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The particle horizon is the greatest distance from which a light signal could have reached the observer since the initial cosmic time $t_i$. Its comoving radius and its radial proper distance at time $t$ areHere proper distance is measured along the spatial slice, rather than by the transverse area of a sphere. The Hubble parameter, deceleration parameter and cosmological density parameter areThe last expression defines the critical density. For blackbody radiation, $p=\rho /3$, so the Friedmann acceleration equation gives $\ddot{a}/a=-\kappa \rho /3$. ConsequentlyThis is the radiation relation between deceleration and density. It does not require spatial flatness.

For the open radiation universe particle horizon, write $x=a/a_0$. The cosmological continuity equation gives $\rho ={\rho }_0{x}^{-4}$, while the present Friedmann equation gives $-k/a_0^2=H_0^2(1-{\Omega }_0)$. ThusTaking the big-bang endpoint $x=0$ and using $dt=dx/\dot{x}$, the present radial proper distance becomesThereforeThe formula applies to $0<{\Omega }_0<1$. Its flat-radiation limit is $H_0{d}_{h0}\to 1$, agreeing with $a\propto t^{1/2}$ and $d_h=2t=H^{-1}$. The divergence as ${\Omega }_0\to 0$ reflects the unbounded past conformal interval of the limiting empty open model; that endpoint is not a radiation-filled universe.

The Horizon problem concerns the nearly uniform temperature of widely separated parts of the Cosmic microwave background. In a purely decelerating Hot Big Bang model, their past light cones at last scattering do not overlap far enough to explain this agreement by thermal contact. During cosmic inflation, accelerated expansion shrinks the comoving Hubble radius. A patch initially small enough for causal communication can be stretched to encompass the later observable universe. Reheating converts the inflationary energy into a hot plasma with correlated initial conditions across that patch. Enough inflation must occur before the observable scales leave the Hubble radius; inflation cannot establish contact between regions that were never initially causally related.

Assume positive fluid density, ${\rho }_0>0$, and write $m=1+3w>0$. A perfect fluid with constant equation of state obeys the cosmological perfect-fluid continuity equation, givingIn the stated units the Friedmann equation is $H^2=\rho /3-k/a^2+\Lambda /3$. Multiplication by $a^2/2$ yields the Friedmann effective potential for a constant-equation-of-state fluid,The allowed region has $V\le 0$. The Friedmann acceleration equation is equivalently $\ddot{a}=-V^{\prime }(a)$, including turning points by continuity. Thus a zero-energy mechanical trajectory reproduces the cosmological evolution, with $a>0$.

For $k=0$, $\Lambda <0$, the potential increases strictly from negative infinity to positive infinity. Its unique zero givesExpansion from the Big Bang stops there, with negative acceleration, then reverses into a Big Crunch. Both the turning point and the final singularity occur in finite proper time: $dt=da/\sqrt{-2V}$ is integrable near a simple turning point and behaves as a constant times $a^{m/2}da$ near zero.

For $\Lambda =0$, $k=+1$, $V$ increases from negative infinity to $1/2$ and crosses zero atThis is again expansion followed by finite-time recollapse. For $k=-1$, the same increasing curve has asymptote $-1/2$ and never meets zero. Expansion continues without a finite maximum; asymptotically $\dot{a}\to 1$ and $a\sim t$, as spatial curvature dominates the diluted fluid.

For $k=0$, $\Lambda >0$, $V$ tends to negative infinity at both ends. It has a maximum atThere is no turning point. Expansion initially decelerates, then accelerates once $a>a_{\ast }$, and approaches de Sitter spacetime expansion $a\propto e^{\sqrt{\Lambda /3}t}$. Thus positive flat dark-energy expansion is unbounded; negative flat dark energy and positive curvature without dark energy recollapse.

The Flatness problem is the extreme closeness to spatial flatness required in the early universe if ordinary decelerating expansion is to produce the observed near-flat universe. From the Friedmann equation, with $c=1$ and curvature constant $k$,During decelerating expansion $|\Omega -1|$ grows. The Friedmann acceleration equation gives $\ddot{a}/a=-4\pi G(\rho +3P)/3$. Thus $\rho +3P<0$ makes an expanding universe accelerate, increases $\dot{a}$ and suppresses $|\Omega -1|$: this is the inflationary solution of the flatness problem.

The slow-roll approximation neglects ${\dot{\varphi }}^2/2$ compared with $V$ and $\ddot{\varphi }$ compared with $3H\dot{\varphi }$. For the positive-field branch, taking $m>0$,Integrating the field equation and $\dot{a}/a=H$ gives the slow-roll solutionThe potential slow-roll parameters are ${ϵ}_V={\eta }_V=2/{\varphi }^2$. They cease to be small when $\varphi$ is of order $1$ in these units; the customary endpoint ${ϵ}_V=1$ is $\boxed{{\varphi }_{\mathrm{end}}\simeq \sqrt{2}}$. Equating the approximate kinetic and potential energies instead gives $\sqrt{2/3}$, another order-one indicator, not a precise endpoint of an approximation that is already failing.

Use units $c=\hslash =k_B=1$, and write $H=\dot{a}/a$ for the Hubble parameter, with dots denoting cosmic time. Differentiating the Friedmann equation givesOn an interval where $H\ne 0$, use the Hubble parameter identity $\dot{H}=\ddot{a}/a-H^2$ and the Friedmann acceleration equation to findSubstitution cancels the curvature term and gives the continuity equationThe cosmological perfect-fluid continuity equation extends by continuity through a regular isolated turning point. Physically it states that the change of energy in a comoving volume is the negative of the pressure work: $d(\rho a^3)=-Pd(a^3)$.

For separately conserved cosmological fluids, each component obeys this equation individually. A constant equation-of-state parameter $w_i=P_i/{\rho }_i$ therefore giveswhere the present scale factor is normalized to $a_0=1$. Thus the component density laws areThe extra factor for radiation in cosmology is the cosmological redshift of each photon's energy; pressureless matter has only number dilution, while the specified dark energy is a cosmological constant.

The critical density at a given expansion rate is the total density that makes the spatial curvature vanish:These cosmological density parameters use the critical density at that same time. Put $E(a)=H(a)/H_0$ and ${\Omega }_{K,0}=-K/H_0^2$. The Friedmann equation yieldsConsequently the fractional-density evolution, rather than just the component-density evolution, isIn particular ${\sum }_i{\Omega }_i=1+K/(a^2{H}^2)$. The simple powers of $a$ alone apply to ${\rho }_i$, or to ${\rho }_i/{\rho }_{\mathrm{crit},0}$, not to the instantaneous cosmological density parameters. At a recollapse turning point $H=0$, those instantaneous ratios are undefined even though the component densities remain finite.

Insert the component-density laws into the Friedmann equation, using ${\rho }_{i,0}=3H_0^2{\Omega }_{i,0}/(8\pi G)$ and $a_0=1$. Multiplying by $a^2$ givesThe radiation in cosmology and pressureless matter contributions add; the printed first formula has lost the plus sign between them. In the Friedmann acceleration equation, $\rho +3P=2{\rho }_r+{\rho }_m-2{\rho }_{\Lambda }$, soEvaluating the first equation today gives $K=H_0^2({\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{\Lambda ,0}-1)$.

Now remove dark energy and assume a positive pressureless matter or radiation in cosmology density, starting on an expanding branch. Let $f(a)=H_0^2({\Omega }_{r,0}/a^2+{\Omega }_{m,0}/a)$. It decreases from infinity to zero, and ${\dot{a}}^2=f(a)-K$, while $\ddot{a}<0$. This immediately distinguishes all three cases of spatial curvature of an FLRW universe.

The early small-$a$ solution has $a\propto t^{1/2}$ if radiation in cosmology is present, and $a\propto t^{2/3}$ for pressureless matter alone. The figure isolates the effect of curvature in three matter-only examples, rather than comparing models with the same present density parameters.

Use units $c=1$ and cosmic time $t$. Define the Hubble parameter $H=\dot{a}/a$, the equation-of-state parameter $\omega =P/\rho$, and the cosmological density parameterHere ${\rho }_{\mathrm{crit}}$ is the critical density. The definition requires $H\ne 0$; assume positive energy density and an expanding branch so that $\log a$ is a valid time coordinate. The Friedmann equation gives $\Omega -1=K/(a^2{H}^2)$. The Friedmann acceleration equation and Hubble parameter identity giveDifferentiating the Friedmann equation and combining with this identity yields the cosmological perfect-fluid continuity equation $\dot{\rho }=-3H(\rho +P)$. ConsequentlyThus the cosmological density parameter flow isThis identity allows time-dependent $\omega$; constancy is needed only for the simple power-law integration in the next part.

The prescribed cosmic string network scaling gives $\dot{\rho }=-2H\rho$. Comparing with the cosmological perfect-fluid continuity equation givesThe averaged pressure is therefore $P=-\rho /3$, provided the network is treated as the stipulated isotropic homogeneous component. The Friedmann acceleration equation gives $\ddot{a}=0$. On an expanding branch let $\rho =C/a^2$ and $B^2=8\pi GC/3-K>0$; thenThis coasting cosmic-string universe has no growing flatness instability, but it does not drive an initially nonflat universe toward flatness. The comoving Hubble radius remains constant, so there is also no shrinking-radius mechanism of the usual cosmic inflation type.

The global causal issue is more precise than saying that a constant comoving Hubble radius solves every Horizon problem. If this ideal coasting solution extends to its zero-size time $t_B$, then ${\int }_{t_B}^{t}d{t}^{\prime }/[B(t^{\prime }-t_B)]$ diverges logarithmically, so it has no finite particle horizon. If string domination starts at a finite earlier time after another cosmological phase, its causal history must instead be evaluated from that earlier phase. The string era alone does not establish that previously disconnected regions had time to equilibrate. A static branch with $B=0$ is a separate possibility; its Hubble-normalized $\Omega$ is undefined and is not covered by the expanding formulas.

Let $M_{\mathrm{Pl}}$ be the reduced Planck mass, $M_{\mathrm{Pl}}^{-2}=8\pi G$ in units $\hslash =c=1$. Combining that equation with the Friedmann acceleration equation givesDifferentiate $3M_{\mathrm{Pl}}^2{H}^2=\rho$ to obtain $\dot{\rho }=-3H(\rho +P)=-3H{\dot{\varphi }}^2$. For the homogeneous canonical scalar field, $\dot{\rho }=\dot{\varphi }\ddot{\varphi }+V_{,\varphi }\dot{\varphi }$, soWhere $\dot{\varphi }\ne 0$, division gives the inflaton equation of motion; continuity extends it to isolated turning points:The Hubble friction term describes loss of kinetic energy into expansion.

There is a small logical caveat if the field is frozen on an entire interval: the homogeneous Einstein constraints alone only give the product equation and do not force $V_{,\varphi }=0$. The frozen scalar fields and the Friedmann constraint counterexample makes this explicit. For a canonical scalar, the field equation also follows directly from its inflaton action in an expanding universe, whose homogeneous Lagrangian is $a^3[{\dot{\varphi }}^2/2-V(\varphi )]$. The Euler-Lagrange equation gives $d(a^3\dot{\varphi })/dt+a^3{V}_{,\varphi }=0$, including the frozen case, and yields the same boxed equation without division by $\dot{\varphi }$.

For separately conserved cosmological fluids, the prerequisite is absence of energy or momentum exchange between components. Each matter action then gives its own local conservation equation ${\nabla }_{\mu }{T}_i^{\mu \nu }=0$. In the homogeneous perfect fluid in general relativity, its time component isThis is the cosmological perfect-fluid continuity equation. Equivalently, a physical volume $V\propto a^3$ following the fluid satisfies $d({\rho }_{i}V)=-P_{i}dV$.

The Einstein field equations instead have the single shared metric on its left and ${\sum }_i{T}_i^{\mu \nu }$ on its right. The Friedmann equation and Friedmann acceleration equation therefore involve total density and total pressure; the same geometry cannot be sourced separately by each component's density alone.

The PDF's separate-conservation statement needs this noninteraction qualification. In an interacting system,where $Q_i$ is the energy-transfer rate into component $i$. Total conservation remains true, but the source-free equation does not hold individually during processes such as annihilation or particle decay.

For a single positive-density component, staticity requires $\rho +3P=0$ from the Friedmann acceleration equation. Among the listed components, only the cosmic string network has $w=-1/3$. Put $a(t)=a_E>0$ and use the Friedmann equation to obtainAll other components are absent. This is the static closed cosmic-string universe.

There is a normalization condition hidden in this expression. String conservation fixes $C_s={\rho }_s{a}^2$, and a static solution exists only whenOnce this condition is imposed, every positive radius $a_E$ is a static solution, with its density determined as above. One cannot independently prescribe an arbitrary string normalization and then choose a radius to compensate: both the string density term and curvature term scale as $a^{-2}$.

Choose a reference epoch with $H_0=\dot{a}(t_0)/a_0\ne 0$ and let $x=a/a_0$. Define the critical density and each cosmological density parameter byFor a closed universe ${\Omega }_{K,0}<0$. Take the dark-energy component to be a cosmological constant, as indicated by its $\Lambda$ label. The constant-equation-of-state density scaling givesSubstitution into the Friedmann equation givesAt the reference epoch, ${\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{s,0}+{\Omega }_{w,0}+{\Omega }_{\Lambda ,0}+{\Omega }_{K,0}=1$. The Friedmann acceleration equation isStrings have ${\rho }_s+3P_s=0$ and make no direct acceleration contribution, although they affect the Friedmann constraint. The normalized density parameters are undefined at a static epoch, where $H_0=0$; the static solution must therefore be described directly with dimensional densities.

First perturb the radius consistently with the conserved string normalization. If $a_E\mapsto a_E(1+ϵ)$ while $C_s={\rho }_s{a}^2$ stays at its static value, the density becomesThe Friedmann equation still has an exact cancellation between the string and curvature terms, and the Friedmann acceleration equation still vanishes. ThusThe homogeneous stability of a static cosmic-string universe is neutral in this restricted sense: there is no restoring force and no exponentially growing displacement. Holding the density fixed while changing the radius would instead change $C_s$ and would not be this pure displacement perturbation.

In cosmic time, all three Friedmann-Lemaitre-Robertson-Walker metrics can be writtenwhereThe closed spatial slice is a three-sphere of radius $a(t)$, soThe complete flat and open spatial slices are noncompact and have

For a homogeneous fluid in the Spatially flat FLRW metric, the equations in part (a) reduce to the cosmological perfect-fluid continuity equationwhile homogeneity makes the spatial relativistic Euler equation automatic. The Einstein field equations give the flat Friedmann equation and Friedmann acceleration equation,equivalently $\ddot{a}/a=-4\pi G(\rho +3P)/3$.

The Friedmann acceleration equation and $H=\dot{a}/a$ give the deceleration parameterAn Einstein-de Sitter universe contains only pressureless matter, so $q_0=1/2$ and cannot explain the observed negative value. For the stated Lambda-CDM model,so its cosmological constant produces the required accelerating universe.

The Friedmann acceleration equation givesFor the scalar field,and henceAn inflationary phase occurs precisely when potential energy dominates enough that

Differentiate the Friedmann equationand use the cosmological perfect-fluid continuity equationAfter dividing by $2H$,Since $\ddot{a}/a=\dot{H}+H^2$, eliminating the curvature term gives the Friedmann acceleration equation

Under the strong energy condition in a Friedmann universe, $\rho +3P\ge 0$, so $\ddot{a}/a\le 0$. Thereforeand wherever $H\ne 0$,

If $H(t_0)>0$, integrating the inequality backward shows that $H^{-1}$ reaches zero no later than $t_0-H(t_0{)}^{-1}$. Thus $H\to +\infty$ and $a\to 0$ at a finite time in the past. If $H(t_0)<0$, the same argument forward in time makes $H\to -\infty$ and $a\to 0$ at a finite future time. The sign of $H$ distinguishes an expanding universe with a past Big Bang from a contracting universe ending in a Big Crunch. Hence a Friedmann universe obeying the strong energy condition cannot remain nonsingular for unlimited proper time in both directions.

The constant $R$ is the curvature radius of the closed spatial slices when $a=1$; at general scale factor their curvature radius is $Ra$ and their scalar curvature is $6/(R^2{a}^2)$.

The cosmological perfect-fluid continuity equation isDuring expansion, $\dot{a}>0$, and thereforeIf $C=\rho (t_0)a(t_0{)}^3$, then throughout the later expanding phaseMultiplying the Friedmann equation by $a^2$ givesThe right-hand side is negative oncewhich is impossible because ${\dot{a}}^2\ge 0$. Thus the scale factor cannot grow without bound.

In fact the expanding phase ends in finite time. Combining the Friedmann and continuity equations yields the Friedmann acceleration equationThe curvature term in the first Friedmann equation also impliesWhile $a$ remains below the finite bound just found, $\ddot{a}$ is bounded above by a strictly negative constant. Hence $\dot{a}$ reaches zero after finite time: the closed universe attains a maximum size and cannot expand forever. This is recollapse of a closed Friedmann universe with nonnegative pressure.