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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 $p=2/{\lambda }^2$.

A graceful exit from inflation ends sustained accelerating expansion and connects it to a hot radiation-dominated universe. Crossing ${ϵ}_H=1$ ends acceleration in a homogeneous expanding model; reheating requires an additional mechanism to transfer inflaton energy into other species. Pure exponential-potential power-law inflation keeps ${ϵ}_H$ constant and its inflaton potential has no finite minimum, so neither an automatic endpoint nor oscillatory reheating follows from that potential alone.

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 $T={\xi }^0$ and ${\xi }^i={\partial }^i\lambda$, and define the physical background Hubble parameter by $H=\dot{a}/(\bar{N}a)$. At first order, the spacetime metric components are $\delta g_{00}=-2{\bar{N}}^2\Phi$, $\delta g_{0i}=a^2{B}_{,i}$ and $\delta g_{ij}=-2a^2(\Psi {\delta }_{ij}+E_{,ij})$. Apply the given passive cosmological gauge transformation component by component. The temporal component changes by $2\bar{N}\dot{\bar{N}}T+2{\bar{N}}^2\dot{T}$; the mixed component changes by ${\bar{N}}^2{T}_{,i}-a^2{\dot{\lambda }}_{,i}$; and the spatial component changes by $-2a\dot{a}T{\delta }_{ij}-2a^2{\lambda }_{,ij}$. HenceThese are the scalar gauge transformations with a background lapse in the paper's sign convention.

Starting in synchronous gauge, preserving both $\Phi =0$ and $B=0$ requiresThe first equation integrates to $\bar{N}T=C(\mathbf{x})$; substituting in the second and integrating givesThe arbitrary lower integration limit is absorbed into $D$. Thus specifying synchronous lapse and shift does not fix the origins of the freely falling clocks or all spatial coordinate labels. This is the residual gauge freedom in synchronous gauge.

During cosmic inflation, choosing these clocks remains arbitrary. For example, the $C$ mode shifts the spatial potential by $HC$ and the density perturbation by $3HC(\bar{\rho }+\bar{P})$. In the de Sitter approximation, $H$ varies little, so a nearly constant metric perturbation can contain a pure coordinate contribution. An inflaton fluctuation or density contrast in synchronous gauge therefore cannot be declared physical solely from its time behavior. Use a uniform-density curvature perturbation or an appropriate field-based gauge-invariant variable, or fix a physical initial slicing. The exact de Sitter case has no evolving homogeneous density clock, which makes density-defined slicing degenerate.

During the standard hot Big Bang, a cold-matter rest frame supplies a convenient additional condition: imposing zero cold-matter velocity removes the nontrivial spatially varying time-shift mode. Before that condition is imposed, the pure-gauge density contrast for constant $w$ is $3H(1+w)C$, which decays as $t^{-1}$ in either radiation or matter domination. In conformal time this is proportional to ${\tau }^{-2}$ during radiation domination and ${\tau }^{-3}$ during matter domination. It can be confused with an independent decaying physical mode unless gauge freedom is fixed. The $D$ mode is a time-independent spatial relabeling and must likewise be fixed by a coordinate or initial-metric convention.

Finally, a scalar density transforms by evaluating the homogeneous scalar at the shifted time:To reach Newtonian gauge from synchronous quantities, impose $\tilde{E}=\tilde{B}=0$. For the nonhomogeneous scalar modes, chooseUsing background stress-energy conservation, $\dot{\bar{\rho }}=-3\bar{N}H(\bar{\rho }+\bar{P})$, gives the synchronous-to-Newtonian density transformation with a lapse:The expression includes the lapse factors because dots refer to the arbitrary time coordinate, not automatically to proper time.

Write $S={\Phi }_0$, $u={\Phi }_{+}$, $v={\Phi }_{-}$, $M=M_p$, $P=uv-\zeta$, $x=|S{|}^2/M^2$ and $q=|u{|}^2+|v{|}^2$. Assume $g\ne 0$ and, initially, $\zeta \ne 0$. The supergravity F-term potential isThe canonical Kähler potential gives the identity Kähler metric. The three Kähler covariant derivatives of a superpotential areSubstituting, including the negative term rather than treating the potential as just a sum of squares, gives the exact answerHere $(1+x{)}^2-3x=1-x+x^2$ explains the first coefficient. It is strictly positive for real $x$, since $1-x+x^2=(x-1/2{)}^2+3/4$. Thus this model's scalar potential is nonnegative everywhere.

In the stated limit $q/M^2\ll 1$, the exponential can be replaced by $e^x[1+O(q/M^2)]$. The other displayed terms retain the effects of the Kähler covariant derivatives of a superpotential. Small waterfall fields alone do not justify discarding $\zeta /M^2$ in those derivatives, or expanding in $x$. If also $|\zeta |/M^2\ll 1$ and $x\ll 1$, the leading result reduces to the global F-term scalar potentialNo gauge sector or gauge coupling is supplied, so there is no specified D-term contribution to add.

The supersymmetric vacuum manifold of a bilinear hybrid superpotential isIndeed $W=0$ and all three Kähler covariant derivatives of a superpotential vanish there, so all supergravity auxiliary fields vanish and supersymmetry is unbroken. The nonnegative scalar potential makes this a global minimum. Conversely zero potential requires $uv=\zeta$, and then $|S{|}^{2}q=0$; for $\zeta \ne 0$ this forces $S=0$. With just the supplied fields this is a complex vacuum manifold before imposing any gauge quotient: the product is fixed but the ratio of $u$ and $v$ is not. More precisely, it is an F-flat vacuum manifold whether or not a gauge symmetry is specified. If one additionally gauges opposite charges, D-flatness typically imposes $|u|=|v|$, giving $|u|=|v|=\sqrt{|\zeta |}$ up to phases and gauge equivalence; that is an extra assumption, not a consequence of $K$ and $W$ alone.

The other familiar branch of F-term hybrid inflation has $u=v=0$. Its nonzero derivative $D_{S}W=-g\zeta (1+x)$ establishes supersymmetry breaking, and its exact scalar potential isThe canonical-Kähler correction to an F-term hybrid valley begins asIn particular the inflaton quadratic mass cancels at this order; the first lift of the global flat valley is quartic in $|S|$.

To identify when this branch is a transverse minimum, rephase the waterfall fields so that $\zeta$ is real positive. To quadratic order in $u,v$, the global scalar potential isThe normalized combinations $(u\pm \bar{v})/\sqrt{2}$ diagonalize this quadratic form, withThus $|S{|}^2>|\zeta |$ gives a valley of transverse minima, while below the threshold the negative eigenvalue produces the waterfall instability in F-term hybrid inflation.

The exact supergravity correction to this stability test is also available without taking $x$ small. Put $b=|\zeta |/M^2$. Expansion of the exact supergravity F-term potential about $u=v=0$ gives equal diagonal coefficients $|g{|}^2{M}^2{e}^x[x+b^2(1+x^2)]$ and off-diagonal magnitude $|g{|}^2{M}^2{e}^{x}b(1+x+x^2)$. HenceFor the usual sub-Planckian symmetry-breaking scale $0<b\ll 1$, the lower eigenvalue is $|g{|}^2{M}^2{e}^x(1-b)[x-b(1+x^2)]$. It is positive for the familiar sub-Planckian valley $b\ll x\ll 1$, and changes sign near $x=b$.

The printed assertion of two full minima requires qualification. The broken branch is a transverse inflationary valley, not generally a second stationary minimum of the complete supergravity potential. In factIts only stationary point in $S$ is the origin. For $0<|\zeta |<M^2$ that point has a negative waterfall eigenvalue $|g{|}^2(|\zeta {|}^2/M^2-|\zeta |)$. A direct counterexample is the path $S=0$, $u=v=a$ with real small $a$, after rephasing $\zeta >0$:The negative quadratic coefficient proves it is not a minimum. No loop corrections or additional stabilizing interactions were specified that could create the requested second stationary vacuum. The mathematically supported interpretation is a SUSY-breaking transverse valley and a zero-energy supersymmetric vacuum minimum manifold. The limit on $u,v$ alone does not fix the parameter range $|\zeta |/M^2$; the familiar waterfall conclusion uses the sub-Planckian range just stated.

In the corresponding global theory the exponential prefactor, Kähler corrections to $D_{i}W$ and universal $-3|W{|}^2/M^2$ term are absent. Its $u=v=0$ valley has exactly constant tree-level energy $|g\zeta {|}^2$ for all $S$, whereas canonical supergravity lifts it by the quartic correction above. The supersymmetric vacuum $S=0$, $uv=\zeta$ and its zero energy persist in both descriptions. If $\zeta =0$, the origin and the $S$ axis instead have vanishing F-terms; the broken-valley interpretation does not apply.

Denote the comoving box volume by $\mathcal{V}$, keeping it distinct from the inflaton potential $V(\varphi )$. For the Canonically rescaled de Sitter scalar mode $u_k=a\delta {\varphi }_k$, the equation in part (b) becomesDirect differentiation verifies the positive-frequency solutionFor $|k\tau |\gg 1$, $u_k$ tends to $e^{-ik\tau }/\sqrt{2k\mathcal{V}}$, the positive-frequency flat-space mode. It therefore selects the incoming Bunch-Davies vacuum. The Bunch-Davies mode normalization in a finite comoving volume is fixed by the Wronskian conditionAfter cosmological horizon exit, $|k\tau |\ll 1$, the mode tends to the time-independent value $iH/\sqrt{2k^3\mathcal{V}}$. The complex mode and its Hermitian conjugate together make the real quantum field.

A perturbation of the inflaton shifts its local clock: $\delta \varphi \simeq \dot{\varphi }\delta t$. Regions with different clock shifts reach the end of inflation at different times and hence undergo different amounts of expansion. The inflaton clock-shift origin of curvature perturbations gives a local expansion difference of magnitude $H\delta t=H\delta \varphi /\dot{\varphi }$. The sign depends on whether $\delta t$ denotes a clock advance or a delay to a fixed end value.

The comoving curvature perturbation has amplitude $\mathcal{R}\simeq H\delta \varphi /\dot{\varphi }$. For adiabatic evolution, superhorizon conservation of single-field comoving curvature carries this amplitude into the later growing density contrast, with the numerical conversion depending on the epoch and the density slice. Thus the approximate density-amplitude statement in the question follows; it does not identify the density contrast and $\mathcal{R}$ in every gauge.

The frozen scalar power per $d\log k$ is $(H/2\pi {)}^2$, so the slow-roll curvature power spectrum isConsequently $⟨{\mathcal{R}}^2⟩=\int (dk/k){\Delta }_{\mathcal{R}}^2(k)$, giving the requested approximate growing-density spectrum. Both $H$ and $\dot{\varphi }$ are evaluated separately when each Fourier mode undergoes cosmological horizon exit.

The inflaton has quantum fluctuations while their wavelengths are shorter than the Hubble radius. Choosing the Bunch-Davies vacuum fixes the short-wavelength positive-frequency modes. During near-de Sitter slow-roll inflation, expansion carries each mode through $k=aH$, leaving a field fluctuation of amplitude approximately $H/(2\pi )$ on spatially flat slices.

The inflaton clock-shift origin of curvature perturbations converts this field displacement into a local time shift of magnitude $|\delta \Phi /\dot{\Phi }|$, and hence into a comoving curvature perturbation of magnitude $H|\delta \Phi /\dot{\Phi }|$. For the single-field adiabatic growing mode on an attractor, the superhorizon conservation of single-field comoving curvature makes this perturbation approximately constant at $k\ll aH$. These vacuum-generated perturbations seed the later density fluctuations.

Use primes for conformal-time derivatives in this question, reserving $V_{,\varphi \varphi }$ for a potential derivative. The inflaton background obeys $(a^2{\bar{\varphi }}^{\prime }{)}^{\prime }+a^4{V}_{,\varphi }=0$, so the linear variation of its action vanishes up to boundary terms. With $\delta \varphi =f/a$, the conformal-time quadratic action for an inflaton perturbation is initiallyThe cross term $-\mathcal{H}(f^2{)}^{\prime }$ integrates to ${\mathcal{H}}^{\prime }{f}^2$. Since ${\mathcal{H}}^{\prime }+{\mathcal{H}}^2=a^{\prime \prime }/a$, this becomesVarying $f$ gives $f^{\prime \prime }-{\nabla }^{2}f+(a^2{V}_{,\varphi \varphi }-a^{\prime \prime }/a)f=0$. The symmetric Fourier transform convention used in the question turns $-{\nabla }^2$ into $k^2$. Dropping the small mass term under the stated slow-roll inflation assumption yieldsFor de Sitter spacetime, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The rescaled field has a canonical kinetic term; $f$ rather than $\delta \varphi$ is therefore the convenient oscillator variable.

The annihilation operators kill the vacuum, and their commutator givesTherefore the scale-invariant inflationary power spectrum has the superhorizon limitThe canonical mode grows as $1/|\tau |$ outside the Hubble radius, while the physical field perturbation $f/a$ freezes. A slowly varying $H$ produces an almost scale-invariant spectrum rather than exact scale invariance.

For a slowly rolling single field, fluctuations in the inflaton clock become the comoving curvature perturbation, $\mathcal{R}\simeq -H\delta \varphi /\dot{\bar{\varphi }}$ up to sign convention. Restoring the reduced Planck mass and using $ϵ={\dot{\bar{\varphi }}}^2/(2M_{\mathrm{pl}}^2{H}^2)$ givesThis conversion presumes nonzero background roll; a strictly constant test scalar in exact de Sitter spacetime does not by itself define a finite comoving curvature perturbation through this formula. The curvature perturbations seed the later density fluctuations.

When metric fluctuations and their Einstein-gravity normalization are restored, the two graviton polarizations have the same massless mode equation. The usual total primordial tensor power spectrum is ${\Delta }_t^2=2H^2/({\pi }^2{M}_{\mathrm{pl}}^2)$, so the single-field leading ratio is $r=16ϵ$. Tensor amplitudes therefore probe the inflationary energy scale, whereas scalar amplitudes also depend on the roll rate. These tensor statements explain the relevance of the mode result; they are not derived from a scalar action with metric perturbations omitted.

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.

A light inflaton behaves approximately as a scalar field in de Sitter space while a wavelength is well inside the Hubble radius. Its Bunch-Davies vacuum has quantum fluctuations. Cosmic inflation stretches each Fourier mode until $k=aH$, after which its physical wavelength exceeds the Hubble radius. The nearly constant growing field mode has a typical fluctuation per logarithmic wavenumber interval $\delta \varphi \sim H/(2\pi )$.

A field fluctuation changes the local position on the rolling background trajectory. Neighboring regions therefore reach the same field value, and the end of inflation, at slightly different times. A clock displacement of magnitude $|\delta \varphi /\dot{\varphi }|$ becomes a difference in local expansion of order $H|\delta \varphi /\dot{\varphi }|$. Equivalently, in a conventional sign choice the comoving curvature perturbation is related to the field fluctuation on spatially flat slices byChanging the sign convention for spatial curvature changes the sign of $\mathcal{R}$, but not its spectrum. This is the inflaton clock-shift origin of curvature perturbations. For a single-field slow-roll attractor there is no independent entropy mode. On a super-Hubble scale, gradient terms are negligible and the superhorizon conservation of single-field comoving curvature preserves the growing adiabatic mode, so fluctuations generated near exit persist as primordial curvature perturbations. This conservation requires the attractor and adiabatic assumptions; a freely chosen non-attractor background would not have the same conclusion.

Divide the primordial tensor power spectrum by the slow-roll curvature power spectrum:Differentiating the scalar-field Friedmann equation and using the inflaton equation gives $\dot{H}=-{\dot{\varphi }}^2/(2M_{\mathrm{pl}}^2)$. Hence $r=16{ϵ}_H\simeq 16ϵ$. For the quadratic model at 50 remaining e-folds,This is above the supplied upper bound, so the model fails the stated tensor constraint, despite its satisfactory scalar tilt and adjustable scalar amplitude. Calibrating $m$ does not reduce $r$, since the coupling cancels from the ratio.

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.