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For $V=C_p(\Phi /M_{\mathrm{Pl}}{)}^p$ and the leading endpoint estimate ${ϵ}_V=1$, the slow-roll approximation gives ${\Phi }_N^2/M_{\mathrm{Pl}}^2=2pN+p^2/2$ and ${ϵ}_V=p/(4N+p)$. Thus ${\mathcal{P}}_{\mathcal{R}}=C_p(2pN+p^2/2{)}^{p/2}(4N+p)/(24{\pi }^2{M}_{\mathrm{Pl}}^{4}p)$ and the scalar spectral index obeys $n_s-1=-2(p+2)/(4N+p)$. The coefficient sets the amplitude, whereas the monomial power and exit e-fold count set the leading tilt. Changing the power without specifying the coefficient leaves the overall normalization undetermined.

For $V=\lambda {\varphi }^n$ on $\varphi >0$, slow-roll approximation gives $ϵ\simeq n^2{M}_{\mathrm{Pl}}^2/(2{\varphi }^2)$ and $\eta \simeq 2n{M}_{\mathrm{Pl}}^2/{\varphi }^2$. The slow-roll e-fold count integrates to $N\simeq ({\varphi }_N^2-{\varphi }_e^2)/(2n{M}_{\mathrm{Pl}}^2)$. For $n=2,3,4$, the loose order-one conditions $ϵ\le 1$ and $|\eta |\le 1$ imply ${\varphi }^2\ge 2n{M}_{\mathrm{Pl}}^2$ in this approximation. Genuine slow roll requires a much stronger hierarchy. A prescribed endpoint ${\varphi }_e^2=2n{M}_{\mathrm{Pl}}^2$ gives ${\varphi }_N^2=2n(N+1)M_{\mathrm{Pl}}^2$ andThe actual end criterion for accelerated expansion is $ϵ=1$, whose leading potential estimate instead gives ${\varphi }_e^2\simeq n^2{M}_{\mathrm{Pl}}^2/2$. The prescribed endpoint is a convention, not an exact slow-roll or inflation boundary.

Write the reduced Planck mass as $M=(8\pi G{)}^{-1/2}=m_{\mathrm{pl}}/\sqrt{8\pi }$, and use natural units. Near the flat maximum, the quartic hilltop inflation trajectory has $V\simeq \alpha {\sigma }^4$ and $V^{\prime }\simeq -12\alpha {\varphi }^3$. The slow-roll approximation gives $H^2\simeq V/(3M^2)$ and $3H\dot{\varphi }\simeq -V^{\prime }$. Therefore the number of e-folds isThe lower field value dominates this integral. ThusThis is far below $\sigma$ for the assumed small symmetry-breaking scale. Strict slow roll actually fails before the minimum: $|M^2{V}^{\prime \prime }/V|\sim 1$ at ${\varphi }^2\sim {\sigma }^4/(36M^2)$. The last approach to the minimum is not described by slow roll. Using that earlier upper limit changes the dominant fifty-e-fold estimate by an order-one number of e-folds, not its leading scale dependence. No extrapolation of slow roll all the way to the minimum is needed.

A comoving wavenumber crosses the inflationary Hubble scale at $k=aH$. Consequently $d\ln k=d\ln a+d\ln H\simeq Hdt$ when $H$ changes slowly. Combining with the slow-roll approximation givesBoth $k$ and $\varphi$ increase along the rolling branch. If $N$ denotes the remaining number of e-folds rather than elapsed expansion, then $d\ln k\simeq -dN$. More precisely $d\ln k=(1-{ϵ}_H)d\ln a$; the correction is beyond the constant-Hubble approximation used here.

Write $V=V_0{e}^{-A\varphi /m_{\mathrm{pl}}}$, where the stated model has $V_0=m_{\mathrm{pl}}^4$, and take $A>0$. Slow roll givesThus $d[e^{A\varphi /(2m_{\mathrm{pl}})}]/dt=A^2\sqrt{V_0}/(\sqrt{96\pi }{m}_{\mathrm{pl}})$. Absorb the integration constant into $b$ to obtainIt follows that $H=16\pi /[A^2(t+b)]$, soThe exponential potential has constant ${ϵ}_V=A^2/(16\pi )$ and ${\eta }_V=A^2/(8\pi )$. Inflation requires $p>1$, whereas reliable slow-roll approximation requires $A^2/(8\pi )\ll 1$. This is the slow-roll normalization for exponential inflation.

The stated field normalization is a slow-roll result, not an exact solution of the original full equations. Keeping the acceleration and kinetic terms yields the same $p$ but $C^2=A^4{V}_0/[2m_{\mathrm{pl}}^2(48\pi -A^2)]$, as follows by substituting $\dot{\varphi }=2m_{\mathrm{pl}}/[A(t+b)]$ into both full equations. It agrees to leading order for small slope and is the exponential-potential power-law inflation solution.

There is no minimum about which this scalar can undergo the damped oscillations normally used for reheating. Its slow-roll parameters are constant, so an inflating exponential solution has no automatic exit. An independent stopping and energy-transfer mechanism is needed.

At the assumed exit, $V_R=m_{\mathrm{pl}}^4{e}^{-30}$. Equating this to the thermal radiation density givesfor $g_{\ast }=10^3$. Corrections from the small kinetic fraction do not change this rough estimate.

Curvature obeys $|\Omega -1|\propto (aH{)}^{-2}$. For constant ${ϵ}_H=A^2/(16\pi )$ and $N=\log (a_R/a_i)$, its suppression during cosmic inflation is $e^{-2(1-{ϵ}_H)N}$. Starting from order-one curvature and requiring present curvature no larger than order one givesAfter instantaneous reheating, entropy conservation implies $a_R/a_0=(T_0/T_R)(g_{\ast s,0}/g_{\ast s,R}{)}^{1/3}$, with $H_R\simeq 1.66\sqrt{g_{\ast }}{T}_R^2/m_{\mathrm{pl}}$. Present scales make the numerator roughly $60$; using the literal approximate units from question2 gives about $63$. A tighter present curvature tolerance adds its corresponding half-logarithm. In slow-roll approximation this is the conventional requirement of roughly sixty e-folds.

Because the field increases toward ${\varphi }_R$, enough expansion requiresThis is the flatness bound on the initial field in exponential inflation. It is an upper bound on the initial field, not a nonzero minimum: starting earlier at a smaller field produces more expansion. If one additionally restricts the initial potential below $m_{\mathrm{pl}}^4$, then ${\varphi }_i\ge 0$ supplies a lower bound, but that is a separate restriction. Without initial-curvature and final-tolerance choices there is no unique numerical initial-field value.

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.

Now use proper-time dots and $H=\dot{a}/a$. Differentiate the slow-roll relation $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$, then substitute $\dot{\bar{\Phi }}\simeq -V_{,\Phi }/(3H)$:Thus the Hubble slow-roll parameter ${ϵ}_H$ agrees with ${ϵ}_V$ to this order. Since ${\mathcal{P}}_{\psi }\propto H_k^2$ and $k=aH$ at horizon exit,Consequently the tensor spectral index isThe final equality is the requested first-order slow-roll approximation; corrections beyond that order also affect the horizon-exit spectrum itself.

The quantum harmonic oscillator has $\hat{H}=({\dot{\hat{q}}}^2+{\omega }^2{\hat{q}}^2)/2$, with $\omega >0$. Using $\hat{a}|0⟩=0$ and the canonical commutation relation givesFor $q=r{e}^{is}$, the Wronskian normalization isConsequentlyEquality requires $\dot{r}=0$ and $r^2=1/(2\omega )$. The minimum-energy normalized oscillator mode is thereforeThe constant phase $\alpha$ is irrelevant. This mode also solves the oscillator equation of motion; an arbitrary squeezed mode would have larger vacuum energy.

For inflation, introduce the Mukhanov-Sasaki variable $v=z\mathcal{R}$. Since $z$ depends only on conformal time,Integrating the cross term by parts gives the canonical bulk actionup to the boundary term $-\frac{1}{2}\int d^{3}x[(z^{\prime }/z)v^2{]}_{\mathrm{boundary}}$. The resulting Euler-Lagrange field equation and its Fourier transform areAt leading order in the slow-roll approximation, retain a small positive, nearly constant $ϵ$ in $z=a\sqrt{2ϵ}$, while approximating $a=-1/(H\tau )$ and $H$ as constant. Then $z^{\prime \prime }/z\simeq 2/{\tau }^2$, soStrict exact de Sitter spacetime would have $ϵ=0$ and would not supply the nonzero curvature kinetic coefficient assumed here; the calculation is the leading quasi-de Sitter limit, not a substitution of zero into $z$.

For $-k\tau \gg 1$, the expansion correction is negligible and each canonical mode is a quantum harmonic oscillator of conformal frequency $k$. The Bunch-Davies vacuum selects its positive-frequency, minimum-energy mode $e^{-ik\tau }/\sqrt{2k}$ in that early subhorizon regime. It does not minimize an instantaneous Hamiltonian after the effective squared frequency has become negative outside the horizon.

A basis of exact solutions to the leading mode equation is $e^{-ik\tau }(1-i/(k\tau ))$ and its complex conjugate. Write the normalized Bogoliubov transformation combination asThe Bunch-Davies vacuum boundary condition sets ${\beta }_k=0$ and ${\alpha }_k=1$ up to phase. HenceSubstitution verifies the equation, and $v_k{v}_k^{\ast \prime }-v_k^{\prime }{v}_k^{\ast }=i$ verifies the Wronskian normalization.

Dividing by $z$ gives the comoving curvature perturbation varianceThis is the dimensional slow-roll curvature power spectrum in the printed normalization, with the reduced Planck mass set to one. The corresponding dimensionless cosmological power spectrum is $k^3{P}_{\mathcal{R}}/(2{\pi }^2)=H^2/(8{\pi }^2ϵ)$, which is independent of $k$ at this order. Restoring the reduced Planck mass divides both power expressions by $M_{\mathrm{Pl}}^2$. Slowly varying background quantities are evaluated near each mode's horizon exit; their variation generates the small departure from exact scale invariance.

The slow-roll approximation neglects $\ddot{\varphi }$ relative to $3H\dot{\varphi }$ and kinetic energy relative to the potential. It is self-consistent on the attractor when the potential slow-roll parameter $ϵ\ll 1$ and the second potential slow-roll parameter $|\eta |\ll 1$. ThusHere, as in the inflaton equations, $M_{\mathrm{pl}}$ denotes the reduced Planck mass, not the unreduced mass used in the thermal calculation. The slow-roll curvature power spectrum becomesso

To differentiate with respect to horizon-exit scale, write $\mathcal{N}=\ln a$, increasing with physical time. Along the slow-roll trajectory,The difference between differentiating with respect to $\ln k$ and $\mathcal{N}$ contributes only at second slow-roll order to the tilt. Since $\ln {\Delta }_{\mathcal{R}}^2=3\ln V-2\ln |V^{\prime }|+\mathrm{constant}$,Therefore the scalar spectral index in potential slow-roll parameters isAll background quantities in these expressions are evaluated when the particular mode exits the Hubble radius.

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.

Distinguish the slow-roll approximation parameter $ϵ$ from the infinitesimal positive contour regulator $\epsilon$. In the interaction picture, the ordered unequal-time Wick contraction isA differentiated vertex field instead supplies $u_k^{\ast \prime }(\tau )$. These conjugates are fixed by the annihilation operators and creation operators and cannot be dropped while retaining the same contour.

Convert the interaction time integral to conformal time. Since $dt=ad\tau$ and each cosmic-time derivative is $a^{-1}{\partial }_{\tau }$, the integrated cubic vertex isThe in-in formalism supplies $-2i$ multiplying the expectation of the Hamiltonian, so its negative sign gives $+2i$ multiplying this vertex. At nonzero external momenta, the connected contractions join each of the three external fields to one vertex field. There are six bijections, not three. Choosing which external field meets the undifferentiated vertex gives three cyclic choices; swapping the two differentiated fields gives a further factor of two. This is the Wick-pairing multiplicity for a zeta zeta-prime-squared vertex.

For example, before the internal momenta are integrated, one cyclic contribution includesThe factors of $2\pi$ cancel to leave one overall $(2\pi {)}^3$ and the external momentum delta. Thus the properly normalized reduced expression isThe vacuum prescription for inflationary in-in integrals damps the early-time oscillations and selects the Bunch-Davies vacuum. Internal self-contractions correspond to disconnected zero-momentum tadpole terms and are excluded from this connected three-point function.

The printed intermediate expression has unconjugated modes and only three literal cyclic terms with coefficient $-2i$. The conjugate form of the result above would use $-4i$, together with the conjugated contour. A literal $-2i$ with only three terms gives half the final printed answer. The six-contraction expression above is the consistent reduction of the supplied Hamiltonian and leads to that final answer.

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.

For the monomial inflation potential on $\varphi >0$, $V_{,\varphi }/V=n/\varphi$ and $V_{,\varphi \varphi }/V=n(n-1)/{\varphi }^2$. Thus, to leading slow-roll approximation,For $n=2,3,4$, $2n\ge n^2/2$. Therefore the loose order-one requirements $ϵ\le 1$ and $|\eta |\le 1$ giveThis is the interpretation under which the printed threshold follows. It is not a sufficient criterion for genuine slow roll, which requires much smaller parameters and hence ${\varphi }^2/M_{\mathrm{Pl}}^2\gg 2n$ for these fixed powers. Nor is its equality the actual end of acceleration for $n=2,3$: there $ϵ\simeq n/4<1$. For $n=4$ the leading estimate has $ϵ\simeq 1$. The potential-based order-one criterion $|{\eta }_V|<1$ also differs from the Hubble-flow criterion, especially at $n=4$. We use the endpoint the paper explicitly stipulates, rather than treating this boundary as a controlled slow-roll approximation.