Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 53 / 4

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.