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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.

The instantaneous-decay approximation keeps the scale factor and physical volume fixed across the decay. Energy conservation and rapid thermalization giveMeanwhile the abundance relation gives $s_{-}={\rho }_{\varphi ,-}/(Ym)\simeq 3M_{\mathrm{Pl}}^2/(Ym{t}_d^2)$. Dividing yields the entropy-injection factorThere is a numerical error in the printed coefficient: part (c) omits the factor $1/3$. The denominator $s_{-}=3M_{\mathrm{Pl}}^2/(Ym{t}_d^2)$ supplies that factor, while the radiation energy contributes $3^{3/4}$ to the numerator. Thus the consistent coefficient is $c_1{3}^{-1/4}{c}_2^{-3/4}$, rather than $c_1{3}^{3/4}{c}_2^{-3/4}$. Both cannot follow from the stated part (b) using the same decay-time approximation.

Its increase is especially transparent without dropping the original radiation. Let $R={\rho }_{\varphi ,-}/{\rho }_{r,-}$. At unchanged $g_{\star }$, $T_{+}^4/T_{-}^4=1+R$, soRelic domination means $R\gg 1$, so the leading estimate is much larger than one. This is entropy production by decay of a dominant relic.

There is no conflict with part (a). The equilibrium gas before the decay was a separately conserved, reversibly expanding system. During decay it receives energy from the nonthermal relic, and the decay products thermalize irreversibly. Its cosmological perfect-fluid continuity equation now has an energy-injection source, so $d({\rho }_{r}V)+P_{r}dV$ does not vanish. The assumptions behind cosmological entropy conservation therefore fail during the event. Moreover the zero-chemical-potential equilibrium expression in part (a) cannot be imposed on the decoupled relic as though it were already part of that same equilibrium radiation bath. The Second law of thermodynamics permits, and here requires, the resulting entropy increase.

In the constant-$H$ de Sitter approximation, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The massive rescaled mode obeysFor a fixed mass in the specified open interval, at sufficiently late superhorizon scales with $|k\tau |\ll 1$, the $k^2$ term is negligible compared with the inverse-square term in the limiting equation. Try $v\propto (-\tau {)}^p$. The indicial equation is $p(p-1)+m_{\chi }^2/H^2-2=0$, givingThus both independent branches of the massive de Sitter superhorizon scalar mode are present inThe coefficients are constant in time and fixed by initial conditions. Writing $-k\tau >0$ avoids ambiguous fractional powers of the negative number $k\tau$; a chosen complex phase in the printed notation can be absorbed into the coefficients.

For the specified mass range, $1/2<\nu <\sqrt{5}/2<3/2$. The physical fluctuation is $\delta {\chi }_{\mathbf{k}}=v_{\mathbf{k}}/a=-H\tau v_{\mathbf{k}}$, so its two branches behave as $(-\tau {)}^{3/2-\nu }$ and $(-\tau {)}^{3/2+\nu }$. Both exponents are strictly positive. Therefore both physical-field branches decay as $\tau \to 0^{-}$:The slower branch of $v$ itself grows because $1/2-\nu <0$, but the expanding scale factor more than cancels that growth. A spatially homogeneous background field of this mass satisfies the corresponding $k=0$ equation and also has these decaying powers.

Write $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{k}^3}$, so $u_k(0)=A_k$. For the cyclic term with the first field undifferentiated, insertion into the in-in formalism giveswhere $K=k_1+k_2+k_3>0$. The two differentiated modes contribute ${\tau }^2$, cancelling the ${\tau }^{-2}$ in $a^2$.

Along the vacuum-selected contour, the lower boundary vanishes. The elementary integrals areand thereforeMultiplying by $4i{M}_{\mathrm{Pl}}^2{ϵ}^2$, taking the real part and summing the three cyclic choices gives the curvature bispectrum from a zeta zeta-prime-squared interaction:This matches the final normalization. The coefficient requires all six connected Wick contractions.

The calculation assumes the Gaussian Bunch-Davies vacuum, tree order in the specified cubic interaction, effectively constant $H$ and $ϵ$ during the integral, the approximate de Sitter scale factor, and observation after the modes freeze as $\tau \to 0^{-}$. The contour tilt is kept until the early boundary has been eliminated; an undamped real-axis integral cannot simply discard that boundary. The reduced Planck mass convention is the one in the supplied De Sitter curvature mode function. Other cubic interactions and nonlinear field redefinitions are outside the specified model, so this is not by itself the complete bispectrum of a general single-field action.

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.

For comoving galaxies their relative comoving vector $\mathbf{\chi }$ is constant, and the physical displacement is $\mathbf{r}(t)=a(t)\mathbf{\chi }$, where $a$ is the scale factor. Differentiating gives the Hubble lawPeculiar velocity relative to this comoving frame is excluded in this relation.

Transport a wave crest between neighbouring comoving observers separated by a small physical distance $d\ell$. Their recession speed is $Hd\ell$, while the light travel time is $dt=d\ell /c$ to first order. The local Doppler shift therefore changes its wavelength by $d\lambda /\lambda =Hdt$. Multiplying these infinitesimal shifts along the ray, or integrating their logarithms, yieldsThe use of the local first-order Doppler law is exact in the limit of infinitesimal steps: the accumulated second-order errors tend to zero. A single special-relativistic recession velocity assigned to widely separated galaxies would not justify this cosmological redshift relation.

Differentiation of the Hubble parameter gives $\dot{H}+H^2=\ddot{a}/a$. The strong energy condition and the acceleration equation therefore implyAssume the present universe is expanding, $H_0>0$, and normalize the scale factor by $a(t_0)=1$. Since $\ddot{a}\le 0$, the past expanding branch has $\dot{a}(t)\ge \dot{a}(t_0)>0$. On this branch,Integrating from a past time $t$ to $t_0$ gives $H(t{)}^{-1}\le H_0^{-1}+t-t_0$. Thus, where the denominator is positive,The printed inequality needs this past-time qualification: for future times with $H>0$, the inequality reverses. Independently, a concave function lies below its tangent, soPositivity of $a$ prevents the past branch extending before the zero of this tangent. If $t_B$ is the beginning of the expanding universe, its age satisfies $\boxed{t_0-t_B\le H_0^{-1}}$. Without the normalization $a(t_0)=1$, the tangent bound instead reads $a(t)\le a(t_0)[1+H_0(t-t_0)]$.

Before decay, the nonrelativistic $\chi$ density redshifts as $a^{-3}$, while the photon density redshifts as $a^{-4}$. Therefore their ratio grows as $a$. Let a minus sign denote the instant just before decay and defineUsing the assumed radiation domination law $a\propto t^{1/2}$ gives $r_d=(t_d/t_{\mathrm{ref}}{)}^{1/2}{\rho }_{\chi }(t_{\mathrm{ref}})/{\rho }_{\gamma }(t_{\mathrm{ref}})$.

Across an instantaneous decay the scale factor does not change. Energy conservation and complete thermalization into a zero-chemical-potential photon bath giveSince blackbody radiation has ${\rho }_{\gamma }\propto T_{\gamma }^4$, instantaneous photon heating by a decaying relic givesThe already decoupled Cosmic neutrino background is not heated. Dividing its unchanged temperature by the new photon temperature therefore yieldsThe relation written with $a_d/a_{\mathrm{ref}}$ is independent of the expansion approximation. Replacing it by the time ratio uses the radiation-dominated background assumed in the question; an appreciable matter contribution would require solving for that background instead. This photon entropy injection is not adiabatic, so the pre-decay photon entropy need not be conserved across the decay.

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

For an infinitesimal test mass $\delta m$ on the surface of a homogeneous sphere, its Newtonian mechanical energy isThe spatially flat case corresponds to zero specific energy. Usingand dividing by $\delta m{R}^2/2$ givesSince $R$ is proportional to the scale factor $a$, this is the Newtonian derivation of the flat Friedmann equation

Now take $a(t_0)=1$ and $\rho ={\rho }_0{a}^{-4}$. PutThe contracting branch of the Friedmann equation isIntegrating $ada=-Cdt$ from $t_0$ givesThereforeon the contracting interval, reaching $a=0$ at $t=t_0+1/(2C)$.

At equality, let ${\rho }_{\mathrm{cc}}={\rho }_{\mathrm{dust}}={\rho }_{\ast }$. The cosmological constant has $P_{\mathrm{cc}}=-{\rho }_{\ast }$, while nonrelativistic matter has $P_{\mathrm{dust}}=0$. ThereforeThe scale factor consequently has $\ddot{a}>0$: the expansion is accelerated at $a=a_{\mathrm{eq}}$.

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

Separate stress-energy conservation and the constant equation-of-state parameter $w$ give the constant-equation-of-state density scalingBecause $w<-1$, the phantom energy density grows with the scale factor, whereas the pressureless matter density decays as $a^{-3}$. Phantom energy therefore dominates at late times. With $\Lambda =0$, the Friedmann equation becomesIntegrating and absorbing the integration constant into the finite time $t^{\ast }$ givesThe exponent is negative, so $a(t)\to \infty$ as $t\to t^{\ast -}$. This is a Big Rip.

A suitable Spatially flat FLRW metric isBoth source and detector are comoving, so cosmological redshift stretches the wave period in proportion to the scale factor. If $f_e$ is the local source frequency at emission and $f_o$ is the frequency measured in Earth's local frame, thenThe same relation applies to gravitational waves propagating on the FLRW background in the geometric-optics regime.

For the metric in vector cosmological perturbation, an orthonormal tetrad is $E_0=a^{-1}{\partial }_{\eta }$, $E_i=a^{-1}(B_i{\partial }_{\eta }+{\partial }_i)$ to first order. A photon has $p^0=(ϵ/a^2)(1+B_i{e}^i)$ and $p^i=(ϵ/a^2)e^i$, so the covariant component is $p_0=-ϵ+O(B^2)$. The covariant geodesic equation gives $d{p}_0/d\eta =(2p^0{)}^{-1}{\partial }_{\eta }{g}_{\alpha \beta }{p}^{\alpha }{p}^{\beta }=ϵe^i{\dot{B}}_i$. The derivative of the scale factor cancels by the null vector condition, proving the redshift formula.

In a flat universe containing separately conserved radiation in cosmology and pressureless matter, the Friedmann equation becomes $(a^{\prime }{)}^2=H_0^2({\Omega }_{r,0}+{\Omega }_{m,0}a)$ in conformal time. Setting $a=0$ at $\tau =0$ gives the stated quadratic-plus-linear scale factor. The linear term controls radiation domination and the quadratic term controls matter domination. The formula presumes $a_0=1$, ${\Omega }_{r,0}+{\Omega }_{m,0}=1$ and no cosmological constant.

A round three-sphere of radius $R$ embedded by $|\mathbf{x}{|}^2+u^2=R^2$ has metric $R^2\{d{r}^2/(1-r^2)+r^{2}d{\Omega }^2\}$ for $r=|\mathbf{x}|/R$. Multiplying by $b(t{)}^2$ gives a scale factor $a=Rb$ and normalized positive spatial curvature of an FLRW universe $K=1$. This radial chart covers a hemisphere; $r=\sin \chi$ gives the global polar description.