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The cosmological principle asserts that, after averaging over sufficiently large scales, the universe is spatially homogeneous and isotropic on constant-cosmic time slices. Homogeneity means no distinguished location; isotropy means no distinguished spatial direction. These are statistical large-scale statements, not assertions that individual galaxies, clusters or the cosmic web are uniform, and they do not require the universe to be unchanging in time.

The nearly isotropic cosmic microwave background is strong evidence for large-scale isotropy about us. Its dominant dipole is explained by our peculiar velocity; the much smaller residual fluctuations provide the seeds and record of structure formation, rather than invalidating the principle. Large-area galaxy and radio-source distributions and deep galaxy surveys support statistical isotropy and a tendency towards statistical homogeneity after smoothing beyond the strongly clustered scales. An approximately isotropic Hubble flow and broadly uniform background radiation supply complementary checks.

There is an important logical distinction: isotropy seen by one observer alone does not prove homogeneity. A spherically symmetric inhomogeneous universe centered on that observer could look isotropic. Combining the observations with the assumption that our location is not privileged, so comparable observers would see comparable large-scale isotropy, motivates the homogeneous FLRW metric. Selection effects, foregrounds, finite survey volumes and the fact that observations lie on our past light cone must be accounted for. The evidence supports a large-scale statistical principle, not exact uniformity on every scale.

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.

Set $L=8\pi G{\rho }_{\Lambda }/3>0$. The Friedmann effective potential for matter, radiation and vacuum now readsWith nonzero nonnegative dust or radiation, $V$ tends to $-\infty$ at both ends. Its stationary point satisfiesEquivalently $V^{\prime }(a)=0$ when $2L=M/a^3+2Q/a^4$; the right-hand side decreases strictly, so there is exactly one stationary point. It is a maximum becauseThe energy line in the preceding sketch therefore gives five broad types of history:

At the critical energy there are also marginal branches within the monotonic classes: expansion from zero can approach $a_E$ asymptotically, or expansion can start asymptotically at $a_E$ and proceed to infinity; contraction gives their reversals. They must not be mistaken for reaching the static solution in finite time. Indeed near the maximum,so $dt\simeq da/[\sqrt{-V^{\prime \prime }(a_E)/2}|a-a_E|]$ diverges logarithmically. The static equilibrium is unstable. Since $V(a_E)<0$, the recollapsing, bouncing and static cases require positive spatial curvature $k>0$.

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.

Neglecting the rapidly oscillating radiation source on subhorizon scales, stress-energy conservation and the trace of the Einstein field equations giveEquivalently in cosmic time, ${\delta }_{m,tt}+2H{\delta }_{m,t}-4\pi G{\rho }_m{\delta }_m=0$. Radiation still contributes to the background expansion, even when its perturbation source is averaged away.

Put $\eta =a/a_{\mathrm{eq}}$ and let ${\rho }_{\mathrm{eq}}$ be the density of either component at matter-radiation equality. Then ${\rho }_m={\rho }_{\mathrm{eq}}{\eta }^{-3}$ and ${\rho }_r={\rho }_{\mathrm{eq}}{\eta }^{-4}$. With $C=(8\pi G/3){\rho }_{\mathrm{eq}}{a}_{\mathrm{eq}}^2$, the Friedmann equation givesApply the chain rule to the density equation:Thus the Mészáros equation isSubstituting $D_{+}=1+3\eta /2$ gives $(2+3\eta )(3/2)-3(1+3\eta /2)=0$, sois indeed a solution. It is approximately constant during radiation domination and grows as $a$ during matter domination. The second independent solution iswhich behaves as $\log (4/\eta )-3$ at small $\eta$. Thus smoothing radiation does not eliminate every logarithmic solution; matching through actual horizon entry fixes the combination excited.

Modes with $k\ll k_{\mathrm{eq}}$ enter after equality and retain the $P(k)\propto k$ shape. Modes with $k\gg k_{\mathrm{eq}}$ enter during radiation domination: their horizon-entry spectrum is $P_H\propto k^{-3}$ and they gain little growth until equality. The late matter power spectrum therefore turns over near $k_{\mathrm{eq}}$ and has a large-scale $k$ branch and a small-scale $k^{-3}$ branch, with the more accurate radiation-era matching givingEquivalently the cold-dark-matter transfer function falls approximately as $(k_{\mathrm{eq}}/k{)}^2$, up to the logarithm. The turnover is the consequence of the Mészáros effect; it is not a change of the primordial scale-invariance condition.

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.

For a radial null geodesic, the FLRW metric givesTwo successive wave crests traverse the same comoving path between comoving observers. Subtracting their path integrals, to first order in their short emission and reception periods, gives $\delta t_0/a(t_0)=\delta t_e/a(t_e)$. Since cosmic time is each observer's proper time and the frequency is the reciprocal period,This neglects peculiar velocities and local gravitational redshift contributions.

For B's light passing through A on its way to Earth, with the comparison event at A being that passage, cosmological redshift factors multiply:This is the intended numerical answer with that reception-event specification. The reception-event dependence of intergalactic redshift is essential: Earth's two redshifts by themselves neither specify the A–B separation nor identify the emission event at B whose light A receives at a chosen time. For arbitrary directions or a different reception time at A, the general answer is $a(t_{A,\mathrm{rec}})/a(t_{B,\mathrm{em}})-1$, with the emission time fixed by the A–B null geodesic; it cannot be deduced from the two supplied numbers alone.

On the sphere, both cosmic time and the comoving coordinate $r$ are constant. The positive spatial induced metric obtained from the FLRW metric is therefore $d{\ell }^2=a^2{r}^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$. Its area element is the square root of its determinant:Integrating over the spherical coordinates givesThus $ar$ is the areal radius. In a curved spatial slice it need not equal the proper radial distance $a{\int }_0^r(1-k{r}^{\prime 2}{)}^{-1/2}d{r}^{\prime }$; the angular part of the FLRW metric, rather than a Euclidean radial-distance assumption, establishes the surface area.

A superhorizon density contrast is gauge dependent. The stated growth laws describe the growing adiabatic mode in the comoving-gauge density contrast (and the corresponding usual synchronous growing mode), not an arbitrary coordinate density perturbation. Work with a flat, single-fluid background with constant $w=p/\rho$, negligible anisotropic stress, $c=1$, and a comoving Fourier wavenumber $K$.

Here is a derivation from the linear gravitational constraints and evolution equation. For an adiabatic mode, the scalar gravitational potential $\Phi$ obeysPrimes denote conformal time derivatives. The Friedmann equation gives $a\propto {\tau }^{2/(1+3w)}$ and $\mathcal{H}=2/[(1+3w)\tau ]$, so the bracket vanishes. On superhorizon scales, $K\tau \ll 1$, the leading solutions are a constant potential and a decaying solution proportional to ${\tau }^{-(5+3w)/(1+3w)}$. Keeping the growing adiabatic branch therefore makes $\Phi$ time independent to leading order.

Combining the time-time and time-space linear Einstein constraints eliminates the coordinate velocity and gives the comoving density constraintThis is the relativistic comoving constraint, not a subhorizon Newtonian approximation. Since $\rho \propto a^{-3(1+w)}$, it implies $\Delta \propto (a^2{H}^2{)}^{-1}\propto a^{1+3w}$. Also $a\propto t^{2/[3(1+w)]}$ in cosmic time, whenceBoth epochs here mean their respective leading growing solution; matching through matter-radiation equality requires the full two-component evolution. The constant superhorizon curvature or potential is consistent with this growth because the comoving density perturbation carries the additional factor $(K/aH{)}^2$.

Vary the scalar field and integrate by parts, taking the variation to vanish at the boundary:Thus the curved-spacetime Klein-Gordon equation is $\square \varphi -V_{,\varphi }=0$. In the flat FRW metric, $\sqrt{-g}=a^3$ and $g^{00}=-1$. A spatially homogeneous field therefore hasConsequentlyDots now denote cosmic time derivatives. The $3H\dot{\varphi }$ term is Hubble friction caused by expansion of the comoving volume.

At linear order a coordinate displacement ${\xi }^{\mu }$ changes a metric perturbation by $q_{\mu \nu }\mapsto q_{\mu \nu }-{\mathcal{L}}_{\xi }{\bar{g}}_{\mu \nu }$. Work in conformal time with ${\bar{g}}_{\mu \nu }=a^2\mathrm{diag}(-1,1,1,1)$, write ${\xi }^0=\alpha$ and lower spatial indices with ${\delta }_{ij}$. Directly evaluating the Lie derivative of a covariant tensor field givesHere primes on $\alpha ,{\xi }_i$ mean conformal-time derivatives, while primes on $q_{0\mu }^{\prime }$ label transformed components. Setting the transformed lapse and shift perturbations to zero requiresThese are first-order time equations. Their explicit local solutions areThus four coordinate functions can impose the four synchronous conditions in any regular perturbative patch. This is synchronous gauge fixing by coordinate displacement. The arbitrary spatial functions $C,D_i$ remain as residual gauge freedom; the conditions alone do not fix the coordinate system completely.

Geometrically, synchronous coordinates follow a congruence of freely falling observers launched normally from a spatial slice, using their proper time. This gives unit lapse and zero shift in cosmic time; conversion to background conformal time gives the displayed convention. “Always possible” is local: geodesic caustics or singularities can prevent a global synchronous chart. Such global obstructions do not invalidate local linear cosmological perturbation theory.

Put $K=2q_0-1>0$ and parametrize the scale factor byOn the expanding branch $0<\theta <\pi$, the equation from part (ii) givesConsequently,Choose the origin of cosmic time at the Big Bang, $y=0$, $\theta =0$. Integration proves the closed matter-dominated Friedmann solution:The same parametrization continues through maximum expansion at $\theta =\pi$ to collapse at $\theta =2\pi$: $dt/d\theta$ remains nonnegative while $dy/d\theta$ changes sign. The maximum scale factor is $y_{\max }=2q_0/K$, reached at $H_0{t}_{\max }=\pi q_0/K^{3/2}$; the total lifetime is twice this time. The squared equation alone does not prescribe the sign of $\dot{y}$, so this continuation supplies the collapsing branch explicitly.

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.

As ${\Omega }_M\downarrow 0$, use $\mathrm{arsinh}z=\log (z+\sqrt{1+z^2})\sim \log (2z)$ to obtainAt fixed Hubble constant, the age of a flat matter-Lambda universe diverges logarithmically. The limiting purely vacuum-dominated de Sitter spacetime has an exponential scale factor and no Big Bang at a finite value of cosmic time in this flat slicing.

For $0<{\Omega }_M<1$, set $x=a^{3/2}$ and choose the expanding branch of the Friedmann equation. ThenSeparating variables and choosing cosmic time $t=0$ at the Big Bang givesThus the flat dust-and-vacuum expansion isSetting $a(t_0)=1$ gives the age of a flat matter-Lambda universe,The two matter-vacuum age endpoint limits follow below.

Retain the homogeneous lapse function $\bar{N}(t)$, and define the physical Hubble parameter $H=\dot{a}/(\bar{N}a)$. At first order the scalar metric perturbations give $\delta g_{00}=-2{\bar{N}}^2\Phi$, $\delta g_{0i}=a^2{B}_{,i}$, and $\delta g_{ij}=-2a^2(\Psi {\delta }_{ij}+E_{,ij})$. Applying the passive coordinate displacement component by component gives the scalar gauge transformations with a background lapseFor example the $0i$ transformation is $\delta {\tilde{g}}_{0i}=a^2{\partial }_{i}B-a^2{\partial }_i\dot{\lambda }+{\bar{N}}^2{\partial }_i{\xi }^0$. The spatial isotropic part acquires $-2a\dot{a}{\xi }^0{\delta }_{ij}$, explaining the positive sign in the transformation of $\Psi$.

Preserving synchronous gauge requires both transformed $\Phi$ and $B$ to remain zero. The first condition is ${\partial }_t(\bar{N}{\xi }^0)=0$, and the second is $\dot{\lambda }={\bar{N}}^2{\xi }^0/a^2$. Integrating them givesThus fixing the lapse function and shift vector perturbations does not exhaust the scalar coordinate freedom. The functions $C,D$ describe residual synchronous-gauge freedom; an integration-origin change in the integral is absorbed into $D$.

To move from synchronous gauge to Newtonian gauge, choose $\lambda =-E_S$ to set $\tilde{E}=0$. Requiring $\tilde{B}=0$ then givesA scalar energy density transforms as $\delta {\rho }_N=\delta {\rho }_S-\dot{\bar{\rho }}{\xi }^0$. Dividing by the background energy density, which changes the contrast only at second order if perturbed in that denominator, yields the synchronous-to-Newtonian density transformation with a lapseThe last form uses background stress-energy conservation, $\dot{\bar{\rho }}=-3\bar{N}H(\bar{\rho }+\bar{P})$. With cosmic time $\bar{N}=1$ this is ${\delta }_N={\delta }_S-3(1+w)a^{2}H{\dot{E}}_S$; with conformal time $\bar{N}=a$ it is ${\delta }_N={\delta }_S-3(1+w)\mathcal{H}{E}_S^{\prime }$, where $\mathcal{H}=a^{\prime }/a$. Keeping these clock conventions distinct is essential to the factors of $a$.

A kinematic redshift is the relativistic Doppler effect arising from relative motion of emitter and receiver. In a local inertial frame, recession decreases the received frequency and approach increases it. A cosmological redshift instead compares photons emitted and received by comoving observers at different cosmic times: expansion of the scale factor stretches their wavelengths. It exists even when both observers have zero peculiar velocity. Both are frequency ratios measured by observers, but a large cosmological redshift should not be inserted into a flat-spacetime radial-speed formula to define the galaxy's unique recession speed.

In the intended comparison of emission and reception epochs, with negligible peculiar velocities, the Earth-observed cosmological redshifts correspond to $a_A=a_0/2$ and $a_B=a_0/10$. A photon emitted at the B epoch and received at the A epoch consequently hasThese factors divide, whereas the redshifts themselves do not subtract. There is a geometrical qualification to this usual answer: the two Earth-observed redshifts alone do not specify an arbitrary actual B-to-A observation. The assumed events must be connected by a null geodesic, as when A is on the same radial photon path between B and Earth. If the galaxies have an unspecified angular separation, or A observes at a different cosmic time, the B emission event reaching A is different and its redshift is not determined by these two numbers alone.

The age of an FLRW universe at an epoch is the elapsed cosmic time since the Big Bang, not the lookback time from today. For positive matter and positive cosmological constant, integrating $dt=da/\dot{a}$ givesPut $u=\sqrt{{\Omega }_{\Lambda ,0}/{\Omega }_{m,0}}{a}^{\prime 3/2}$; then $du=(3/2)\sqrt{{\Omega }_{\Lambda ,0}/{\Omega }_{m,0}}\sqrt{a^{\prime }}d{a}^{\prime }$. The integral becomes the age of a flat matter-Lambda universe:Choose $0<\theta <\pi /2$ such that $\tan \theta =\sqrt{{\Omega }_{m,0}/{\Omega }_{\Lambda ,0}}{a}^{-3/2}$. The upper-limit value of $u$ is $\cot \theta$, andConsequentlyThe integration constant is fixed by $t\to 0$ as $z\to \infty$, where $\theta \to \pi /2$. Equivalently, differentiating $u=\cot \theta$ changes the last integral to $-\int d\theta /\sin \theta$, recovering the supplied trigonometric identity. The formula assumes positive ${\Omega }_{\Lambda ,0}$ and ${\Omega }_{m,0}$; its zero-vacuum limit is regular even though this parametrization is singular.

Normalize the scale factor by $x=a/a_0$. Constant-equation-of-state density scaling for pressureless matter gives $\rho ={\rho }_0{x}^{-3}$. Multiplying the Friedmann equation by $x^2$ and substituting the expressions from part (i),Take the expanding branch $\dot{x}>0$ and set the origin of cosmic time at the Big Bang, $x=0$. The age of a dust universe without a cosmological constant is consequently