Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 (Cosmology)

For a component with constant barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation givesHence ${\rho }_i={\rho }_{i,0}{a}^{-3(1+w_i)}$. Choose the present scale factor to be $a_0=1$, so the cosmological redshift obeys $1+z=a^{-1}$, and define the present cosmological density parameter bySubstitution in the spatially flat Friedmann equation then yields

The Friedmann acceleration equation and $H=\dot{a}/a$ give the deceleration parameterAn Einstein-de Sitter universe contains only pressureless matter, so $q_0=1/2$ and cannot explain the observed negative value. For the stated Lambda-CDM model,so its cosmological constant produces the required accelerating universe.

The age of an FLRW universe follows from $dt=da/(aH)$. Neglecting radiation and spatial curvature givesFor an Einstein-de Sitter universe, ${\Omega }_{m,0}=1$ and ${\Omega }_{\Lambda ,0}=0$, so

The measured Hubble time would give the Einstein-de Sitter agewhich is less than the measured $12\mathrm{Gyr}$ age of stars that must themselves be younger than the universe. The measurements are therefore incompatible with an Einstein-de Sitter universe. For the stated spatially flat Lambda-CDM model, direct evaluation of the preceding integral givesand hence $t_0\simeq 13.5\mathrm{Gyr}$. The period of accelerating expansion caused by the cosmological constant therefore allows an age consistent with the stellar lower bound.

For a spatially homogeneous canonical scalar field, spatial derivatives vanish. Reading the time and spatial components of its energy-momentum tensor in the comoving frame givesThus the kinetic term contributes equally to energy density and pressure, whereas the scalar potential contributes negative pressure.

Substitution of these expressions into the cosmological perfect-fluid continuity equation givesContinuity through a turning point therefore yields the homogeneous Klein-Gordon equationDuring slow-roll inflation, the acceleration and kinetic-energy terms are negligible. In terms of the reduced Planck mass, the approximate evolution is

This is the potential of Starobinsky inflation. Put $b=\sqrt{2/3}/M_{\mathrm{Pl}}$ and $y=e^{-b\varphi }$, so $V=V_0(1-y{)}^2$. Its potential slow-roll parameter and second potential slow-roll parameter areAt large positive $\varphi$, $y\ll 1$, so both ${ϵ}_V$ and $|{\eta }_V|$ are small and slow-roll inflation is possible. Near the minimum, a Taylor expansion gives $1-y\simeq b\varphi$ and henceThey are large for $|\varphi |\ll M_{\mathrm{Pl}}$, so that region cannot sustain slow roll.

Near the minimum, the Taylor expansion of the potential isNeglecting Hubble friction during one short oscillation reduces the Klein-Gordon equation to the harmonic oscillator equationThe virial theorem gives $⟨{\dot{\varphi }}^2/2⟩=⟨V⟩$, so $⟨P_{\varphi }⟩=0$. Restoring the slow cosmological damping in the averaged cosmological perfect-fluid continuity equation givesThe coherently oscillating inflaton therefore behaves as pressureless matter.

The matter-like scaling is special to a quadratic minimum. More generally, the effective equation of state of an oscillating scalar field in $V\propto |\varphi {|}^n$ isFor example, a quartic minimum has $⟨w⟩=1/3$ and redshifts like radiation in cosmology. Potentials without a stable minimum, oscillations whose period is not short relative to the Hubble time, and significant decay of the inflaton into other particles also invalidate the matter-like argument.

At neutrino decoupling, neutrinos, electrons, positrons, and photons share one temperature. The neutrinos subsequently free stream, so $T_{\nu }\propto a^{-1}$. In the still-coupled electromagnetic plasma, the effective entropy degrees of freedom change during electron-positron annihilation in cosmology fromto $g_{\ast s,\mathrm{after}}=2$. Separate cosmological entropy conservation in that plasma gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$, while $T_{\nu }a$ remains constant. Consequently the Cosmic neutrino background temperature obeys

For one species with two internal states, the relic-neutrino energy density obtained from the frozen Fermi-Dirac distribution isIn the relativistic limit, set $x=ap/T_{\nu }$ and use the standard Riemann zeta function integralThis gives

At $T_{\nu }/a\ll m_{\nu }$, the Taylor expansion of the relativistic energy is $\sqrt{p^2+m_{\nu }^2}=m_{\nu }+p^2/(2m_{\nu })+O(p^4/m_{\nu }^3)$. The number density isThe ratio of the next momentum moment to this one isThereforeso $\boxed{F=15\zeta (5)/(2\zeta (3))\simeq 6.47}$.

Comparing the correction in part c with the nonrelativistic kinetic energy $m_{\nu }⟨v^2⟩/2$ gives the characteristic late-time speedThus the late-time relic-neutrino speed redshifts as $a^{-1}$.

The Cosmic neutrino background began free streaming at neutrino decoupling, long before cosmological recombination, so its directional flux can retain information about density fluctuations from epochs inaccessible to the Cosmic microwave background. Earlier decoupling does not by itself guarantee a larger comoving radial distance. A sufficiently massive neutrino eventually becomes nonrelativistic, and its travelled distance iswhich may be smaller than the photon distance because $v(a)<c$. Hence the cosmic neutrino background last-scattering surface need not lie beyond the Cosmic microwave background last-scattering surface; it generally does for neutrinos that remain sufficiently relativistic for sufficiently long.

The creation and annihilation operators obeyThe two-point correlation function in their vacuum has Fourier amplitude $|f_k{|}^2/a^2$. After cosmological horizon exit, $|k\tau |\ll 1$, and the stated Bunch-Davies vacuum mode and $a=-1/(H\tau )$ giveThe dimensionless power spectrum is consequently

The exact mode functions satisfy the Wronskian normalizationUsing the Dirac delta function generated by the ladder-operator commutator therefore gives the exact canonical commutation relationThis remains true on superhorizon scales, so the exact operators do not literally commute. A claim of classical behavior must instead compare this fixed commutator with the growing statistical fluctuations, or appeal to quantum decoherence.

At leading order in $|k\tau |$, the inflationary Fourier mode and its derivative areThus the leading parts of $\hat{f}$ and $\hat{\pi }$ contain the same quadrature of the creation and annihilation operators and satisfyTheir commutator consequently vanishes in this leading approximation. The exact nonzero result in part b is carried by the subleading decaying mode. Relative to the rapidly growing anticommutator, its effect is suppressed by powers of $|k\tau |$, which is the squeezing captured by the classicality parameter of a cosmological perturbation. The perturbation can therefore be treated as an effectively classical stochastic field on superhorizon scales even though its exact quantum commutator is unchanged.

The supplied superhorizon evolution equation isFor an adiabatic cosmological perturbation, $\delta P=({\bar{P}}^{\prime }/{\bar{\rho }}^{\prime })\delta \rho$, so $\delta P_{\mathrm{nad}}=0$ and the comoving curvature perturbation $\mathcal{R}$ is conserved on superhorizon scales.

For photons and cold dark matter, put $A={\bar{\rho }}_{\gamma }$, $C={\bar{\rho }}_c$, and use $\bar{P}=A/3$, $A^{\prime }=-4\mathcal{H}A$, and $C^{\prime }=-3\mathcal{H}C$. ThenThe cosmological entropy perturbation $S={\delta }_c-3{\delta }_{\gamma }/4$ implies ${\delta }_c=S+3{\delta }_{\gamma }/4$, so the adiabatic part cancels andSince $\bar{\rho }+\bar{P}=C+4A/3$, the curvature evolves asIn the sign convention requested in the question, $d\mathcal{R}/d\ln a=-fS$, this means