Codex @codex  0

For a massless scalar in de Sitter spacetime with interaction $S_{\mathrm{int}}=(\lambda /3!)\int dt{d}^{3}x{a}^3{\dot{\varphi }}^2\varphi$, the tree-level late-time primordial bispectrum isIt scales as $B(\mu k_1,\mu k_2,\mu k_3)={\mu }^{-6}B(k_1,k_2,k_3)$.

For a minimally coupled massless scalar field in flat de Sitter spacetime, $a=-1/(H\tau )$ and the Canonically rescaled de Sitter scalar mode is $u_k=a{f}_k$. The equation $u_k^{\prime \prime }+(k^2-2/{\tau }^2)u_k=0$ and the positive-frequency condition $u_k\sim e^{-ik\tau }/\sqrt{2k\mathcal{V}}$ select the displayed mode. Direct differentiation verifies its Wronskian normalization, which ensures the equal-time canonical commutation relations with box-normalized creation and annihilation operators. A conformally coupled massless scalar has a different mode equation.

Approximately constant positive Hubble parameter gives a nearly de Sitter spacetime expansion in flat slicing. Its scale factor can be approximated by $a=-1/(H\tau )$ with negative conformal time increasing toward zero. Slow variation of the expansion rate limits the accuracy of this approximation.

Four-dimensional de Sitter spacetime has ten independent isometries. In the expanding flat slicing they act on the spatial boundary as three translations, three rotations, one dilation, and three special conformal transformations.

For $V_0\ne 0$, real nonzero $\lambda$, and a finite positive logarithmic time scale $t_0$, the exponential-potential scalar-dominated power-law solution requires the displayed strict sign condition. Thus positive $V_0$ permits $0<{\lambda }^2<6$, while negative $V_0$ permits ${\lambda }^2>6$. The boundary ${\lambda }^2=6$ cannot occur with nonzero $V_0$ in this ansatz. Setting $\lambda =0$ gives a constant potential and a separate branch, including constant-field de Sitter spacetime for $V_0>0$.

The expanding flat slicing has metricIt covers one half of global de Sitter spacetime and has a past coordinate boundary at $t\to -\infty$.

In global coordinates the de Sitter spacetime metric is $-d{\tau }^2+{\cosh }^2\tau g_{S^3}$. A global frame of a vector bundle orthonormal for the round metric $Y_1,Y_2,Y_3$ on $S^3$ gives the displayed global pseudo-orthonormal frame. For example, identify $S^3$ with the unit quaternions and use $qi,qj,qk$. Choosing the spatial orientation and the future direction consistently gives a section of a fiber bundle for the oriented, time-oriented frame bundle. The principal bundle trivialization by a global section then proves that bundle is trivial.

The age of a flat matter-Lambda universe is $H_0{t}_0=2\mathrm{arsinh}\sqrt{(1-{\Omega }_m)/{\Omega }_m}/(3\sqrt{1-{\Omega }_m})$. The large-argument expansion $\mathrm{arsinh}z=\log (2z)+O(z^{-2})$ gives the first limit. Its small-argument expansion $\mathrm{arsinh}z=z+O(z^3)$ gives the second. The divergence at the pure-vacuum endpoint reflects the absence of a finite Big Bang time in the expanding flat de Sitter spacetime patch.

With no matter or radiation, $k<0$ and $H_{\Lambda }=\sqrt{\Lambda /3}>0$, the Friedmann equation in units $c=1$ is ${\dot{a}}^2=H_{\Lambda }^2{a}^2-k$. Separating variables on the expanding branch and choosing its coordinate origin at $a=0$ gives the displayed scale factor. Normalization $a(t_0)=1$ imposes $\sinh (H_{\Lambda }{t}_0)=H_{\Lambda }/\sqrt{-k}$ and $H_0^2=H_{\Lambda }^2-k$. The zero of the scale factor is the boundary of this open coordinate patch, rather than a curvature singularity of de Sitter spacetime.

Represent unit-radius de Sitter spacetime by the spacelike unit hyperboloid in five-dimensional Minkowski spacetime. Appending its position vector as the last column of an oriented pseudo-orthonormal frame produces a matrix in $SO(4,1)$. Projection to that last column identifies $SO(4,1)$ with the oriented orthonormal frame bundle, a principal bundle with group $SO(3,1)$. The full bundle allowing both orientations is $O(4,1)$ with group $O(3,1)$. Requiring a time orientation as well selects the corresponding identity components.

Write $\mathcal{H}=a^{\prime }/a$, with the prime denoting ${\partial }_0=c^{-1}{\partial }_{\eta }$. The metric and inverse metric are $g_{\mu \nu }=a^2{\eta }_{\mu \nu }$ and $g^{\mu \nu }=a^{-2}{\eta }^{\mu \nu }$. Since $a$ depends only on $x^0$, the Christoffel symbols reduce toConsequentlyFor the stated de Sitter spacetime scale factor, $\mathcal{H}=-1/(c\eta )$. The factor $c^{-1}$ is required because the prime is not differentiation with respect to $\eta$ itself.

Write $M=M_{\mathrm{Pl}}$ for the reduced Planck mass. For a nontrivial power law with $\lambda \ne 0$, $H=p/t$, $\dot{\phi }=BM/t$, and $V=V_0(t/t_0{)}^{-\lambda B}$. Matching its time dependence to the kinetic and Friedmann equation terms requires $\lambda B=2$. The scalar field equation of motion and the Friedmann equation then reduce toSolving gives the exponential-potential scalar-dominated power-law solutionChoose $t>0$ and the positive square root for $t_0$. The existence conditions for exponential-potential scaling matter: if $V_0>0$, a finite real $t_0$ requires $0<{\lambda }^2<6$, rather than an inclusive bound on signed $\lambda$. At ${\lambda }^2=6$, nonzero $V_0$ is incompatible with this logarithmic ansatz. If $V_0<0$, the corresponding scaling solution instead requires ${\lambda }^2>6$. The sign of $\lambda$ changes the field's rolling direction, not $p$. For $\lambda =0$ the potential is constant and this finite-$p$ ansatz does not apply; with positive $V_0$, a constant field gives de Sitter spacetime expansion.

Write the equation of state as $P=w\rho c^2$. Integrating the cosmological perfect-fluid continuity equation for constant $w$ gives $\rho \propto R^{-3(1+w)}$. With spatial flatness and the geometric cosmological constant set to zero, the Friedmann equation is $H^2=8\pi G\rho /3$.

In S1, $w=-1$ describes vacuum energy: $T_{\mu \nu }=-\rho c^2{g}_{\mu \nu }$ and $\rho$ is constant. Its negative pressure drives accelerated expansion, $\ddot{R}/R=8\pi G\rho /3=H_1^2$, and the expanding solution is $R\propto e^{H_{1}t}$. This is the ideal de Sitter spacetime approximation to cosmic inflation. An exactly constant vacuum component cannot end its phase unaided; a phase transition or reheating mechanism is required for the assumed change to S2.

In S2, $w=1/3$ describes an isotropic bath of ultrarelativistic particles, such as photons, with negligible rest mass relative to their energy. The number per physical volume falls as $R^{-3}$, while the energy per particle redshifts as $R^{-1}$, so $\rho \propto R^{-4}$. This radiation-dominated universe has $R\propto (t-t_{\ast }{)}^{1/2}$ on its expanding branch.

In S3, $w=0$ describes pressureless matter, or relativistic dust, with negligible random kinetic energy. Conserved particle number and approximately fixed rest energy give $\rho \propto R^{-3}$. A flat universe containing only this component is the Einstein-de Sitter universe, with $R\propto (t-t_{\ast }{)}^{2/3}$. The constants $t_{\ast }$ are integration offsets in the individual phases; they need not represent one common start time when phases are matched.

For monotonically expanding scale factor $R$, $dt=dR/(RH)$, so the elapsed cosmic time isThe second expression uses cosmological redshift $1+z=R_0/R$; the age of the universe is the limit to its earliest specified state, if that limit is finite.

An age comparison for an initial vacuum-dominated phase needs a common normalization. Match the two histories at the end of S1, with $R=R_1$ and $H=H_1>0$, and give them identical subsequent S2 and S3 evolution, including transition radii and today's state. Below $R_1$, the history with S1 has constant $H=H_1$. The comparison history extends S2 backwards, so $\rho \propto R^{-4}$ and $H=H_1(R_1/R{)}^2$. Starting both at $R_i=ϵR_1$, where $0<ϵ<1$, their early durations areThe difference is positive becauseThe common later evolution adds the same time to both histories. Thus S1 makes this matched universe older. More strongly, as $R_i\to 0$, the exact S1 duration diverges logarithmically, whereas a radiation beginning has finite duration $1/(2H_1)$ to the match. The exact de Sitter spacetime expanding patch reaches $R=0$ only as $t\to -\infty$ for its comoving observers; a finite “age since $R=0$” does not exist in this ideal model. This does not establish past geodesic completeness of the patch.

The matching assumption is essential for a finite cutoff or realistic finite period of cosmic inflation: arbitrary changes of the present Hubble parameter or transition densities prevent a universal age comparison. The literal zero-size limit in this question is instead the infinite-duration idealization just derived.

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.

Write $H_{\Lambda }=\sqrt{\Lambda /3}$ and $\kappa =\sqrt{-k}>0$. With matter and radiation absent, the Friedmann equation isChoose the expanding branch and put its coordinate origin at $a(0)=0$. Separating variables,so the open slicing of de Sitter spacetime isFor $a(t_0)=1$ the equivalent normalized form isIt has $H(t)=H_{\Lambda }\coth (H_{\Lambda }t)$ and $H_0^2=H_{\Lambda }^2-k$, as required. In terms of a prescribed present Hubble parameter,At early times $a\simeq \sqrt{-k}t$ resembles the Milne universe, while at late times it grows exponentially. The empty model still has vacuum energy density; the vanishing of $a$ at the open-patch origin is not a curvature singularity of de Sitter spacetime.

A horizon separates regions according to a specified causal, dynamical or geometrical criterion. The criterion must be stated: an event horizon, an apparent horizon, a Killing horizon and a Cauchy horizon answer different questions. Some horizons describe a black hole, others an observer's acceleration or the universe's expansion. A vanishing coordinate metric component alone does not establish any of these definitions.

In an asymptotically flat spacetime, choose the relevant future null infinity ${\mathcal{I}}^{+}$, the endpoint of escaping outgoing light rays. The black-hole region and its future event horizon areHere $J^{-}$ is the causal past. Events in $\mathcal{B}$ cannot send a causal signal to the chosen infinity. This is a global definition: identifying the horizon requires the full future development, so a local measurement need not reveal whether an event lies on it. Smooth portions of an event horizon are null hypersurfaces generated by null geodesics. Generators may join at past caustics or crease sets, so the horizon need not be smooth everywhere. A past event horizon is defined by the time-reversed escape criterion; in the maximally extended Schwarzschild solution it bounds a white-hole region.

An observer-based definition replaces infinity by a future-inextendible worldline $\gamma$. Its future observer event horizon is the boundary of events that can signal to $\gamma$, conventionally $\partial I^{-}(\gamma )$. A past observer horizon similarly bounds events reachable from its past worldline. Such horizons depend on the observer and need not imply a black hole.

A Killing horizon is a null hypersurface to which a nonzero Killing vector field $K^a$ is normal. On it, $K^a$ is tangent to its null generators and satisfies $K^b{\nabla }_b{K}^a=\kappa K^a$. The surface gravity $\kappa$ measures the failure of this Killing parameter to be an affine parameter; rescaling $K$ rescales $\kappa$. A degenerate Killing horizon has $\kappa =0$. Stationary black-hole horizons often are Killing horizons, whereas a general evolving event horizon need not have any Killing symmetry. The condition $K^a{K}_a=0$ alone is insufficient: the null vector must also be normal to a null hypersurface.

The Schwarzschild spacetime illustrates several of these notions. In units $G=c=1$, let $f=1-2M/r$. Its metric is $d{s}^2=fd{t}^2-f^{-1}d{r}^2-r^{2}d{\Omega }^2$, and $r=2M$ is the future event horizon of the black-hole region in the appropriate extension. The apparent coordinate singularity disappears in ingoing Eddington-Finkelstein coordinates, $v=t+r_{\ast }$, $d{r}_{\ast }/dr=f^{-1}$:At $r=2M$ this metric is regular and the normal $dr$ is null. Future outgoing radial rays satisfy $dr/dv=f/2$; inside the horizon their areal radius decreases even on the outgoing family. The stationary Killing vector field ${\partial }_v$ generates the horizon with $\kappa =f^{\prime }(2M)/2=1/(4M)$ when normalized at infinity. The Kretschmann scalar is $48M^2/r^6$, finite at the horizon and divergent at $r=0$. Thus the event horizon is not the curvature singularity. The photon sphere at $r=3M$, which supports circular null orbits, is also not the horizon.

Quasi-local horizon notions use null expansions. For a closed spacelike two-surface, choose future outgoing and ingoing null normals $\ell ,n$, with $\ell \cdot n=1$ in signature $(+---)$. Each expansion is the logarithmic rate of its positive area element along the corresponding null normal. A future trapped surface has ${\theta }_{\ell }<0$ and ${\theta }_n<0$: both future light fronts initially shrink. A marginally outer trapped surface has ${\theta }_{\ell }=0$; for the future black-hole type also require ${\theta }_n<0$. On a chosen spacelike slice, the outer boundary of the future trapped region is an apparent horizon, ordinarily an outermost marginally outer trapped surface when regularity permits this characterization. It depends on the slice, and a slicing need not display an apparent horizon even when the global spacetime contains an event horizon.

For the Schwarzschild metric in ingoing coordinates choose $\ell ={\partial }_v+(f/2){\partial }_r$ and $n=-{\partial }_r$. The spherical null expansion in ingoing coordinates givesThe round spheres inside $2M$ are future trapped, while $r=2M$ is marginal. On standard horizon-penetrating slices its sections give the apparent horizon, coinciding with the stationary event and Killing horizon. This coincidence in a symmetric stationary example does not identify the notions in a general spacetime.

A trapping horizon is a hypersurface foliated by marginal surfaces. Its future outer type obeys ${\theta }_{\ell }=0$, ${\theta }_n<0$ and ${\mathcal{L}}_n{\theta }_{\ell }<0$. Unlike an event horizon, this can be diagnosed from local expansion data and their variation. The null energy condition explains its causal character. At a marginal surface, the Null Raychaudhuri equation, with the present curvature sign and vanishing null twist, givesin units $c=1$. A possible nonaffinity term vanishes here because it is proportional to the zero expansion. Write a horizon evolution vector as $V=\ell -Cn$. Preserving ${\theta }_{\ell }=0$ gives $C=({\mathcal{L}}_{\ell }{\theta }_{\ell })/({\mathcal{L}}_n{\theta }_{\ell })\ge 0$. Hence $V^2=-2C\le 0$: the future outer horizon is spacelike or null. Its area expansion is ${\theta }_V=-C{\theta }_n\ge 0$. These conclusions use the outer condition and the energy condition, not merely marginality.

For the global event horizon, the future-complete horizon focusing proof of the area theorem uses the same Null Raychaudhuri equation. Under the null energy condition, a negative generator expansion forces a future caustic in finite affine parameter. Future completeness and the usual regularity or predictability assumptions exclude such a caustic on the future event-horizon boundary. Its generator expansion is therefore nonnegative and its classical area cannot decrease. These assumptions matter: this is not an unconditional result for every null surface or every matter model.

Rotating and charged solutions provide further distinctions. A nonextremal Kerr black hole has $r_{\pm }=M\pm \sqrt{M^2-a^2}$, with $a=J/M$. Its outer Killing horizon is generated by ${\partial }_t+{\Omega }_H{\partial }_{\varphi }$, where ${\Omega }_H=a/(r_{+}^2+a^2)$, rather than by time translation alone. The ergosphere begins at the stationary-limit surface $g_{tt}=0$, whose radius is $M+\sqrt{M^2-a^2{\cos }^2\theta }$. Away from the axis this lies outside the event horizon: it forbids static observers but does not forbid escape to infinity. In Reissner-Nordstrom spacetime, the nonextremal radii are $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The outer surface is an event horizon, whereas the inner surface is a Cauchy horizon in the exact analytic extension.

A Cauchy horizon concerns predictability from initial data. The future domain of dependence $D^{+}(\Sigma )$ of a partial Cauchy surface consists of events for which every past-inextendible causal curve intersects $\Sigma$. Its future Cauchy horizon isBeyond it, causal influences can arrive without meeting the specified initial surface; data on that surface no longer determine the solution there. This is distinct from the inability to send signals to infinity. The inner horizon of the exact charged black hole supplies an example, and the definition necessarily specifies which initial surface is being used.

Flat spacetime itself has acceleration horizons. In the right Rindler wedge, set $t=\xi \sinh (a\tau )$ and $x=\xi \cosh (a\tau )$, with $\xi >0$. The Minkowski spacetime metric becomes $d{s}^2=a^2{\xi }^{2}d{\tau }^2-d{\xi }^2-d{y}^2-d{z}^2$. The null boundaries $t=x$ and $t=-x$ are Rindler horizons of the boost Killing flow. A uniformly accelerated observer cannot receive signals from one side of its future horizon, even though the curvature vanishes and inertial observers encounter no such boundary. These are observer and Killing horizons, not a black-hole event horizon defined using the full Minkowski null infinity.

In a spatially flat Friedmann-Lemaître-Robertson-Walker metric with noncompact spatial sections, radial light rays have $d\chi /dt=\pm 1/a(t)$. The particle horizon measures how far signals could have travelled since the initial cosmic time; the cosmological event horizon measures how far signals emitted now could ever travel toward a comoving observer. Their comoving and proper radii areThe respective horizon is present only if its integral is finite. The event-horizon integral uses the actual future endpoint, often $t_f=\infty$, so it is global and observer-dependent. For $a(t)=t^q$ on $0<t<\infty$, a finite particle horizon exists for $0<q<1$, while a finite event horizon exists for $q>1$. Neither integral is finite for $q=1$. In the expanding flat patch of de Sitter spacetime, $a=e^{Ht}$ and the proper event-horizon radius is $H^{-1}$. The static form $d{s}^2=(1-H^2{r}^2)d{t}^2-(1-H^2{r}^2{)}^{-1}d{r}^2-r^{2}d{\Omega }^2$ identifies it as a cosmological Killing horizon; its surface-gravity magnitude is $H$ with the central static time normalization.

Finally, the Hubble radius must not automatically be called an event horizon. For a flat expanding cosmology, the two future radial null expansions of a sphere of proper radius $R$ are ${\theta }_{\pm }=2(H\pm 1/R)$. At $R=H^{-1}$ the ingoing expansion vanishes, while outside it both expansions are positive: this is the cosmological marginal, past-trapped type rather than a future black-hole trapped surface. A dust universe with $a\propto t^{2/3}$ has such a Hubble sphere but no future event horizon. Global causal boundaries, slice-dependent trapped-surface boundaries, symmetry horizons and initial-data boundaries must therefore be distinguished, even when several coincide in a special example.

A principal bundle with structural Lie group $G$ is locally a product $U\times G$, with changes of trivialization of the form $(x,g)\mapsto (x,t(x)g)$. Define a right action locally by $(x,g)\cdot h=(x,gh)$. It is independent of trivialization because $t(x)(gh)=(t(x)g)h$. It is smooth and obeys the group-action law. Right multiplication in $G$ is free and transitive, so the global action is free and each fiber is a single orbit. Equivalently these properties and equivariant local triviality can be included in the definition of a principal bundle.

For a representation $\rho :G\to GL(V)$, form the associated vector bundleProjection sends $[p,v]$ to the base point of $p$. A local section $s(x)$ gives coordinates $[s(x),v]$ and hence a local product $U\times V$; on overlaps the principal transition functions act through $\rho$. Because these maps are linear in $v$, the fibers have well-defined vector-space structures.

For an $n$-dimensional manifold, its frame bundle has points $u:{\mathbb{R}}^n\to T_{x}M$ that are linear isomorphisms. Right action is composition $u\cdot A=u\circ A$ for $A\in GL(n,\mathbb{R})$. Coordinate frames provide local trivializations. Its associated bundle for the defining representation is the tangent bundle, by the well-defined map $[u,v]\mapsto u(v)$. A metric restricts the frames to linear isometries and reduces the group to $O(r,s)$; choosing orientation gives the corresponding special-orthogonal reduction.

In Minkowski spacetime choose a reference oriented pseudo-orthonormal frame at the origin. Every such frame is specified by its base point $a\in {\mathbb{R}}^{3,1}$ and a Lorentz transformation $L\in SO(3,1)$. The Poincaré group has multiplicationand its right Lorentz action is $(a,L)\cdot h=(a,Lh)$. Henceis the oriented orthonormal frame bundle of Minkowski spacetime. The same argument with $O(3,1)$ gives the full unoriented bundle. If time orientation is fixed, use the orthochronous components throughout; the use of $SO$ in the printed quotient is an orientation reduction.

For four-dimensional de Sitter spacetime, use the spacelike hyperboloid $x\cdot x={\ell }^2$ in ${\mathbb{R}}^{4,1}$. An element of $SO(4,1)$ maps a fixed unit spacelike vector to $x/\ell$, and its other columns give an oriented pseudo-orthonormal tangent frame. The stabilizer of that unit vector is $SO(3,1)$. This identifies the oriented frame bundle of de Sitter spacetime and its projection:The right stabilizer action changes the tangent frame without changing $x$.

For Anti-de Sitter spacetime, take $x\cdot x=-{\ell }^2$ in ${\mathbb{R}}^{3,2}$. A unit timelike normal has orthogonal complement of signature $(3,1)$, so its stabilizer is again $SO(3,1)$. Appending this normal to a tangent frame realizes the oriented orthonormal frame bundle of anti-de Sitter spacetime asConnected components or orientation/time-orientation restrictions must be chosen consistently. This quotient is the anti-de Sitter quadric, which has a periodic timelike coordinate. If “anti-de Sitter spacetime” denotes its universal cover, pull the frame bundle back to that cover, rather than identify the covered spacetime with the unmodified quotient.

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.

Cosmic inflation addresses the horizon problem by extending the causal past of the observed region, the flatness problem by damping the curvature contribution, and the cosmological monopole problem by diluting unwanted relics. A canonical inflaton can drive this expansion when its potential energy density exceeds twice its kinetic energy density.

Use the unreduced Planck mass $m_P=G^{-1/2}$, consistent with the coefficient in the Friedmann equation, and take $V_0>0$, $c\ne 0$. Seek the exponential-potential power-law inflation solution $\varphi =A\log (\beta t)$ and $a=a_{\ast }(t/t_{\ast }{)}^p$. The kinetic energy density is proportional to $t^{-2}$. The inflaton potential must have the same scaling, soSubstitution into the inflaton equation of motion givesThe Friedmann equation now reduces toConsequently the nontrivial positive-potential solution isA positive $V_0$ requires $p>1/3$. Accelerating expansion occurs exactly for $p>1$, givingThis is an exact solution, not a use of the slow-roll approximation. The first Hubble slow-roll parameter and equation-of-state parameter are constant:If the reduced Planck mass $M=m_P/\sqrt{8\pi }$ is used instead, the dimensionless exponential slope is $\lambda =c/\sqrt{8\pi }$ and the exponent becomes $p=2/{\lambda }^2$; these are the same solution.

For $p>1$, the comoving Hubble radius decreases as $(aH{)}^{-1}\propto t^{1-p}$. A region initially in causal contact can grow beyond that radius before the hot radiation-dominated universe begins. In the ideal power law extrapolated all the way to $t=0$, the particle horizon integral ${\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$ diverges, so the solution has no finite past particle horizon. A finite, physically trustworthy inflationary interval instead requires a sufficiently large number of e-folds to enlarge an initially causal patch.

For a small curvature contribution, the Friedmann equation gives $|\Omega -1|=|k|/(a^2{H}^2)\propto t^{2-2p}\to 0$, solving the flatness problem. Conserved nonrelativistic magnetic monopoles have number density proportional to $a^{-3}$, while the inflaton energy density is proportional to $t^{-2}$; their energy-density ratio therefore falls as $t^{2-3p}$. This gives the required dilution, provided later reheating does not recreate them. Inflationary shear damping similarly suppresses anisotropic expansion.

The main failure is no automatic graceful exit or reheating. The first Hubble slow-roll parameter never reaches one, and the exponential inflaton potential has no finite minimum about which the field can oscillate. Additional dynamics are needed for a graceful exit from inflation and transfer to a hot radiation-dominated universe. The ideal solution also retains a curvature singularity at $t=0$; solving the horizon problem does not itself explain the beginning of the universe. For $c=0$, the constant inflaton potential instead permits constant-field de Sitter spacetime expansion; the logarithmic field solution is inapplicable.