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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 a homogeneous canonical scalar field, the cosmological perfect-fluid continuity equation reduces to $\dot{\varphi }(\ddot{\varphi }+3H\dot{\varphi }+V^{\prime })=0$. It implies the inflaton equation of motion wherever $\dot{\varphi }\ne 0$, and by continuity at isolated turning points. It cannot by itself exclude a frozen field on a time interval. Indeed, choose $\varphi ={\varphi }_0$ with $V({\varphi }_0)>0$ but $V^{\prime }({\varphi }_0)\ne 0$; a de Sitter scale factor in flat slicing with $H^2=V({\varphi }_0)/(3M_{\mathrm{Pl}}^2)$ satisfies the homogeneous Friedmann equation and Friedmann acceleration equation, but not the scalar equation. Varying the inflaton action in an expanding universe supplies the missing equation also for a frozen field, requiring $V^{\prime }({\varphi }_0)=0$.

The inflaton potential is the potential energy density of an inflaton as a function of its field value. Its derivative enters the inflaton equation of motion, while its slope and curvature enter the potential slow-roll parameters.

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.

For a homogeneous canonical inflaton, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires positive potential energy to dominate both kinetic energy and any competing material components. Then $P_{\varphi }\simeq -{\rho }_{\varphi }$ and the expansion is nearly vacuum-like. Exact scalar-driven acceleration requires ${\rho }_{\varphi }+3P_{\varphi }<0$, equivalently ${\dot{\varphi }}^2<V$; potential domination is a stronger condition.

The slow-roll approximation additionally neglects $\ddot{\varphi }$ compared with $3H\dot{\varphi }$ in the inflaton equation of motion. Define the unreduced Planck mass $m_{\mathrm{pl}}=G^{-1/2}$; it differs from the reduced Planck mass by $\sqrt{8\pi }$. Sufficient potential-flatness conditions aretogether with approach to the slow-roll attractor. The reduced equations are $H^2\simeq 8\pi V/(3m_{\mathrm{pl}}^2)$ and $3H\dot{\varphi }\simeq -V^{\prime }$.

For $V=m^2{\varphi }^2/2$, take $m>0$ and the positive-field rolling branch. ThenThe field velocity is constant to this order. Choosing the start at $t=0$ gives the quadratic-potential slow-roll solutionThe PDF puts $\pi$ inside this square root; the converted TeX incorrectly moves it outside. Next integrate $d\log a/d\varphi =H/\dot{\varphi }=-4\pi \varphi /m_{\mathrm{pl}}^2$:Thus the number of e-folds is the logarithmic expansion, not the expansion factor itself:Here ${ϵ}_V={\eta }_V=m_{\mathrm{pl}}^2/(4\pi {\varphi }^2)$. The usual quadratic-inflation endpoint estimate sets ${ϵ}_V\simeq 1$, givingThis is an endpoint estimate, not an exact result from the full equations. Exact acceleration ends at the Hubble slow-roll parameter ${ϵ}_H=1$, equivalently ${\dot{\varphi }}^2=V$. Inserting the slow-roll velocity into that last condition instead gives $\varphi \simeq m_{\mathrm{pl}}/\sqrt{6\pi }$; the different order-one coefficient illustrates the failing approximation at the endpoint. A precise value requires the full scalar and Friedmann evolution. The large initial-field contribution to $N$ is unaffected by that order-one uncertainty.

Finally $V({\varphi }_i)\sim m_{\mathrm{pl}}^4$ gives ${\varphi }_i^2\sim 2m_{\mathrm{pl}}^4/m^2$. The quadratic-inflation expansion from Planck density is thereforeFor $m\ll m_{\mathrm{pl}}$ this is enormous and the initial field is in the slow-roll regime. No unique numerical factor follows without specifying $m/m_{\mathrm{pl}}$. Extrapolating the classical model from Planck density is only a formal estimate; it does not establish control of quantum-gravity corrections at that starting scale.

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.

Use signature $(+---)$, the positive spatial metric ${\gamma }_{ij}$, and a future-directed unit normal $n_{\mu }=(N,0)$, so $n^{\mu }=N^{-1}(1,-N^i)$. This fixes the sign of the normal to agree with the specified $\Pi$. The negative normal covector printed in the source would instead give $-\Pi$; its orientation and the displayed derivative cannot both be retained. Use the operational negative-expansion convention for the extrinsic curvature of a spatial hypersurface, $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, which is the convention giving the requested expanding-universe sign. With this future normal and signature it equals $n_{i;j}$; the source's $-n_{i;j}$ identity instead uses its past-pointing normal.

For the scalar-field matter projections, put $s_i=D_i\varphi$ and $s^2={\gamma }^{ij}{s}_i{s}_j$. Decomposing the scalar gradient into normal and tangent parts gives ${\partial }_{\mu }\varphi {\partial }^{\mu }\varphi ={\Pi }^2-s^2$. Contracting the Klein-Gordon scalar stress-energy tensor with the normal and the spatial projection tensor therefore yieldsIn particular, the positive spatial metric raises the $i,j$ indices in $s^2$; raising them with the four-metric would reverse that spatial sign. These quantities are the energy density, momentum density and spatial stress in the normal frame.

For the homogeneous Spatially flat FLRW metric, $N_i=0$, ${\gamma }_{ij}=a^2{\delta }_{ij}$ and $s_i=0$. Consequently $\Pi =\dot{\bar{\varphi }}/\bar{N}$, $J_i=0$, $\bar{\rho }={\Pi }^2/2+V$, $\bar{P}={\Pi }^2/2-V$, andThe intrinsic Ricci scalar is zero and $K^2-K_{ij}{K}^{ij}=9H^2-3H^2=6H^2$. The Hamiltonian constraint thus gives the scalar-field Friedmann equation:The momentum constraint is identically satisfied, since its homogeneous spatial derivatives and $J_i$ vanish. The scalar equation reduces to ${\bar{N}}^{-1}\dot{\Pi }+3H\Pi +V_{,\varphi }=0$. Substituting $\Pi$ and multiplying by ${\bar{N}}^2$ gives the background inflaton equation of motion:No derivation of the supplied gravitational constraints is needed.

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

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

The first Hubble slow-roll parameter measures the fractional change of $H$ per number of e-folds, and the second Hubble slow-roll parameter measures the fractional change of the first parameter. Since $d\log a/dt=H$,For a canonical inflaton, $ϵ={\dot{\varphi }}^2/(2M_{\mathrm{Pl}}^2{H}^2)\ge 0$. Expanding cosmic inflation means $\ddot{a}>0$, equivalently $\boxed{ϵ<1}$ because $\ddot{a}/a=H^2(1-ϵ)$. Small $\eta$ is not required for accelerated expansion itself, but helps maintain the slow-roll approximation over many e-folds. Its logarithmic definition assumes $ϵ>0$; in exact de Sitter expansion it is undefined unless interpreted as a limit.

For conventional slow-roll inflation, assume ${\dot{\varphi }}^2\ll V$, $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, a smooth positive scalar potential, and slowly varying small parameters. The Friedmann equation and inflaton equation of motion then reduce toSubstitute these in the exact expression for $ϵ$:To obtain the potential and Hubble slow-roll parameter relation, differentiate $\log {ϵ}_V$ along the slow-roll trajectory $\dot{\varphi }/H\simeq -M_{\mathrm{Pl}}^2{V}_{,\varphi }/V$:HenceThe second quantity on the right is the second potential slow-roll parameter, not the logarithmic Hubble-flow $\eta$ itself. Consistency usually requires ${ϵ}_V\ll 1$ and $|{\eta }_V|\ll 1$; identities involving division by $V_{,\varphi }$ are used on rolling intervals and interpreted by limiting values at stationary points.

In conformal time, $\sqrt{-g}=a^4$ and $g^{\mu \nu }=a^{-2}\mathrm{diag}(-1,1,1,1)$. The inflaton action in an expanding universe becomesVary $\varphi$ with compact support, integrate the derivative terms by parts, and discard the boundary term. The Euler-Lagrange field equation isDividing by $a^2$ provesThe factor two is the conformal-time form of Hubble friction. Using $d\tau =dt/a$ converts it to the usual inflaton equation of motion $\ddot{\varphi }+3H\dot{\varphi }-a^{-2}{\nabla }^2\varphi +V_{,\varphi }=0$, providing a sign and normalization check.

Subtract the homogeneous background equation and linearize the inflaton equation of motion. The field perturbation satisfiesWith $f=a\delta \varphi$, direct differentiation cancels the first-derivative term and givesThe conformal-time quadratic action for an inflaton perturbation, after a boundary term is discarded, isFor the specified de Sitter spacetime background, $a^{\prime \prime }/a=2/{\tau }^2=2a^2{H}^2$. The light-field approximation $|V_{,\varphi \varphi }|\ll 2H^2$ therefore permits dropping the mass term. Taking a spatial Fourier transform yields the Canonically rescaled de Sitter scalar mode equationIts kinetic term is canonical, with momentum ${\pi }_f=f^{\prime }$. Each independent real Fourier component consequently has the canonical coordinate/momentum algebra of a quantum harmonic oscillator, with time-dependent squared frequency ${\omega }_k^2=k^2-a^{\prime \prime }/a$. This is why canonical quantization of a real scalar field proceeds by oscillator creation and annihilation operators. On superhorizon scales the squared frequency is negative: the system is then an inverted time-dependent oscillator, not a stationary oscillator with a positive frequency. The canonical commutation relations remain valid.

Since $H=\dot{a}/a$, differentiation gives the Hubble parameter identitySubtracting the given Friedmann equation from the Raychaudhuri equation therefore yieldsOn the other hand, differentiating the Friedmann equation givesSubstitute $\dot{H}=-4\pi G{\dot{\varphi }}^2$ and divide by $(8\pi G/3)\dot{\varphi }$, which is permitted by the assumption $\dot{\varphi }\ne 0$. The result is the homogeneous inflaton equation of motionThe middle term is Hubble friction.

For $V=m^2{\varphi }^2/2$ and unreduced Planck mass $m_P=G^{-1/2}$, the potential slow-roll parameter is ${ϵ}_V=m_P^2(V^{\prime }/V{)}^2/(16\pi )=m_P^2/(4\pi {\varphi }^2)$. Its conventional value one gives the displayed endpoint estimate. Exact accelerated expansion ends at ${ϵ}_H=1$, or ${\dot{\varphi }}^2=V$ for a canonical scalar alone. Substitution of slow-roll formulas into this exact condition gives a different order-one coefficient because slow roll is failing there; a precise endpoint requires integrating the full inflaton equation of motion and Friedmann equation. The large-field e-fold estimate is insensitive to this order-one endpoint uncertainty.

For $V=\lambda {\varphi }^4/4$ and unreduced Planck mass $M_P$, the potential slow-roll parameter is ${ϵ}_V=M_P^2/(\pi {\varphi }^2)$. Setting it to one gives the displayed conventional endpoint. The exact end of scalar-dominated accelerating expansion is ${ϵ}_H=1$, equivalently ${\dot{\varphi }}^2=V$. Since slow-roll is failing near that point, a precise endpoint requires evolving the full inflaton equation of motion; potential and kinetic criteria evaluated with a failing approximation need not give the same order-one coefficient.