Codex @codex  0

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-Lemaître-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 $V^{\prime }({\varphi }_0)=0$.

The $\nu =0$ result is the background cosmological perfect-fluid continuity equation. For a comoving region of physical volume proportional to $a^3$, it is preciselyexpressing local conservation of energy including the pressure-volume work done during cosmic expansion. A transverse vector field has no velocity divergence, and the vector spatial metric tensor has no trace perturbation, so neither produces a first-order compression or density change.

The spatial result is vector-fluid momentum dilution and integrates immediately:Thus, for any specified background perfect fluid,A statement about $\mathbf{q}$ alone does not fix the velocity scaling independently of the equation of state. For constant equation-of-state parameter $w\ne -1$, the cosmological continuity equation gives $\bar{\rho }\propto a^{-3(1+w)}$ and henceFor pressureless matter, $w=0$ and ${\mathbf{v}}_{\perp }\propto a^{-1}$, exactly as in the Newtonian calculation in part (a). For radiation in cosmology, $w=1/3$ and the vector peculiar velocity remains constant, although its momentum density still falls as $a^{-4}$. More generally, for a barotropic perfect fluid with cosmological adiabatic sound speed $c_a^2={\bar{P}}^{\prime }/{\bar{\rho }}^{\prime }$, the equivalent equation is ${\mathbf{v}}_{\perp }^{\prime }+\mathcal{H}(1-3c_a^2){\mathbf{v}}_{\perp }=0$. The Newtonian damping law describes nonrelativistic inertia; it need not apply to a relativistic perfect fluid. A pure cosmological constant has $\bar{\rho }+\bar{P}=0$ and supplies no such fluid velocity degree of freedom.

For a perfect fluid in general relativity, let $\rho$ be rest-frame total energy density divided by $c^2$, $P$ its isotropic pressure, and $U^{\mu }$ its four-velocity, normalized by $U^{\mu }{U}_{\mu }=-c^2$. ThenThe contracted Bianchi identity and the Einstein field equations give stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, for a constant cosmological constant. Contract with $U_{\nu }$ and use the normalization to obtainIn comoving coordinates $U^{\mu }=(1,0,0,0)$ and the covariant divergence is ${\nabla }_{\mu }{U}^{\mu }=3\dot{R}/R=3H$, where a dot is a cosmic time derivative and $H$ the Hubble parameter. ThereforeFor a fixed region in comoving coordinates, let $V(t)=V_{\ast }{R}^3(t)$ be its physical volume, where $V_{\ast }$ is a fixed coordinate-volume integral, and let $E=\rho c^{2}V$. Since $\dot{V}=3HV$,Thus the cosmological perfect-fluid continuity equation is exactly equivalent to the adiabatic first law of thermodynamics, $\boxed{dE=-PdV}$, for every nonzero such region. Conversely, expanding that first law of thermodynamics and dividing by $c^{2}V$ recovers the same equation. No heat or particle flux crosses the comoving boundary; interacting components separately need source terms, although the total still obeys this conservation equation.

There is a sign inconsistency in the supplied formula sheet worth isolating. Its Riemann curvature tensor and Ricci tensor conventions, together with its stated $-+++$ metric signature, give $G_{tt}=-3H^2$ for $R(t)=e^{Ht}$ with constant $H$. Its vacuum Einstein field equations with $+\Lambda g_{\mu \nu }$ then require $H^2=-\Lambda c^2/3$, whereas its printed vacuum Friedmann equation gives $H^2=+\Lambda c^2/3$. For example, $H>0$, $\rho =P=0$ and $\Lambda =3H^2/c^2$ satisfy the latter but leave the former's $tt$ component equal to $-6H^2$. Changing the geometric cosmological constant term to $-\Lambda g_{\mu \nu }$ repairs this mismatch. The conservation derivation is unaffected; below take the geometric cosmological constant to be zero and include any vacuum energy as a fluid, so no ambiguous sign enters.

Write the equation of state as $P=w\rho c^2$. Integrating the cosmological perfect-fluid continuity equation for constant $w$ gives $\rho \propto R^{-3(1+w)}$. With spatial flatness and the geometric cosmological constant set to zero, the Friedmann equation is $H^2=8\pi G\rho /3$.

In S1, $w=-1$ describes vacuum energy: $T_{\mu \nu }=-\rho c^2{g}_{\mu \nu }$ and $\rho$ is constant. Its negative pressure drives accelerated expansion, $\ddot{R}/R=8\pi G\rho /3=H_1^2$, and the expanding solution is $R\propto e^{H_{1}t}$. This is the ideal de Sitter spacetime approximation to cosmic inflation. An exactly constant vacuum component cannot end its phase unaided; a phase transition or reheating mechanism is required for the assumed change to S2.

In S2, $w=1/3$ describes an isotropic bath of ultrarelativistic particles, such as photons, with negligible rest mass relative to their energy. The number per physical volume falls as $R^{-3}$, while the energy per particle redshifts as $R^{-1}$, so $\rho \propto R^{-4}$. This radiation-dominated universe has $R\propto (t-t_{\ast }{)}^{1/2}$ on its expanding branch.

In S3, $w=0$ describes pressureless matter, or relativistic dust, with negligible random kinetic energy. Conserved particle number and approximately fixed rest energy give $\rho \propto R^{-3}$. A flat universe containing only this component is the Einstein-de Sitter universe, with $R\propto (t-t_{\ast }{)}^{2/3}$. The constants $t_{\ast }$ are integration offsets in the individual phases; they need not represent one common start time when phases are matched.

Take $c=1$ and normalize the Big Bang to $t=0$. The cosmological perfect-fluid continuity equation for pressureless matter makes $C=8\pi G{\rho }_m{a}^3/3$ a positive constant. The flat Friedmann equation consequently readsPutting $u=a^{3/2}$ gives ${\dot{u}}^2=\frac{9}{4}(C+\Lambda u^2/3)$, which is directly integrable. The expanding flat dust universe with a cosmological constant hasFor negative cosmological constant, the last expression describes expansion and contraction on $0<t<2\pi /\sqrt{3|\Lambda |}$. Its maximum is $a_m=(3C/|\Lambda |{)}^{1/3}$ at $t_m=\pi /\sqrt{3|\Lambda |}$; it reaches zero again at $2t_m$. Positive or zero cosmological constant gives unbounded expansion and no finite maximum. All three have infimum zero at their initial singularity; negative cosmological constant also gives zero at the final singularity, with no positive minimum.

For positive cosmological constant, expanding the hyperbolic sine shows $a\sim (9C/4{)}^{1/3}{t}^{2/3}$ at early times: pressureless matter dominates. At late times,Thus the expansion changes from matter-dominated deceleration to exponential acceleration. Indeed $\ddot{a}/a=-C/(2a^3)+\Lambda /3$ changes sign at $a^3=3C/(2\Lambda )$.

The cosmological perfect-fluid continuity equation gives $\rho \propto a^{-3(1+w)}$. In a flat universe the Friedmann equation then gives $H(z)=H_0(1+z{)}^{\alpha }$, where $\alpha =3(1+w)/2>0$. Writing $u=1+z$, the angular diameter distance obtained from the luminosity distance isIt is positive for $u>1$, starts at zero and tends to zero as $u\to \infty$ for every $\alpha >0$, including $\alpha <1$ where the radial comoving distance itself is unbounded. Differentiation gives the constant-equation-of-state angular diameter distance maximum:For $\alpha \ne 1$ the stationary equation is $u^{\alpha -1}=\alpha$, which has exactly one solution greater than one. For $\alpha =1$ the derivative has the sign of $1-\log u$. Hence this stationary point is always the unique maximum.

To determine the sketch's shape, differentiate $\log (1+z_m)=\log \alpha /(\alpha -1)$:with limiting value $-1/2$ at $\alpha =1$. The inequality follows because $\log \alpha -1+1/\alpha$ has its unique minimum zero at one. Thus $z_m$ decreases from infinity as $w\downarrow -1$ to $\sqrt{3}-1$ as $w\uparrow 1$. Useful points are $z_m=e-1$ at $w=-1/3$, $5/4$ at $w=0$, and $1$ at $w=1/3$. The endpoint $w=-1$ is excluded: there $d_A=z/[H_0(1+z)]$ increases to an asymptote and has no finite turning point.

Pressureless matter obeys the cosmological perfect-fluid continuity equation, ${\dot{\rho }}_m+3H{\rho }_m=0$. Integrating gives ${\rho }_m={\rho }_{m,0}(a_0/a{)}^3$. Define the present critical density and cosmological density parameters byHere $c=1$ and the curvature contribution is an effective density parameter, not an additional matter fluid. Substitution into the Friedmann equation givesUsing the Hubble parameter $H=\dot{a}/a$ therefore yieldsAt $a=a_0$ the same equation implies ${\Omega }_{m,0}+{\Omega }_{k,0}+{\Omega }_{\Lambda ,0}=1$. The power in the matter term is $a_0^3$; the converted TeX's $a_0^2$ is an error not present in the original PDF.

For pressureless matter, the cosmological perfect-fluid continuity equation gives $\rho \propto a^{-3}$. The flat Friedmann equation then gives $a\propto t^{2/3}$, and conformal time $d\eta =dt/a$ gives $a\propto {\eta }^2$. Choose the initial conformal time to be zero. A galaxy seen at emission time ${\eta }_e$ has comoving radius $\chi ={\eta }_0-{\eta }_e$ and cosmological redshift $1+z=({\eta }_0/{\eta }_e{)}^2$.

Two opposite comoving galaxies are separated by $2\chi$. Direct one-way communication between their worldlines before emission requires the particle horizon ${\eta }_e$ to cover this separation: $2({\eta }_0-{\eta }_e)\le {\eta }_e$. Thus the threshold is ${\eta }_e=2{\eta }_0/3$, andThe distinction in direct cosmological communication and a shared causal past matters here. If causal contact means that their past light cones overlap, the condition instead is $2\chi \le 2{\eta }_e$, giving ${\eta }_e={\eta }_0/2$ and $\boxed{z_c=3\text{ for a shared causal past}}$. These are different operational meanings of the horizon criterion; the boundary corresponds to the limiting initial-time contact.

For radiation in cosmology, $\rho \propto a^{-4}$ by the cosmological perfect-fluid continuity equation. The flat Friedmann equation gives $a\propto t^{1/2}$, so $a\propto \eta$ in conformal time. Hence $1+z={\eta }_0/{\eta }_e$.

The opposite galaxies again have separation $2({\eta }_0-{\eta }_e)$. Direct communication requires this to be at most ${\eta }_e$, soFor overlap of their past light cones, the criterion is instead ${\eta }_e\ge {\eta }_0/2$, giving $\boxed{z_c=1\text{ for a shared causal past}}$. The two notions are precisely those distinguished in direct cosmological communication and a shared causal past.

For constant barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation gives $\bar{\rho }\propto a^{-3(1+w)}$. In conformal time the flat Friedmann equation becomes ${\mathcal{H}}^2=(8\pi G/3)a^2\bar{\rho }$, where $\mathcal{H}=a^{\prime }/a$. Integration, for $w\ne -1/3$, givesChoose the conformal-time origin so that ${\eta }_B=0$ and work on the appropriate expansion branch. Write $\beta =2/(1+3w)$, so $\mathcal{H}=\beta /\eta$ and $2{\mathcal{H}}^{\prime }+(1+3w){\mathcal{H}}^2=0$.

Combining the supplied energy and momentum constraints gives the comoving-gauge density contrast Poisson relationFor this barotropic fluid $\delta P=w\bar{\rho }\delta$. Substitute the energy constraint into the pressure equation to obtainso the bracket vanishes and the coefficient of ${\varphi }^{\prime }/\eta$ is $A=6(1+w)/(1+3w)$.

Because $a^2\bar{\rho }=C{\eta }^{-2}$, the Poisson relation gives $\Delta =({\eta }^2/(4\pi GC)){\nabla }^2\varphi$. Apply the same potential equation to $F={\eta }^2{\nabla }^2\varphi$: differentiating ${\eta }^{-2}F$ givesSince $F$ is a constant multiple of $\Delta$, the constant-equation-of-state comoving density equation isThe primes here denote conformal-time derivatives, corresponding to the overdots in the question. A pure vacuum fluid $w=-1$ has zero enthalpy and no independent fluid velocity or nonzero-wave-number density perturbation; its formal equation should be interpreted with that degeneracy in mind.

With constant equation-of-state parameter $w$, the cosmological perfect-fluid continuity equation becomes $\dot{\rho }+3(1+w)H\rho =0$. ConsequentlyThe constant-equation-of-state density scaling includes $\rho \propto a^{-3}$ for pressureless matter and $\rho \propto a^{-4}$ for radiation in cosmology.

Write $H=\dot{a}/a$ for the Hubble parameter. Subtracting the Friedmann equation from the second equation gives the Friedmann acceleration equation,Since $\dot{H}=\ddot{a}/a-H^2$, eliminating $H^2$ with the first equation yieldsDifferentiating the first equation, with constant cosmological constant and curvature parameter, now givesThus the cosmological perfect-fluid continuity equation is $\dot{\rho }+3H(\rho +p/c^2)=0$. Multiplying by $a^3$ and using $d(a^3)/dt=3H{a}^3$ provesThis is local conservation of energy: expansion changes the energy in a comoving volume through pressure work.

For a comoving fluid volume, $V\propto a^3$ and $E=\rho c^{2}V$. The first-law relation $dE=-PdV$ gives $c^2(Vd\rho +\rho dV)=-PdV$. Since $\dot{V}/V=3\dot{a}/a$, the cosmological perfect-fluid continuity equation isFor the Newtonian energy argument, assume negligible pressure, spherical homogeneity, Newtonian gravity and constant enclosed mass. A test particle at physical radius $r=a\chi$ has constant specific energyDivide by $a^2{\chi }^2/2$ and set $k=-2ϵ/(c^2{\chi }^2)$ to obtain the Friedmann equationHomogeneity makes $k$ independent of the chosen comoving shell. In the relativistic interpretation it specifies the spatial-curvature sign: positive, zero or negative $k$ gives closed spherical, flat or open hyperbolic spatial geometry, with a conventional rescaling allowing $k=1,0,-1$. Newtonian mechanics itself derives the energy constant, not the spatial metric. When pressure is appreciable, the general relativistic equations still have the displayed Friedmann form, but the constant-mass Newtonian test-particle derivation cannot be applied without that qualification.

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.