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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.

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.

The slow-roll equations turn the slow-roll curvature power spectrum intoThe spectral tilt is the logarithmic derivative of this dimensionless spectrum. Apply the horizon-exit logarithmic derivative in slow roll:With the potential slow-roll parameter $ϵ$ above and the second potential slow-roll parameter $\eta =V_{,\varphi \varphi }/V$,This is the first-order departure of the scalar spectral index from unity, in reduced Planck units.

The slow-roll equations give $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$ and $\dot{\Phi }\simeq -V^{\prime }/(3H)$. Substituting in the stated slow-roll curvature power spectrum yieldsWith the potential slow-roll parameter ${ϵ}_V=M_{\mathrm{Pl}}^2{V}^{\prime 2}/(2V^2)$,All quantities are evaluated at horizon exit. The two forms have the same normalization because $(\sqrt{8\pi }{)}^4=64{\pi }^2$.

At horizon exit $k=aH$, $d\log k/dt=H(1-{ϵ}_H)$, where ${ϵ}_H=-\dot{H}/H^2$ is the Hubble slow-roll parameter. To leading slow-roll order,The slow-roll curvature power spectrum is proportional to $V^3/V^{\prime 2}$. Therefore the scalar spectral index in potential slow-roll parameters isUsing the potential slow-roll parameter and the second potential slow-roll parameter givesThe neglected horizon-exit derivative correction contributes at higher slow-roll order.

Apply the horizon-exit derivative from part (d) to the potential slow-roll parameter and the second potential slow-roll parameter. The potential slow-roll flow equations arewhere the third potential slow-roll parameter is ${\xi }_V^2=M_{\mathrm{Pl}}^4{V}^{\prime }{V}^{\prime \prime \prime }/V^2$. For example, differentiating ${\eta }_V$ gives $-M_{\mathrm{Pl}}^4(V^{\prime }/V)(V^{\prime \prime \prime }/V-V^{\prime \prime }{V}^{\prime }/V^2)$, which is the second expression. Differentiating $n_s-1\simeq -6{ϵ}_V+2{\eta }_V$ now gives the running of the scalar spectral indexThis is the leading second-order slow-roll running. The notation ${\xi }_V^2$ denotes a signed derivative combination and does not force its value to be nonnegative.

Derive both cases from $V=C_p(\Phi /M_{\mathrm{Pl}}{)}^p$, using the positive-field branch. The potential slow-roll parameter and second potential slow-roll parameter areTaking the leading inflation-end estimate ${ϵ}_V=1$, the remaining number of e-folds isThe corresponding monomial-inflation curvature amplitude and tilt areIn particular,The quartic case is redder. Its small-$k$ matter envelope is $k^{0.9508}$ instead of $k^{0.9669}$; its high-$k$ envelope is $k^{-3.0492}{\ln }^{2}k$ instead of $k^{-3.0331}{\ln }^{2}k$. With the late cosmology fixed, the turnover and baryonic acoustic scale do not move. If the spectra are matched in amplitude at a pivot $k_{\ast }$, their ratio is approximatelyThis gives more large-scale power and less small-scale power relative to the pivot, with about a $7\%$ relative tilt across two decades in wavenumber.

The quartic coefficient is not specified, so changing the power of the potential does not by itself specify the overall amplitude. If $C_2=V_0$ and the quartic coefficient is $C_4$, the formula above gives, at the same $N$,Consequently choosing the same dimensionful coefficient would give a very different normalization, whereas choosing $C_4/V_0\simeq 1/496$ matches the pivot amplitude. These are different normalization choices, not a universal amplitude prediction from a quartic shape alone.

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.

The identityshows that accelerating expansion has $H>0$ and $ϵ<1$, whereas accelerated contraction has $H<0$, $\ddot{a}<0$, and $ϵ>1$.

For the monomial inflation potential $V=\lambda {\varphi }^p$, the two potential slow-roll parameters areThus ${ϵ}_V<1$ requires $|\varphi |>p{M}_{\mathrm{Pl}}/\sqrt{2}$, while $|{\eta }_V|<1$ requires $|\varphi |>\sqrt{p|p-1|}{M}_{\mathrm{Pl}}$.

At ${\varphi }_{\ast }$, the specified velocity gives kinetic energyHence ${\rho }_{\varphi }=3V$, $P_{\varphi }=V$, and the first Hubble slow-roll parameter isThe universe is therefore decelerating, despite the flatness of the potential: the chosen initial kinetic energy violates the slow-roll approximation.

This is the potential of Starobinsky inflation. Put $b=\sqrt{2/3}/M_{\mathrm{Pl}}$ and $y=e^{-b\varphi }$, so $V=V_0(1-y{)}^2$. Its potential slow-roll parameter and second potential slow-roll parameter areAt large positive $\varphi$, $y\ll 1$, so both ${ϵ}_V$ and $|{\eta }_V|$ are small and slow-roll inflation is possible. Near the minimum, a Taylor expansion gives $1-y\simeq b\varphi$ and henceThey are large for $|\varphi |\ll M_{\mathrm{Pl}}$, so that region cannot sustain slow roll.

For a monotonic inflaton trajectory, the potential slow-roll parameter satisfiesThe remaining slow-roll e-fold count is the integral from the present field value to its end value along the direction of motion:

The potential slow-roll parameter isFrom the slow-roll approximation,ThereforeThus $ϵ\ll 1$ implies ${\dot{\varphi }}^2\ll V$, consistently with potential-energy domination.

The same equations giveOn the branch $V^{\prime }/V>0$, the definition of $ϵ$ turns this into

For the hilltop inflation potential,The two potential slow-roll parameters are consequentlyand

Use the geodesic radial comoving coordinate, in which the Friedmann-Lemaître-Robertson-Walker metric has the formOn a slice of constant cosmic time, the radial proper-length element is $R(t)\mathrm{d}\chi$. ThereforeIf one starts from a curvature coordinate with radial element $R\mathrm{d}r/\sqrt{1-k{r}^2}$, then $\chi ={\int }_0^r\mathrm{d}{r}^{\prime }/\sqrt{1-k{r}^{\prime 2}}$; the curvature coordinate $r$ itself is not generally the $\chi$ used above.

For a comoving galaxy, $\dot{\chi }=0$, so $\dot{D}=\dot{R}\chi =HD$. The definitions areThis recession law concerns proper distances on common cosmic-time slices. It is not a local special-relativistic relative velocity measurement; a large recession rate is not a signal exceeding the local speed of light.

For a comoving volume $V=V_0{R}^3$, total fluid energy is $E=\rho c^{2}V$. Adiabatic energy conservation gives $\mathrm{d}E+P\mathrm{d}V=0$. Expanding,henceHere $\rho$ is mass-equivalent energy density. Radiation loses energy both by dilution and by pressure work; pressureless matter only dilutes in this description.

In the time-unit convention for the cosmological constant, it represents a uniform vacuum contribution with pressure equal to minus its energy density. There is a units convention in the printed field equations: their term is $\Lambda /3$, so the symbol $\Lambda$ there has units of inverse time squared. If the geometrical constant is instead given in inverse length squared, this symbol equals $c^2{\Lambda }_{\mathrm{geom}}$. The curvature parameter $k$ labels constant-curvature spatial slices; with a conventional scale-factor normalization it is $+1,0,-1$ for positively curved, flat and negatively curved slices. It fixes local curvature sign, not every possible global spatial topology.

For a flat fluid-only model without a cosmological constant and with $P=w\rho c^2$, the two field equations giveThe two printed cases and their scale-factor laws are worked under the literal inline labels below.

Inflation is accelerated expansion, $\ddot{R}>0$, equivalently a decreasing comoving Hubble radius $(RH{)}^{-1}$ on an expanding branch. An approximately constant positive vacuum-like energy can give nearly exponential expansion. Enough inflation can put regions now widely separated inside a common earlier causal patch, suppress $|k|/(RH{)}^2$ and dilute unwanted relics such as monopoles. These are the horizon, flatness and relic-abundance motivations. Reheating is needed to populate the subsequent hot universe. Quantum fluctuations can seed primordial perturbations, but inflation alone is not a universal solution to the initial singularity or a proof that all initial conditions inflate.

Now adopt the printed natural units $c=\hslash =1$. For the Homogeneous massive-scalar equation in an expanding universe,The continuity equation givesWhere $\dot{\varphi }\ne 0$, this implies the expanding-universe Klein-Gordon equationA regular canonical-field solution extends it through isolated turning points by continuity. Purely dividing the conservation equation does not determine a trajectory on an interval where $\dot{\varphi }$ is identically zero; canonical field dynamics, or variation of its action, supplies the equation there. In particular a constant nonzero massive field is not an exact freely evolving solution.

With $\Lambda =k=0$,ThereforeThis inflation condition for a homogeneous massive scalar means a potential-dominated field has sufficiently negative pressure to accelerate expansion. In quadratic chaotic inflation, a large initial field with $V=m^2{\varphi }^2/2$ can enter slow roll, with $3H\dot{\varphi }\simeq -m^2\varphi$ and $H^2\simeq 4\pi G{m}^2{\varphi }^2/3$. The potential slow-roll parameter is ${ϵ}_V=1/(4\pi G{\varphi }^2)$, so a large field relative to the gravitational scale can support many e-folds. The field eventually rolls far enough that this regime ends and oscillation and reheating become relevant. Large potential energy with arbitrary initial kinetic energy does not by itself guarantee an inflationary interval.

Differentiating the flat Friedmann equation and then using the scalar equation,For the expanding branch,This Nonincreasing Hubble rate for a canonical inflaton decreases strictly where the field rolls and can be momentarily stationary at a turning point; the statement “always decreasing” means nonincreasing. This does not conflict with acceleration, because $\ddot{R}/R=H^2+\dot{H}$ remains positive when ${ϵ}_H=-\dot{H}/H^2<1$.

At horizon exit, larger comoving wavelengths leave earlier, when $H$ is larger. For a canonical field in slow roll, with reduced Planck mass ${\overline{M}}_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$,The decreasing expansion rate gives a red tensor tilt at this order. The scalar curvature spectrum also contains the conversion from field fluctuations to curvature:Here $N=\ln R$ counts elapsed e-folds. Thus the Decreasing Hubble rate does not alone fix scalar tilt. Evolution of ${ϵ}_H$ matters. For the quadratic slow-roll model, ${\eta }_V={ϵ}_V$ and $n_s-1\simeq -6{ϵ}_V+2{\eta }_V=-4{ϵ}_V<0$. If $N_{\mathrm{rem}}$ e-folds remain until the end, leading estimates give $n_s\simeq 1-2/N_{\mathrm{rem}}$ and tensor-to-scalar ratio $r_{\mathrm{TS}}\simeq 8/N_{\mathrm{rem}}$, with small end-of-inflation corrections.

The relevant contemporary comparison is a nearly scale-invariant but red scalar spectrum. The February2015 Planck analysis reported $n_s=0.968\pm 0.006$ for its temperature, low-polarization and lensing combination, and an upper limit $r_{0.002}<0.11$ at95percent confidence. The scalar tilt is qualitatively consistent with a slowly evolving inflaton, but that result does not establish the quadratic potential; its relatively large tensor prediction was already disfavored against lower-tensor alternatives in that analysis.^[ref]

At horizon exit during single-field slow-roll inflation, $d/d\log k\simeq -M_{\mathrm{Pl}}^2(V^{\prime }/V)d/d\Phi$. Applying this operator to the potential slow-roll parameter and the second potential slow-roll parameter yields the displayed leading-order flow equations. Corrections to the horizon-exit derivative contribute only at higher slow-roll order here.

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.

The quadratic potential has equal potential slow-roll parameters, ${ϵ}_V={\eta }_V=1/(2N+1)$. Consequently $n_s=1-4/(2N+1)$ and $r=16/(2N+1)$ at leading order. The relation $r=4(1-n_s)$ is independent of the inflaton mass, so a satisfactory scalar tilt does not by itself guarantee a satisfactory tensor prediction.

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.

For $V(\varphi )=\lambda {\varphi }^4/4$ with $\lambda >0$, use natural units and the unreduced Planck mass $M_P=G^{-1/2}$. The slow-roll equations give $H\simeq \sqrt{2\pi \lambda /3}{\varphi }^2/M_P$ and $\dot{\varphi }\simeq -M_P\sqrt{\lambda /(6\pi )}\varphi$. Integrating gives the displayed exponential field decay, valid on either sign branch, andThe potential slow-roll parameters require ${\varphi }^2\gg M_P^2$. The expression is valid only during slow-roll inflation; its formal late-time saturation does not describe the subsequent reheating or radiation era.