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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 a flat universe with a single constant equation-of-state parameter $w>-1$, $H(z)=H_0(1+z{)}^{\alpha }$. Its angular diameter distance is $H_0^{-1}[(1+z{)}^{-\alpha }-(1+z{)}^{-1}]/(1-\alpha )$, with logarithmic limit $H_0^{-1}\log (1+z)/(1+z)$ at $\alpha =1$. Differentiation gives the displayed unique maximum, interpreted as $z_m=e-1$ at $\alpha =1$. The distance vanishes at both endpoints $z=0$ and $z\to \infty$. As $w\downarrow -1$, the maximum moves to infinite cosmological redshift.

For a perfect fluid in general relativity with constant equation-of-state parameter $w$, use conformal time $dt=ad\tau$ and the conformal Hubble parameter $\mathcal{H}=a^{\prime }/a=aH$. The Friedmann acceleration equation gives ${\mathcal{H}}^{\prime }=a\ddot{a}=-4\pi G{a}^2(1+3w)\rho /3$. The Friedmann equation eliminates $\rho$ and gives this Riccati equation. For positive energy density, $\alpha \ne 0$ and appropriately shifted conformal time, its usual branches are $\cot (\alpha \tau )$ for $k=1$, $1/(\alpha \tau )$ for $k=0$, and $\coth (\alpha \tau )$ for $k=-1$. Their scale factors are proportional to ${\sin }^{1/\alpha }(\alpha \tau )$, $(\alpha \tau {)}^{1/\alpha }$ and ${\sinh }^{1/\alpha }(\alpha \tau )$ respectively on intervals where the bases are positive. At $\alpha =0$ the conformal Hubble parameter is constant and the separate solution is $a=A{e}^{h\tau }$.

In natural units, conservation of the stress-energy tensor for a homogeneous perfect fluid in an FLRW metric gives the displayed equation for its energy density and pressure. It is equivalent to $d(\rho a^3)=-Pd(a^3)$. For a constant equation-of-state parameter $w=P/\rho$, separation gives $\rho \propto a^{-3(1+w)}$. If $\rho$ instead denotes mass density, the pressure term is $P/c^2$.

This is local stress-energy conservation; it does not in general define a globally conserved total energy for an expanding universe. The adiabatic work in a comoving volume makes this distinction explicit.

An isotropic, noninteracting network of domain walls with energy density proportional to $a^{-1}$ has effective equation-of-state parameter $w=-2/3$. This follows from constant-equation-of-state density scaling $\rho \propto a^{-3(1+w)}$.

Dark energy is a smooth component with sufficiently negative pressure to drive accelerated cosmic expansion. The supernova discovery of accelerating expansion measures an expansion history through luminosity distance; interpreting that history as a component of the stress tensor additionally assumes general relativity. A cosmological constant is one possibility, whereas dynamical dark energy permits a changing equation-of-state parameter.

Use natural units, the unreduced Planck mass $m_P=G^{-1/2}$, $V_0>0$ and $c\ne 0$. In a flat universe dominated by a canonical inflaton, $\varphi =(2m_P/c)\log (\beta t)$ and $a\propto t^p$ solve the inflaton equation of motion and Friedmann equation exactly when $p=16\pi /c^2$ and ${\beta }^2=V_0{c}^2/[2m_P^2(3p-1)]$. The positive-potential scaling solution requires $p>1/3$, while cosmic inflation requires $p>1$. Its constant equation-of-state parameter is $w=-1+2/(3p)$. In terms of the reduced Planck mass $M=m_P/\sqrt{8\pi }$, the exponential slope is $\lambda =c/\sqrt{8\pi }$ and the same exponent is $p=2/{\lambda }^2$.

For $V=V_0{e}^{-\lambda \phi /M_{\mathrm{Pl}}}$, $\lambda \ne 0$, the displayed ansatz satisfies the flat Friedmann equation and scalar field equation of motion when $V_0{t}_0^2=2M_{\mathrm{Pl}}^2(6-{\lambda }^2)/{\lambda }^4$. Its equation-of-state parameter is $w=-1+{\lambda }^2/3$ and its deceleration parameter is $q={\lambda }^2/2-1$. It gives exponential-potential power-law inflation only for $0<{\lambda }^2<2$; the same algebraic family includes decelerating expansion outside that range when its potential normalization is real.

For positive constant cosmological constant, zero cosmological spatial curvature and a separate positive-density fluid with constant equation-of-state parameter $w>-1$, set $H_{\Lambda }=\sqrt{\Lambda /3}$ in the time-unit convention. Eliminating the fluid density gives $\dot{H}=-3(1+w)(H^2-H_{\Lambda }^2)/2$. On the expanding Big Bang branch,The origin is the Big Bang. The fluid's early-time power-law expansion approaches exponential vacuum domination at late times.

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.

Throughout this question use natural units with $c=1$. For zero potential and nonzero kinetic energy, the homogeneous canonical scalar field has $P=\rho ={\dot{\phi }}^2/2$, so $\boxed{w=1}$. Its equation of motion becomes $d(a^3\dot{\phi })/dt=0$, giving $\dot{\phi }\propto a^{-3}$ and thereforeThe same scaling follows from cosmological energy conservation with constant equation-of-state parameter $w=1$. If the field is exactly constant, both $P$ and $\rho$ vanish and their ratio is undefined; that is the separate empty solution.

For nonrelativistic particles the energy density is dominated by rest energy, while their isotropic kinetic pressure is approximately $p_m=nm⟨v^2⟩/3$. ThereforeThe matter pressure is negligible in this approximation, including in the gravitational combination ${\rho }_m+3p_m/c^2$. For isotropic radiation in cosmology, $E=pc$ implies one third of the energy density in each diagonal spatial stress:

Because the two fluids do not exchange energy or momentum, they are separately conserved cosmological fluids: ${\nabla }_{\alpha }{T}_i^{\alpha \beta }=0$ for each, rather than merely conservation of their sum. In a comoving volume proportional to $a^3$, the first law of thermodynamics givesEquivalently this follows by projecting the individual stress-energy conservation equation along the common comoving four-velocity. Differentiating and dividing by ${\rho }_i{c}^2{a}^3$ yieldsFor constant equation-of-state parameter $w_i$,The extra power for radiation in cosmology represents the cosmological redshift of photon energies in addition to number dilution. Differentiating the total Friedmann equation gives only the total continuity equation; the separate equations also use the stated absence of interactions between components.

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.

The Friedmann acceleration equation gives the gravitational deceleration from a uniform pressureless matter density, plus the acceleration from a cosmological constant. Here $\rho$ is the matter mass density, $G$ is the gravitational constant, $H=\dot{a}/a$ is the Hubble parameter, and $k$ is the constant spatial-curvature coefficient in the chosen normalization of $a$. The Friedmann equation is the first integral relating expansion to density, curvature and vacuum energy. With $c=1$, the matter, curvature and cosmological-constant contributions to $H^2$ are $8\pi G\rho /3$, $-k/a^2$ and $\Lambda /3$. The cosmological constant has effective density $\Lambda /(8\pi G)$ and equation-of-state parameter $w=-1$; its negative pressure explains its positive acceleration contribution when $\Lambda >0$.

For the Newtonian derivation with $\Lambda =0$, follow a test mass on the edge of a uniform sphere of physical radius $R=a\chi$. Conservation of the enclosed matter gives $M=4\pi \rho R^3/3$ constant. The shell theorem removes the force of external spherical shells and givesIntegrating the force equation once gives the conserved specific mechanical energyDivide by $R^2/2=a^2{\chi }^2/2$ to obtainHomologous motion makes $\epsilon /{\chi }^2$ constant between shells. Newtonian mechanics supplies this energy integration constant; its interpretation as intrinsic spatial curvature belongs to the relativistic FRW metric.

An outward acceleration $\Lambda \mathbf{r}/3$ comes from the Newtonian potential per unit masssince $-\nabla {\Phi }_{\Lambda }=\Lambda \mathbf{r}/3$. With conventional dimensional $\Lambda$, restore $c^2$ in this potential and the corresponding acceleration and Friedmann terms.

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 $w<-1/3$.

Use natural units and the unreduced Planck mass $M_P=G^{-1/2}$; it is $\sqrt{8\pi }$ times the reduced Planck mass. A homogeneous canonical inflaton has ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires ${\dot{\varphi }}_i^2/2\ll V({\varphi }_i)$, with other matter and the curvature term negligible in the Friedmann equation. The equation-of-state parameter is then near $-1$, and the Friedmann acceleration equation gives accelerating expansion.

For sustained slow-roll inflation, Hubble friction must dominate the inflaton acceleration: $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, with the scalar potential changing only slightly in one Hubble time. In terms of the potential slow-roll parameter and second potential slow-roll parameter, this requiresFor the quartic scalar potential,Thus $|{\varphi }_i|\gg M_P$ supplies a broad slow-roll range when $\lambda \ll 1$. At an initial density of order the Planck density, potential domination givesso both potential slow-roll parameters are small for small $\lambda$. Small initial kinetic energy is a separate condition; the height of the scalar potential alone does not enforce it. These classical equations describe the post-Planck evolution under the model assumptions, rather than resolving quantum gravity at the initial endpoint.

Take ${\varphi }_i>0$; the negative branch follows by the symmetry $\varphi \mapsto -\varphi$. Dropping kinetic energy from the Friedmann equation and acceleration from the inflaton equation of motion givesWriting $b=M_P\sqrt{\lambda /(6\pi )}$, the quartic-potential slow-roll solution isIndeed, $\dot{\varphi }\simeq -b\varphi$, and integrating $H=\dot{a}/a$ gives $\log (a/a_i)=\pi ({\varphi }_i^2-{\varphi }^2)/M_P^2$. These expressions apply during slow-roll inflation; their formal late-time saturation must not be used beyond the end of that approximation.

The exact end of accelerating expansion is ${ϵ}_H=-\dot{H}/H^2=1$, equivalently ${\dot{\varphi }}^2=V$ for the scalar-dominated flat model. The quartic-inflation endpoint estimate uses the leading potential condition ${ϵ}_V\simeq 1$, givingThe numerical coefficient is an endpoint estimate: the slow-roll approximation breaks down there. Imposing $|{\eta }_V|=1$ estimates when slow roll loses accuracy and gives a different order-one coefficient, not a different exact acceleration condition.

The slow-roll e-fold count isThe quartic-inflation expansion from Planck density combines this initial estimate with the conventional endpoint to giveFor small $\lambda$ the dominant term is enormous; the order-one endpoint uncertainty does not affect that conclusion.

To estimate the reheating temperature, assume prompt, efficient reheating converts the end-of-inflation energy into a thermal radiation bath. With the same endpoint convention, $V_e\simeq \lambda M_P^4/(4{\pi }^2)$, and equating this to the radiation energy density givesThe quartic-inflation prompt-reheating estimate fixes the scaling; the endpoint kinetic energy changes the coefficient by order one. Without an inflaton coupling and its thermalization history, $V$ alone cannot specify the actual reheating temperature.

Finally, during cosmic inflation the comoving Hubble radius decreases:An initially causally connected patch is enlarged enough that the present observable region can lie within it. Its portions may subsequently be far outside one another's Hubble radii while retaining their common earlier thermal history. This solves the horizon problem if the number of e-folds is sufficiently large, conventionally of order sixty for the relevant thermal history; $N\sim 2\pi /\sqrt{\lambda }$ can readily exceed that. The necessary count depends on reheating and later expansion. The causal mechanism and the conversion back to a hot universe are discussed in David Tong's account of inflation.

With constant equation-of-state parameter $w$, the cosmological perfect-fluid continuity equation becomes $\dot{\rho }+3(1+w)H\rho =0$. ConsequentlyThe constant-equation-of-state density scaling includes $\rho \propto a^{-3}$ for pressureless matter and $\rho \propto a^{-4}$ for radiation in cosmology.

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.

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.

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.