the electromagnetic bath contains two photon polarizations and four electron/positron spin states. After neutrino decoupling, its entropy degrees of freedom are initiallyand after annihilation only the two photon polarizations remain. Electromagnetic entropy conservation therefore givesThe decoupled Cosmic neutrino background receives none of this entropy; its distribution temperature obeys $a{T}_{\nu }=\text{constant}$, and $T_{\nu }=T_{\gamma }$ before annihilation. ThusThis is the temperature of a decoupled relativistic relic in the instantaneous-decoupling approximation. The ratio remains the same afterwards in the absence of additional photon heating. At late times $T_{\nu }$ denotes the redshifted relic distribution temperature even when neutrino masses become relevant; it need not be a kinetic-equilibrium temperature.

Before decay, the nonrelativistic $\chi$ density redshifts as $a^{-3}$, while the photon density redshifts as $a^{-4}$. Therefore their ratio grows as $a$. Let a minus sign denote the instant just before decay and defineUsing the assumed radiation domination law $a\propto t^{1/2}$ gives $r_d=(t_d/t_{\mathrm{ref}}{)}^{1/2}{\rho }_{\chi }(t_{\mathrm{ref}})/{\rho }_{\gamma }(t_{\mathrm{ref}})$.

Across an instantaneous decay the scale factor does not change. Energy conservation and complete thermalization into a zero-chemical-potential photon bath giveSince blackbody radiation has ${\rho }_{\gamma }\propto T_{\gamma }^4$, instantaneous photon heating by a decaying relic givesThe already decoupled Cosmic neutrino background is not heated. Dividing its unchanged temperature by the new photon temperature therefore yieldsThe relation written with $a_d/a_{\mathrm{ref}}$ is independent of the expansion approximation. Replacing it by the time ratio uses the radiation-dominated background assumed in the question; an appreciable matter contribution would require solving for that background instead. This photon entropy injection is not adiabatic, so the pre-decay photon entropy need not be conserved across the decay.

The photon number density in a thermal zero-chemical-potential bath scales as $T_{\gamma }^3$. The heating factor from the previous calculation therefore increases photon number at fixed physical volume by $(1+r_d{)}^{3/4}$. The decay is nonbaryonic and preserves baryon number, giving baryon-to-photon dilution by entropy injection:If the decay occurs after the light-element abundances have frozen out in Big Bang nucleosynthesis but before the epoch probed by the cosmic microwave background, abundance measurements infer the earlier ${\eta }_{\mathrm{BBN}}$, whereas the CMB measures the later ${\eta }_{\mathrm{CMB}}$. In this simple scenario,A significant discrepancy in this direction could indicate photon entropy injection by $\chi$. Several light-element abundances help distinguish a changed baryon density from a changed expansion history. If decay occurs during nucleosynthesis, the inference instead requires time-dependent $\eta$ and a modified expansion history; the before/after comparison cannot automatically be applied to every decay time allowed in the question.

In conformal time, $\sqrt{-g}=a^4$ and $g^{\mu \nu }=a^{-2}\mathrm{diag}(-1,1,1,1)$. The inflaton action in an expanding universe becomesVary $\varphi$ with compact support, integrate the derivative terms by parts, and discard the boundary term. The Euler-Lagrange field equation isDividing by $a^2$ provesThe factor two is the conformal-time form of Hubble friction. Using $d\tau =dt/a$ converts it to the usual inflaton equation of motion $\ddot{\varphi }+3H\dot{\varphi }-a^{-2}{\nabla }^2\varphi +V_{,\varphi }=0$, providing a sign and normalization check.

Subtract the homogeneous background equation and linearize the inflaton equation of motion. The field perturbation satisfiesWith $f=a\delta \varphi$, direct differentiation cancels the first-derivative term and givesThe conformal-time quadratic action for an inflaton perturbation, after a boundary term is discarded, isFor the specified de Sitter spacetime background, $a^{\prime \prime }/a=2/{\tau }^2=2a^2{H}^2$. The light-field approximation $|V_{,\varphi \varphi }|\ll 2H^2$ therefore permits dropping the mass term. Taking a spatial Fourier transform yields the Canonically rescaled de Sitter scalar mode equationIts kinetic term is canonical, with momentum ${\pi }_f=f^{\prime }$. Each independent real Fourier component consequently has the canonical coordinate/momentum algebra of a quantum harmonic oscillator, with time-dependent squared frequency ${\omega }_k^2=k^2-a^{\prime \prime }/a$. This is why canonical quantization of a real scalar field proceeds by oscillator creation and annihilation operators. On superhorizon scales the squared frequency is negative: the system is then an inverted time-dependent oscillator, not a stationary oscillator with a positive frequency. The canonical commutation relations remain valid.

Use the Fourier normalization of the PDF. The creation and annihilation operators obeyThere is no $(2\pi {)}^3$ in this commutator because the expansion uses $(2\pi {)}^{-3/2}$. The Bunch-Davies vacuum satisfies ${\hat{a}}_{\mathbf{k}}|0⟩=0$. Write $u_k=f_k^{\ast }$ for the coefficient of the annihilation operator. Only the contraction $⟨0|{\hat{a}}_{\mathbf{k}}{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}|0⟩={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$ contributes, soThe given mode hasSince $a^{-2}=H^2{\tau }^2$, the equal-time two-point function of a de Sitter scalar therefore has dimensional power spectrumComparing with the dimensionless power convention in the hint givesEquivalently, angular integration expresses the correlation as ${\int }_0^{\infty }(dk/k){\Delta }_{\delta \varphi }^2\sin (kr)/(kr)$, where $r=|\mathbf{r}|$.

On superhorizon scales, $k\ll aH$ is $|k\tau |\ll 1$, andThis scale-invariant inflationary power spectrum gives equal variance per logarithmic wavenumber interval. It is exactly scale invariant in the massless constant-$H$ approximation here; slow evolution of $H$ and small field mass produce a small spectral tilt. Such quantum primordial perturbations supply the initial fluctuations that subsequently seed cosmological structure after conversion to curvature and density perturbations. The curvature conversion itself requires the metric perturbations excluded from this calculation.

The continuum correlation is understood with the usual infrared regulator, such as finite volume or finite inflationary duration. The strictly massless infinite-volume model has a logarithmic infrared divergence. Coincident products also require ultraviolet regularization or smearing; the finite mode spectrum is unaffected by these qualifications.

Under the stated linear field expansion, the Bunch-Davies vacuum is a centered Gaussian state. By Wick theorem, vacuum correlators reduce to pair contractions; an odd product has no complete pairing. In particular,This holds for arbitrary spatial arguments as a regulated correlator or distribution. One can also use number parity: $\mathcal{P}=(-1{)}^{\hat{N}}$ leaves the vacuum unchanged and sends every creation or annihilation operator, hence the linear field $\hat{f}$, to its negative. An odd product consequently has expectation equal to its own negative. Thus vanishing odd correlators of a free Gaussian field givesFor a coincident product, impose a momentum regulator or smear the field before taking the expectation; the zero result holds for every such symmetry-preserving regulator. It is not a claim that the unregulated local power is an ordinary finite random variable. Interactions, a non-Gaussian state, or inclusion of a nonzero background mean can invalidate the odd-moment conclusion. Within this free perturbation calculation, the primordial bispectrum vanishes.