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For a perfect fluid in general relativity with constant equation-of-state parameter $w$, use conformal time $dt=ad\tau$ and the conformal Hubble parameter $\mathcal{H}=a^{\prime }/a=aH$. The Friedmann acceleration equation gives ${\mathcal{H}}^{\prime }=a\ddot{a}=-4\pi G{a}^2(1+3w)\rho /3$. The Friedmann equation eliminates $\rho$ and gives this Riccati equation. For positive energy density, $\alpha \ne 0$ and appropriately shifted conformal time, its usual branches are $\cot (\alpha \tau )$ for $k=1$, $1/(\alpha \tau )$ for $k=0$, and $\coth (\alpha \tau )$ for $k=-1$. Their scale factors are proportional to ${\sin }^{1/\alpha }(\alpha \tau )$, $(\alpha \tau {)}^{1/\alpha }$ and ${\sinh }^{1/\alpha }(\alpha \tau )$ respectively on intervals where the bases are positive. At $\alpha =0$ the conformal Hubble parameter is constant and the separate solution is $a=A{e}^{h\tau }$.

Use natural units and conformal time $dt=ad\tau$. The conformal Hubble parameter is $\mathcal{H}=a^{\prime }/a=aH$. For the barotropic equation of state $P=(\gamma -1)\rho$, the Friedmann acceleration equation givesThe Friedmann equation gives $8\pi G{a}^2\rho /3={\mathcal{H}}^2+k$. Hence the constant-equation-of-state conformal Riccati equation isFor $\alpha \ne 0$, separation and integration of this Riccati equation yield, after choosing the origin of conformal time,The open positive-energy density branch uses the hyperbolic cotangent, rather than the cotangent. Integrating $a^{\prime }/a=\mathcal{H}$ giveswhere the positive constants are fixed by the energy density normalization. These formulas apply on branches where the bases are positive. For $\alpha >0$, take the Big Bang at $\tau =0$. The cosmic time is thenThe integrals also give the general answer after selecting a suitable reference time when a finite Big Bang endpoint does not exist. The exceptional coasting fluid $\gamma =2/3$ has $\alpha =0$, so $\mathcal{H}=h$ is constant: $a=A{e}^{h\tau }$ and $t=(A/h)e^{h\tau }+t_{\ast }$. If $h=0$, instead $a=A$ and $t=A\tau +t_{\ast }$. A flat fluid with $\gamma =0$ is the constant-density de Sitter spacetime case, for which the flat time integral is logarithmic.

During matter domination, $\gamma =1$ and $\alpha =1/2$. Absorbing numerical factors into separate positive constants gives the closed matter-dominated Friedmann solution, the flat Einstein-de Sitter universe and the open matter-dominated Friedmann solution:To compare the age of a dust universe without a cosmological constant, specify the normalization. At fixed present Hubble parameter $H_0$, with $x=a/a_0$ and present matter cosmological density parameter ${\Omega }_m>0$, the Friedmann equation givesThe integrand decreases strictly with ${\Omega }_m$ for $0<x<1$. Therefore age ordering of dust universes at fixed Hubble parameter gives, on the expanding branches,The final bound is approached by the empty Milne universe. This compares models with the same $H_0$; fixing instead the present scale factor and energy density gives a different comparison.

Use units $c=1$ and take the initially expanding branch with ordinary positive matter or radiation density. In the Friedmann-Lemaître-Robertson-Walker metric, $k=+1,0,-1$ respectively describe positive, zero and negative spatial curvature. Separately conserved pressureless matter has ${\rho }_M\propto a^{-3}$ and radiation in cosmology has ${\rho }_R\propto a^{-4}$, so either dominates the curvature and vacuum terms sufficiently far back on a hot Big Bang branch.

With negligible cosmological constant, a spatially flat radiation universe has $a\propto t^{1/2}$ and a flat matter universe has $a\propto t^{2/3}$; both expand forever while decelerating. An open universe also expands forever and eventually becomes approximately curvature dominated, with $a\propto t$. A closed universe containing only ordinary matter/radiation reaches a maximum size and recollapses. For example, dust gives ${\dot{a}}^2=C/a-k$ with $C>0$, so only $k=+1$ admits the positive turning point $a=C$.

Curvature alone does not determine the fate when the cosmological constant is retained. Positive $\Lambda$ can cause late accelerated expansion and prevent even a closed universe from recollapsing; whether a closed expanding branch turns around depends on the relative matter and vacuum terms. Negative $\Lambda$ can cause recollapse even for flat or open spatial sections. These statements follow by inspecting ${\dot{a}}^2=C/a-k+\Lambda a^2/3$ for dust; the curvature sign is a geometrical classification, not by itself a universal destiny criterion.

Define the critical density ${\rho }_c=3H^2/(8\pi G)$ and cosmological density parameters ${\Omega }_M={\rho }_M/{\rho }_c$, ${\Omega }_R={\rho }_R/{\rho }_c$, ${\Omega }_{\Lambda }=\Lambda /(3H^2)$ and ${\Omega }_k=-k/(a^2{H}^2)$. The Friedmann equation gives ${\Omega }_M+{\Omega }_R+{\Omega }_{\Lambda }+{\Omega }_k=1$. Dividing the Friedmann acceleration equation by $-H^2$ givesConsequentlyThe deceleration parameter measures the sign and size of the acceleration relative to $a{H}^2$: $q>0$ means deceleration, $q<0$ accelerated expansion and $q=0$ instantaneously zero acceleration. In particular a falling Hubble parameter alone does not imply deceleration, since $q=-1-\dot{H}/H^2$.

For conformal time $d\eta =dt/a$, the conformal Hubble parameter is $\mathcal{H}=aH$. Using matter conservation and differentiating ${\Omega }_M\propto {\rho }_M/H^2$,This is the matter density flow with a cosmological constant. Neglecting ${\Omega }_{\Lambda }$ during matter domination reduces it to the requested relationThat last equation is not exact with nonzero $\Lambda$. For example, a flat dust-plus-vacuum state with $({\Omega }_M,{\Omega }_{\Lambda })=(0.9,0.1)$ gives ${\Omega }_M^{\prime }/\mathcal{H}=-0.27$, whereas the matter-only expression gives $-0.09$.

For the flatness problem, let ${\Omega }_{\mathrm{tot}}={\Omega }_M+{\Omega }_R+{\Omega }_{\Lambda }$. Curvature satisfies ${\Omega }_{\mathrm{tot}}-1=k/(aH{)}^2$, soIn decelerating near-flat radiation and matter eras, deviations therefore grow as $a^2$ and $a$ respectively. Maintaining small curvature at a late epoch requires very small early deviations unless a prior mechanism suppresses them. Cosmic inflation, with $q\simeq -1$, drives the deviation down approximately as $a^{-2}$. If vacuum energy is present, flatness means ${\Omega }_{\mathrm{tot}}=1$, not ${\Omega }_M=1$.

Write $\mathcal{H}=a^{\prime }/a$ for the conformal Hubble parameter and ${\theta }_N=i\mathbf{k}\cdot {\mathbf{v}}_N$ for the velocity divergence of a Fourier mode. Work in the synchronous gauge comoving with the cold dark matter, so ${\mathbf{v}}_C=0$. This remains consistent with its pressureless momentum equation ${\mathbf{v}}_C^{\prime }+\mathcal{H}{\mathbf{v}}_C=0$. Its density equation then gives ${\delta }_C^{\prime }=-h^{\prime }/2$.

For the string component, $1+3w_S=0$. Thus it makes no contribution to the trace gravitational source in the supplied perfect-fluid model. Substitute $h^{\prime }=-2{\delta }_C^{\prime }$ and $h^{\prime \prime }=-2{\delta }_C^{\prime \prime }$ in that trace equation to obtainThe absence of a pressure-gradient term reflects the zero cosmological sound speed of the cold matter.

For the $w_S=-1/3$ coasting fluid, the continuity and momentum equations becomeThe second equation follows by taking $i\mathbf{k}\cdot$ the vector equation; in particular, $i\mathbf{k}\cdot i\mathbf{k}=-k^2$, which fixes the pressure sign. Differentiate the first equation and use ${\theta }_S={\delta }_C^{\prime }-3{\delta }_S^{\prime }/2$:Eliminating ${\delta }_C^{\prime \prime }$ with the cold-matter equation givesThese are equations for the specified idealized barotropic string fluid. A cosmic string network with additional anisotropic stress need not obey the same pressure response.

Write $\delta ={\delta }_{C,\mathbf{k}}$ and $\mathcal{H}=a^{\prime }/a$, the conformal Hubble parameter. The flat Friedmann equation givesOutside the horizon, the stated adiabatic cosmological perturbation relation is ${\delta }_R=(4/3)\delta$. It expresses equality of ${\delta }_i/(1+w_i)$ for dust and radiation in the density convention of the supplied equations. During radiation domination, $a\propto \tau$, so $\mathcal{H}=1/\tau$ and ${\rho }_C$ is negligible. Consequently $8\pi G{a}^2{\rho }_R=3/{\tau }^2$, and the linear cosmological density perturbation equation becomesThe trial solution $\delta \propto {\tau }^q$ gives $q(q-1)+q=4$, hence $q=2,-2$.

During matter domination, $a\propto {\tau }^2$, so $\mathcal{H}=2/\tau$. Now the radiation source is negligible and $4\pi G{a}^2{\rho }_C=(3/2){\mathcal{H}}^2=6/{\tau }^2$. The equation becomeswhose exponents satisfy $q(q-1)+2q=6$, or $(q-2)(q+3)=0$. Thus the growing solutions in both eras areThe other modes are ${\tau }^{-2}$ in the radiation limit and ${\tau }^{-3}$ in the matter limit. These are statements in the density and gauge convention of the supplied evolution equation; a density contrast outside the horizon is not a gauge-independent observable by itself.

Write $\rho =\bar{\rho }(1+\delta )$, take a constant barotropic equation-of-state parameter $w$, and keep first-order terms in $\delta$, $v^i$ and $h_{ij}$. In synchronous gauge, the inverse metric is $g^{00}=-a^{-2}$, $g^{0i}=0$, $g^{ij}=a^{-2}({\delta }^{ij}-h^{ij})$. The normalization $u^{\mu }{u}_{\mu }=-1$ gives $u^0=a^{-1}+O(v^2)$ and $u^i=a^{-1}{v}^i$. Substitution in the perfect fluid stress tensor gives the linear perfect-fluid stress tensor in synchronous gauge:In the first component the pressure contributions cancel; in the last, the velocity product is second order and the inverse-metric perturbation supplies the minus sign.

Use projections of stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, to retain all connection terms efficiently. Let $\mathcal{H}=a^{\prime }/a$ be the conformal Hubble parameter and $h={\delta }^{ij}{h}_{ij}$. Since $\sqrt{-g}=a^4(1+h/2)$, the fluid expansion isThe energy projection is $u^{\mu }{\partial }_{\mu }\rho +(\rho +P)\vartheta =0$. Its background is ${\bar{\rho }}^{\prime }+3\mathcal{H}(1+w)\bar{\rho }=0$. Expanding and subtracting that background leaves the synchronous perfect-fluid density equationFor momentum, the spatial projection is $(\rho +P)u^{\nu }{\nabla }_{\nu }{u}_i+({\delta }_i{}^{\nu }+u_i{u}^{\nu }){\partial }_{\nu }P=0$. The given connections yield $u^{\nu }{\nabla }_{\nu }{u}_i=v_i^{\prime }+\mathcal{H}{v}_i$ at first order. The pressure projection contains both ${\partial }_i\delta P$ and $v_i{\bar{P}}^{\prime }$; retaining the latter is essential. With $\delta P=w\bar{\rho }\delta$ and ${\bar{P}}^{\prime }=-3w\mathcal{H}(1+w)\bar{\rho }$, division by $\bar{\rho }(1+w)$ gives the Cosmological Euler equation in synchronous gauge,For Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}}$, these results areThe velocity equation assumes nonzero enthalpy $\bar{\rho }+\bar{P}$, so a cosmological constant with $w=-1$ is not a propagating fluid-velocity case. A time-dependent $w$ would also introduce additional terms and is not the constant-barotropic system used here.

For a single barotropic perfect fluid in general relativity, the perturbations obey $\delta P=w\delta \rho =w\bar{\rho }\delta$. This adiabatic closure is needed: constant background $P/\rho$ alone would not eliminate an independent entropy perturbation. The absence of scalar anisotropic stress allows the common potential used in Newtonian gauge in cosmology.

Substituting the density constraint into the pressure equation gives the gravitational potential evolution of a barotropic fluidSince $a\propto {\tau }^{2/(1+3w)}$, the conformal Hubble parameter is $\mathcal{H}=2/[(1+3w)\tau ]$. The bracket cancels identically, leavingFor a Fourier transform mode, ${\nabla }^2$ becomes $-k^2$.

During radiation domination, $w=1/3$ and ${\Phi }_k^{\prime \prime }+4{\Phi }_k^{\prime }/\tau +(k^2/3){\Phi }_k=0$. Set $x=k\tau /\sqrt{3}$ and write $\Phi =u/x$. The resulting equation is $u_{xx}+2u_x/x+(1-2/x^2)u=0$, so the two Spherical Bessel functions in the hint giveAt $x\ll 1$, the two solutions approach a constant and a mode proportional to ${\tau }^{-3}$. The regular adiabatic mode, normalized to its primordial potential, isAfter entry into the sound horizon, $x\gg 1$, the potential oscillates at cosmological sound speed $1/\sqrt{3}$ with envelope $x^{-2}\propto a^{-2}$. The Hubble radius and sound horizon differ by the sound-speed factor; outside the Hubble radius the regular potential is constant, while well inside it radiation supports acoustic oscillations.

During matter domination, $w=0$ gives ${\Phi }_k^{\prime \prime }+6{\Phi }_k^{\prime }/\tau =0$ at every wavenumber. HenceThe growing density mode has a constant potential both outside and inside the Hubble radius; the other potential mode decays. Pressureless matter has zero cosmological sound speed, so horizon entry does not produce the radiation acoustic decay. These formulas cover both independent solutions, while the subsequent sketches select the regular adiabatic growing mode.

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.

The same invariant interval can be written as $g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }={\tilde{g}}_{\alpha \beta }d{\tilde{x}}^{\alpha }d{\tilde{x}}^{\beta }$. Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and ${\xi }^{\mu }=(T,{\partial }^{i}L)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background ${\bar{g}}_{00}=-a^2$, ${\bar{g}}_{ij}=a^2{\delta }_{ij}$, the Lie derivative components areHere primes denote conformal time and $\mathcal{H}$ is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.

The $00$ component is $\delta g_{00}=-2a^{2}A$, immediately giving the lapse function perturbation transformation. The $0i$ component is $a^2{\partial }_{i}B$, giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is $6a^{2}C$, so tracing the spatial Lie derivative gives the transformation of $C$. Its trace-free scalar part is $2a^2({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, giving the transformation of $E$. ThusAs usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.

At first order, tensor cosmological perturbations are spatial transverse-traceless tensors. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In Fourier space, the latter terms carry a factor $k_i$ or $k_j$. The transverse-traceless projector removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore the tensor perturbation is gauge invariant at linear order around the homogeneous background. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.

The tight-coupling approximation requires the positive scattering rate $\kappa =-\dot{\tau }$ to exceed both $k$ and the conformal Hubble parameter. The photon-baryon velocity slip and photon quadrupole are then small, and the leading fluid motion has $v_b=v_{\gamma }=v$. One must combine the equations before setting the slip to zero: a small slip times a large collision rate can exert a finite force.

Add the photon Euler equation to $R$ times the baryon Euler equation with Thomson drag. The collisions cancel and, neglecting ${\Theta }_2$ at this order, the result isSince $\dot{R}=\mathcal{H}R$, the common velocity obeysThe inertia of the baryons reduces the photon-baryon sound speed to $c_s^2=1/[3(1+R)]$.

Differentiate the photon continuity equation and substitute this velocity equation:Thus the photon-baryon acoustic oscillator isThe terms on the right drive the oscillation gravitationally; the first-derivative term comes from changing baryon inertia. Diffusion damping enters only beyond this leading tight-coupling approximation.

Throughout this question dots denote conformal time derivatives, and $\mathcal{H}=\dot{a}/a$ is the conformal Hubble parameter. The monopole of the photon Boltzmann hierarchy gives ${\dot{\Theta }}_0+k{\Theta }_1/3=\dot{\varphi }$. With the specified velocity convention, the photon continuity equation is thereforeThe dipole equation is ${\dot{\Theta }}_1+k(2{\Theta }_2/5-{\Theta }_0)=\dot{\tau }({\Theta }_1+v_b)+k\psi$. Thus the photon Euler equation isIn particular, $\dot{\tau }<0$ makes the scattering term damp the photon-baryon velocity slip; it does not amplify it.

The linear momentum density of the photon-baryon fluid is weighted by enthalpy, not merely energy density:Elastic Thomson scattering exchanges momentum internally. Its photon force density is $(4{\bar{\rho }}_{\gamma }/3)\dot{\tau }(v_{\gamma }-v_b)$, so the baryon force density is its negative. Dividing by ${\bar{\rho }}_b$ gives the baryon Euler equation with Thomson dragThis $R$ is the baryon loading parameter, the baryon-to-photon inertia ratio. The two collision terms cancel exactly when the equations are weighted by their enthalpies. Background conservation gives ${\bar{\rho }}_b\propto a^{-3}$ and ${\bar{\rho }}_{\gamma }\propto a^{-4}$, so $\dot{R}=\mathcal{H}R$.

The printed $\mathcal{H}$ is the conformal Hubble parameter, $\mathcal{H}=aH=\dot{a}$; ${\mathcal{H}}^{-1}$ is the comoving Hubble radius, rather than the physical radius $H^{-1}$. Since the derivative requested is with respect to cosmic time,Thus the cosmic-time derivative of the comoving Hubble radius hasFor an expanding universe, the same factor $1+3\omega$ determines whether a small departure from flatness grows and whether the comoving Hubble radius grows. Ordinary matter with $\omega >-1/3$ therefore has both signs associated with the conventional flatness problem and horizon problem. Scales whose physical wavelengths now exceed the Hubble radius were even farther outside it, in relative terms, earlier in such an era, rather than being brought inside for causal equilibration. Accelerated expansion with $\omega <-1/3$ reverses both signs.

There is a qualification to the word “always”: an actual horizon problem depends on the complete past history, not just this local sign. In a flat, constant-$\omega$ hot Big Bang model with $\omega >-1/3$, $a\propto (t-t_B{)}^p$ with $p=2/[3(1+\omega )]<1$, so the comoving particle horizon ${\int }_{t_B}^{t}d{t}^{\prime }/a(t^{\prime })$ is finite. The conventional causal problem then accompanies the flatness instability. An earlier accelerated era or a different past boundary can change the causal conclusion even when the present-era comoving Hubble radius is growing. Thus the requested correspondence is valid in that usual expanding hot Big Bang setting, not a universal logical equivalence between global horizons and local stability. At $\omega =-1/3$ both local effects are marginal.

In linear cosmological perturbation theory, an unsourced vector cosmological perturbation of a perfect fluid in synchronous gauge obeys stress-energy conservation in the form ${\mathbf{q}}^{\prime }+4\mathcal{H}\mathbf{q}=0$, where $\mathbf{q}=(\bar{\rho }+\bar{P}){\mathbf{v}}_{\perp }$ is the momentum density, $\mathcal{H}$ the conformal Hubble parameter, and the prime a conformal time derivative. Its conserved quantity is $a^4\mathbf{q}$. For constant equation-of-state parameter $w\ne -1$, the background cosmological continuity equation gives ${\mathbf{v}}_{\perp }\propto a^{3w-1}$. Thus a pressureless matter vector velocity falls as $a^{-1}$, while a radiation vector velocity is constant. This statement assumes vanishing vector anisotropic stress and no momentum exchange with another component.