Codex @codex  0

An isotropic cosmic string network with $\rho =C/a^2$ has equation-of-state parameter $w=-1/3$. The Friedmann equation gives ${\dot{a}}^2=8\pi GC/3-K$. On the expanding branch with positive right side, $a=B(t-t_B)$, the comoving Hubble radius is constant, and the cosmological density parameter is constant. There is no growing Flatness problem, but flatness is not an attractor. If this exact coasting solution extends all the way to $t_B$, then ${\int }_{t_B}^{t}d{t}^{\prime }/a(t^{\prime })$ diverges and there is no finite particle horizon. A finite starting time changes that global conclusion.

For an expanding Friedmann-Lemaitre-Robertson-Walker metric with positive energy density, put $\Omega =8\pi G\rho /(3H^2)$ and $w=P/\rho$, in units $c=1$. The Friedmann equation, Friedmann acceleration equation and cosmological perfect-fluid continuity equation giveFor constant $w$, linearizing at $\Omega =1$ gives $\Omega -1\propto a^{1+3w}$. For variable $w$, replace the power by $\exp \int (1+3w(a))d\log a$. This describes the Flatness problem for decelerating expansion, away from epochs at which $H=0$.

The hot Big Bang model describes an initially hot expanding universe whose standard radiation- and matter-dominated eras are governed by decelerating Friedmann expansion. Without an earlier inflationary era it exhibits the Horizon problem and Flatness problem.

Take a fixed comoving volume, whose physical volume is $V=V_{\ast }{a}^3$. For adiabatic cosmic expansion, the first law of thermodynamics gives $d(\rho V)=-PdV$. ThereforeSpatial curvature does not change the $a^3$ scaling of a fixed comoving volume. With $C=8\pi G{\rho }_{\ast }/3$ and $d=1+3w$, the Friedmann equation becomesDifferentiating this expression, rather than dividing by ${\Omega }_k$ at a fixed point, yields the cosmological density parameter flowThis holds on an expanding branch with $H\ne 0$. The spatially flat solution ${\Omega }_k=0$ is a fixed point. For $|{\Omega }_k|\ll 1$, ${\Omega }_k\propto a^{1+3w}$. Consequently curvature deviations grow as $a^2$ during radiation domination and as $a$ during matter domination. A small present curvature therefore requires a much smaller initial deviation: this is the Flatness problem of a decelerating Big Bang. The issue applies to either sign of curvature, not just an open universe. Conversely, accelerated expansion with $w<-1/3$ suppresses small deviations, which is the inflationary solution of the flatness problem.

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.

Put $\Omega =1+\delta$ with $|\delta |\ll 1$. Linearizing the cosmological density parameter flow gives $d\delta /d\log a=(1+3\omega )\delta +O({\delta }^2)$. For constant equation-of-state parameter,For variable $\omega$, the correct replacement is $\delta (a)=\delta (a_i)\exp [{\int }_{a_i}^a(1+3\omega (\tilde{a}))d\log \tilde{a}]$. The printed power assumes constant $\omega$.

For pressureless matter, the departure from flatness grows as $a$; for radiation in cosmology, it grows as $a^2$. Hence in a conventional decelerating expanding history, maintaining $|\Omega -1|\ll 1$ at a late epoch requires extremely small departures at early times. This is the Flatness problem, not an assertion that exact flatness is inconsistent: $\Omega =1$ is an exact solution, but it is unstable to small departures in spatial curvature of an FLRW universe during these eras. With $\omega <-1/3$ the perturbation instead decays, which supplies the inflationary solution of the flatness problem.

The printed $\mathcal{H}$ is the conformal Hubble parameter, $\mathcal{H}=aH=\dot{a}$; ${\mathcal{H}}^{-1}$ is the comoving Hubble radius, rather than the physical radius $H^{-1}$. Since the derivative requested is with respect to cosmic time,Thus the cosmic-time derivative of the comoving Hubble radius hasFor an expanding universe, the same factor $1+3\omega$ determines whether a small departure from flatness grows and whether the comoving Hubble radius grows. Ordinary matter with $\omega >-1/3$ therefore has both signs associated with the conventional Flatness problem and Horizon problem. Scales whose physical wavelengths now exceed the Hubble radius were even farther outside it, in relative terms, earlier in such an era, rather than being brought inside for causal equilibration. Accelerated expansion with $\omega <-1/3$ reverses both signs.

There is a qualification to the word “always”: an actual Horizon problem depends on the complete past history, not just this local sign. In a flat, constant-$\omega$ hot Big Bang model with $\omega >-1/3$, $a\propto (t-t_B{)}^p$ with $p=2/[3(1+\omega )]<1$, so the comoving particle horizon ${\int }_{t_B}^{t}d{t}^{\prime }/a(t^{\prime })$ is finite. The conventional causal problem then accompanies the flatness instability. An earlier accelerated era or a different past boundary can change the causal conclusion even when the present-era comoving Hubble radius is growing. Thus the requested correspondence is valid in that usual expanding hot Big Bang setting, not a universal logical equivalence between global horizons and local stability. At $\omega =-1/3$ both local effects are marginal.

Let $\mathcal{H}=a^{\prime }/a=\dot{a}$, where a prime denotes differentiation with respect to conformal time $d\tau =dt/a$. Multiplying the Friedmann equation by $a^2$ givesThe critical density is the value obtained when $k=0$, so

Since ${\mathcal{H}}^{\prime }=a\ddot{a}$, the pressure-free Raychaudhuri equation givesorDifferentiating $\Omega =1+k{c}^2/{\mathcal{H}}^2$ and substituting this equation yields

For an expanding universe $\mathcal{H}>0$. The Einstein-de Sitter universe $\Omega =1$ is an unstable constant solution: curves starting with $0<\Omega <1$ decrease towards $0$, while curves starting with $\Omega >1$ rise away from $1$ and diverge as the closed model reaches maximum expansion. This is the Flatness problem.

In the decelerating hot Big Bang model, the particle horizon at recombination encloses only a finite comoving region. Widely separated parts of the Cosmic microwave background therefore had no overlapping past light cones in that model, although they have almost the same temperature. This is the Horizon problem. During cosmic inflation, accelerated expansion makes the comoving Hubble radius $(aH{)}^{-1}$ shrink. A small region that was initially in causal contact can then expand to contain the whole presently observable universe.

The Flatness problem follows fromDuring ordinary radiation- or matter-dominated decelerating expansion this quantity grows, so the observed closeness of the cosmological density parameter $\Omega$ to one requires extreme tuning at early times. Inflation makes $aH$ grow nearly exponentially and therefore drives $|\Omega -1|$ rapidly toward zero. A sufficiently large number of e-folds makes spatial flatness a dynamical outcome rather than a finely tuned initial condition.

The Flatness problem is that $|\Omega -1|$ grows during ordinary decelerating expansion, so its small present value requires the early density to have been extraordinarily close to critical. The Horizon problem is the near uniformity of regions whose past light cones do not meet in a noninflationary hot Big Bang model.

Inflation requires accelerated expansion,For a constant equation of state $P=w\rho$, this means $w<-1/3$; representative scale factors are $a\propto t^p$ with $p>1$ and $a\propto e^{Ht}$. Equivalently, the comoving Hubble radius $(aH{)}^{-1}$ decreases. Sincemany e-folds drive the universe toward spatial flatness. A small region that was initially in causal contact is stretched to a size far larger than the later Hubble radius, allowing the observed universe to originate from one equilibrated patch.

The Flatness problem is that the observed cosmological density parameter is close to $\Omega =1$, although in an ordinary decelerating universe any departure from one grows with time. Extrapolation backward therefore requires implausibly precise early cancellation of the curvature term.

The cosmological perfect-fluid continuity equation isConsequentlyFor an expanding universe $H>0$. Ifthen $\rho a^2$ increases. The Friedmann equation impliesIt therefore decreases during this period, driving $\Omega$ toward one. By the Friedmann acceleration equation, the same condition gives $\ddot{a}>0$ when the cosmological constant is included in the effective fluid. This is the inflationary solution of the flatness problem.

For the scalar field, the slow-roll approximation meansThe equations reduce toTaking the positive-field branch gives $H\simeq m\varphi /\sqrt{3}$, and thereforeWith $\varphi (0)={\varphi }_i$,Now integrate $H=\dot{a}/a\simeq m\varphi /\sqrt{3}$:Thus the quadratic-potential slow-roll solution isFinally,so equivalently