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Consider a species in thermal equilibrium, with energy density and pressure depending on temperature and with vanishing chemical potential, as in the supplied thermodynamic relation. Put $U=\rho (T)V$. The first law of thermodynamics givesSince $dP/dT=(\rho +P)/T$,The differential is therefore exact:After fixing the irrelevant additive entropy zero, the entropy density at zero chemical potential isMore precisely the first law leaves an additive constant in $S$; extensivity fixes that constant for the usual entropy-density normalization. A nonzero chemical potential would instead require $Ts=\rho +P-\mu n$, so the stated formula is not a universal identity for a decoupled massive species with conserved particle number.

For a reversible thermodynamic process that is an adiabatic process in a comoving volume, $\dot{V}=3HV$ and the cosmological perfect-fluid continuity equation gives $\dot{\rho }=-3H(\rho +P)$. Differentiating the extensive expression explicitly, or using the exact differential above, givesThis is cosmological entropy conservation for the closed equilibrium gas with no entropy-producing energy injection.

For radiation in cosmology, $P=\rho /3$, so $s=4\rho /(3T)$. Comparing $s=c_1{g}_{\star }{T}^3$ and $\rho =c_2{g}_{\star }{T}^4$ yields the coefficient ratioWhen all relativistic species share the same temperature and count, the familiar normalizations are $c_1=2{\pi }^2/45$ and $c_2={\pi }^2/30$. More generally the energy and entropy effective counts can differ; the common $g_{\star }$ here uses the assumptions stated in the question.

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.

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.

The prescribed cosmic string network scaling gives $\dot{\rho }=-2H\rho$. Comparing with the cosmological perfect-fluid continuity equation givesThe averaged pressure is therefore $P=-\rho /3$, provided the network is treated as the stipulated isotropic homogeneous component. The Friedmann acceleration equation gives $\ddot{a}=0$. On an expanding branch let $\rho =C/a^2$ and $B^2=8\pi GC/3-K>0$; thenThis coasting cosmic-string universe has no growing flatness instability, but it does not drive an initially nonflat universe toward flatness. The comoving Hubble radius remains constant, so there is also no shrinking-radius mechanism of the usual cosmic inflation type.

The global causal issue is more precise than saying that a constant comoving Hubble radius solves every Horizon problem. If this ideal coasting solution extends to its zero-size time $t_B$, then ${\int }_{t_B}^{t}d{t}^{\prime }/[B(t^{\prime }-t_B)]$ diverges logarithmically, so it has no finite particle horizon. If string domination starts at a finite earlier time after another cosmological phase, its causal history must instead be evaluated from that earlier phase. The string era alone does not establish that previously disconnected regions had time to equilibrate. A static branch with $B=0$ is a separate possibility; its Hubble-normalized $\Omega$ is undefined and is not covered by the expanding formulas.

Take a homogeneous comoving cell of physical volume $V_c=a^3{V}_0$ and total entropy $S=s{V}_c$. At vanishing chemical potential, the reversible first law of thermodynamics readsFor an extensive equilibrium system, the thermodynamic Euler relation is $E=TS-P{V}_c+\mu N$. At $\mu =0$ this givesThe supplied pressure identity verifies consistency directly: $dP=(\rho +P)dT/T$ impliesThe thermodynamic Euler relation fixes the additive normalization that a differential identity alone would leave undetermined.

For adiabatic cosmological expansion, the cosmological perfect-fluid continuity equation gives $\dot{\rho }+3H(\rho +P)=0$, while ${\dot{V}}_c=3H{V}_c$. HenceThis proves cosmological entropy conservation in the stated perfect-fluid approximation. Kinetic equilibrium alone does not guarantee adiabatic evolution: entropy injection, dissipative stresses or out-of-equilibrium reactions would require additional source terms. Those effects are implicitly excluded in this calculation.

For separately conserved cosmological fluids, the prerequisite is absence of energy or momentum exchange between components. Each matter action then gives its own local conservation equation ${\nabla }_{\mu }{T}_i^{\mu \nu }=0$. In the homogeneous perfect fluid in general relativity, its time component isThis is the cosmological perfect-fluid continuity equation. Equivalently, a physical volume $V\propto a^3$ following the fluid satisfies $d({\rho }_{i}V)=-P_{i}dV$.

The Einstein field equations instead have the single shared metric on its left and ${\sum }_i{T}_i^{\mu \nu }$ on its right. The Friedmann equation and Friedmann acceleration equation therefore involve total density and total pressure; the same geometry cannot be sourced separately by each component's density alone.

The PDF's separate-conservation statement needs this noninteraction qualification. In an interacting system,where $Q_i$ is the energy-transfer rate into component $i$. Total conservation remains true, but the source-free equation does not hold individually during processes such as annihilation or particle decay.

For a constant equation-of-state parameter $w=P/\rho$, the cosmological perfect-fluid continuity equation integrates to $\rho \propto a^{-3(1+w)}$. Comparing the given powers yieldsThis constant-equation-of-state density scaling describes a cosmic string network and a cosmological domain wall network. An intuitive check is that the energy in a fixed comoving cell grows in proportion to the total string length, $a$, or total wall area, $a^2$, while the physical cell volume grows as $a^3$. Thus the densities scale as $a^{-2}$ and $a^{-1}$ respectively. These effective equations of state use the noninteracting, statistically isotropic network assumptions of the problem.

Let $V$ be the physical volume of a region with fixed comoving coordinates, so $V\propto a^3$, and write $U=\rho V$. The first law of thermodynamics is $dU=TdS-PdV+\mu dN$. Use extensive quantities: under replication, $U(\lambda S,\lambda V,\lambda N)=\lambda U(S,V,N)$. The Euler theorem for homogeneous functions givesSetting $\mu =0$ yieldsThis is the entropy form of the Gibbs-Duhem equation; it fixes the normalization by extensivity, rather than adding an arbitrary volume-independent entropy constant.

For equilibrium expansion of the same region, the first law and the cosmological perfect-fluid continuity equation implyTherefore $\boxed{\dot{S}=0}$, establishing cosmological entropy conservation for adiabatic equilibrium evolution. Here “comoving volume” means a patch moving with the cosmological fluid; its physical volume $V$ changes. The corresponding coordinate volume is constant. The local TeX's $\dot{V}/P$ in the continuity-equation hint is a transcription error: the PDF correctly has $\dot{V}/V$.

The cosmological perfect-fluid continuity equation and the constant barotropic equation of state $P=w\rho$ giveOn every interval where $a$ is monotone, separation of variables yieldsHence the constant-equation-of-state density scaling iswhere $\mu$ is a constant. The same expression extends through a regular turning point by continuity.

In cosmic time, all three Friedmann-Lemaitre-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 $\ddot{a}/a=-4\pi G(\rho +3P)/3$.

Take a fixed comoving region, whose physical volume is $V=V_0{a}^3$. Its energy is $E=\rho V$, so the first-law relation $dE=-PdV$ givesDifferentiating with respect to cosmic time and using the Hubble parameter $H=\dot{a}/a$ yields the cosmological perfect-fluid continuity equation

For the barotropic equation of state $P=w\rho$, separation gives the constant-equation-of-state density scalingSubstituting this into the flat Friedmann equation and integrating the expanding branch gives, for $w\ne -1$,For $w=-1$, the density and Hubble parameter are constant and instead $a(t)\propto e^{Ht}$.

Conformal time is defined byAt radiation–string equality, let each component have density ${\rho }_{\mathrm{eq}}$ at $a=a_{\mathrm{eq}}$. Since radiation in cosmology has $w=1/3$ and a cosmic string network has $w=-1/3$,ConsequentlyThusOn the expanding branch, integration givesChoosing the Big Bang to occur at $\tau =0$ gives the radiation--cosmic-string Friedmann solution

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

Substitution of these expressions into the cosmological perfect-fluid continuity equation givesContinuity through a turning point therefore yields the homogeneous Klein-Gordon equationDuring slow-roll inflation, the acceleration and kinetic-energy terms are negligible. In terms of the reduced Planck mass, the approximate evolution is

Near the minimum, the Taylor expansion of the potential isNeglecting Hubble friction during one short oscillation reduces the Klein-Gordon equation to the harmonic oscillator equationThe virial theorem gives $⟨{\dot{\varphi }}^2/2⟩=⟨V⟩$, so $⟨P_{\varphi }⟩=0$. Restoring the slow cosmological damping in the averaged cosmological perfect-fluid continuity equation givesThe coherently oscillating inflaton therefore behaves as pressureless matter.

The cosmological perfect-fluid continuity equation isFor constant equation-of-state parameter $w=p/\rho$, integration giveswhere $a_0=1$. Define the present cosmological density parameterThe Friedmann equation then givesSpatial curvature may be included as an effective $w=-1/3$ component. Once the parameters are specified, the scale factor follows from the first-order equation $\dot{a}=aH(a)$, or equivalently

Differentiate the Friedmann equationand use the cosmological perfect-fluid continuity equationAfter dividing by $2H$,Since $\ddot{a}/a=\dot{H}+H^2$, eliminating the curvature term gives the Friedmann acceleration equation

Under the strong energy condition in a Friedmann universe, $\rho +3P\ge 0$, so $\ddot{a}/a\le 0$. Thereforeand wherever $H\ne 0$,

If $H(t_0)>0$, integrating the inequality backward shows that $H^{-1}$ reaches zero no later than $t_0-H(t_0{)}^{-1}$. Thus $H\to +\infty$ and $a\to 0$ at a finite time in the past. If $H(t_0)<0$, the same argument forward in time makes $H\to -\infty$ and $a\to 0$ at a finite future time. The sign of $H$ distinguishes an expanding universe with a past Big Bang from a contracting universe ending in a Big Crunch. Hence a Friedmann universe obeying the strong energy condition cannot remain nonsingular for unlimited proper time in both directions.

The constant $R$ is the curvature radius of the closed spatial slices when $a=1$; at general scale factor their curvature radius is $Ra$ and their scalar curvature is $6/(R^2{a}^2)$.

The cosmological perfect-fluid continuity equation isDuring expansion, $\dot{a}>0$, and thereforeIf $C=\rho (t_0)a(t_0{)}^3$, then throughout the later expanding phaseMultiplying the Friedmann equation by $a^2$ givesThe right-hand side is negative oncewhich is impossible because ${\dot{a}}^2\ge 0$. Thus the scale factor cannot grow without bound.

In fact the expanding phase ends in finite time. Combining the Friedmann and continuity equations yields the Friedmann acceleration equationThe curvature term in the first Friedmann equation also impliesWhile $a$ remains below the finite bound just found, $\ddot{a}$ is bounded above by a strictly negative constant. Hence $\dot{a}$ reaches zero after finite time: the closed universe attains a maximum size and cannot expand forever. This is recollapse of a closed Friedmann universe with nonnegative pressure.

For a component with the barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation isUsing $H=\dot{a}/a$ makes this a separable differential equation:Taking $a_0=1$ gives the constant-equation-of-state density scalingThe spatially flat Friedmann equation and the cosmological density parameters therefore giveAlong a radial light ray, $d\chi =dt/a$. Since the redshift-time relation is $dt=-dz/[(1+z)H]$ and $a=(1+z{)}^{-1}$,Hence the comoving radial distance is

Write $A=8\pi G/(3c^2)$ and $H=\dot{a}/a$. Differentiating the Friedmann equation givesThe cosmological perfect-fluid continuity equation says $\dot{\rho }=-3H(\rho +P)$, so, first where $H\ne 0$ and then everywhere by continuity,Using $K{c}^2/a^2=A\rho -H^2$ from the original equation givesSince $\dot{H}+H^2=\ddot{a}/a$ for the Hubble parameter, this is the Friedmann acceleration equation.

For a comoving region, $E=\rho V$ and $V=a^3{V}_0$. The work law givesso differentiation with respect to time yieldsSince $\dot{V}/V=3\dot{a}/a$, division by $V$ gives the cosmological perfect-fluid continuity equation