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The regional mass density is $3H_0^2{\Omega }_m/(8\pi G{a}^3)$, while the Einstein-de Sitter universe has ${\rho }_{\mathrm{bg}}=1/(6\pi G{t}^2)$. Substituting the parametric spherical collapse from part (a) yieldsAt spherical-collapse turnaround, $\theta =\pi$, and henceFor an isolated uniform sphere, the Newtonian gravitational potential energy is $U=-3G{M}^2/(5R)$. At spherical-collapse turnaround the coherent kinetic energy is zero, so conservation of energy fixes $E=U_m$. At virial equilibrium, the virial theorem gives $2T_v+U_v=0$, hence $E=U_v/2$. Equating these energies proves the virial radius from turnaround energy:This uses the same mass and gravitational structure coefficient before and after collapse and neglects boundary pressure, mass loss and external energy exchange.

If virial equilibrium is assigned to $2t_m$, the sphere's density is eight times its turnaround density, whereas the background mass density has decreased by four. ConsequentlyThe often quoted value $18{\pi }^2$ is a density ratio; the density contrast defined as $(\rho -{\rho }_{\mathrm{bg}})/{\rho }_{\mathrm{bg}}$ is smaller by one. This finite virialized estimate and the infinite cold-collapse contrast in part (b) describe different physical endpoints.

A useful description of the low-column Lyman-alpha forest at $z\simeq 3$ is a power law $f_X(N,z)=A(z)N^{-\beta }$, with $\beta$ around $1.4$–$1.5$ over the ordinary unsaturated forest range. The same column dependence holds for $f_z$ at a fixed redshift. It is not a universal unbroken power law: saturation, increased neutral fraction and changing absorber populations produce breaks. Lyman limit systems become optically thick at $N\simeq 10^{17.2}{\mathrm{cm}}^{-2}$ because $N{\sigma }_{912}\simeq 1$ with ${\sigma }_{912}\simeq 6.3\times 10^{-18}{\mathrm{cm}}^2$. Damped Lyman-alpha systems have $N\gtrsim 2\times 10^{20}{\mathrm{cm}}^{-2}$ and strong radiative damping wings. They are rare in number but contain substantial neutral mass. The highest-column distribution steepens or cuts off rather than extending indefinitely.

This sketch shows representative shapes and transition scales, not an observational fit; its ordinate is distribution per unit column, not counts per logarithmic bin. In an Einstein-de Sitter universe, $H=H_0(1+z{)}^{3/2}$, so a fixed comoving population with fixed physical cross-section above a fixed column threshold has $d\mathcal{N}/dz\propto (1+z{)}^{1/2}$. ThusThe observed redshift evolution of the Lyman-alpha forest is much stronger for commonly selected forest lines. For example, Kim, Cristiani and D'Odorico's absorption survey finds an index about $2.19$ in $d\mathcal{N}/dz\propto (1+z{)}^{\gamma }$ for $10^{13.64}<N/{\mathrm{cm}}^{-2}<10^{16}$ over $1.5<z<4$, corresponding to an increase by approximately $2^{2.19}\simeq 4.56$. This number is selection-dependent. Expansion changes the gas density, and the evolving cosmic ionizing background changes its neutral fraction through photoionization equilibrium, moving systems across a fixed neutral-column threshold. Structure also evolves. A non-evolving population with fixed physical cross-section is therefore not an adequate physical model of the observed forest.

For a homogeneous isotropic metric, radial light propagation covers comoving distance $\int cdt/a(t)$. The physical particle horizon is $d_p(t)=a(t){\int }_{t_i}^{t}cd{t}^{\prime }/a(t^{\prime })$, when this is finite. A future event horizon exists when $d_e(t)=a(t){\int }_t^{t_f}cd{t}^{\prime }/a(t^{\prime })$ is finite. The Hubble radius is the instantaneous scale $c/H$, with $H=\dot{a}/a$; it is not generally the same as either causal horizon.

For an Einstein-de Sitter universe, $H=2/(3t)$, so $c/H=3ct/2$, while $d_p=3ct$ from a beginning at zero. The future integral diverges, so there is no event horizon. For an eternal-past de Sitter spacetime, $H$ is constant and the particle-horizon integral from $-\infty$ diverges. Its event horizon and Hubble radius both equal $c/H$. Thus a Hubble radius is not automatically a particle horizon, even in these simple isotropic models.

For an Einstein-de Sitter universe, $H=\dot{a}/a=2/(3t)$. The flat matter-only Friedmann equation $H^2=8\pi G{\rho }_m/3$ gives ${\rho }_m=1/(6\pi G{t}^2)$. With zero sound speed the perturbation equation is thereforeSubstituting $t^s$ yields $s(s-1)+4s/3-2/3=0$, or $(3s-2)(s+1)=0$. Hencewith growing mode proportional to the scale factor and decaying mode proportional to $t^{-1}$.

For $q_0=1/2$, the expanding Friedmann equation reduces to $\dot{y}=H_0{y}^{-1/2}$. Separating variables and placing the Big Bang at $t=0$ givesThus the flat pressureless Einstein-de Sitter universe hasIn particular its present age is $2/(3H_0)$.

At the present epoch, define $s=1+z$ and $F(z)=\dot{z}/H_0$. For flat matter plus a cosmological constant, neglecting radiation,In an Einstein-de Sitter universe, $E=s^{3/2}$, henceThe redshift drift is zero at $z=0$, has initial slope $F^{\prime }(0)=-1/2$, and decreases monotonically because $F^{\prime }=1-3\sqrt{s}/2<0$. Its second derivative is $-3/(4\sqrt{s})<0$, so the sketch bends downward, with $F\sim -z^{3/2}$ at large positive redshift.

Use the observed wavelength in $d\lambda /\lambda =dv/c$. Since ${\lambda }_{\mathrm{obs}}=(1+z){\lambda }_{\mathrm{rest}}$,For an Einstein-de Sitter universe at $z=3$, $H(z)=H_0(1+z{)}^{3/2}=8H_0$, and the redshift drift is $\dot{z}=4H_0-8H_0=-4H_0$. With the rounded numerical values supplied in the paper,Therefore the spectroscopic velocity drift accumulated over ten years isThe negative sign means a shift toward shorter observed wavelength. Omitting the factor $1+z$ would overestimate the apparent velocity shift by four.

As ${\Omega }_M\uparrow 1$, use $\mathrm{arsinh}z\sim z$ near zero. ConsequentlyThe flat dust-and-vacuum expansion tends to $a(t)=(3H_{0}t/2{)}^{2/3}$, the Einstein-de Sitter universe.

Use the stipulated common normalization $A=H_0{\Omega }_{m,0}$ and plot $R=A{r}_h/c$. The Einstein-de Sitter universe has $H_{0,E}=A$, whereas the $({\Omega }_{m,0},{\Omega }_{\Lambda ,0})=(0.3,0.7)$ model has $H_{0,\Lambda }=A/0.3$. Their comoving particle horizons are thereforeBoth decrease with increasing $z$, or increase with cosmic time. The Einstein-de Sitter universe curve diverges at $z=-1$ and has $R_E(0)=2$, $R_E(2)=2/\sqrt{3}$. The matter-plus-vacuum curve remains finite at $z=-1$: the integrand approaches a constant there, and its high-redshift tail is integrable. Moreover it lies below the Einstein-de Sitter curve for this particular normalization, since, with $u=1+z^{\prime }>0$,The high-redshift ratio of the two curves tends to $\sqrt{0.3}$, not one. Holding $H_0\sqrt{{\Omega }_{m,0}}$ fixed would give a different comparison; it is $H_0{\Omega }_{m,0}$ that the original question specifies.

A positive cosmological constant eventually makes the scale factor grow exponentially. The future integral ${\int }_t^{\infty }cd{t}^{\prime }/a(t^{\prime })$ then converges, so the remaining conformal time is finite. The comoving particle horizon approaches a finite limiting visibility radius; it does not shrink. The remaining distance to that limit is the comoving radius of the cosmological event horizon. Thus a signal emitted sufficiently far away today can never reach us, even though the proper particle-horizon radius continues to grow. This is the total conformal lifetime of a flat matter-Lambda universe.

For an Einstein-de Sitter universe, the Hubble parameter is $H(z)=H_0(1+z{)}^{3/2}$. The comoving particle horizon becomesThusIt tends to zero towards the initial singularity and increases without bound as $z\to -1$ in the infinite future. At the present time it is $2c/H_0=3c{t}_0$, since the Einstein-de Sitter universe has age $t_0=2/(3H_0)$.

Normalize the Einstein-de Sitter universe by $a(t_0)=1$, giving $a(t)=(t/t_0{)}^{2/3}$. A photon emitted inward today at the current comoving particle horizon starts at ${\chi }_0=3c{t}_0$. By time $t$ it has traversed the comoving distanceAt arrival at Earth this equals ${\chi }_0$, hence $(t/t_0{)}^{1/3}=2$ andThe emitting point is at the horizon location today, not a photon already arriving today from the initial singularity. The calculation of photon arrival from an Einstein-de Sitter particle horizon follows the future inward null geodesic from that event.

In an Einstein-de Sitter universe, $H(z)=100h(1+z{)}^{3/2}\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$. At the cloud's approximate epoch $z_{\ast }=3$, $H=800h\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, and $\delta z=0.1$ corresponds to $\delta v\simeq 7500\mathrm{km}{\mathrm{s}}^{-1}$. ThereforeChoosing the midpoint $z_{\ast }=3.05$ instead gives $9.1h^{-1}\mathrm{Mpc}$ proper. Thus the intended approximate proper distance is about $9h^{-1}\mathrm{Mpc}$; the small variation reflects the epoch choice in the local expansion, not a factor-of-four physical discrepancy. For example the exact comoving separation is $2c{H}_0^{-1}[(1+z_c{)}^{-1/2}-(1+z_q{)}^{-1/2}]$, whose conversion at the midpoint agrees with that estimate.

At $z=0$, spatial flatness gives $\tan {\theta }_0=\sqrt{(1-{\Omega }_{\Lambda ,0})/{\Omega }_{\Lambda ,0}}$, hence $\cos {\theta }_0=\sqrt{{\Omega }_{\Lambda ,0}}$ and $\sin {\theta }_0=\sqrt{1-{\Omega }_{\Lambda ,0}}$. ThereforeFor ${\Omega }_{\Lambda ,0}\to 0$, $\mathrm{arsinh}u\sim u$ in the previous integral gives $t_0\to 2/(3H_0)$, the Einstein-de Sitter universe result. For ${\Omega }_{m,0}=0$, the positive-vacuum flat slicing has no finite-age Big Bang, explaining the other endpoint divergence.

In an Einstein-de Sitter universe, $H(a)=H_0{a}^{-3/2}$, so direct integration of $dt=da/(aH)$ gives $t(a)=2a^{3/2}/(3H_0)$. At $z=8$, $a=1/9$ and $a^{3/2}=1/27$. The fraction of the present age of an FLRW universe and the available formation interval are thereforeThis is only about $3.7\%$ of the present age. The available interval is $t(8)$ itself, not $t_0-t(8)$, which would instead be the lookback time.

For ${\Omega }_{m,0}=0.3$ and ${\Omega }_{\Lambda ,0}=0.7$, at the same Hubble constant, the exact integral givesThus the formation interval is longer, by about a factor of $1.83$. There is also a direct comparison: for $0<a<1$, $0.3a^{-3}+0.7<a^{-3}$, so $H(a)$ is smaller than its Einstein-de Sitter universe value at fixed $a$, and integrating $da/(aH)$ gives a greater age. At $z=8$ the leading matter-era result is $t\simeq 2/[3H_0\sqrt{0.3}(1+z{)}^{3/2}]$; the smaller matter density, with the same $H_0$, explains nearly all the increase.

In an Einstein-de Sitter universe, $E(z)=(1+z{)}^{3/2}$. Under the non-evolution assumptions of the previous part, the incidence above a fixed neutral-hydrogen column density threshold consequently scales as $(1+z{)}^{1/2}$. Between the two specified epochs,The higher-redshift rate would be about $41\%$ larger.

Observed Lyman-alpha forest incidence above substantial column thresholds rises much more rapidly. For example, the survey at Kim and collaborators' measurements finds an exponent $2.19\pm 0.27$ for $10^{13.64}\lesssim N_{\mathrm{HI}}\lesssim 10^{16}{\mathrm{cm}}^{-2}$ over this redshift range, giving a representative increase $2^{2.19}\simeq 4.56$, rather than $\sqrt{2}$. The empirical exponent is threshold dependent. The forest therefore cannot be treated as a conserved population of identical fixed-size neutral clouds: gas expansion, structure evolution and the cosmic ionizing background change neutral fractions, absorption strengths and the effective area above the threshold. In photoionization equilibrium, $n_{\mathrm{HI}}\Gamma \simeq \alpha (T)n_e{n}_p$, so changing density or ionization rate changes a counted neutral column even without creating or destroying its baryons.

Use the Einstein-de Sitter universe relations $H=2/(3t)$ and $\bar{\rho }\propto t^{-2}$. Write $H_{\mathrm{ta}}=H(t_{\mathrm{ta}})$ and $R_{\mathrm{ta}}=R(t_{\mathrm{ta}})$. Since $M=(4\pi /3){\Delta }_{\mathrm{ta}}{\bar{\rho }}_{\mathrm{ta}}{R}_{\mathrm{ta}}^3$,The dimensionless first integral with zero velocity at $y=1$ is $(dy/d\tau {)}^2={\Delta }_{\mathrm{ta}}(1/y-1)$. On the expanding branch this givesEvaluating the integral gives the inverse-trigonometric expression in the source. Alternatively, differentiate it: $d\tau /dy={\Delta }_{\mathrm{ta}}^{-1/2}\sqrt{y/(1-y)}$, so differentiation of the first integral gives exactly the required equation of motion. At $y=1$, the integral equals $\pi /2$. Thus ${\tau }_{\mathrm{ta}}=2/3$ fixes $\sqrt{{\Delta }_{\mathrm{ta}}}=3\pi /4$. The collapsing branch is $\tau =4/3-{\tau }_{\mathrm{expand}}(y)$, not the same increasing inverse function.

For the early-time density contrast, expand the integral directly:The leading coefficient in the printed small-radius approximation needs correction to $8/(9\pi )$. The printed coefficient is inconsistent with both the exact integral and approach to the homogeneous background. Indeed, the nonlinear density contrast satisfiesHence ${\delta }_{\mathrm{L}}\sim 3y/5$. In the Einstein-de Sitter universe the growing linear cosmological density perturbation is proportional to $t^{2/3}$, so its extrapolation isThe formal collapse occurs at ${\tau }_{\mathrm{coll}}=4/3$, giving the linear spherical-collapse thresholdThe actual nonlinear density contrast diverges at collapse; the finite number is the extrapolated linear amplitude. Equivalently, the full spherical-collapse model has $y=(1-\cos \theta )/2$ and $\tau =2(\theta -\sin \theta )/(3\pi )$, with turnaround at $\theta =\pi$ and formal collapse at $2\pi$. Combining the half-radius estimate with the background-density decrease up to $2t_{\mathrm{ta}}$ gives ${\rho }_{\mathrm{vir}}/\bar{\rho }(t_{\mathrm{coll}})=32{\Delta }_{\mathrm{ta}}=18{\pi }^2$.