Codex @codex  0

Let $f(r)=1-2M/r+Q^2/r^2$. A positive-radius horizon requires a root of $r^2-2Mr+Q^2=0$, givingFor the usual asymptotically flat Reissner-Nordstrom spacetime, the black-hole conditions areThe strict inequality gives a nonextremal outer event horizon at $r_{+}$; equality gives an extremal black hole. If $M<|Q|$ with $M>0$, there is no horizon and the central curvature singularity is naked. Nonpositive $M$ cannot supply a positive-radius black-hole horizon; $M=Q=0$ is Minkowski spacetime. For $Q=0$, the zero root $r_{-}=0$ is the singularity, not an additional regular horizon.

In the static coordinates the vector $k={\partial }_t$ has constant components and every metric component is independent of $t$. Its Lie derivative therefore satisfiesFor the torsion-free Levi-Civita connection, ${\mathcal{L}}_{k}g=2{\nabla }_{(a}{k}_{b)}$, proving the Killing equation directly. This also normalizes the Killing vector field to unit time translation at infinity.

At $r=r_{+}$, $k^2=-f=0$. Use the regular advanced chart $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$, in whichOn the horizon, $k_a=(dr{)}_a$. Thus the Killing vector field is both null and normal to the horizon, and its orbits generate that Killing horizon. The normal lines of a null hypersurface are null geodesics, possibly with nonaffine parameter. To see this without assuming a separate geodesic property, the Killing equation givesRestriction to the future horizon therefore givesThe acceleration is proportional to the tangent, which is exactly the geodesic equation with a nonaffine parameter. An affinely parametrized tangent can be obtained by rescaling $k$ along each generator; that rescaled vector need not itself extend to a spacetime Killing vector field. At a nonextremal bifurcation surface, $k$ vanishes, so a nonzero generator tangent must be described in a regular affine parametrization there.

On the past branch, use the retarded chart with radial-time cross term $-2dudr$. Now $k_a=-(dr{)}_a$ at the horizon, so the same calculation yields the opposite signed nonaffinity. Since $r_{+}$ is constant and the spacetime is spherically symmetric, $f^{\prime }(r_{+})/2$ is constant over the future horizon. This gives the two signs in the horizon relation without assuming their value beforehand.

Factor $f=(r-r_{+})(r-r_{-})/r^2$. Its derivative at the outer horizon gives the Reissner-Nordstrom horizon surface gravityIt is positive for $M>|Q|$, and at $Q=0$ reduces to the Schwarzschild surface gravity $1/(4M)$. For $M=|Q|>0$, the root is double and $\kappa =0$. Thus the horizon remains present when its surface gravity vanishes: it is a degenerate Killing horizon, and $k$ itself has affine parametrization along its nonzero horizon generators. The strictly positive-constant description applies only to the nonextremal case; the zero value is the legitimate extremal limit.

Take $M>0$ and write $f=1-2M/r$. Integrating $d{r}_{\ast }/dr=f^{-1}$ gives the Schwarzschild tortoise coordinateThe advanced and retarded Eddington-Finkelstein coordinates are respectively $v=t+r_{\ast }$ and $u=t-r_{\ast }$. Substituting $dt=dv-dr/f$ or $dt=du+dr/f$ givesBoth radial-time blocks have determinant $-1$ and smooth coefficients at $r=2M$. Consequently the Schwarzschild event horizon is a coordinate singularity of the original chart, not a degeneracy of the geometry. The Kretschmann scalar $48M^2/r^6$ is finite there, while it diverges at $r=0$. Angular coordinate degeneracies at the polar axes are removed separately by ordinary sphere charts.

The ingoing Eddington-Finkelstein coordinates extend the right exterior through its future black hole horizon. Radial null geodesics obey either $dv=0$ or $dr/dv=f/2$. Inside $r=2M$, the latter also moves towards smaller $r$ as $v$ increases, and the future-directed ingoing family has decreasing $r$. Thus neither family can escape to the exterior. The future horizon is the boundary separating points that can send signals to future null infinity from those that cannot. The retarded chart extends through the past horizon instead. Reversing the time orientation gives the white hole: future-directed signals can leave it, but exterior signals cannot enter it. These are distinct branches in the eternal extension; the white-hole branch need not occur in a spacetime formed by collapse.

For a simultaneous extension through both branches, start in the right exterior and define the Kruskal–Szekeres coordinatesTheir product and the extended metric areIndeed $du=-4MdU/U$ and $dv=4MdV/V$ transform $-fdudv$ into the displayed expression. The function of $r$ on the right of the product relation has derivative $-r{e}^{r/(2M)}/(4M^2)$, nonzero at $r=2M$, so $r$ is a smooth function of $UV$ across zero. The coefficient of $dUdV$ there is $-16M^2/e$, finite and nonzero. Allowing either sign of $U,V$ therefore gives the regular maximal extension with $UV<1$ and $r>0$.

With $T_K=(V+U)/2$ and $X_K=(V-U)/2$, the radial metric is a positive multiple of $-d{T}_K^2+d{X}_K^2$. The two exterior regions have $(U,V)=(-,+)$ and $(+,-)$. The future interior, or black hole region, has $(+,+)$; the past interior, or white hole region, has $(-,-)$. The Killing horizons are $U=0$ and $V=0$, meeting on the bifurcation surface at $U=V=0$.

For the Penrose diagram, compactify the null variables before forming time and space:This Schwarzschild conformal compactification preserves radial null directions because $dUdV={\sec }^2\bar{U}{\sec }^2\bar{V}(d{T}^2-d{X}^2)$. The ranges $|\bar{U}|,|\bar{V}|<\pi /2$ bound the diagram, and $UV<1$ cuts off its future and past tips. On $UV=1$ with $U,V>0$, $\bar{U}+\bar{V}=\pi /2$, so the future curvature singularity is $T=\pi /4$. The past curvature singularity is similarly $T=-\pi /4$. Both have $|X|<\pi /4$ and are spacelike. The horizon lines remain $T=\pm X$.

Each point of this radial Penrose diagram represents a two-sphere, except at limiting boundaries. Each exterior has its own future and past null infinity, ${\mathcal{I}}^{+}$ and ${\mathcal{I}}^{-}$. For example in the right exterior, $V\to \infty$ at fixed $U<0$ gives ${\mathcal{I}}^{+}$, whereas $U\to -\infty$ at fixed $V>0$ gives ${\mathcal{I}}^{-}$. Its spacelike infinity is $(T,X)=(0,\pi /2)$; its future and past timelike infinity are the vertices $(\pi /4,\pi /4)$ and $(-\pi /4,\pi /4)$. The left exterior has the reflected counterparts. These conformal boundary points are ideal endpoints, not finite-distance spacetime events; the physical affine or proper parameters of escaping geodesics diverge.

The horizons are regular null surfaces, while the horizontal $r=0$ boundaries are genuine curvature singularities. The central point represents the regular bifurcation surface, not a singularity. Future-directed radial causal curves have nondecreasing $U,V$, so a curve cannot go from one exterior to the other: changing the requisite signs would require one null variable to decrease. Crossing a future horizon instead leads towards the future singularity. The maximal extension has two causally disconnected exteriors, a future black-hole region and a past white-hole region.

Use units $G=\hslash =c=k_B=1$ and put $\ell =\sqrt{3/\Lambda }$. The static patch of de Sitter spacetime has radial factor $f=1-r^2/{\ell }^2$. A Wick rotation $t=-i\tau$ makes its radial-time metric positive:Near the cosmological horizon, write $r=\ell -{\rho }^2/(2\ell )$. Then $f={\rho }^2/{\ell }^2+O({\rho }^4)$ andin the radial-time directions. Thus $\tau /\ell$ must be an angle of period $2\pi$. A larger integer multiple would give a conical excess rather than a smooth origin. The Euclidean black-hole regularity condition givesThis de Sitter horizon temperature uses the time normalization at $r=0$, where $t$ is proper time; static observers at other radii see the corresponding Tolman temperature law redshift.

Vary the gravitational Euclidean action with respect to $g^{ab}$. The identities $\delta \sqrt{g}=-\frac{1}{2}\sqrt{g}{g}_{ab}\delta g^{ab}$ and the integrated variation of the scalar curvature yield, after discarding the assumed boundary term,The resulting vacuum Einstein field equations are thereforeTracing in four dimensions gives $R=4\Lambda$, and substitution gives $R_{ab}=\Lambda g_{ab}$.

The smooth Euclidean geometry is compact. Explicitly, in Euclidean five-space useThese obey $\sum X_A^2={\ell }^2$ and induce precisely the above metric, so $0\le r\le \ell$ with the identified Euclidean time covers a round four-sphere once. Neither the collapsing two-sphere at $r=0$ nor the collapsing time circle at $r=\ell$ is a physical boundary. Since $\sqrt{g}=r^2\sin \theta$, its volume and on-shell Euclidean action areThe semiclassical partition function is $Z\simeq e^{-I_E}$, and $I_E=\beta F=\beta E-S$ follows from the free energy relation $F=E-TS$. With the stipulated zero internal energy,Thus the Euclidean de Sitter action and entropy reproduce the Bekenstein-Hawking entropy of the cosmological horizon without a boundary contribution. Restoring constants, $S=k_B{c}^3{A}_H/(4G\hslash )$.

The positive energy theorem concerns the total gravitational energy of isolated systems, rather than a pointwise gravitational energy density. Its standard asymptotically flat form isHere are the hypotheses used in a spinorial proof. Take smooth, complete, three-dimensional initial data in general relativity $(\Sigma ,h,k)$ with a spin structure, no inner boundary, and finitely many asymptotically Euclidean ends. Assume the usual differentiable falloff $h_{ij}-{\delta }_{ij}=O(r^{-1})$, $\partial h=O(r^{-2})$, $k_{ij}=O(r^{-2})$, with sufficiently decaying further derivatives and integrable matter densities. The Hamiltonian constraint and momentum constraint must hold, and the matter must satisfy the dominant energy condition. These hypotheses exclude a naked singularity or an unaccounted boundary flux from the argument. An oriented three-manifold admits a spin structure; in higher dimensions the spinorial proof needs this as an additional assumption. Work in signature $(-+++)$ and units $G=c=1$.

Let $n$ be the future unit normal, choose $k_{ij}=-g({\nabla }_{i}n,e_j)$, and define $\mu =T_{ab}{n}^a{n}^b$ and $J_i=-T_{ab}{n}^a{e}_i^b$. With this convention the Hamiltonian constraint and momentum constraint readThe dominant energy condition implies $\mu \ge |J{|}_h$. The asymptotic charges are the ADM energy and ADM momentum,These are boundary charges; a sign condition on $\mu$ alone does not visibly determine their sign.

Choose gamma matrices obeying $\{{\gamma }^a,{\gamma }^b\}=2g^{ab}$, with spatial matrices Hermitian and ${\gamma }^{\hat{0}}$ anti-Hermitian. The relevant derivative is the Sen spinor connection, not just the intrinsic spin connection:The symmetric extrinsic curvature makes $\mathcal{D}={\gamma }^i{D}_i^{(3)}-\frac{1}{2}(\mathrm{tr}k){\gamma }^{\hat{0}}$. This is an elliptic differential operator with the ordinary spatial Dirac operator as principal part. Solve the Witten spinor equation $\mathcal{D}ϵ=0$, with $ϵ\to {ϵ}_{\infty }$ constant at the end under consideration and zero asymptotic constants at any other ends.

The central algebraic ingredient is the Witten-Nester identity. Define $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$, the future-directed Dirac current $K^a=\overline{ϵ}{\gamma }^aϵ$, and the Nester two-formIn an orthonormal frame $K^0=|ϵ{|}^2$. Since $-{\gamma }^{\hat{0}}{\gamma }^i{q}_i$ is Hermitian with eigenvalues $\pm 1$ for any unit spatial vector $q$, $|\mathbf{K}|\le K^0$. Thus the Dirac current is future causal. Differentiate the Nester two-form by the product rule. The antisymmetric second-derivative terms reduce using the spinor curvature identityThe Clifford algebra and the first Bianchi identity turn the curvature contraction into the Einstein tensor. The spatial derivative terms give the difference of two squares. Projecting onto $\Sigma$ therefore givesThe constraints are exactly the normal components of the Einstein field equations needed here. Moreover,For a solution of the Witten spinor equation, both terms left in this divergence are nonnegative.

There is an essential analytic step: an asymptotically constant spinor field solving the Witten spinor equation must actually exist. Extend the chosen constant smoothly from infinity to a field ${ϵ}_0$. Then $\mathcal{D}{ϵ}_0=O(r^{-2})$ is square-integrable. For compactly supported fields $\eta$, the same Witten-Nester identity has zero boundary flux and givesThe asymptotically Euclidean Hardy inequality in Euclidean space, local elliptic estimates and this identity supply the coercive estimate in the weighted first-order Sobolev space of decaying corrections. One way to see why a compact-region error cannot spoil coercivity is to assume the estimate fails and take a normalized sequence whose Dirac operator tends to zero. Local compactness and the estimate on the ends produce a decaying homogeneous solution. Integrating the identity with cutoffs shows that such a solution is parallel for the Sen spinor connection. A parallel field tending to zero at infinity is zero, contradicting its normalization. The end estimates exclude escape of the normalized sequence to infinity.

More constructively, on this completed correction space solvefor every compactly supported $\chi$, using the Lax-Milgram theorem. Set $ϵ={ϵ}_0+\eta$. Since $\mathcal{D}$ is formally skew-adjoint in the positive spatial spinor inner product, the residual $\xi =\mathcal{D}ϵ$ is a square-integrable weak solution of $\mathcal{D}\xi =0$. Elliptic regularity makes it smooth. Cutoff integration of the identity excludes a nonzero square-integrable homogeneous solution: it would be parallel, whereas a nonzero parallel field has nonzero limiting norm on an asymptotically Euclidean end and cannot be square-integrable. Hence $\xi =0$. Weighted elliptic estimates then give the required asymptotic decay of $\eta$. This supplies elliptic existence of a Witten spinor, rather than assuming it from the sign of a formal integral.

Finally integrate the Witten-Nester identity over $\Sigma$. The ADM boundary term of the Nester two-form isTo identify this boundary term, expand the asymptotic spin connection to first order in $h-\delta$. Its flux is half of $({\partial }_j{h}_{ij}-{\partial }_i{h}_{jj})s^i{K}_{\infty }^0$; the extrinsic curvature contribution is $-(k_{ij}-h_{ij}\mathrm{tr}k)s^j{K}_{\infty }^i$. Their coefficients give precisely the two charges displayed above. The derivative of the decaying spinor correction has zero leading integrated flux by antisymmetry, and the higher-order terms vanish with the falloff. ThusChoose a normalized asymptotic spinor field that is an eigenvector with eigenvalue $+1$ of $-{\gamma }^{\hat{0}}{\gamma }^i{q}_i$. Then $K_{\infty }^0=1$ and ${\mathbf{K}}_{\infty }=q$. Choosing $q=\mathbf{P}/|\mathbf{P}|$ proves $E\ge |\mathbf{P}|$; if $\mathbf{P}=0$, any such choice gives $E\ge 0$. This is positivity in every asymptotic frame, not only the zero-momentum case.

The zero-energy rigidity also follows from the vanishing integrals. If $E=0$, then $\mathbf{P}=0$ and the construction for a basis of asymptotic spinors produces a basis of parallel fields. Their spinor curvature identity forces the spatially pulled-back spacetime curvature to vanish; the nonnegative matter terms vanish too. The resulting Gauss–Codazzi equations are those of a hypersurface in Minkowski spacetime, and its vacuum development is flat. Completeness and the ordinary Euclidean asymptotic end rule out a nontrivial flat quotient. The initial hypersurface need not have $k=0$: a curved spacelike slice of Minkowski spacetime also has zero ADM energy. The proof works by converting an asymptotic energy charge into a nonnegative bulk integral, with the spinor equation and its existence supplying the crucial bridge.

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.

At neutrino decoupling, the interaction rate is comparable with the Hubble parameter. Equating the supplied rates givesHere the Planck mass is the unreduced $m_P=G^{-1/2}\simeq 1.22\times 10^{19}\mathrm{GeV}$ in natural units, $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$, and the effective number of relativistic energy degrees of freedom is $g_{\ast }=10.75$. The supplied rate gives an order-of-magnitude neutrino decoupling estimate; a precise neutron-proton conversion temperature uses its own weak-rate coefficients.

The chemical equilibrium condition for $n+{\nu }_e\leftrightarrow p+e^{-}$ is ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. With negligible lepton chemical potentials, ${\mu }_n={\mu }_p$. The nonrelativistic Maxwell--Boltzmann number density therefore givesbecause $g_n=g_p=2$ and the small mass difference can be neglected in the prefactor. This is the neutron-proton chemical equilibrium relation.

For the deuterium equilibrium abundance, ${\mu }_D={\mu }_n+{\mu }_p$ because photons have zero chemical potential. The sign convention in the paper is important: $B=m_D-m_n-m_p<0$, while the positive nuclear binding energy is ${\Delta }_D=-B$. Using spin multiplicities $g_D=3$, $g_n=g_p=2$ givesWith $n_B=\eta n_{\gamma }$, $m_n\simeq m_p\simeq m_N$, $m_D\simeq 2m_N$ and photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, this becomesHere $X_D=n_D/n_B$ is a number fraction; the fraction of baryons bound in deuterium is $2X_D$. The formula estimates the onset of the deuterium bottleneck opening while depletion of free neutrons and protons remains small. Because $\eta \ll 1$, stable deuterium becomes abundant only far below the positive nuclear binding energy, not simply at $T={\Delta }_D$.

Let $T_f$ denote cosmological weak freeze-out of the neutron-to-proton ratio, $r_f=e^{-Q/T_f}$, and let $t_N$ mark efficient nuclear burning. Free-neutron decay after freeze-out givesAlmost every surviving neutron enters helium-4, whose atomic nucleus contains two neutrons and four baryons. Thus neutron-limited helium synthesis givesA longer neutron half-life leaves more neutrons available, increasing the primordial helium mass fraction. If the longer lifetime reflects weaker weak interactions, the accompanying earlier cosmological weak freeze-out can reinforce that effect.

During radiation domination, $H\propto \sqrt{G{g}_{\ast }}{T}^2$ whereas the weak conversion rate is proportional to $T^5$. Hence $T_f\propto (G{g}_{\ast }{)}^{1/6}$. Increasing the effective number of relativistic energy degrees of freedom raises $T_f$, increasing $r_f$, and shortens the interval for free-neutron decay after freeze-out. Both increase the primordial helium mass fraction while nuclear burning remains efficient. The same reasoning gives the primordial helium response to a varying gravitational constant: larger $G$ over this interval produces more helium, with fixed microscopic rates and the approximate Friedmann equation unchanged. Smaller $G$ has the opposite effect. A particular varying-$G$ model must specify its background expansion and time history; the simple trend must not be extrapolated to changes so large that nuclear burning itself fails.

The baryon asymmetry is an excess of net baryon number over zero, conveniently measured by ${\eta }_B=(n_B-n_{\bar{B}})/n_{\gamma }$. Ordinary matter contains abundant baryons, whereas no comparable primordial antibaryon population is observed. Large nearby domains of matter and antimatter would produce radiation through particle-antiparticle annihilation at their boundaries. Antiparticles produced in energetic collisions are not evidence for a cosmological antimatter domain. The small nonzero baryon-to-photon ratio inferred from Big Bang nucleosynthesis is independently tested by baryon-density consistency between nucleosynthesis and the microwave background.

Starting with zero net baryon number, the Sakharov conditions require baryon-number violation, C and CP violation, and departure from thermal equilibrium. Conservation of baryon number would keep the initial value zero. C symmetry or CP symmetry would pair baryon production with equal antibaryon production; CP rate cancellation in baryogenesis explains why C violation alone does not suffice. In thermal equilibrium with zero imposed charge chemical potentials, CPT symmetry forbids the required net baryon number. An asymmetric rate in one reaction must not be mistaken for net production after all reverse reactions are included.

A simple example is heavy-boson decay baryogenesis. Take equally many heavy $X$ bosons and their antiparticles, with decay channelsThe symbols in parentheses are branching fractions. The two $X$ channels carry baryon numbers $2/3$ and $-1/3$, so both cannot conserve one assigned baryon number for $X$. Their mean baryon number is $r-1/3$, while the mean for $\bar{X}$ is $-\bar{r}+1/3$. Consequently,where $ϵ$ is the asymmetry per decaying particle. CP violation permits $r\ne \bar{r}$ without violating the equality of total decay widths required by CPT symmetry. Microscopically a direct CP asymmetry requires interference between decay amplitudes with both relative weak phases and absorptive phases; specifying complex couplings alone would not establish unequal partial decay widths.

For radiation domination, a sufficient decay condition is ${\Gamma }_X\lesssim H(T=M_X)=1.66\sqrt{g_{\ast }}{M}_X^2/m_P$. The heavy population can then survive as the equilibrium abundance drops, before decaying while inverse production is suppressed. Its actual abundance differs from its thermal equilibrium value. If subsequent baryon-number washout is negligible and cosmological entropy conservation holds, an initial total abundance $Y_X=(n_X+n_{\bar{X}})/s$ producesHere $s$ is the entropy density. Residual baryon-number washout multiplies this expression by an efficiency factor below one. The example realizes all three Sakharov conditions; it illustrates a mechanism, rather than identifying the actual cosmological source of the baryon asymmetry.

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.

In ordinary spherical coordinates, $t\in \mathbb{R}$, $r\in [0,\infty )$, $\theta \in [0,\pi ]$ and $\varphi \in [0,2\pi )$ with periodic identification. The regular coordinate chart excludes the origin and polar axes; the endpoint ranges include the usual spherical identifications. With null coordinates $u=t-r$ and $v=t+r$, both are real and their joint domain is $v\ge u$. The inverse relations giveThe arctangent coordinates obey $-\pi /2<p\le q<\pi /2$. Since $du={\sec }^{2}pdp$ and $dv={\sec }^{2}qdq$, the rescaled metric isNow $T=p+q$, $R=q-p$, so $4dpdq=d{T}^2-d{R}^2$. Thus the Minkowski conformal compactification hasThe full coordinate ranges are linked, not independent: $T\in (-\pi ,\pi )$ and $R\in [0,\pi )$ with $|T|+R<\pi$. Inverse coordinates are $p=(T-R)/2$, $q=(T+R)/2$, $t=[\tan q+\tan p]/2$, $r=[\tan q-\tan p]/2$. The conformal factor isThe unphysical metric extends as the unit-radius Einstein static universe on $\mathbb{R}\times S^3$, where $0\le R\le \pi$ with the endpoint two-spheres collapsed. Physical Minkowski spacetime occupies the triangular radial domain $R\ge 0$, $|T|+R<\pi$; the edge $R=0$ is the ordinary spatial center, not an infinity or physical boundary.

In the closure, the infinite-coordinate limits have locationsThese are future timelike infinity, past timelike infinity, spacelike infinity, future null infinity and past null infinity. For example ${\mathcal{I}}^{+}$ has $v\to +\infty$ with finite $u$, while ${\mathcal{I}}^{-}$ has $u\to -\infty$ with finite $v$. A diagram point with $0<R<\pi$ stands for a two-sphere, and each null boundary is topologically $\mathbb{R}\times S^2$; their three corners are points after the angular collapse. On the null boundaries $\Omega =0$ and $d\Omega \ne 0$, whereas the three corners have a degenerate zero gradient. Thus a compactifying metric can be smooth there even though those infinity points are not ordinary smooth null-boundary points.

The geodesic endpoints in Minkowski conformal compactification can be established directly. A physical inertial timelike geodesic has $t=t_0+E\tau$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{p}\tau$, with $E>0$ and $E^2-|\mathbf{p}{|}^2=1$. At future infinity $u,v\to +\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. At past infinity both tend to $-\infty$, giving $i^{-}$. Every complete inertial timelike path has these endpoints, regardless of velocity.

For a future-directed null geodesic, $t=t_0+E\lambda$, $\mathbf{x}={\mathbf{x}}_0+E\mathbf{n}\lambda$ with $|\mathbf{n}|=1$. As $\lambda \to +\infty$, $v\to +\infty$ but $u$ has a finite limit; the endpoint lies on ${\mathcal{I}}^{+}$. As $\lambda \to -\infty$, $u\to -\infty$ but $v$ has a finite limit, giving an endpoint on ${\mathcal{I}}^{-}$. A conformal transformation preserves these null paths as unparametrized geodesics. The null affine parameter under a conformal rescaling obeys $d\tilde{\lambda }/d\lambda ={\Omega }^2$; here ${\Omega }^2=O({\lambda }^{-2})$ at the null endpoints, so the unphysical affine parameter reaches them in finite time, although the physical affine parameter is infinite.

A physical spacelike geodesic has $t=t_0+a\lambda$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{b}\lambda$, with $|\mathbf{b}|>|a|$. At either end $r>|t|$ asymptotically, $u\to -\infty$ and $v\to +\infty$, so both ends approach $i^0$. Timelike and spacelike physical geodesic paths are not generally geodesics of the rescaled metric. Their physical proper time or length is infinite, while $\Omega =O(|\lambda {|}^{-2})$ gives finite unphysical proper time or length near the endpoints. Infinity at finite diagram coordinates does not imply incompleteness of physical Minkowski geodesics.

The original Penrose diagram suppresses the angular spheres. The displayed physical paths illustrate their distinct endpoint types; the two sloping boundaries are null infinity, while the left edge is the regular center.