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For a homogeneous cosmic fluid, the physical volume bounded by fixed comoving coordinates obeys $V\propto a^3$. The cosmological continuity equation therefore gives $d(\rho c^{2}V)=-PdV$. This is the first law of thermodynamics with conserved entropy and no additional chemical work. The internal energy of that volume need not be constant: a component with negative pressure gains internal energy as the universe expands. Local stress-energy conservation does not imply a global conserved energy for an arbitrary expanding spacetime.

For a closed Friedmann equation normalized to $k{c}^2=1$, constant equation-of-state parameter $w$, and conformal time $\tau$, let $q=(1+3w)/2$. The cosmological continuity equation and Friedmann acceleration equation give ${\mathcal{H}}^{\prime }=-q(1+{\mathcal{H}}^2)$, where $\mathcal{H}=a^{\prime }/a$. For $q\ne 0$ this integrates to the stated scale factor on any interval where the sine is positive, with ${\alpha }^{2q}=8\pi G{\rho }_0/3$. For $w=-1/3$ the separate solution is $a=A{e}^{H_0\tau }$ with $H_0^2+1=8\pi G{\rho }_0/3$; the sine-power formula is inapplicable.

During matter domination in a flat FLRW metric, the cosmological continuity equation gives $\rho \propto a^{-3}$, and the Friedmann equation gives $a\propto {\tau }^2$ and $8\pi G\rho a^2=12/{\tau }^2$. The CDM density equation in a matter-radiation universe, with radiation neglected, is therefore ${\delta }^{\prime \prime }+2{\delta }^{\prime }/\tau -6\delta /{\tau }^2=0$. Its Euler-Cauchy equation powers satisfy $(p-2)(p+3)=0$, proving the displayed general solution. The growing density contrast is proportional to the scale factor, while the decaying mode is proportional to $a^{-3/2}$. This all-scale form uses the CDM-comoving synchronous gauge; other density slices require their corresponding gauge transformation.

The cosmological principle permits a Friedmann-Lemaître-Robertson-Walker metric. The Friedmann equation, with vanishing cosmological constant, isFor separately conserved pressureless matter, the cosmological continuity equation gives ${\bar{\rho }}_m={\bar{\rho }}_{m0}{a}^{-3}$: its rest energy is conserved while its physical volume grows as $a^3$. Define the present critical density and matter density parameter byEvaluating the Friedmann equation today gives $-k=H_0^2(1-\Omega )>0$. Thus the model has negative spatial curvature of an FLRW universe; the second contribution to $H^2$ represents that spatial curvature of an FLRW universe, rather than another material fluid. Taking the expanding branch givesHere $0<\Omega <1$. At early times, $a\ll \Omega /(1-\Omega )$, matter domination gives $\dot{a}\simeq H_0\sqrt{\Omega }{a}^{-1/2}$, hence, measuring cosmic time from the Big Bang,At late times the $a^{-2}$ term dominates the Friedmann equation. This curvature-dominated universe has $\dot{a}\to H_0\sqrt{1-\Omega }$ and thereforeThe late expansion approaches the coasting Milne model. The early limit belongs to this idealized matter-only model; including radiation in cosmology would alter the sufficiently early evolution.

Define the momentum density $q_i=(\bar{\rho }+\bar{P})v_{\perp i}$, lowering its spatial label with ${\delta }_{ij}$, so the mixed stress-energy tensor obeys $T^i{}_0=q_i$ and $T^0{}_i=-q_i$. All products of $J_{ij}$ or its derivatives with $q_i$ are second order and must be dropped in linear cosmological perturbation theory.

For $\nu =0$, the three terms in stress-energy conservation areHere the trace $J_{ii}^{\prime }$ vanishes, ${\Gamma }_{00}^i={\Gamma }_{i0}^0=0$, and ${\partial }_i{q}_i=(\bar{\rho }+\bar{P}){\partial }_i{v}_{\perp }^i=0$. ThusFor $\nu =i$, spatial pressure gradients vanish because this is a purely vector cosmological perturbation. The first two terms areIn the remaining contraction, the two background-connection times momentum density terms cancel:The other terms are ${\Gamma }_{0i}^0\bar{\rho }=0$ and $-\bar{P}{\Gamma }_{ji}^j=0$. ConsequentlyExpanding this conservation law and using the background cosmological continuity equation gives the equivalent peculiar velocity equation, when $\bar{\rho }+\bar{P}\ne 0$,

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.

Set $H=\dot{a}/a$, the Hubble parameter. The first supplied component of the Einstein field equations gives the Friedmann acceleration equationSubstitution into the second supplied equation givesTherefore the first Friedmann equation isDifferentiate this equation with respect to cosmic time:Insert the Friedmann acceleration equation on the left and cancel $8\pi Ga/3$, which is nonzero for $a>0$. This givesThis is the cosmological continuity equation, expressing stress-energy conservation for the perfect fluid in general relativity. No division by $\dot{a}$ was made, so the derivation remains valid at a cosmological bounce or an instant of maximum expansion.

For the equation of state $p=-\varrho$, the cosmological continuity equation gives $a\dot{\varrho }=0$. Hence $\varrho ={\varrho }_0$ is constant. First take the physically intended positive energy density and defineThe Friedmann equation and the original Friedmann acceleration equation becomeSolve the second equation as $a=A{e}^{H_{\Lambda }t}+B{e}^{-H_{\Lambda }t}$. The first equation then requires $4H_{\Lambda }^{2}AB=k$. Choosing the time origin separately in each solution gives all positive branches:For $k=-1$, there is also the time reverse $a=H_{\Lambda }^{-1}\sinh [H_{\Lambda }(t_B-t)]$ on $t<t_B$, which contracts to a future zero. For $k=+1$, the cosmological bounce joins contraction to expansion at the positive minimum $a(t_b)=H_{\Lambda }^{-1}$. Retaining $\ddot{a}=H_{\Lambda }^{2}a$ at the minimum excludes a spurious permanently static solution that would satisfy the first-order constraint alone.

These are vacuum-energy FLRW solutions: the expanding flat branch, closed slicing of de Sitter spacetime, and open slicing of de Sitter spacetime. To compare them at the same energy density, the figure sets $T=H_{\Lambda }(t-t_c)$, takes $A_0=H_{\Lambda }^{-1}$, and aligns the open boundary and closed minimum at $T=0$. Thus the three plotted functions are $H_{\Lambda }a=e^T$, $\cosh T$, and $\sinh T$, the last only for $T>0$.

Only the expanding $k=-1$ branch has a finite past zero. The expanding $k=0$ branch tends to zero only as $t\to -\infty$, and the $k=+1$ branch is bounded below. All three describe portions of de Sitter spacetime, with Ricci scalar $R=12H_{\Lambda }^2$ and Kretschmann scalar $24H_{\Lambda }^4$. In particular the $k=-1$ zero is a boundary of that slicing, not a curvature singularity. The inequality in part (d) does not apply, because $\varrho +3p=-2{\varrho }_0<0$.

For zero energy density, the remaining solutions are $a=A_0>0$ for $k=0$, and the expanding or contracting Milne model $a=\pm (t-t_B)>0$ for $k=-1$; no positive real solution exists for $k=+1$. If negative energy density is admitted mathematically, write ${\alpha }^2=-8\pi G{\varrho }_0/3>0$. Only $k=-1$ is possible, with $a={\alpha }^{-1}\sin [\alpha (t-t_B)]$ for $0<\alpha (t-t_B)<\pi$, an open slicing of anti-de Sitter spacetime. These limiting and negative-density cases do not alter the positive-density comparison above.

An isotropic source's bolometric luminosity is distributed over a wavefront of physical area $4\pi a_0^2{r}^2$. Each photon loses an energy factor $(1+z{)}^{-1}$ through cosmological redshift, and successive photons arrive with a further rate factor $(1+z{)}^{-1}$ through cosmological time dilation. Therefore the radiative flux and luminosity distance areThese are bolometric formulas; a fixed observed frequency would require the shifted emission spectrum as well.

Normalize the scale factor by $a_0=1$. Summing the radiative flux over the source counts on an FLRW past light cone makes the transverse shell-area factors cancel:Thus ${\mathcal{F}}_0$ is power per effective detector area, summed over the whole sky; an all-sky detector of effective area $A_{\det }$ receives $A_{\det }{\mathcal{F}}_0$. The corresponding intensity per solid angle is ${\mathcal{F}}_0/(4\pi )$. Restoring the speed of light multiplies these flux integrals by $c$.

Spatially uniform $n(t)$ by itself does not specify its time evolution. For a persistent population comoving with the expansion, with neither creation nor destruction, the cosmological continuity equation for source number is $\dot{n}+3Hn=0$. Its conserved comoving number density is $n_0=a^{3}n$, so the cosmological bolometric background from conserved sources becomes $L{n}_0\int adt$. For the Einstein-de Sitter universe,The last equality uses $H_0=2/(3t_0)$. The early-time improper integral converges despite the diverging physical source density. If instead the proper number density were maintained at a constant value $n_0$ by continuous source creation, the same geometry would give $3L{n}_0{t}_0/11$. For a completely unspecified source history the general integral above is the determined answer; constancy of individual luminosity alone does not imply number conservation.

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.

For a flat universe containing only the stated fluid, the cosmological continuity equation is $\dot{\rho }+3H(\rho +P)=0$. A constant equation-of-state parameter $w$ therefore gives $\rho ={\rho }_0(a/a_0{)}^{-3(1+w)}$. Combine this with the Friedmann equation $H^2=8\pi G\rho /3$ to findOn the expanding branch, specifying $H(t_0)=H_0>0$ yieldsThe solution is used only where its denominator is positive. For $w\ne -1$, integration also givesFor $w>-1$, choosing the zero of time at the initial singularity gives $H=2/[3(1+w)t]$ and $\rho =1/[6\pi G(1+w{)}^2{t}^2]$. For $w=-1$, $H=H_0$, $\rho$ is constant and $a=a_0{e}^{H_0(t-t_0)}$. For $w<-1$, the displayed initial-value solution instead grows towards a future finite-time singularity; it is not the ordinary positive-time Big Bang branch. With positive density, accelerated expansion requires $w<-1/3$.

Adding and subtracting the canonical scalar field density and pressure givesA real scalar with positive kinetic energy consequently needs $w\ge -1$. On the expanding scalar-dominated branch, divide $\dot{\varphi }$ by $H$ and use the Friedmann equation:Choose the positive rolling sign. Integration and elimination of $a$ using ${\rho }_{\varphi }={\rho }_{\varphi }^0(a/a_0{)}^{-3(1+w)}$ give the scalar-field reconstruction for a constant equation of state:Changing the rolling sign reverses the exponential slope. At $w=-1$, $\varphi$ is constant and $V={\rho }_{\varphi }^0$; elimination by an invertible $\varphi (a)$ then degenerates. The cosmological continuity equation gives $\ddot{\varphi }+3H\dot{\varphi }+V_{,\varphi }=0$ whenever $\dot{\varphi }\ne 0$, so the reconstruction satisfies the field equation as well as the Friedmann equation.

Compare the reconstructed exponent with $V_0{e}^{-\lambda \sqrt{8\pi G}\varphi }$. In this constant-$w$, scalar-dominated scaling solution,Strict ${\rho }_{\varphi }+P_{\varphi }>0$ requires $w>-1$, and acceleration requires $w<-1/3$. HenceThe accelerating branch has $V_0>0$. The zero-slope limit is a constant potential with $\rho +P=0$, so it does not meet the strict inequality. This result characterizes the exponential-potential power-law inflation solution; an arbitrary trajectory with the same potential need not have constant $w$ throughout its evolution.

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.

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

During radiation domination, the cosmological continuity equation gives ${\rho }_R\propto a^{-4}$. The flat Friedmann equation in conformal time is $(a^{\prime }/a{)}^2=8\pi G{a}^2{\rho }_R/3$. Hence $a^{\prime 2}$ is constant, and choosing the Big Bang at $\tau =0$ yieldsFor the CDM density equation in a matter-radiation universe, neglect ${\rho }_C{\delta }_C$ while retaining ${\delta }_C$ itself. The adiabatic initial conditions give ${\delta }_R=4{\delta }_C/3$, soThis Euler-Cauchy equation has powers ${\tau }^p$ with $p(p-1)+p-4=0$, namely $p=2,-2$. Selecting the growing superhorizon adiabatic CDM mode in radiation domination gives

Differentiate the Friedmann equation and use the cosmological continuity equation to obtain $\dot{H}+H^2=-(4\pi G/3)(\rho +3P/c^2)$. The relation extends to turning points by continuity. Radiation obeys ${\rho }_R={\rho }_{R0}{a}^{-4}$, while dark energy with $P_{\Lambda }=-{\rho }_{\Lambda }{c}^2$ has constant density. Substitution at $a(t_0)=1$ gives $k{c}^2=\beta H_0^2$, and at arbitrary $a$ givesPut $x=a^2$. For $\beta >4$, the upward-opening polynomial $P(x)=x^2-\beta x+\beta$ has roots $x_{\pm }=(\beta \pm \sqrt{\beta (\beta -4)})/2$, with $1<x_{-}<2<x_{+}$. The expanding Big Bang branch reaches $x_{-}$ in finite time. There $H=0$ but $\dot{H}<0$, so expansion turns to contraction; it returns to $a=0$ in finite time, a Big Crunch. A sketch of $a^4{H}^2/H_0^2=P(x)$ has the forbidden negative interval between the roots, while $a^4\dot{H}/H_0^2=\beta (x-2)$ is a line crossing zero at two.

For $\beta =4$, $P(x)=(x-2{)}^2$. The branch starting at zero has $\dot{x}=2H_0(2-x)$, givingAt small $t$, $a\sim 2\sqrt{H_{0}t}$, the radiation-dominated scale factor. At large $t$, $a=\sqrt{2}[1-\frac{1}{2}{e}^{-2H_{0}t}+O(e^{-4H_{0}t})]$. It approaches an Einstein static universe with equal radiation and dark-energy densities, rather than recollapsing or entering unbounded expansion.

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 ${\Omega }_m\simeq 0.315$.

For a homogeneous scalar field, every spatial derivative vanishes. The stress-energy tensor consequently has no momentum density or directional stress:The only spatial tensor left is ${\delta }^{ij}$, so the stress is isotropic. This is the perfect fluid form in its comoving frame. Now expand the time component of stress-energy conservation using the covariant derivative:The first term is $\dot{\rho }$. In the second, only $\lambda =0$ contributes, and ${\sum }_i{\Gamma }^i{}_{i0}=3H$, giving $3H\rho$. The third isHere the quoted Christoffel symbols include their lower-index symmetry ${\Gamma }^i{}_{j0}={\Gamma }^i{}_{0j}$. Therefore the cosmological continuity equation isAs a check using the scalar field equation, $\dot{\rho }=\dot{\varphi }(\ddot{\varphi }+V_{,\varphi })=-3H{\dot{\varphi }}^2=-3H(\rho +P)$, with exactly the same sign.

With zero scalar potential and a homogeneous scalar field, the preceding perfect fluid identification givesThis equation of state is that of a stiff fluid. The field equation integrates to $a^3\dot{\varphi }=K$, since ${\partial }_t(a^3\dot{\varphi })=0$. Hence $\rho =K^2/(2a^6)$, agreeing with the cosmological continuity equation for $w=1$. If this nonzero component dominates a flat universe, the Friedmann equation giveson the expanding branch. Integrating once more, $a^3=3\sqrt{4\pi G/3}|K|(t-t_B)$. Choosing the Big Bang at $t_B=0$ yieldsThe homogeneous free scalar as a stiff fluid construction requires $K\ne 0$; a constant zero-potential field has zero energy density and cannot dominate this expansion.

For constant equation-of-state parameter $w$, the cosmological continuity equation integrates to $\rho ={\rho }_0{a}^{-3(1+w)}$ using today's normalization. In conformal time, $\mathcal{H}=a^{\prime }/a=\dot{a}$, so the Friedmann equation becomesSince ${\mathcal{H}}^{\prime }=a\ddot{a}$, the Friedmann acceleration equation givesFor $k{c}^2=1$, put $q=(1+3w)/2$. When $q\ne 0$, integration gives $\mathcal{H}=\cot (q(\tau -{\tau }_0))$. Integrating $a^{\prime }/a=\mathcal{H}$ then yields the closed constant-equation-of-state Friedmann scale factorChoose a connected interval with positive sine; the origin of conformal time can absorb ${\tau }_0$. For $w>-1/3$ this is the expansion and recollapse branch between consecutive zeros of the sine. Negative $q$ instead gives the corresponding minimum-radius branch on its positive-sine interval. The printed power formula excludes $w=-1/3$: in that exceptional case $\mathcal{H}=H_0$ is constant and $a=A{e}^{H_0\tau }$, with $H_0^2+1=8\pi G{\rho }_0/3$. A real solution requires the latter constant to be at least one; equality permits a static solution.

For pressure-free matter, $w=0$, so $\beta =1/2$. Set the big-bang origins to zero. The closed matter-dominated Friedmann solution follows by ${\sin }^2(\beta \tau )=(1-\cos 2\beta \tau )/2$ and integrating $dt=ad\tau$:The entire cycle is $0\le 2\beta \tau \le 2\pi$; maximum expansion has $a=\alpha$ at $2\beta \tau =\pi$, and the big crunch occurs at $t=\alpha \pi /(2\beta )$.