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A coasting fluid has pressure $p=-\rho /3$. The cosmological perfect-fluid continuity equation makes its density scale as $a^{-2}$. If it alone fills a flat expanding universe, the scale factor grows linearly in physical time. Its background density has the same scale-factor dependence as a spatial-curvature term, although the fluid does not itself create spatial curvature.

For an expanding Friedmann-Lemaitre-Robertson-Walker metric with positive energy density, put $\Omega =8\pi G\rho /(3H^2)$ and $w=P/\rho$, in units $c=1$. The Friedmann equation, Friedmann acceleration equation and cosmological perfect-fluid continuity equation giveFor constant $w$, linearizing at $\Omega =1$ gives $\Omega -1\propto a^{1+3w}$. For variable $w$, replace the power by $\exp \int (1+3w(a))d\log a$. This describes the Flatness problem for decelerating expansion, away from epochs at which $H=0$.

For an extensive equilibrium system, the thermodynamic Euler relation gives $E=TS-PV+\mu N$. At zero chemical potential, divide by $V$ to obtain $s=(\rho +P)/T$. Equivalently, $dP=(\rho +P)dT/T$ makesThe first law of thermodynamics identifies this as $dS$; the thermodynamic Euler relation fixes the entropy normalization. For adiabatic expansion satisfying the cosmological perfect-fluid continuity equation, the numerator on the right vanishes, proving cosmological entropy conservation.

For a homogeneous canonical scalar field, the cosmological perfect-fluid continuity equation reduces to $\dot{\varphi }(\ddot{\varphi }+3H\dot{\varphi }+V^{\prime })=0$. It implies the inflaton equation of motion wherever $\dot{\varphi }\ne 0$, and by continuity at isolated turning points. It cannot by itself exclude a frozen field on a time interval. Indeed, choose $\varphi ={\varphi }_0$ with $V({\varphi }_0)>0$ but $V^{\prime }({\varphi }_0)\ne 0$; a de Sitter scale factor in flat slicing with $H^2=V({\varphi }_0)/(3M_{\mathrm{Pl}}^2)$ satisfies the homogeneous Friedmann equation and Friedmann acceleration equation, but not the scalar equation. Varying the inflaton action in an expanding universe supplies the missing equation also for a frozen field, requiring $V^{\prime }({\varphi }_0)=0$.

Take $w$ constant and $w\ne -1$, with a barotropic perturbation $\delta P=w\delta \rho$. Define the density contrast $\delta =\delta \rho /\bar{\rho }$ and use the scalar velocity potential convention $v_i(\mathbf{k})=i{k}_i\theta (\mathbf{k})$, so ${\partial }_i{v}^i=-k^2\theta$. This $\theta$ is a potential, not the sometimes-used velocity-divergence variable. In synchronous gauge in cosmology, the inverse metric is $g^{00}=-a^{-2}$ and $g^{ij}=a^{-2}({\delta }^{ij}-h^{ij})$ to first order. Discarding products of perturbations in the perfect fluid in general relativity givesFor the linearization, each $h_{ij}$ must be small: the PDF's determinant condition alone is insufficient, since $h=\mathrm{diag}(L,0,0)$ has zero determinant even for large $L$.

Write $\mathcal{H}=a^{\prime }/a$. The time component of covariant conservation of stress-energy isHere ${\Gamma }_{\mu 0}^{\mu }=4\mathcal{H}+h^{\prime }/2$. To first order, ${\Gamma }_{ij}^0{T}^{ij}=a^{-2}w\bar{\rho }[3\mathcal{H}(1+\delta )+h^{\prime }/2]$: the two metric-$h$ contributions cancel. Multiplying the conservation equation by $a^2$, including ${\Gamma }_{00}^0{T}^{00}$, givesThe background cosmological perfect-fluid continuity equation, ${\bar{\rho }}^{\prime }=-3\mathcal{H}(1+w)\bar{\rho }$, cancels the first and third terms. ThusThe $h^{\prime }$ term accounts for perturbation of the spatial volume expansion in synchronous gauge in cosmology. The PDF's correct connection ${\Gamma }_{0j}^i=\mathcal{H}{\delta }_j^i+h^{\prime i}{}_j/2$ is garbled in the TeX aid.

Assume positive fluid density, ${\rho }_0>0$, and write $m=1+3w>0$. A perfect fluid with constant equation of state obeys the cosmological perfect-fluid continuity equation, givingIn the stated units the Friedmann equation is $H^2=\rho /3-k/a^2+\Lambda /3$. Multiplication by $a^2/2$ yields the Friedmann effective potential for a constant-equation-of-state fluid,The allowed region has $V\le 0$. The Friedmann acceleration equation is equivalently $\ddot{a}=-V^{\prime }(a)$, including turning points by continuity. Thus a zero-energy mechanical trajectory reproduces the cosmological evolution, with $a>0$.

For $k=0$, $\Lambda <0$, the potential increases strictly from negative infinity to positive infinity. Its unique zero givesExpansion from the Big Bang stops there, with negative acceleration, then reverses into a Big Crunch. Both the turning point and the final singularity occur in finite proper time: $dt=da/\sqrt{-2V}$ is integrable near a simple turning point and behaves as a constant times $a^{m/2}da$ near zero.

For $\Lambda =0$, $k=+1$, $V$ increases from negative infinity to $1/2$ and crosses zero atThis is again expansion followed by finite-time recollapse. For $k=-1$, the same increasing curve has asymptote $-1/2$ and never meets zero. Expansion continues without a finite maximum; asymptotically $\dot{a}\to 1$ and $a\sim t$, as spatial curvature dominates the diluted fluid.

For $k=0$, $\Lambda >0$, $V$ tends to negative infinity at both ends. It has a maximum atThere is no turning point. Expansion initially decelerates, then accelerates once $a>a_{\ast }$, and approaches de Sitter spacetime expansion $a\propto e^{\sqrt{\Lambda /3}t}$. Thus positive flat dark-energy expansion is unbounded; negative flat dark energy and positive curvature without dark energy recollapse.

The cosmological perfect-fluid continuity equation gives ${\rho }_r={\rho }_{r,0}{a}^{-4}$ and ${\rho }_m={\rho }_{m,0}{a}^{-3}$. In conformal time, $\dot{a}=a^{\prime }/a$ and $H=a^{\prime }/a^2$. Thus the flat Friedmann equation isChoose the expanding branch and put the Big Bang at $\tau =0$. Integration givesInverting it supplies the radiation-matter scale factor in conformal time:The early linear term gives radiation domination, while the late quadratic term gives matter domination. The solution presumes the stated radiation-plus-matter model, without a cosmological constant.

At matter-radiation equality, $a_{\mathrm{eq}}={\Omega }_{r,0}/{\Omega }_{m,0}$. Since ${\Omega }_{r,0}+{\Omega }_{m,0}=1$,In units $c=1$, the comoving particle horizon at equality is ${\tau }_{\mathrm{eq}}$; its physical size is $a_{\mathrm{eq}}{\tau }_{\mathrm{eq}}$. The angular diameter distance to the equality surface is $a_{\mathrm{eq}}({\tau }_0-{\tau }_{\mathrm{eq}})$. Therefore the horizon angle at matter-radiation equality corresponding to one horizon length isFor ${\Omega }_{r,0}=10^{-4}$, this is $4.20\times 10^{-3}$ radians, or $0.241^{\circ }$, about $14.4$ arcminutes. If quoting the full diameter of a horizon-sized patch, the result is twice this, $0.481^{\circ }$. The small-angle approximation is excellent. Expanding in $\sqrt{{\Omega }_{r,0}}\ll 1$ would give $4.14\times 10^{-3}$ radians; retaining both cosmic components near equality avoids an inaccurate pure-matter extrapolation there. The scale factor cancels between physical size and angular diameter distance, and $H_0$ cancels from the ratio.

If cosmological recombination occurs only a modest expansion after equality, its causal patch still subtends a small part of the sky. Widely separated regions of the observed Cosmic microwave background have nearly the same temperature even though their past particle horizons did not overlap in the ordinary decelerating history. Local thermalization after the Big Bang therefore cannot explain this large-angle uniformity. This is the Horizon problem; inflation supplies an earlier connected region that can grow to encompass the observed sky.

Put $\Omega ={\Omega }_{m,0}$ and $b=1-\Omega$. Integrating the cosmological perfect-fluid continuity equation for the separately conserved cosmological fluids givesIn particular, ${\rho }_X{a}^2$ is constant. At the present epoch the flat Friedmann equation implies ${\Omega }_{X,0}=b$. Nonnegative fluid densities therefore require $0\le \Omega \le 1$.

The conformal time relation $dt=ad\tau$ gives the conformal Hubble parameter $\mathcal{H}=aH$. ConsequentlyOn the expanding branch, divide by $\mathcal{H}$ and use $a^{\prime }=a\mathcal{H}$:Eliminating the matter term yields the conformal Riccati equation for matter and a coasting fluid,The nonnegative square root fixes the convenient parameter convention; only ${\alpha }^2$ enters the differential equation.

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 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.

Because density is a scalar field, $\tilde{\rho }(\tilde{x})=\rho (x)$. Write both sides as background plus perturbation and expand the background time shift:The coordinate shift acting on an already first-order perturbation contributes only at second order. Thus the density perturbation gauge transformation isThe uniform-density curvature perturbation in the paper's sign convention isThe derivative on ${\bar{\rho }}^{\prime }$ is essential and is present in the PDF; the TeX transcription drops it. The scalar-curvature combination transforms aswhile the density term changes by $-\mathcal{H}T$. Hence the two time-slicing changes cancel:On a uniform-density slice, $\delta \rho =0$ and this variable is the signed spatial-curvature perturbation. The construction assumes ${\bar{\rho }}^{\prime }\ne 0$; a pure constant-density cosmological constant does not define such a time slicing.

In Newtonian gauge in cosmology, $E=B=0$ and $C=-\Phi$. The background cosmological perfect-fluid continuity equation gives ${\bar{\rho }}^{\prime }=-3\mathcal{H}(\bar{\rho }+\bar{P})$, so, writing $D=\bar{\rho }+\bar{P}$,Set the cosmological adiabatic sound speed $c_a^2={\bar{P}}^{\prime }/{\bar{\rho }}^{\prime }$. Then $D^{\prime }=-3\mathcal{H}D(1+c_a^2)$. Differentiate the previous expression and insert the perturbed energy-conservation equation:Consequently the curvature evolution equation in this sign convention isThe last equality defines the total energy-frame velocity by $\mathbf{q}=D\mathbf{v}$. For adiabatic cosmological perturbations, the non-adiabatic pressure perturbation vanishes. On superhorizon scales, the stated suppression of the velocity divergence makes the remaining gradient term negligible. Thus $\zeta$ is conserved to leading order in $k/\mathcal{H}$. The positive sign of the pressure source follows from the paper's definition of $\zeta$ and must not be replaced by a formula using the opposite curvature convention.

Superhorizon conservation of uniform-density curvature lets one carry a primordial perturbation from its inflationary generation through otherwise complicated eras and predict the initial conditions for later Cosmic microwave background anisotropy and structure growth. It relies on adiabaticity: an isocurvature perturbation can source curvature evolution, so the same conclusion is not automatic in an arbitrary multifield model.