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Normalize the scale factor to $a(t_0)=1$, use $c=1$, and define the instantaneous critical density by ${\rho }_{\mathrm{cr}}(t)=3H(t{)}^2/(8\pi G)$. Separately conserved pressureless matter and radiation in cosmology obey the cosmological continuity equation,The present cosmological density parameters areThe cosmological constant has equivalent energy density ${\rho }_{\Lambda }=\Lambda /(8\pi G)$. Substituting into the Friedmann equation yieldsAt the present epoch this gives $k=H_0^2({\Omega }_0-1)$, or curvature contribution ${\Omega }_{K0}=-k/H_0^2=1-{\Omega }_0$. Dividing each component by $H_0^2$ provesSpatial curvature is not included in the total physical-density parameter ${\Omega }_0$. In particular, positive vacuum density does not have the radiation or matter scaling.

Use $c=1$ and signature $+---$. The positively curved FLRW metric in the usual radial chart isSet $r=\sin \chi$ and introduce conformal time by $dt=ad\eta$. Then $d{r}^2/(1-r^2)=d{\chi }^2$ in the chart, and the spherical polar coordinate $\chi \in [0,\pi ]$ extends the spatial description to the full three-sphere. ThusThe $r$ coordinate itself is not one-to-one across the equator, which is why the new polar coordinate is useful.

For $\Lambda =0$ and $k=1$, the Friedmann equation is ${\dot{a}}^2=(8\pi G/3)\rho a^2-1$. Since $a^{\prime }=a\dot{a}$, it becomesFor pressureless matter, the background continuity equation gives $\rho a^3={\rho }_0{a}_0^3$. Define the conserved normalization $M=(4\pi /3){\rho }_0{a}_0^3$. It makes the equation $(a^{\prime }{)}^2=2GMa-a^2$. The closed matter-dominated Friedmann solution is the cycloidIndeed $a^{\prime }=GM\sin \eta$ and $2GMa-a^2=(GM{)}^2{\sin }^2\eta$. Integrating $dt/d\eta =a$ gives the second expression, with the Big Bang at $t=\eta =0$. It expands to $a=2GM$ at $\eta =\pi$ and recollapses at $\eta =2\pi$. Here $M$ is the constant appearing in the Friedmann equation; it is not the total mass of the entire closed spatial slice, whose volume is $2{\pi }^2{a}^3$.

Use cosmic time $t$ here; derivatives with respect to conformal time will be used from Question 2 onward. Conservation of pressureless matter and radiation in cosmology makes ${\rho }_m{a}^3$ and ${\rho }_r{a}^4$ constant. DefineThe Friedmann equation is a mechanical energy equation,For nonnegative dust and radiation, at least one nonzero, $V$ tends to $-\infty$ as $a\to 0$ and to $+\infty$ as $a\to \infty$, whileThus every energy line meets the potential once. An expanding solution reaches this turning point and then contracts: $a_{tt}=-V^{\prime }(a)/2<0$. The Friedmann effective potential for matter, radiation and vacuum is sketched below, alongside the positive-vacuum case.

To bound the full cosmic time from the initial zero of $a$ to its final zero, use the Friedmann acceleration equation:Put maximum expansion at $t=0$, with $a(0)=a_{\max }$ and $a_t(0)=0$. The forced-oscillator identity isUntil the first zero and for $0\le t\le \pi /(2\omega )$, the kernel is nonnegative and the bracket is nonpositive. Hence $a(t)\le a_{\max }\cos (\omega t)$, so contraction cannot remain at positive $a$ beyond $\pi /(2\omega )$. Applying the same argument backward from maximum expansion bounds the expanding half. Therefore the negative-vacuum Friedmann lifetime bound isAt either endpoint, nonzero dust or radiation has divergent density. If both are absent, the same estimate instead bounds the open Anti-de Sitter spacetime coordinate patch; its coordinate zeros are not a finite lifetime of the maximally extended vacuum spacetime.

Without radiation, the Friedmann acceleration equation changes sign when ${\rho }_m=2{\rho }_{\Lambda }$. The cosmological redshift satisfies $1+z=a_0/a$, and conservation of pressureless matter gives ${\rho }_m(z)={\rho }_{m,0}(1+z{)}^3$. HenceThe expanding spatially flat model is decelerating above this redshift and accelerating below it. Positive ${\rho }_{\Lambda }$ is required for such a transition; a negative value gives no accelerated expanding epoch. If the displayed redshift is negative, the transition lies in our future.

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 ${\Omega }_M=1$.

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 $a^3=3C/(2\Lambda )$.

Normalize the region's reference scale factor to one. Pressureless matter has $\rho ={\rho }_0{a}^{-3}$, so the closed Friedmann equation isThe expanding branch can reach only $a\le a_m$, where $a_m={\Omega }_m/({\Omega }_m-1)$. Substitute $a=a_m(1-\cos \theta )/2$. Since $da/d\theta =a_m\sin \theta /2$ and $\dot{a}=H_0\sqrt{({\Omega }_m-1)(a_m/a-1)}$, cancellation on $0<\theta <\pi$ givesIntegration from the Big Bang establishesThe same parametrization continues through spherical-collapse turnaround at $\theta =\pi$ onto the contracting branch up to $\theta =2\pi$. Its maximum-expansion time is $t_m=\pi {\Omega }_m/[2H_0({\Omega }_m-1{)}^{3/2}]$.

Define neutral-hydrogen column density along a ray by $N=\int n_{\mathrm{HI}}d\ell$, in atoms per unit area. The neutral-hydrogen column-density distribution per redshift isso $f_{z}dNdz$ is the expected number of absorbers in those intervals along one sightline. A logarithmic column bin instead has distribution $N\ln (10)f_z$; the bin convention must be specified.

Let $n_c$ be a constant comoving number density and ${\sigma }_p$ a constant physical interception area. The corresponding physical number density is $n_p=n_c(1+z{)}^3$. Proper light-path length satisfies $d\ell =c|dt|=cdz/[(1+z)H(z)]$. The expected intersection count, or first-order probability in infinitesimal $dz$, is thereforeThe Friedmann equation, the constant-equation-of-state density scaling for pressureless matter, and the present cosmological density parameters giveCombining them yields the absorber incidence and comoving number density relationHere the constant cross-section is physical, as required for this result; a fixed comoving cross-section would instead decrease in physical area as $(1+z{)}^{-2}$. It is convenient to define absorption distance by $dX/dz=(H_0/H(z))(1+z{)}^2$ and $f_X=f_z/(dX/dz)$, so a population with these non-evolution assumptions has constant incidence per $X$.

Use natural units with $c=1$, write $H=\dot{a}/a$, and measure cosmic time from the Big Bang. The cosmological continuity equation follows by differentiating the Friedmann equation and using the Friedmann acceleration equation; for pressureless matter it gives $\dot{\rho }+3H\rho =0$, hence $\rho ={\rho }_0{a}^{-3}$ when the present scale factor is $a_0=1$. The present critical density and cosmological density parameter giveConsequently, with $C=H_0^2{\Omega }_0$, the Friedmann equation becomes ${\dot{a}}^2=C/a-k$.

Introduce conformal time through $dt=ad\tau$. A prime denotes a derivative with respect to $\tau$, so $a^{\prime }=a\dot{a}$ andSet $u=\sqrt{k}\tau$, with $u=0$ at the Big Bang. On the expanding branch the substitution $a=C(1-\cos u)/(2k)$ satisfies this equation. Integrating $dt=adu/\sqrt{k}$, with $t(0)=0$, gives the closed matter-dominated Friedmann solution:The parametrization continues smoothly through the maximum scale factor, reached at $u=\pi$, onto the contracting branch. Its curve in the $(t,a)$ plane is a rescaled cycloid.

For a presently expanding universe, $H_0>0$ selects $0<u_0<\pi$. The normalization $a(u_0)=1$ givesSubstitution yields the age of a closed dust universe:The first later zero of the scale factor occurs at $u=2\pi$, so the lifetime of a closed dust universe gives the Big Crunch timeThis is the total lifetime measured from the Big Bang; the remaining cosmic time is $t_{\mathrm{BC}}-t_0$. As a check, the limit ${\Omega }_0\downarrow 1$ gives $H_0{t}_0\to 2/3$, while the Big Crunch recedes to infinite time, as expected for a flat matter-dominated universe with zero cosmological constant.

Use geometrized units $G=c=1$ until converting the clock readings. By the Birkhoff theorem, the spherical vacuum exterior is a portion of the Schwarzschild spacetime with fixed mass $M$:The exterior is not flat and is not radiating merely because the surface moves. When the surface crosses $r=2M$, this exterior must be represented in regular Ingoing Eddington-Finkelstein coordinates, rather than treating the Schwarzschild-coordinate divergence as a physical barrier.

The Oppenheimer-Snyder model has a homogeneous pressureless matter interior. Since its initial expansion vanishes but its density is positive, the Friedmann equation requires positive spatial curvature. Choose a comoving radial angle $\chi$ and the boundary's proper time $\tau$, with the stopwatch set to zero at release. The interior FLRW metric and density areThe density law follows from the dust stress-energy tensor and its covariant conservation. The Friedmann equation and the initial rest condition giveIts contracting branch has the parameterizationIndeed $d\tau /d\eta =a$ and $\dot{a}=-\sin \eta /(1+\cos \eta )$ satisfy the displayed equation; $\eta =0$ is the turning point and $\eta =\pi$ is the final curvature singularity.

For homogeneous dust-ball matching, the boundary has areal radius $R=a\sin {\chi }_0$. Matching its mass to the exterior givesThis $M$ is the gravitational mass parameter, not the integral of rest density over the curved proper-volume element. The matching can also be checked directly. The outward angular extrinsic curvature inside is $K_{\theta \theta }=R\cos {\chi }_0$; outside it is $R\sqrt{1-2M/R+{\dot{R}}^2}$. The boundary motion hasso these curvatures agree. The induced metric is $-d{\tau }^2+R^{2}d{\Omega }^2$ on both sides and $K_{\tau \tau }=0$ because the pressure-free boundary is geodesic. No artificial surface stress layer is required.

The observer moving with this boundary meets the event horizon when $R=2M$. From $R=R_0(1+\cos \eta )/2$, the proper-time horizon crossing in homogeneous dust collapse occurs atThe singularity is reached at ${\tau }_s=\pi a_{\max }/2$, so the remaining clock time isThese are finite proper times, even though the external static time diverges at horizon crossing.

Convert the supplied rounded units consistently: $M=10^{38}$ Planck masses and $R_0=6.25\times 10^{39}$ Planck lengths. Thus $2M/R_0=0.032$, ${\chi }_0\simeq 0.17985$, and $a_{\max }\simeq 3.4939\times 10^{40}$ in Planck units. Multiplying geometric times by $5\times 10^{-44}$ seconds givesAt the accuracy justified by the rounded constants, these are about three milliseconds to horizon crossing and about seven microseconds more to the singularity. For $R_0\gg 2M$, expanding ${\eta }_H=\pi -2{\chi }_0$ gives $\Delta \tau \simeq (2/3)a_{\max }{\chi }_0^3\simeq 4M/3$, independently confirming the microsecond scale. The exact expression above retains the finite release-radius correction.

Strictly, this boundary clock reading is when the observer crosses the horizon, not when the horizon first exists anywhere. The event horizon inside an Oppenheimer-Snyder cloud is an outgoing radial null ray. Since $d\tau =ad\eta$, the interior metric is conformal to $-d{\eta }^2+d{\chi }^2+{\sin }^2\chi d{\Omega }^2$, and that ray has $d\chi /d\eta =1$. Tracing it back from $({\eta }_H,{\chi }_0)$ givesFor this cloud it starts at the centre after release, at comoving time ${\tau }_{\mathrm{birth}}\simeq 2.722\times 10^{-3}$ seconds. This answers the literal global-formation interpretation as well; it is not an event on the outer observer's worldline. The distinction matters because an event horizon is defined by the entire future causal escape problem, not by a locally detectable signal of formation.

For the observer at infinity, use the boundary's conserved Killing energy $E$ in the exterior:As $\tau \uparrow {\tau }_H$, $\dot{R}\to -E$ and $R-2M\simeq E({\tau }_H-\tau )$. Consequentlywhere ${\tau }_{\ast }>0$ fixes a harmless dimensionless logarithm. For actual outgoing light signals the relevant retarded time is $u=t-r_{\ast }(R)$. Since $r_{\ast }\simeq 2M\log (R-2M)+O(1)$,Therefore she never receives a signal showing the surface cross the horizon at any finite asymptotic time. Radiation emitted ever closer to crossing arrives ever later and is increasingly redshifted and diluted; photons emitted at or inside the event horizon cannot reach her. The formal crossing time at infinity is infinite, not the finite boundary stopwatch reading. Nor can the earlier global birth of the horizon be observed directly as a local flash from its central birth event.

Write $y=a/a_0$. Conservation of pressureless matter gives ${\rho }_m={\rho }_{m,0}{y}^{-3}$. The Friedmann equation, retaining spatial curvature, isSince $\dot{y}=yH$ and ${\Omega }_{m,0}=2q_0$, multiplication by $y^2$ givesIn the curvature convention $H^2=8\pi G{\rho }_m/3-k{c}^2/a^2$, this also fixes $-k{c}^2/(a_0^2{H}_0^2)=1-2q_0$.

For pressureless matter, the Friedmann acceleration equation with vanishing cosmological constant isEvaluating at the present cosmic time and using the definition of the deceleration parameter givesThe denominator is the critical density, so the cosmological density parameter satisfies the dust relationThis uses pressureless matter and $\Lambda =0$; spatial flatness has not been assumed.

Let $H=\dot{a}/a$ be the Hubble parameter, and normalize the scale factor by $a(t_0)=1$. The cosmological continuity equation gives ${\rho }_M={\rho }_{M0}{a}^{-3}$ for pressureless matter, whereas the cosmological constant has constant energy density ${\rho }_{\Lambda }$. The spatially flat Friedmann equation therefore becomesThe present Hubble constant is $H_0=H(t_0)$. In terms of the present critical density ${\rho }_{\mathrm{crit},0}=3H_0^2/(8\pi G)$, the matter cosmological density parameter is ${\Omega }_M={\rho }_{M0}/{\rho }_{\mathrm{crit},0}$. Spatial flatness gives ${\rho }_{\Lambda }/{\rho }_{\mathrm{crit},0}=1-{\Omega }_M$, so

Use units with $c=1$. For the flat FRW metric, the Friedmann equation is $H^2=8\pi G\rho /3+\Lambda /3$, where $H=\dot{a}/a$ is the Hubble parameter. The cosmological continuity equation for a separately conserved component with constant equation of state $P=w\rho$ gives $\rho \propto a^{-3(1+w)}$. Consequently radiation in cosmology scales as $a^{-4}$, pressureless matter as $a^{-3}$, and cosmological constant energy is constant. With $a_0=1$, define the present critical density and cosmological density parameters byThe Hubble constant $H_0$ measures the expansion rate today. Here ${\Omega }_r$ includes photons and relativistic neutrinos, ${\Omega }_m$ includes baryons and cold dark matter, and ${\Omega }_{\Lambda }$ measures the vacuum contribution. Spatial flatness implies ${\Omega }_r+{\Omega }_m+{\Omega }_{\Lambda }=1$. ThusSince conformal time satisfies $d\tau =dt/a$, we have $a^{\prime }=a\dot{a}=a^{2}H$. Multiplication by $a^4$ provesFor an exam-level estimate, $H_0=100h\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ with $h\simeq 0.7$, ${\Omega }_m\simeq 0.3$, ${\Omega }_{\Lambda }\simeq 0.7$, and the specified radiation in cosmology gives ${\Omega }_r=4.2\times 10^{-5}{h}^{-2}\simeq 8.6\times 10^{-5}$. These are rounded illustrative flat-universe values; precision cosmological density parameters depend on the fitted model. For comparison, the base flat-model fit in Planck 2018 cosmological parameters gives $H_0\simeq 67.4\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and ${\Omega }_m\simeq 0.315$.

In the printed convention $\rho$ is energy density, since the gravitational source is $8\pi G\rho /(3c^2)$. For nonrelativistic pressureless matter, ${\rho }_m={\rho }_{m,0}(1+z{)}^3$. Define the cosmological density parameters byThe Friedmann equation then givesLet $n_c$ be the constant comoving number density of absorbers, and let ${\sigma }_p$ be their fixed proper interception area. Their proper number density is $n_p(z)=n_c(1+z{)}^3$, with $a_0=1$. The redshift-time relation gives a local proper photon-path incrementFor an infinitesimal path, the expected number of interceptions, and to first order the probability of an interception, is $d\mathcal{N}=n_p{\sigma }_{p}d\ell$. Thus absorber incidence and comoving number density implyTwo factors survive: proper density brings three powers of $1+z$, whereas proper path length brings the inverse of one power. A fixed comoving area instead of the assumed fixed proper area would change the result. The formula is an incidence rate, not a probability bounded by one over a finite redshift interval; multiple interceptions can occur.

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 pressureless matter, $w=0$, the constant-equation-of-state comoving density equation reduces to ${\Delta }^{\prime \prime }+2{\Delta }^{\prime }/\eta -6\Delta /{\eta }^2=0$. Substitution of $\Delta \propto {\eta }^s$ gives $s^2+s-6=(s-2)(s+3)=0$. Since $a\propto {\eta }^2$,In the growing mode, $a^2\bar{\rho }\Delta$ is constant, so the Poisson constraint makes $\varphi$ constant for a nonzero Fourier mode. The momentum constraint and the flat Friedmann equation giveThis is the growing-mode velocity potential in matter domination, in the question's convention where the peculiar velocity is $\nabla v$.

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.