Scalar-Field Inflation and Slow Roll

A canonical scalar field can sustain inflation when potential energy dominates its motion and Hubble damping keeps the field close to a slowly evolving trajectory. Slow roll is an approximation to the exact scalar–gravity system, not a separate equation of motion. This page derives both descriptions, identifies the conditions under which they agree, and follows a quadratic example through the point where accelerated expansion actually ends.

Required background. The Einstein Equation supplies Einstein Equations for the homogeneous constraint and acceleration law. Classical Field Modes and Canonical Variables supplies Flat-Spacetime Quantum Field Theory for the canonical scalar action and field normalization.

Helpful background. Relativistic Matter and Compact Objects supplies the stress–energy interpretation, while Relativistic Cosmology supplies the FLRW dynamics and e-fold variable.

The exact scalar–gravity system

Consider one minimally coupled scalar field with action

$$S_\phi=\int d^4x\sqrt{-g}\left[ -\frac12 g^{\mu\nu}\partial_\mu\phi\,\partial_\nu\phi-V(\phi) \right].$$

For a homogeneous field in a spatially flat FLRW spacetime, with a dot denoting the cosmic-time derivative $d/dt$ and $H=\dot a/a$,

$$\rho_\phi=\frac12\dot\phi^2+V, \qquad p_\phi=\frac12\dot\phi^2-V.$$

The exact equations are

$$H^2=\frac{8\pi G}{3} \left(\frac12\dot\phi^2+V\right), \qquad \dot H=-4\pi G\dot\phi^2,$$ $$\ddot\phi+3H\dot\phi+V_{,\phi}=0.$$

The Klein–Gordon equation follows either by varying the action or by inserting the scalar density and pressure into energy conservation. Differentiating the Friedmann constraint and using the Klein–Gordon equation recovers the stated $\dot H$ equation, so the three equations are mutually consistent rather than independent evolution laws.

Define the reduced Planck mass only as an abbreviation, $M\equiv(8\pi G)^{-1/2}$, and set

$$\epsilon_H\equiv-\frac{\dot H}{H^2} =\frac{4\pi G\dot\phi^2}{H^2} =\frac{\dot\phi^2}{2M^2H^2}.$$

Inflation ends at the exact crossing $\epsilon_H=1$. For a scalar-dominated background this is equivalent to $\dot\phi^2=V$ and to $p_\phi=-\rho_\phi/3$. The scalar equations and this exact inflation criterion are given in Ellis, Vennin, and Wands 2025/2026, §§23.2–23.2.1, printed pp. 3–5, eqs. (23.7)–(23.17) (Official PDF).

Slow roll as a controlled approximation

If kinetic energy and field acceleration are both small in the relevant senses,

$$\dot\phi^2\ll V, \qquad \lvert\ddot\phi\rvert\ll\lvert3H\dot\phi\rvert, \qquad \lvert\ddot\phi\rvert\ll\lvert V_{,\phi}\rvert,$$

then the exact system reduces to

$$H^2\simeq\frac{8\pi G}{3}V=\frac{V}{3M^2}, \qquad 3H\dot\phi\simeq-V_{,\phi}.$$

The potential slow-roll parameters are

$$\epsilon_V\equiv \frac{1}{16\pi G}\left(\frac{V_{,\phi}}{V}\right)^2 =\frac{M^2}{2}\left(\frac{V_{,\phi}}{V}\right)^2,$$ $$\eta_V\equiv \frac{1}{8\pi G}\frac{V_{,\phi\phi}}{V} =M^2\frac{V_{,\phi\phi}}{V}.$$

When $\epsilon_V\ll1$ and $\lvert\eta_V\rvert\ll1$ along a trajectory that has reached the slow-roll branch,

$$\epsilon_H\simeq\epsilon_V, \qquad \epsilon_2\equiv\frac{d\ln\epsilon_H}{dN} \simeq4\epsilon_V-2\eta_V.$$

These potential conditions diagnose a slowly varying solution; they do not guarantee that arbitrary initial data reach it before overshooting the flat region. Locally, if $H$ and $V_{,\phi}$ vary slowly, a velocity displacement from the branch obeys $\delta\dot\phi\propto a^{-3}$. This is the origin of the homogeneous attractor statement. It is weakened near a sharp feature, a turn in multifield space, a noncanonical kinetic term, or the end of inflation. Baumann develops the exact and slow-roll systems in Baumann 2012, §§6.1–6.3, printed pp. 31–34, eqs. (61)–(90) (Open PDF).

E-folds along a monotonic trajectory

Let $N$ count expansion forward, $dN=Hdt$. For a field rolling toward smaller $\phi$ with $V_{,\phi}\gt 0$, the number of e-folds from $\phi_i$ to the end is

$$N_{i\to\mathrm{end}} =\int_{t_i}^{t_{\mathrm{end}}}Hdt \simeq\frac{1}{M^2} \int_{\phi_{\mathrm{end}}}^{\phi_i} \frac{V}{V_{,\phi}}\,d\phi =8\pi G\int_{\phi_{\mathrm{end}}}^{\phi_i} \frac{V}{V_{,\phi}}\,d\phi.$$

The limits and sign must be changed for another rolling direction. The integral is trustworthy only over the portion of the trajectory on which slow roll holds; replacing the exact end $\epsilon_H=1$ by $\epsilon_V=1$ is part of the approximation.

For

$$V(\phi)=\frac12m^2\phi^2,$$

one obtains

$$\epsilon_V=\eta_V=\frac{2M^2}{\phi^2}, \qquad \phi_{\mathrm{end}}^{\rm(SR)}=\sqrt2\,M,$$ $$N^{\rm(SR)} =\frac{\phi_i^2-(\phi_{\mathrm{end}}^{\rm(SR)})^2}{4M^2}.$$

Thus $\phi_i=\sqrt{242}\,M$ gives $N^{\rm(SR)}=60$.

The two smallness conditions have different jobs. Small $\epsilon_V$ keeps the kinetic contribution subdominant and allows accelerated expansion; small $\lvert\eta_V\rvert$ keeps the slope from changing too much while the field moves. A flat potential with a field moving too fast can therefore violate slow roll even though its potential parameters are small. The numerical example below starts with the approximate attractor velocity so that it tests the accuracy of the approximation and its endpoint, rather than an arbitrary initial transient.

Numerical check through inflation end

The slow-roll endpoint can be checked without making a slow-roll approximation in the evolution. Measure the field in units of $M$ and time in units of $m^{-1}$—equivalently, set $M=m=1$ in the dimensionless equations—and choose

$$\phi_i=\sqrt{242}, \qquad \dot\phi_i=-\sqrt{\frac23},$$

where the initial velocity is the leading slow-roll value, compute $H_i$ from the exact Friedmann constraint, and integrate

$$\dot\phi=\pi_\phi, \qquad \dot\pi_\phi=-3H\pi_\phi-\phi, \qquad \dot N=H$$

until $\epsilon_H=1$. An adaptive fifth-order Runge–Kutta integration with relative tolerance $10^{-10}$ and absolute tolerance $10^{-12}$ gives

$$\phi_{\mathrm{end}}=1.00934270\,M, \qquad N_{i\to\mathrm{end}}=61.05140832.$$

Changing the maximum dimensionless time step from $0.1$ to $0.01$ changes the reported $N$ by less than $3\times10^{-11}$. The slow-roll estimate is lower by $1.0514$ e-folds, or $1.75\%$. This is not a failure of the attractor: most of the evolution follows it closely. The difference accumulates where the field leaves slow roll, and the exact endpoint lies below the proxy $\sqrt2\,M$.

The mass $m$ sets the time and energy scales but cancels from this dimensionless trajectory. In other potentials inflation may end through a waterfall instability, loss of effective-field-theory control, or a transition in another field before the single-field condition $\epsilon_H=1$ is reached.

What the approximation does not say

Slow roll is not synonymous with inflation. The exact condition $\epsilon_H\lt 1$ can hold briefly even when $\lvert\ddot\phi\rvert$ is not negligible, while a potential with small local $\epsilon_V$ and $\lvert\eta_V\rvert$ need not yield many e-folds if its flat region is too narrow. Conversely, a field that begins with substantial kinetic energy may first undergo a decelerating phase and join the slow-roll branch only after Hubble damping has removed enough of its velocity.

Nor is $\dot\phi\simeq-V_{,\phi}/(3H)$ a universal attractor formula. Near an exceptionally flat segment, the potential force can become negligible and the exact equation gives $\dot\phi\propto a^{-3}$. During this “ultra-slow-roll” interval $\epsilon_H$ decreases rapidly, the usual potential hierarchy does not organize the evolution, and the nominal decaying perturbation mode can become important. The exact background equations remain valid; it is the reduced first-order description that must be replaced.

Exercises

For the quadratic potential, verify the slow-roll e-fold result and compare the potential and Hubble parameters at the numerical initial point above.

Solution

Since $V/V_{,\phi}=\phi/2$,

$$N^{\rm(SR)} =\frac{1}{M^2}\int_{\sqrt2M}^{\sqrt{242}M}\frac{\phi}{2}\,d\phi =\frac{242-2}{4}=60.$$

The potential parameter is $\epsilon_{V,i}=2/242=0.00826446$. With $M=m=1$ and $\dot\phi_i^2=2/3$, the exact constraint gives $H_i^2=(\dot\phi_i^2+\phi_i^2)/6=40.4444$, so

$$\epsilon_{H,i} =\frac{\dot\phi_i^2}{2H_i^2} =0.00824176.$$

Their $0.28\%$ relative difference is consistent with the point being well inside, but not exactly on, the leading slow-roll approximation.

References

• Baumann, D., “TASI Lectures on Inflation,” arXiv:0907.5424v2 (2012), Open PDF.
• Ellis, J., V. Vennin, and D. Wands, “Inflation,” in F. Takahashi et al. (Particle Data Group), Review of Particle Physics, International Journal of Modern Physics A 41, 2630011 (2026), revised August 2025, Official PDF.