Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 310 (Cosmology)

A perfect fluid in general relativity has stress-energy tensorwhere $u^{\mu }{u}_{\mu }=-1$. Besides the metric tensor, one needs the energy density $\rho$, the pressure $P$, and three independent components of the normalized four-velocity $u^{\mu }$: five scalar functions of spacetime in total.

The four equations of stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, split parallel and perpendicular to $u^{\mu }$ intoAn equation of state, such as a barotropic equation of state $P=P(\rho )$, supplies the fifth relation needed to close the evolution.

In cosmic time, all three Friedmann-Lemaître-Robertson-Walker metrics can be writtenwhereThe closed spatial slice is a three-sphere of radius $a(t)$, soThe complete flat and open spatial slices are noncompact and have

For a homogeneous fluid in the Spatially flat FLRW metric, the equations in part (a) reduce to the cosmological perfect-fluid continuity equationwhile homogeneity makes the spatial relativistic Euler equation automatic. The Einstein field equations give the flat Friedmann equation and Friedmann acceleration equation,equivalently $\ddot{a}/a=-4\pi G(\rho +3P)/3$.

Normalize the present scale factor to $a(t_0)=1$. In a flat universe containing radiation in cosmology, pressureless matter, and a cosmological constant, the Friedmann equation iswhere ${\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{\Lambda ,0}=1$. Since $dt=da/(aH)$, the age of an FLRW universe isFor a matter-only flat Einstein-de Sitter universe,Taking the present value $H_0\simeq 70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, for which the Hubble time is $H_0^{-1}\simeq 14.0\mathrm{Gyr}$, gives

The identityshows that accelerating expansion has $H>0$ and $ϵ<1$, whereas accelerated contraction has $H<0$, $\ddot{a}<0$, and $ϵ>1$.

For the monomial inflation potential $V=\lambda {\varphi }^p$, the two potential slow-roll parameters areThus ${ϵ}_V<1$ requires $|\varphi |>p{M}_{\mathrm{Pl}}/\sqrt{2}$, while $|{\eta }_V|<1$ requires $|\varphi |>\sqrt{p|p-1|}{M}_{\mathrm{Pl}}$.

At ${\varphi }_{\ast }$, the specified velocity gives kinetic energyHence ${\rho }_{\varphi }=3V$, $P_{\varphi }=V$, and the first Hubble slow-roll parameter isThe universe is therefore decelerating, despite the flatness of the potential: the chosen initial kinetic energy violates the slow-roll approximation.

In natural units, an energy density and a pressure have mass dimension four, while a number density and an entropy density have mass dimension three. Since $m$ and $T$ each have mass dimension one,

The reaction maintaining chemical equilibrium isFor $T<m_e$ and negligible chemical potentials, the electron and Positron are dilute and obey the nonrelativistic equilibrium number density. Including two spin states for each particle,The photon number density is $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, so the electron-positron equilibrium pair abundance below the electron mass isIf $n_e$ denotes only one of the charge-conjugate species, this expression is divided by two.

Let $Y_{\mathrm{He}}\simeq 0.25$ be the present primordial helium mass fraction. If $n_{\mathrm{He}}$ is the helium-nucleus density and $n_H$ the hydrogen-nucleus density, thenSince $Y_{\mathrm{He}}=4n_{\mathrm{He}}/n_b$,This retains the correction from neutrons bound in helium. Charge neutrality requires one electron per proton, including bound electrons, so with the baryon-to-photon ratio $\eta =n_b/n_{\gamma }\simeq 6\times 10^{-10}$,

Pair equilibrium with negligible chemical potentials cannot continue once its predicted electron abundance falls below the residual abundance required by charge neutrality. Put $x=m_e/T$ and equate part (b), to logarithmic accuracy, to the order-$10^{-10}$ result of part (c):Taking a logarithm and retaining the numerical factor singled out in the hint givesThe solution is $x\simeq 28$, soThus the zero-chemical-potential electron-positron equilibrium description must fail before roughly $20\mathrm{keV}$; thereafter the small asymmetric electron population survives.

For each Fourier mode, the cosmological Poisson equation givesBecause nonrelativistic matter has ${\bar{\rho }}_m\propto a^{-3}$, this implies $\Phi \propto {\Delta }_m/a$. Write $\Phi =C{\Delta }_m/a$ and use $\mathcal{H}=a^{\prime }/a$. Direct differentiation in the stated potential equation givesThe supplied Friedmann equation identity therefore yieldsSince $d/d\tau =ad/dt$, this becomes the linear growth equation

During matter domination, $a\propto t^{2/3}$, $H=2/(3t)$, and $4\pi G{\bar{\rho }}_m=2/(3t^2)$. Substituting ${\Delta }_m\propto t^s$ into the linear growth equation giveswhose roots are $s=2/3$ and $s=-1$. ThereforeThe growing one of the matter-dominated density-perturbation modes is

Increasing the dark energy density makes dark-energy domination begin earlier. The larger Hubble parameter strengthens the $2H{\dot{\Delta }}_m$ friction term while the clustering source $4\pi G{\bar{\rho }}_m{\Delta }_m$ continues to dilute, so less large-scale structure of the universe can grow from a fixed primordial amplitude.

Deep in cosmological-constant domination, $H\simeq H_{\Lambda }$ is constant and ${\bar{\rho }}_m$ is negligible. The growth equation reduces toand henceThe surviving mode is constant: matter growth freezes.

The background remains matter dominated, so $H=2/(3t)$ and $4\pi G{\bar{\rho }}_m=2/(3t^2)$. For the reduced gravitational clustering strength, substitute ${\Delta }_m\propto t^s$ into the modified equation:The growing root is the matter-era growth with reduced gravitational clustering strengthIt satisfies $0<s(f)<2/3$ for $0<f<1$. Between $t_f$ and the onset $t_{\Lambda }$ of dark-energy domination, standard growth supplies $(t_{\Lambda }/t_f{)}^{2/3}$ whereas the modified model supplies $(t_{\Lambda }/t_f{)}^{s(f)}$. Subsequent growth freezes in both, so

A larger cosmological constant changes both the expansion history and the growth history: it alters cosmological distances, the baryon acoustic oscillation scale as observed in angle and redshift, and the time at which growth begins to freeze. The new interaction in part (d) leaves the background Friedmann equation unchanged but changes clustering abruptly from $t_f$ onward.

Redshift-resolved measurements of the linear growth factor, weak gravitational lensing, galaxy clustering, and cosmological distance measures can therefore compare geometry with growth. Equal present amplitudes do not imply equal $D(z)$, and only the higher-dark-energy model changes the distance-redshift relation. This breaks the degeneracy.

With the symmetric Fourier normalization displayed in the question, the creation and annihilation operators obey the canonical commutation relationSince $\delta \hat{\varphi }=\hat{f}/a$, its vacuum two-point correlation function isComparison with the definition in the question givesUsing $|f_k{|}^2=(1+1/(k^2{\tau }^2))/(2k)$ and $a=-1/(H\tau )$,On superhorizon scales, $k\ll aH$ or $|k\tau |\ll 1$, so the scale-invariant inflationary power spectrum freezes to

A perturbation $\delta \varphi$ shifts the time at which the rolling inflaton reaches a fixed value by $\delta t\simeq \delta \varphi /\dot{\varphi }$. The corresponding perturbation in local expansion is $\delta N=H\delta t$, which is the comoving curvature perturbation up to the stated sign convention:ThereforeUsing ${\dot{\varphi }}^2=2ϵH^2{M}_{\mathrm{Pl}}^2$ gives the slow-roll curvature power spectrumFor each $k$, the slowly varying quantities on the right are evaluated at cosmological horizon exit, $k=aH$.

At cosmological horizon exit, $d\log k/dN=1-ϵ\simeq 1$. Differentiating the slow-roll curvature power spectrum and using the Hubble slow-roll parameters givesThe supplied relations implyConsequently the scalar spectral index is

For $V=\lambda {\varphi }^4$,Thus $n_s-1=-6{ϵ}_V+2{\eta }_V=-3{ϵ}_V$. Combining this with the tensor-to-scalar ratio $r=16{ϵ}_V$ gives the quartic-inflation tilt-tensor relationThe measured tilt $n_s-1=-0.04$ would require $r\simeq 0.213$, whereas $r=0.001$ predicts $n_s-1\simeq -1.88\times 10^{-4}$. Therefore quartic slow-roll inflation is inconsistent with the stated measurements.