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An adiabatic mode corresponds locally to a common displacement along the background history. Its density changes obey $\delta {\rho }_i/{\dot{\bar{\rho }}}_i=\delta t$ for every component. The cosmological perfect-fluid continuity equation then implies that ${\Delta }_i/(1+w_i)$ is common to all fluids. Thus, in the common comoving gauge in cosmology relevant to the growing superhorizon mode,Equivalently, the relative entropy perturbation ${\Delta }_c-(3/4){\Delta }_r$ vanishes. A common time-slicing shift preserves this relation because it changes both normalized density perturbations by the same amount.

The flat background Friedmann equation gives $4\pi G{a}^2\bar{\rho }=(3/2){\mathcal{H}}^2$. During radiation domination, $\mathcal{H}=1/\tau$ and the matter background is negligible. Substituting the adiabatic relation into the density equation therefore givesWith ${\Delta }_c\propto {\tau }^p$, the indicial equation is $p^2=4$, so the modes are ${\tau }^2$ and ${\tau }^{-2}$.

During matter domination, $\mathcal{H}=2/\tau$ and the radiation background is negligible. The same calculation giveswhose indicial equation is $p^2+p-6=(p-2)(p+3)=0$. Its modes are ${\tau }^2$ and ${\tau }^{-3}$. Thus the superhorizon growing density mode of adiabatic matter and radiation satisfies ${\Delta }_c\propto {\tau }^2$ in both eras. The common conformal-time exponent does not imply the same scale-factor dependence: it is ${\Delta }_c\propto a^2$ in the radiation era and ${\Delta }_c\propto a$ in the matter era.

Use signature $(+---)$, the positive spatial metric ${\gamma }_{ij}$, and a future-directed unit normal $n_{\mu }=(N,0)$, so $n^{\mu }=N^{-1}(1,-N^i)$. This fixes the sign of the normal to agree with the specified $\Pi$. The negative normal covector printed in the source would instead give $-\Pi$; its orientation and the displayed derivative cannot both be retained. Use the operational negative-expansion convention for the extrinsic curvature of a spatial hypersurface, $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, which is the convention giving the requested expanding-universe sign. With this future normal and signature it equals $n_{i;j}$; the source's $-n_{i;j}$ identity instead uses its past-pointing normal.

For the scalar-field matter projections, put $s_i=D_i\varphi$ and $s^2={\gamma }^{ij}{s}_i{s}_j$. Decomposing the scalar gradient into normal and tangent parts gives ${\partial }_{\mu }\varphi {\partial }^{\mu }\varphi ={\Pi }^2-s^2$. Contracting the Klein-Gordon scalar stress-energy tensor with the normal and the spatial projection tensor therefore yieldsIn particular, the positive spatial metric raises the $i,j$ indices in $s^2$; raising them with the four-metric would reverse that spatial sign. These quantities are the energy density, momentum density and spatial stress in the normal frame.

For the homogeneous Spatially flat FLRW metric, $N_i=0$, ${\gamma }_{ij}=a^2{\delta }_{ij}$ and $s_i=0$. Consequently $\Pi =\dot{\bar{\varphi }}/\bar{N}$, $J_i=0$, $\bar{\rho }={\Pi }^2/2+V$, $\bar{P}={\Pi }^2/2-V$, andThe intrinsic Ricci scalar is zero and $K^2-K_{ij}{K}^{ij}=9H^2-3H^2=6H^2$. The Hamiltonian constraint thus gives the scalar-field Friedmann equation:The momentum constraint is identically satisfied, since its homogeneous spatial derivatives and $J_i$ vanish. The scalar equation reduces to ${\bar{N}}^{-1}\dot{\Pi }+3H\Pi +V_{,\varphi }=0$. Substituting $\Pi$ and multiplying by ${\bar{N}}^2$ gives the background inflaton equation of motion:No derivation of the supplied gravitational constraints is needed.

For the growing adiabatic mode in an epoch of constant $w$, the superhorizon potential is constant. The Friedmann equation gives $4\pi G{a}^2(\bar{\rho }+\bar{P})=\frac{3}{2}(1+w){\mathcal{H}}^2$, so the given relation impliesThus ${\varphi }_{\mathrm{rad}}=-2\mathcal{R}/3$ and ${\varphi }_{\mathrm{mat}}=-3\mathcal{R}/5$. Photon continuity, neglecting its gradient term on superhorizon scales, gives $({\Theta }_0-\varphi {)}^{\prime }=0$. The radiation-era initial condition fixes ${\Theta }_0-\varphi =\mathcal{R}$. Through the matter-radiation transition it follows that ${\Theta }_{0\ast }=\mathcal{R}+{\varphi }_{\mathrm{mat}}=2\mathcal{R}/5$. Since $\psi =\varphi$, the matched large-scale emission perturbation isKeeping the radiation value of ${\Theta }_0$ unchanged across the transition would give the wrong coefficient. This is the Sachs-Wolfe radiation-to-matter matching for the paper's sign convention for comoving curvature perturbation.

Neglect the Integrated Sachs-Wolfe effect, the superhorizon-suppressed Doppler term, observer monopole, and any observer kinematic dipole. Put ${\mathbf{x}}_0=0$ by statistical homogeneity, and defineThe plane-wave expansion on ${\mathbf{x}}_{\ast }=-{\chi }_{\ast }\mathbf{e}$ gives angular multipolesStatistical isotropy and orthonormal spherical harmonics then imply $⟨a_{\ell m}{a}_{{\ell }^{\prime }{m}^{\prime }}^{\ast }⟩={\delta }_{\ell {\ell }^{\prime }}{\delta }_{m{m}^{\prime }}{C}_{\ell }$. Radial integration supplies $(4\pi {)}^2(2{\pi }^2)/(2\pi {)}^3=4\pi$, giving the large-angle angular power spectrumThis assumes adiabatic growing modes, matter domination at emission, negligible anisotropic stress, and wavenumbers that are outside the Hubble radius then. It treats recombination as instantaneous, omits late potential evolution and rescattering, and omits lensing at this linear order. The formal primordial contribution is defined for $\ell >0$; the observed dipole has a large additional kinematic contribution, so the clean large-angle primordial comparison is usually $\ell \ge 2$. A scale-invariant spectrum also yields the Sachs-Wolfe plateau $\ell (\ell +1)C_{\ell }=2\pi {\mathcal{P}}_{\mathcal{R}}/25$ for $\ell \ge 1$.

Set $y=a^2$. Multiplication of the Friedmann equation by $4a^4$ givesFor the expanding branch emerging from $y(0)=0$, $\dot{y}(0)=2\sqrt{\Gamma }$. Differentiating where $\dot{y}\ne 0$ gives $\ddot{y}=(4\Lambda /3)y-2$, and continuity extends it through a regular turning point. Put $\kappa =\sqrt{4\Lambda /3}$. Solving this linear equation with both initial data givesInitially $a(t)\sim (2\sqrt{\Gamma }t{)}^{1/2}$. If $\lambda >1$, $y$ stays positive and strictly increasing; at late times $a\sim [3(\lambda -1)/(4\Lambda ){]}^{1/2}{e}^{\kappa t/2}$. This universe expands forever.

If $0<\lambda <1$, expansion stops when $\tanh \kappa t_{\max }=\lambda$, givingThe physical solution expands from zero and then symmetrically recollapses to zero; negative $a^2$ beyond this interval is not a continuation of a real scale factor.

The flat matter-dominated Friedmann equation gives $\rho =1/(6\pi G{t}^2)$ when $a=t^{2/3}$. With $P=C{\rho }^{4/3}$,For linear density modes with polytropic pressure, substitution gives the Euler-Cauchy equationA power law has exponent satisfying $n(n-1)+4n/3+{\bar{c}}_s^2{k}^2-2/3=0$. Thus, for distinct roots,For $k\ll k_J$, the two modes approach $t^{2/3}$ and $t^{-1}$: gravity produces a growing linear cosmological density perturbation and a decaying mode. For $k\gg k_J$, put $\nu =\sqrt{{\bar{c}}_s^2{k}^2-25/36}$; the real solution isThese are pressure-supported oscillations with a decaying envelope. For the repeated-root cosmological density mode $k=k_J$, the complete solution is $t^{-1/6}(A_k+B_k\log t)$. The specified $k_J=5/(6{\bar{c}}_s)$ separates real exponents from logarithmic oscillations; the growing exponent actually crosses zero at $k=\sqrt{2/3}/{\bar{c}}_s$, a slightly smaller value.

The Flatness problem is the extreme closeness to spatial flatness required in the early universe if ordinary decelerating expansion is to produce the observed near-flat universe. From the Friedmann equation, with $c=1$ and curvature constant $k$,During decelerating expansion $|\Omega -1|$ grows. The Friedmann acceleration equation gives $\ddot{a}/a=-4\pi G(\rho +3P)/3$. Thus $\rho +3P<0$ makes an expanding universe accelerate, increases $\dot{a}$ and suppresses $|\Omega -1|$: this is the inflationary solution of the flatness problem.

The slow-roll approximation neglects ${\dot{\varphi }}^2/2$ compared with $V$ and $\ddot{\varphi }$ compared with $3H\dot{\varphi }$. For the positive-field branch, taking $m>0$,Integrating the field equation and $\dot{a}/a=H$ gives the slow-roll solutionThe potential slow-roll parameters are ${ϵ}_V={\eta }_V=2/{\varphi }^2$. They cease to be small when $\varphi$ is of order $1$ in these units; the customary endpoint ${ϵ}_V=1$ is $\boxed{{\varphi }_{\mathrm{end}}\simeq \sqrt{2}}$. Equating the approximate kinetic and potential energies instead gives $\sqrt{2/3}$, another order-one indicator, not a precise endpoint of an approximation that is already failing.

The effective number of relativistic energy degrees of freedom count two photon polarizations, four electron/positron spin states, and six neutrino/antineutrino states, with the Fermi-Dirac distribution factor $7/8$ for fermions. Treating these species as relativistic and at a common temperature gives $\boxed{g_{\ast }=2+(7/8)(4+6)=10.75}$.

The mass-equivalent radiation mass density is $\rho =({\pi }^2/30)g_{\ast }(k_{B}T{)}^4/({\hslash }^3{c}^5)$. Therefore the Friedmann equation givesAt $T_0=10^{10}\mathrm{K}$, $H_0\simeq 0.503{\mathrm{s}}^{-1}$. Equating the specified weak interaction rate to $H_0(T/T_0{)}^2$ gives $(T_{\ast }/T_0{)}^3=H_0/(1{\mathrm{s}}^{-1})$ and henceThe approximate equilibrium neutron-to-proton ratio, neglecting chemical potentials, is $r_{\ast }\simeq \exp [-(m_n-m_p)c^2/(k_B{T}_{\ast })]$, about $0.152$ using the mass difference $1.293\mathrm{MeV}$. This is a freeze-out estimate rather than a detailed reaction-network computation.

After weak cosmological weak freeze-out, free neutrons undergo beta-minus decay; neutrinos decouple, and electron-positron annihilation heats the photons relative to the neutrinos. Initially the abundant energetic photons destroy deuterium almost as soon as it forms. Once the deuterium bottleneck ends, nuclear reactions rapidly pass through deuterium, tritium and helium-three to helium-4. Reactions eventually become too slow compared with cosmic expansion; nearly all surviving neutrons are bound in helium-four. Its mass fraction is approximately $Y_4=2r/(1+r)$, where $r$ is evaluated when nuclear assembly begins.

Increasing $g_{\ast }$ increases the Hubble parameter, raises $T_{\ast }$, increases the freeze-out neutron-to-proton ratio, and leaves less time for free-neutron decay after freeze-out. Both effects increase $Y_4$; decreasing $g_{\ast }$ reverses them. Most deuterium is only an intermediate product and is burned into helium, so its residual abundance and its absolute change are tiny compared with the helium mass fraction. This does not imply a negligible fractional change in residual deuterium. Faster expansion also shortens the time available to burn it, and its relative abundance can be appreciably sensitive to extra radiation. The distinction matters when interpreting the final wording of the question; joint deuterium and helium abundances constrain extra relativistic species, as discussed in the Particle Data Group nucleosynthesis review.

Use units $c=\hslash =k_B=1$, and write $H=\dot{a}/a$ for the Hubble parameter, with dots denoting cosmic time. Differentiating the Friedmann equation givesOn an interval where $H\ne 0$, use the Hubble parameter identity $\dot{H}=\ddot{a}/a-H^2$ and the Friedmann acceleration equation to findSubstitution cancels the curvature term and gives the continuity equationThe cosmological perfect-fluid continuity equation extends by continuity through a regular isolated turning point. Physically it states that the change of energy in a comoving volume is the negative of the pressure work: $d(\rho a^3)=-Pd(a^3)$.

For separately conserved cosmological fluids, each component obeys this equation individually. A constant equation-of-state parameter $w_i=P_i/{\rho }_i$ therefore giveswhere the present scale factor is normalized to $a_0=1$. Thus the component density laws areThe extra factor for radiation in cosmology is the cosmological redshift of each photon's energy; pressureless matter has only number dilution, while the specified dark energy is a cosmological constant.

The critical density at a given expansion rate is the total density that makes the spatial curvature vanish:These cosmological density parameters use the critical density at that same time. Put $E(a)=H(a)/H_0$ and ${\Omega }_{K,0}=-K/H_0^2$. The Friedmann equation yieldsConsequently the fractional-density evolution, rather than just the component-density evolution, isIn particular ${\sum }_i{\Omega }_i=1+K/(a^2{H}^2)$. The simple powers of $a$ alone apply to ${\rho }_i$, or to ${\rho }_i/{\rho }_{\mathrm{crit},0}$, not to the instantaneous cosmological density parameters. At a recollapse turning point $H=0$, those instantaneous ratios are undefined even though the component densities remain finite.

Insert the component-density laws into the Friedmann equation, using ${\rho }_{i,0}=3H_0^2{\Omega }_{i,0}/(8\pi G)$ and $a_0=1$. Multiplying by $a^2$ givesThe radiation in cosmology and pressureless matter contributions add; the printed first formula has lost the plus sign between them. In the Friedmann acceleration equation, $\rho +3P=2{\rho }_r+{\rho }_m-2{\rho }_{\Lambda }$, soEvaluating the first equation today gives $K=H_0^2({\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{\Lambda ,0}-1)$.

Now remove dark energy and assume a positive pressureless matter or radiation in cosmology density, starting on an expanding branch. Let $f(a)=H_0^2({\Omega }_{r,0}/a^2+{\Omega }_{m,0}/a)$. It decreases from infinity to zero, and ${\dot{a}}^2=f(a)-K$, while $\ddot{a}<0$. This immediately distinguishes all three cases of spatial curvature of an FLRW universe.

The early small-$a$ solution has $a\propto t^{1/2}$ if radiation in cosmology is present, and $a\propto t^{2/3}$ for pressureless matter alone. The figure isolates the effect of curvature in three matter-only examples, rather than comparing models with the same present density parameters.

For the open matter-dominated Friedmann solution, write $\Omega ={\Omega }_{m,0}$, so $0<\Omega <1$. The present Friedmann equation gives $-K=H_0^2(1-\Omega )$. With $A=H_0^2\Omega$ and $B=H_0^2(1-\Omega )>0$, the expanding equation becomesChoose the hyperbolic substitutionThen $da=(A/(2B))\sinh \eta d\eta$ and $\sqrt{a/(a+A/B)}=\tanh (\eta /2)$. Their product givesTake the Big Bang to be $t=0$, $\eta =0$, and integrate. The parametric solution isThe parameter is determined explicitly byThus $\eta$ is conformal time multiplied by $\sqrt{-K}$, with its origin at the Big Bang; it is not equal to unscaled cosmic time. In particular today ${\eta }_0=\mathrm{arcosh}(2/\Omega -1)$. Small $\eta$ gives $a\propto {\eta }^2$ and $t\propto {\eta }^3$, hence $a\propto t^{2/3}$. Large $\eta$ gives $a/t\to H_0\sqrt{1-\Omega }=\sqrt{-K}$, in agreement with the curvature-dominated universe limit.

Let $T_{-}$ be the residual gas temperature immediately before decay. This is the $T_d$ that enters the target abundance formula, not the higher reheating temperature after decay. Before the decay, separately conserved relic number and gas entropy give the conserved cosmological relic abundance $Y=n_{\varphi }/s_{-}$. The cold nonrelativistic relic has negligible kinetic energy, henceAssume its density dominates over that of the residual gas and that curvature and dark energy are negligible in the Friedmann equation. Using the reduced Planck mass $M_{\mathrm{Pl}}^2=(8\pi G{)}^{-1}$ gives $3M_{\mathrm{Pl}}^2{H}_d^2\simeq {\rho }_{\varphi }(t_d^{-})$. Under the allowed sudden decay of a dominant nonrelativistic relic convention $H_d\simeq \Gamma =t_d^{-1}$, the temperature immediately before decay isThe coefficient inherits the approximation $H_d\simeq 1/t_d$: the result is a sudden-decay estimate, not an exact age relation for a matter-only universe. If $t_d$ were instead the exact cosmic age in an uninterrupted spatially flat matter era, $H_d=2/(3t_d)$ would replace the numerator $3M_{\mathrm{Pl}}^2$ by $4M_{\mathrm{Pl}}^2/3$.

The assumptions also require thermal decoupling in cosmology and negligible prior decay so that $Y$ remains conserved, a fixed relativistic count $g_{\star }$, and rapid thermalization after the energy release. In the same approximation, the reheating temperature $T_{+}$ follows from $c_2{g}_{\star }{T}_{+}^4\simeq 3M_{\mathrm{Pl}}^2/t_d^2$:Its absence of $Y$ at fixed decay rate makes clear why the two temperatures must be distinguished.

Use units $c=1$ and cosmic time $t$. Define the Hubble parameter $H=\dot{a}/a$, the equation-of-state parameter $\omega =P/\rho$, and the cosmological density parameterHere ${\rho }_{\mathrm{crit}}$ is the critical density. The definition requires $H\ne 0$; assume positive energy density and an expanding branch so that $\log a$ is a valid time coordinate. The Friedmann equation gives $\Omega -1=K/(a^2{H}^2)$. The Friedmann acceleration equation and Hubble parameter identity giveDifferentiating the Friedmann equation and combining with this identity yields the cosmological perfect-fluid continuity equation $\dot{\rho }=-3H(\rho +P)$. ConsequentlyThus the cosmological density parameter flow isThis identity allows time-dependent $\omega$; constancy is needed only for the simple power-law integration in the next part.

Normalize the scale factor so that $a_0=1$. With separately conserved pressureless matter and radiation in cosmology, their energy densities scale as $a^{-3}$ and $a^{-4}$. Hence the flat Friedmann equation isEquality of the two densities defines matter-radiation equality, with $a_{\mathrm{eq}}={\Omega }_{r,0}/{\Omega }_{m,0}$. Define the dimensionless variableSubstituting $a=a_{\mathrm{eq}}y$ givesBoth reference density fractions are positive. In the two-component flat model their sum at the reference epoch is one; no dark-energy component is being included in this background calculation.

Relativistic pressure gives the radiation fluid a sound speed $c_s^2=1/3$. Well inside the sound horizon, $k\gg \mathcal{H}$, radiation linear cosmological density perturbations undergo rapid pressure-supported oscillations instead of sustained gravitational growth. Their potential decays. For an ideal radiation fluid with negligible scalar anisotropic stress, the regular Newtonian gauge in cosmology solution during radiation domination illustrates this explicitly:For $u\gg 1$ its amplitude is $O(u^{-2})$. Thus radiation in cosmology remains essential to the background Friedmann equation, while its rapidly oscillating gravitational forcing can be smoothed out when following the slowly varying cold dark matter density mode. This is the approximation behind the Mészáros equation.

The important limitation is that “subhorizon” alone is not a pointwise theorem that $|{\Phi }_r|\ll |{\Phi }_m|$. Radiation can dominate the instantaneous potential during early radiation domination, and the forcing near horizon entry seeds the later logarithmic matter mode. We neglect that transient and rapidly oscillating forcing in the subsequent coarse-grained growth equation, not radiation's contribution to the expansion rate. Close to horizon entry, or for accurate oscillatory evolution, the coupled radiation and matter perturbation equations must be retained. For the slow matter potential, the metric time derivatives in part (a) are also suppressed relative to its spatial Laplacian by $O((\mathcal{H}/k{)}^2)$. The approximation smooths radiation perturbations while retaining radiation in cosmology in the background cosmic expansion.

The first Hubble slow-roll parameter measures the fractional change of $H$ per number of e-folds, and the second Hubble slow-roll parameter measures the fractional change of the first parameter. Since $d\log a/dt=H$,For a canonical inflaton, $ϵ={\dot{\varphi }}^2/(2M_{\mathrm{Pl}}^2{H}^2)\ge 0$. Expanding cosmic inflation means $\ddot{a}>0$, equivalently $\boxed{ϵ<1}$ because $\ddot{a}/a=H^2(1-ϵ)$. Small $\eta$ is not required for accelerated expansion itself, but helps maintain the slow-roll approximation over many e-folds. Its logarithmic definition assumes $ϵ>0$; in exact de Sitter expansion it is undefined unless interpreted as a limit.

For conventional slow-roll inflation, assume ${\dot{\varphi }}^2\ll V$, $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, a smooth positive scalar potential, and slowly varying small parameters. The Friedmann equation and inflaton equation of motion then reduce toSubstitute these in the exact expression for $ϵ$:To obtain the potential and Hubble slow-roll parameter relation, differentiate $\log {ϵ}_V$ along the slow-roll trajectory $\dot{\varphi }/H\simeq -M_{\mathrm{Pl}}^2{V}_{,\varphi }/V$:HenceThe second quantity on the right is the second potential slow-roll parameter, not the logarithmic Hubble-flow $\eta$ itself. Consistency usually requires ${ϵ}_V\ll 1$ and $|{\eta }_V|\ll 1$; identities involving division by $V_{,\varphi }$ are used on rolling intervals and interpreted by limiting values at stationary points.

The result above gives $K=-3H+\kappa$. Since the shear is trace-free and already first order, it cannot contribute linearly to the quadratic tensor contraction $K_{ij}{K}^{ij}=\mathrm{tr}(K^2)$. Thereforeup to second-order scalar cosmological perturbations. Combining these with the supplied spatial Ricci scalar gives the Hamiltonian constraintIts background part is the flat Friedmann equation $6H^2=16\pi G\bar{\rho }$. Subtracting that equation proves the linearized cosmological Hamiltonian constraint:Here $\rho$ is the energy density measured by the hypersurface normal, as required by the normal-normal projection of the Einstein field equations.

Differentiate the Friedmann equation and use energy conservation to obtain $\dot{H}=-4\pi G(\rho +P/c^2)$ on the expanding branch, where $H\ne 0$. Since $\ddot{a}/a=\dot{H}+H^2$,For pressureless matter, conservation gives ${\rho }_M={\rho }_{M0}{a}^{-3}$. In terms of the specified cosmological density parameters, flatness implies ${\Omega }_M+{\Omega }_{\Lambda }=1$, andSet $b=a^{3/2}$ and take ${\Omega }_M>0$. Then $\dot{b}=\frac{3}{2}{H}_0\sqrt{{\Omega }_M+{\Omega }_{\Lambda }{b}^2}$. Integrating with the Big Bang at $t=0$ givesFor another time origin replace $t$ by $t-t_B$. The cosmological constant has effective density ${\rho }_{\Lambda }=3H_0^2{\Omega }_{\Lambda }/(8\pi G)$. Equality with the matter density means $a^3={\Omega }_M/{\Omega }_{\Lambda }$, soIf ${\Omega }_M=0$, the expanding solution is pure exponential de Sitter scale factor in flat slicing and there is no finite matter-density equality time or dust Big Bang; the stated hyperbolic-sine formula presumes a nonzero dust component.

For the Einstein static universe with $P=0$ and ${\rho }_0>0$, the acceleration equation gives $\Lambda =4\pi {\rho }_0$. The first Friedmann equation then gives $k/a_0^2=4\pi {\rho }_0$, soKeep $\Lambda ,k$ fixed and use the specified perturbation closure $\delta P=\delta \rho /3$. Linearization of the constraint gives $\delta \rho =-3{\rho }_0\delta a/a_0$. In the acceleration equation the background terms cancel, leavingThusThe growing mode establishes linear instability of the static universe; a specially chosen decaying perturbation does not remove instability. This uses the pressure-density relation explicitly supplied for the perturbation. If zero density is allowed, there is also the degenerate vacuum static case $k=0$, $\Lambda =0$, which is not the positive-density static universe considered here.