Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 53 (Cosmology)

The particle horizon is the greatest distance from which a light signal could have reached the observer since the initial cosmic time $t_i$. Its comoving radius and its radial proper distance at time $t$ areHere proper distance is measured along the spatial slice, rather than by the transverse area of a sphere. The Hubble parameter, deceleration parameter and cosmological density parameter areThe last expression defines the critical density. For blackbody radiation, $p=\rho /3$, so the Friedmann acceleration equation gives $\ddot{a}/a=-\kappa \rho /3$. ConsequentlyThis is the radiation relation between deceleration and density. It does not require spatial flatness.

For the open radiation universe particle horizon, write $x=a/a_0$. The cosmological continuity equation gives $\rho ={\rho }_0{x}^{-4}$, while the present Friedmann equation gives $-k/a_0^2=H_0^2(1-{\Omega }_0)$. ThusTaking the big-bang endpoint $x=0$ and using $dt=dx/\dot{x}$, the present radial proper distance becomesThereforeThe formula applies to $0<{\Omega }_0<1$. Its flat-radiation limit is $H_0{d}_{h0}\to 1$, agreeing with $a\propto t^{1/2}$ and $d_h=2t=H^{-1}$. The divergence as ${\Omega }_0\to 0$ reflects the unbounded past conformal interval of the limiting empty open model; that endpoint is not a radiation-filled universe.

The horizon problem concerns the nearly uniform temperature of widely separated parts of the cosmic microwave background. In a purely decelerating Hot Big Bang model, their past light cones at last scattering do not overlap far enough to explain this agreement by thermal contact. During cosmic inflation, accelerated expansion shrinks the comoving Hubble radius. A patch initially small enough for causal communication can be stretched to encompass the later observable universe. Reheating converts the inflationary energy into a hot plasma with correlated initial conditions across that patch. Enough inflation must occur before the observable scales leave the Hubble radius; inflation cannot establish contact between regions that were never initially causally related.

For a quantitative thermal-relic prediction, assume one ordinary weakly interacting neutrino flavor: one populated neutrino helicity and one antineutrino helicity, negligible chemical potential, early thermal equilibrium, and relativistic decoupling around the MeV epoch. Equal neutrino and antineutrino populations imply no new lepton asymmetry. The Fermi-Dirac distribution is maintained while the weak interaction rate exceeds the expansion rate. After neutrino decoupling, the distribution free streams, its momentum scale falls as $a^{-1}$, and its comoving number is conserved. A mass of $1\mathrm{keV}$ is negligible at Big Bang nucleosynthesis temperatures, but the relic later becomes nonrelativistic. Its energy density then changes from radiation-like $a^{-4}$ scaling to matter-like $a^{-3}$ scaling. Stability prevents its removal by decay; decoupling while relativistic prevents a subsequent equilibrium Boltzmann suppression of its number.

Before electron-positron annihilation in cosmology, the pair has two spin states, just as photons do, and the fermionic number integral gives $n_{\nu +\bar{\nu }}/n_{\gamma }=3/4$ at equal temperature. The factor follows by expanding $(e^x+1{)}^{-1}$ into an alternating exponential series: its $x^2$ integral is $3/4$ of the bosonic one.

After neutrino decoupling, cosmological entropy conservation in the photon-electron plasma changes its relativistic entropy count from $2+(7/8)4=11/2$ to $2$. The decoupled neutrinos do not share this heating. HenceThese counts include both the new neutrino and its antineutrino; another factor of two would double count them. This is the number density of a relativistic neutrino relative to photons, evaluated after photon heating.

They are nonrelativistic today, so the relic-neutrino energy density is dominated by rest mass:The number follows from the supplied critical density, without a Hubble-unit convention. It is conditional on the thermal history stated above; a sterile, colder or nonthermally produced state need not have this abundance.

At weak freeze-out, photons, electrons, positrons and all four neutrino flavors are relativistic and approximately at the same temperature. The effective number of relativistic energy degrees of freedom changes fromThe expansion rate at fixed temperature is larger by $\sqrt{50/43}\simeq 1.078$. Equating the supplied weak interaction time with the supplied age gives $T_f^3=\sqrt{g_{\ast }}$ when $T_f$ is measured in MeV. ThereforeEarlier cosmological weak freeze-out at a higher temperature leaves more neutrons. With negligible chemical potentials, the neutron-to-proton ratio in equilibrium is $r_f=\exp (-Q/T_f)$, where $Q=m_n-m_p\simeq 1.293\mathrm{MeV}$. If almost all surviving neutrons enter helium-4, the primordial helium mass fraction before neutron decay is $Y_f=2r_f/(1+r_f)$. Literal use of the given rate normalizations yieldsEquivalently, the leading response is $\delta Y_f/Y_f=[Q/(T_f(1+r_f))]\delta T_f/T_f$. This is a roughly one-percentage-point absolute increase in the helium fraction, rather than a one-percent relative increase.

The deuterium bottleneck delays nuclear burning, allowing free-neutron decay after freeze-out. If its end temperature is $T_N$, and ${\tau }_n$ is the neutron mean life, a sharper instantaneous-freeze-out estimate conserves total baryon number and givesThis formula evolves the neutron fraction, not just the neutron-to-proton ratio, because each neutron decay also creates a proton. After electron annihilation, let $r_T=(4/11{)}^{1/3}$. The radiation counts relevant to the age at fixed $T_N$ areConsequently $t_N^{(4)}/t_N^{(3)}\simeq \sqrt{3.36/3.82}\simeq 0.939$. Less time is available for neutron decay, adding to the helium increase. As an explicit illustrative completion, $T_N=0.07\mathrm{MeV}$ and ${\tau }_n\simeq 880\mathrm{s}$ give an increase of approximately $0.012$ in this toy model. The precise helium fraction requires the nuclear-burning temperature, baryon abundance and properly normalized weak rates; the question supplies only approximate time laws. In particular, the large no-decay fractions in the table must not be interpreted as precise predictions for the observed primordial abundance. The robust effect is extra relativistic energy raises the primordial helium fraction, with an absolute change of order $10^{-2}$ for one thermally populated flavor.

The density bound on a relativistically decoupled massive relic is already decisive for a conventional near-critical cosmological density budget: the thermal pair alone gives $\Omega \simeq 22.4$. The same additional relativistic pair increases the effective number of neutrino species by one during nucleosynthesis and raises the primordial helium mass fraction, so its expansion effect supplies another constraint. Thus a stable, fully thermalized MeV-decoupling neutrino pair of this mass is incompatible with the usual cosmological density budget.

This conclusion is not a prohibition on every particle with that mass. Earlier decoupling followed by entropy dilution, a reduced production abundance, or interactions that change the assumed thermal history can alter both tests. At fixed mass, requiring this pair to contribute no more than the supplied critical density would demand a number density at most $1/22.4$ of the standard value, or a thermal temperature at most $(1/22.4{)}^{1/3}\simeq 0.355$ of the standard neutrino temperature. No numerical abundance or exclusion independent of these assumptions follows from the mass and equal particle-antiparticle counts alone.

A general scalar cosmological perturbation has four scalar metric functions: a lapse perturbation, a scalar shift, a spatial curvature perturbation and a scalar spatial shear. The two scalar functions in a cosmological gauge transformation can set the shift and shear to zero. For nondegenerate spatial modes this gives conformal Newtonian gauge, leaving the two potentials $\psi$ and $\varphi$. The scalar traceless spatial Einstein equation makes their difference proportional to scalar anisotropic stress. In general relativity, with no scalar anisotropic stress and the usual boundary conditions, $\psi =\varphi$. Pressure alone does not spoil equality; free-streaming relativistic species can do so through their anisotropic stress.

Write $\Psi =(\varphi +\psi )/2$, the Weyl lensing potential. A conformal rescaling leaves the unparameterized null geodesics unchanged, although the affine parameter changes. To first order, the conformal optical metric for scalar perturbations hasThe only nonzero connection coefficients at this order areHere spatial indices are raised with ${\delta }^{ij}$. An affine parameter $\lambda$ for this rescaled metric therefore obeysThe first line is also ${\eta }_{,\lambda \lambda }+2{\eta }_{,\lambda }d\Psi /d\lambda +2x_{,\lambda }^i{\eta }_{,\lambda }{\partial }_i\Psi =0$, since the derivative of the potential is along the ray.

Put $v^i=d{x}^i/d\eta$. The chain rule gives $x_{,\lambda \lambda }^i={\eta }_{,\lambda }^{2}d{v}^i/d\eta +v^i{\eta }_{,\lambda \lambda }$. Eliminating the affine acceleration givesThe extra spatial-gradient term inside the parentheses does not replace the spatial term already contained in the total derivative.

At reception, the null condition is $|v{|}^2=1+4{\Psi }_0$. With the specified unit propagation vector $e$, this impliesThe zero-order trajectory is $x^i=e^i(\eta -{\eta }_0)$. In every term already containing a first-order potential, we may substitute $v=e$. Defining the transverse projector $P_{\perp }^{ij}={\delta }^{ij}-e^i{e}^j$, the first-order geodesic equation becomesIntegrating with the reception position and velocity gives the first-order photon trajectory in scalar perturbations:Its first derivative at reception is $e^i+2e^i{\Psi }_0$, and differentiating twice gives the previous equation. For a first-order calculation, evaluating the integrals on the unperturbed ray is sufficient; shifting their evaluation points by the first-order displacement changes them only at second order. This is the Born approximation in cosmological lensing.

To fix the angular-deflection sign, define ${\chi }_s={\eta }_0-{\eta }_s>0$, $n_{\mathrm{obs}}=-e$, and $n_{\mathrm{src}}=x_s/|x_s|$. The latter is the unlensed coordinate direction to the source, while the former is its apparent direction measured by the observer. The longitudinal integral changes the radial distance, not the direction at this order. The transverse displacement isDividing by ${\chi }_s$ yieldsThis is the angular deflection kernel for cosmological lensing. It depends on the sum of the two metric potentials; setting them equal reduces it to the familiar twice-the-Newtonian-potential expression. With angular derivatives, the lensing potential is $2{\int }_0^{{\chi }_s}[({\chi }_s-\chi )/({\chi }_s\chi )]\Psi d\chi$, and its angular gradient is $\alpha$.

Gravitational lensing displaces apparent source positions and produces magnification and shear through angular derivatives of the deflection. Weak gravitational lensing of galaxy shapes probes the foreground mass distribution, including dark matter. Lensing also remaps the cosmic microwave background, smoothing its acoustic features and converting part of its E-mode polarization into B modes. Strong deflections can produce multiple images, though a single first-order ray calculation does not describe that regime completely.

The broad turnover near $k\sim 10^{-2}{\mathrm{Mpc}}^{-1}$ marks the matter-radiation equality scale. Modes entering the Hubble radius after matter-radiation equality grow efficiently during matter domination. Modes entering earlier have their cold-matter growth delayed by radiation domination: their subhorizon growth is logarithmic rather than proportional to $a$. This is the Mészáros effect. The smaller undulations on the declining side are baryon acoustic oscillations inherited from the photon-baryon fluid, with an oscillatory wavenumber spacing set approximately by $\pi$ divided by the sound horizon. They are damped at sufficiently short scales by photon diffusion and the transfer of the baryon perturbation into the total matter field. The plotted spectrum is a linear prediction; nonlinear evolution changes the actual sufficiently small-scale spectrum at redshift zero.

For an adiabatic growing mode, the comoving curvature perturbation transfers to the matter overdensity asup to an irrelevant overall sign convention for $\mathcal{R}$. Normalize the cosmological transfer function by $T\to 1$ at small $k$ and the linear growth factor by $D\to a$ in matter domination. The relation distinguishes a dimensional matter power spectrum from a dimensionless cosmological power spectrum: the dimensional curvature spectrum is proportional to $k^{-3}{\mathcal{P}}_{\mathcal{R}}(k)$. ThereforeFor locally constant scalar spectral index $n_s$, the asymptotic cold-dark-matter power spectrum isbecause the cold-dark-matter transfer function tends to one at small $k$ and scales as $\ln (k/k_{\mathrm{eq}})/(k/k_{\mathrm{eq}}{)}^2$ at large $k$. The asymptotes refer to the smooth envelope, not the individual baryonic wiggles. The large-$k$ local logarithmic slope is $n_s-4+2/\ln (k/k_{\mathrm{eq}})$. For the quadratic potential at $N\simeq 60$, derived below, $n_s\simeq 0.967$, so the limiting powers are approximately $k^{0.967}$ and $k^{-3.033}{\ln }^{2}k$. Multiplying by $k^3$ to form dimensionless matter power would change both exponents by three; that is not the plotted quantity.

Take the usual smooth late dark energy, in particular a cosmological constant, and hold the primordial perturbations fixed. Increasing its physical density while holding the matter and radiation densities fixed raises $H_0$ in a flat universe. It lowers ${\Omega }_{m0}$, but leaves the physical product ${\Omega }_{m0}{H}_0^2=\kappa {\rho }_{m0}/3$ fixed. It also leaves the early equality ratio $a_{\mathrm{eq}}={\rho }_{r0}/{\rho }_{m0}$ and the equality wavenumber in ${\mathrm{Mpc}}^{-1}$ essentially unchanged, provided dark energy remains negligible at equality. The early cosmological transfer function, including the acoustic scale, is then unchanged.

More dark energy causes accelerated expansion to start earlier, reducing the time for matter growth. The suppression of matter growth by smooth accelerated expansion lowers $D(1)$ and hence the present spectrum. For the subhorizon scales shown,This is approximately a downward amplitude change with the same turnover and wiggle locations in physical inverse-megaparsec units. For matter plus a cosmological constant, an explicit early-normalized growth solution iswith radiation neglected in this late-time growth equation. A numerical factor needs the old and new dark-energy densities. A change to axes expressed using $h$ would additionally relabel wavenumbers and volumes; the printed axes use ${\mathrm{Mpc}}^{-1}$ and ${\mathrm{Mpc}}^3$.

The supplied slow-roll curvature power spectrum simplifies toAt a fixed number of e-folds before the end, multiplying the potential by a constant does not change $V^{\prime }/V$, $V^{\prime \prime }/V$ or the slow-roll field value. It multiplies ${\mathcal{P}}_{\mathcal{R}}$ by that constant. Thus, for an unchanged subsequent cosmology,The scalar spectral index, transfer-function shape, turnover and acoustic features are unchanged at this order. Root-mean-square density fluctuations increase by $\sqrt{2}$. The comparison uses the same exit e-fold count; a different specified reheating history could shift the detailed mapping between $k$ and that count.

Derive both cases from $V=C_p(\Phi /M_{\mathrm{Pl}}{)}^p$, using the positive-field branch. The potential slow-roll parameter and second potential slow-roll parameter areTaking the leading inflation-end estimate ${ϵ}_V=1$, the remaining number of e-folds isThe corresponding monomial-inflation curvature amplitude and tilt areIn particular,The quartic case is redder. Its small-$k$ matter envelope is $k^{0.9508}$ instead of $k^{0.9669}$; its high-$k$ envelope is $k^{-3.0492}{\ln }^{2}k$ instead of $k^{-3.0331}{\ln }^{2}k$. With the late cosmology fixed, the turnover and baryonic acoustic scale do not move. If the spectra are matched in amplitude at a pivot $k_{\ast }$, their ratio is approximatelyThis gives more large-scale power and less small-scale power relative to the pivot, with about a $7\%$ relative tilt across two decades in wavenumber.

The quartic coefficient is not specified, so changing the power of the potential does not by itself specify the overall amplitude. If $C_2=V_0$ and the quartic coefficient is $C_4$, the formula above gives, at the same $N$,Consequently choosing the same dimensionful coefficient would give a very different normalization, whereas choosing $C_4/V_0\simeq 1/496$ matches the pivot amplitude. These are different normalization choices, not a universal amplitude prediction from a quartic shape alone.

For the quadratic potential, the e-fold integral above gives ${\Phi }_N/M_{\mathrm{Pl}}\simeq \sqrt{4N+2}$. At $N\simeq 60$, the field is ${\Phi }_{\ast }/M_{\mathrm{Pl}}\simeq \sqrt{242}\simeq 15.6$. Since $k=aH$ at exit and $N$ counts down,To leading slow-roll order, $|\Delta N|\simeq \Delta \ln k$. The quadratic-inflation field interval from a wavenumber interval is thereforeThe corresponding field endpoints are approximately $\sqrt{232}{M}_{\mathrm{Pl}}$ and $\sqrt{252}{M}_{\mathrm{Pl}}$, or $15.2$ to $15.9$ reduced Planck masses. Longer wavelengths exit earlier at the larger field value. Using $\ln 100=4.605$ instead of the supplied rounded logarithmic interval gives a span of about $0.59M_{\mathrm{Pl}}$; retaining $1-{ϵ}_H$ changes either estimate by less than one percent. The observable interval samples a small part of the large background field value, rather than the entire excursion to the end of inflation.