Slow-Roll Inflation from a Scalar Field

Derivation

Slow-Roll Inflation from a Scalar Field

D-343 Home PU-308 Threads fields · energy · symmetry · chance Depends on The Friedmann Equations from Einstein's Equations, scalar-field-stress-energy, Particle Horizon, Horizon and Flatness Problems

Statement

A single homogeneous scalar field \(\phi\) (the inflaton) minimally coupled to gravity, with potential \(V(\phi)\), obeys the flat-FLRW Friedmann and Klein–Gordon equations \(H^2=\tfrac{1}{3M_{\mathrm{Pl}}^2}\!\left(\tfrac{1}{2}\dot\phi^2+V\right)\) and \(\ddot\phi+3H\dot\phi+V'=0\). When its potential energy dominates its kinetic energy and its acceleration is negligible on the Hubble time, the field enters a slow-roll attractor on which \(H\approx\mathrm{const}\) and the scale factor grows quasi-exponentially, \(a(t)\propto e^{\int H\,dt}\). The regime is governed by the dimensionless slow-roll parameters \(\varepsilon\equiv\tfrac{M_{\mathrm{Pl}}^2}{2}\!\left(V'/V\right)^2\) and \(\eta\equiv M_{\mathrm{Pl}}^2\,V''/V\); \(\varepsilon\ll 1,\ |\eta|\ll 1\) guarantees \(\ddot a>0\) and produces the many e-folds \(N=\int H\,dt\gtrsim 60\) that solve the horizon and flatness problems.

Why it matters

Inflation is the only known mechanism that makes the observed universe a generic outcome rather than a fine-tuned initial condition. A brief epoch of accelerated expansion stretches a single causally connected, nearly flat patch to encompass the entire observable universe, simultaneously erasing the horizon and flatness problems and diluting unwanted relics. Slow-roll is the concrete dynamical realisation: it shows how ordinary field theory, with only a flat enough potential, produces \(\ddot a>0\) for a finite number of e-folds and then gracefully ends.

The same slow-roll parameters that control the background expansion also fix the amplitude and tilt of the primordial density perturbations, so this derivation is the backbone connecting a Lagrangian \(V(\phi)\) to the CMB power spectrum measured by WMAP and Planck. Getting the background right is the prerequisite for every prediction inflation makes.

Assumptions

Homogeneous, isotropic background (flat FLRW metric).If dropped, spatial gradients \((\nabla\phi)^2/a^2\) and anisotropic shear enter the stress tensor; the field equations become PDEs and the clean \(H(\phi)\) attractor is lost.

Single, minimally coupled, canonical scalar field.If dropped (multiple fields, non-canonical kinetic term, or \(\xi R\phi^2\) coupling), the sound speed and the effective potential shift, and isocurvature or entropy modes appear alongside the adiabatic one.

Potential energy dominates: \(\tfrac{1}{2}\dot\phi^2\ll V\).If dropped, \(w=p/\rho\) rises above \(-\tfrac{1}{3}\) and expansion is no longer accelerating; you recover ordinary scalar-field cosmology, not inflation.

Friction dominates inertia: \(|\ddot\phi|\ll 3H|\dot\phi|\).If dropped, the field oscillates or overshoots rather than tracking the attractor; the slow-roll relation \(3H\dot\phi\approx -V'\) fails and \(H\) is no longer nearly constant.

Sub-Planckian energy density, \(V\ll M_{\mathrm{Pl}}^4\).If dropped, quantum-gravity corrections to the Friedmann equation and to \(V(\phi)\) are uncontrolled and the classical treatment breaks down.

Derivation

\[ \rho=\tfrac{1}{2}\dot\phi^2+V(\phi),\qquad p=\tfrac{1}{2}\dot\phi^2-V(\phi) \]

Energy density and pressure of a homogeneous canonical scalar from the scalar-field stress–energy tensor \(T_{\mu\nu}=\partial_\mu\phi\,\partial_\nu\phi-g_{\mu\nu}\mathcal{L}\), evaluated for \(\phi=\phi(t)\) (prior result). A

\[ H^2=\left(\frac{\dot a}{a}\right)^2=\frac{1}{3M_{\mathrm{Pl}}^2}\left(\tfrac{1}{2}\dot\phi^2+V\right) \]

Substitute \(\rho\) into the flat Friedmann equation \(H^2=\rho/3M_{\mathrm{Pl}}^2\) with \(M_{\mathrm{Pl}}^2\equiv(8\pi G)^{-1}\) (Friedmann equations from GR, prior result). A

\[ \ddot\phi+3H\dot\phi+V'(\phi)=0 \]

Equation of motion for \(\phi\): either the covariant conservation \(\dot\rho+3H(\rho+p)=0\) applied to step 1, or the Euler–Lagrange equation for the FLRW action. The \(3H\dot\phi\) term is Hubble friction. Here \('\equiv d/d\phi\). B

\[ \frac{\ddot a}{a}=\dot H+H^2=-\frac{1}{6M_{\mathrm{Pl}}^2}\left(\rho+3p\right)=-\frac{1}{3M_{\mathrm{Pl}}^2}\left(\dot\phi^2-V\right) \]

Second Friedmann (acceleration) equation with \(\rho+3p=2\dot\phi^2-2V\). Accelerated expansion \(\ddot a>0\) requires \(\dot\phi^2<V\), i.e. potential domination. B

\[ \tfrac{1}{2}\dot\phi^2\ll V \;\Longrightarrow\; H^2\approx\frac{V(\phi)}{3M_{\mathrm{Pl}}^2} \]

First slow-roll approximation: impose potential dominance (Assumption 3) in step 2. \(H\) is now a slowly varying function of \(\phi\) alone. A

\[ |\ddot\phi|\ll 3H|\dot\phi| \;\Longrightarrow\; 3H\dot\phi\approx -V'(\phi) \]

Second slow-roll approximation: drop the inertial term in step 3 (Assumption 4). This is a first-order (overdamped) equation — the slow-roll attractor. A

\[ \dot\phi\approx-\frac{V'}{3H}=-\frac{V'\,M_{\mathrm{Pl}}}{\sqrt{3V}} \]

Solve step 6 for \(\dot\phi\) and substitute \(H\) from step 5. The field velocity is fixed algebraically by the local slope of the potential — no free initial velocity survives on the attractor. B

\[ \frac{\tfrac{1}{2}\dot\phi^2}{V}\approx\frac{1}{2V}\frac{(V')^2 M_{\mathrm{Pl}}^2}{3V}=\frac{M_{\mathrm{Pl}}^2}{6}\left(\frac{V'}{V}\right)^2=\frac{\varepsilon}{3} \]

Insert step 7 into the kinetic-to-potential ratio and define the first slow-roll parameter \(\varepsilon\equiv\tfrac{M_{\mathrm{Pl}}^2}{2}(V'/V)^2\). Consistency of Assumption 3 therefore requires \(\varepsilon\ll 1\). C

\[ \varepsilon\equiv\frac{M_{\mathrm{Pl}}^2}{2}\left(\frac{V'}{V}\right)^2,\qquad \eta\equiv M_{\mathrm{Pl}}^2\,\frac{V''}{V} \]

Differentiate the attractor relation \(3H\dot\phi=-V'\) and use step 5 to show that the self-consistency of Assumption 4 (small \(\ddot\phi\)) requires \(|\eta|\ll 1\); this defines the second (potential) slow-roll parameter. Both are properties of the shape of \(V\), not of initial data. C

\[ -\frac{\dot H}{H^2}=\varepsilon_H\approx\varepsilon\ll 1 \;\Longrightarrow\; \frac{\ddot a}{a}=H^2(1-\varepsilon_H)>0 \]

Compute \(\dot H=-\tfrac{1}{2}\dot\phi^2/M_{\mathrm{Pl}}^2\) (differentiate step 2 and use step 3), divide by \(H^2\), and use step 8. Since \(\varepsilon<1\), \(\ddot a>0\): expansion accelerates. To leading order the Hubble slow-roll \(\varepsilon_H\) equals the potential \(\varepsilon\). C

\[ N\equiv\ln\frac{a_{\mathrm{end}}}{a_i}=\int_{t_i}^{t_{\mathrm{end}}}\!H\,dt=\int_{\phi_{\mathrm{end}}}^{\phi_i}\frac{H}{\dot\phi}\,d\phi\approx\frac{1}{M_{\mathrm{Pl}}^2}\int_{\phi_{\mathrm{end}}}^{\phi_i}\frac{V}{V'}\,d\phi \]

Define the number of e-folds and change the integration variable from \(t\) to \(\phi\) using \(dt=d\phi/\dot\phi\) with step 7. Inflation ends when \(\varepsilon(\phi_{\mathrm{end}})\simeq 1\). B

\[ \varepsilon\ll1 \;\Rightarrow\; H\approx\text{const}\;\Rightarrow\; a(t)\approx a_i\,e^{H(t-t_i)} \]

With \(\dot H/H^2=-\varepsilon\) small, \(H\) is nearly constant over many Hubble times, so integrating \(\dot a/a=H\) gives quasi-de Sitter (exponential) expansion — the sought result. A

Result

\[ \boxed{\;H^2\approx\frac{V}{3M_{\mathrm{Pl}}^2},\quad 3H\dot\phi\approx-V',\quad \varepsilon=\frac{M_{\mathrm{Pl}}^2}{2}\!\left(\frac{V'}{V}\right)^2\ll1,\ |\eta|=\left|M_{\mathrm{Pl}}^2\frac{V''}{V}\right|\ll1\;\Rightarrow\; a\propto e^{Ht},\ N\approx\frac{1}{M_{\mathrm{Pl}}^2}\!\int\frac{V}{V'}d\phi\;}\]

Reading. When the potential is flat enough that both \(\varepsilon\) and \(|\eta|\) are small, the inflaton crawls down \(V(\phi)\) at a terminal velocity fixed by the slope, the energy density stays nearly constant, and the scale factor grows exponentially. Inflation persists until \(\varepsilon\to1\); the accumulated growth is \(e^N\) with \(N\) the field-space integral of \(V/V'\). Requiring the observable universe to have been in causal contact needs \(N\gtrsim 50\)–\(60\), which flat potentials supply easily.

Units check. In natural units \(\hbar=c=1\), \([\phi]=[M_{\mathrm{Pl}}]=\)mass, \([V]=\)mass\(^4\), \([V']=\)mass\(^3\), \([V'']=\)mass\(^2\). Then \(V'/V\) has units mass\(^{-1}\), so \(M_{\mathrm{Pl}}^2(V'/V)^2\) is (mass\(^2\))(mass\(^{-2}\)) = dimensionless — good for \(\varepsilon\); and \(M_{\mathrm{Pl}}^2\,V''/V=\)(mass\(^2\))(mass\(^2\)/mass\(^4\)) = dimensionless — good for \(\eta\). \(H^2\sim V/M_{\mathrm{Pl}}^2=\)mass\(^4\)/mass\(^2=\)mass\(^2\), so \([H]=\)mass \(=\)time\(^{-1}\); and \(N=\int H\,dt\) is dimensionless, as an e-fold count must be.

Limiting cases

Exact de Sitter (\(\varepsilon,\eta\to0\)): \(V=\)const, \(\dot\phi\to0\), \(H=\)const exactly, \(a=a_i e^{Ht}\); a cosmological constant. Inflation never ends — a limit to approach, not sit in.
End of inflation (\(\varepsilon\to1\)): \(\ddot a\to0\); the slow-roll approximation collapses, the field picks up kinetic energy and begins to oscillate about the minimum, initiating reheating.
Kinetic domination (\(\tfrac12\dot\phi^2\gg V\)): \(w\to+1\), \(a\propto t^{1/3}\) (stiff fluid); no acceleration — the opposite regime, relevant to pre-inflationary or kination epochs.
Massive free field, \(V=\tfrac12 m^2\phi^2\): \(\varepsilon=\eta=2M_{\mathrm{Pl}}^2/\phi^2\); slow-roll holds for \(\phi\gg M_{\mathrm{Pl}}\) and \(N\approx\phi_i^2/4M_{\mathrm{Pl}}^2\). The archetypal large-field model.
Small \(\eta\), large \(\varepsilon\): steep but not concave potential; expansion accelerates only briefly — insufficient e-folds, a reminder that both parameters matter.

Breaks when

Kinetic energy becomes comparable to potential (\(\varepsilon\gtrsim1\)). Assumption 3 fails, \(w>-\tfrac13\), \(\ddot a<0\): inflation ends. The slow-roll formulae for \(H\), \(\dot\phi\) and \(N\) all lose validity here, which is exactly why \(\varepsilon=1\) marks the endpoint.
The potential is steep or sharply curved (\(|\eta|\gtrsim1\)). Hubble friction can no longer overdamp the motion; \(\ddot\phi\) is not negligible, the field overshoots the attractor, and \(3H\dot\phi\approx-V'\) is wrong. Features, cliffs, or inflection points in \(V\) break the leading-order treatment even when \(\varepsilon\) is momentarily small.
Ultra-slow-roll / near-inflection regions (\(V'\to0\) with \(V''\neq0\)). The relation \(3H\dot\phi=-V'\) gives \(\dot\phi\to0\), but the true solution has \(\dot\phi\propto a^{-3}\) coasting; the standard \(\varepsilon\)-based attractor is non-attractor here and one must integrate the full second-order equation.
Multi-field or non-canonical dynamics. With more than one light direction, or a field-dependent kinetic term / sound speed \(c_s\neq1\), the single-field \(\varepsilon,\eta\) no longer capture the evolution; turning trajectories and isocurvature modes appear.
Trans-Planckian densities (\(V\gtrsim M_{\mathrm{Pl}}^4\)) or \(H\gtrsim M_{\mathrm{Pl}}\). The classical Friedmann equation and the effective potential receive uncontrolled quantum-gravity corrections; the derivation's classical footing is gone.

Failure modes

Confusing \(\varepsilon\) and \(\eta\) with the Hubble slow-roll parameters. \(\varepsilon_V=\tfrac{M_{\mathrm{Pl}}^2}{2}(V'/V)^2\) and \(\eta_V=M_{\mathrm{Pl}}^2V''/V\) are potential parameters; \(\varepsilon_H=-\dot H/H^2\) and \(\eta_H\) are exact kinematic ones. They agree only to leading order. Using the wrong pair in the spectral-index formula \(n_s-1=2\eta_V-6\varepsilon_V\) gives a wrong sign or factor.
Dropping the \(M_{\mathrm{Pl}}^2\) factors or using \(8\pi G\) instead of \((8\pi G)^{-1}\). \(\varepsilon\) and \(\eta\) come out with the wrong dimensions or wrong magnitude by factors of \(8\pi\). Fix \(M_{\mathrm{Pl}}^2=(8\pi G)^{-1}=2.4\times10^{18}\,\)GeV squared (reduced Planck mass) and keep it explicit.
Believing slow-roll means \(\dot\phi=0\). The field does move — \(\dot\phi=-V'/3H\neq0\); it is \(\ddot\phi\) and the kinetic energy relative to \(V\) that are small. Setting \(\dot\phi=0\) gives exact de Sitter and no way to end inflation or count e-folds.
Computing \(N\) with the wrong limits or sign. \(N\) counts e-folds from horizon exit to the end of inflation; integrating from \(\phi_{\mathrm{end}}\) to \(\phi_i\) (small to large field for a large-field model) keeps \(N>0\). Flipping the limits gives a spurious negative e-fold count.
Assuming \(\varepsilon=1\) is where slow-roll first fails. Slow-roll can break earlier if \(|\eta|\) hits 1 while \(\varepsilon\) is still small (e.g. near an inflection); \(\varepsilon=1\) is where acceleration ends, not necessarily where the approximation first breaks.
Using \(V\), not \(V'\), in the e-fold denominator. The integrand is \(V/V'\), not \(V\); a common slip that makes \(N\) dimensionally wrong (it must be dimensionless).

Discussion

The physical heart of slow-roll is overdamping. The inflaton's equation of motion \(\ddot\phi+3H\dot\phi+V'=0\) is a Newtonian particle rolling in the potential \(V\) with a friction coefficient \(3H\) set by the expansion itself. When that friction is large — which happens automatically because \(H\) is large when \(V\) is large — the particle quickly forgets its initial velocity and reaches a terminal drift \(\dot\phi=-V'/3H\), just as a marble in honey. Because the drift is slow, most of the energy stays as potential energy, \(H\) barely changes, and the universe sees an almost-constant vacuum energy: quasi-de Sitter. The genius is that the very expansion the field sources provides the friction that keeps it slow — a self-sustaining accelerated epoch.

The slow-roll parameters translate this into geometry. \(\varepsilon\) measures the fractional change of \(H\) per e-fold (\(\varepsilon=-\dot H/H^2\)); it is the steepness of \(V\), and \(\ddot a>0\) is precisely \(\varepsilon<1\). \(\eta\) measures the curvature of \(V\) and controls how long slow-roll can persist — a small \(\varepsilon\) alone is not enough if the potential's curvature drives the field off the attractor. Together they parameterise how close the dynamics is to exact de Sitter, and, remarkably, the same two numbers evaluated at horizon exit fix the observables: the scalar spectral tilt \(n_s-1=2\eta-6\varepsilon\) and the tensor-to-scalar ratio \(r=16\varepsilon\). A measurement of the CMB is thus a measurement of the shape of \(V\) tens of e-folds before the end of inflation.

This is how inflation solves the classic puzzles. The horizon problem — why causally disconnected CMB patches share a temperature — dissolves because \(N\gtrsim60\) e-folds inflate one tiny causal patch to more than the observable universe, so everything we see was once in contact. The flatness problem — why \(|\Omega-1|\) is so small today despite growing with time — dissolves because \(|\Omega-1|\propto1/(aH)^2\) is driven toward zero as \(a\) grows exponentially at nearly constant \(H\); inflation is an attractor toward spatial flatness. Both fixes require the same accelerated expansion, and slow-roll is the minimal field theory that delivers it while still allowing a graceful exit at \(\varepsilon\to1\).

At the deepest level, slow-roll sits at an intersection of the four threads of this unit. It is a fields story (a scalar field's dynamics), an energy story (vacuum-like potential energy gravitating with \(w\approx-1\)), a symmetry story (the approximate shift symmetry \(\phi\to\phi+\text{const}\) that protects a flat potential is exactly what makes \(\varepsilon,\eta\) naturally small, and its mild breaking sets the tilt), and a chance story (the quantum fluctuations of \(\phi\), frozen at horizon exit with amplitude \(\sim H/2\pi\), seed all cosmic structure — deterministic slow-roll converts vacuum randomness into galaxies). The eta-problem — that generic UV corrections push \(\eta\) toward 1 and spoil inflation — is the statement that this shift symmetry is hard to protect, and is one of the central tensions between inflation and quantum gravity.

Common misconceptions. Inflation is not an explosion into pre-existing space and does not require the inflaton to sit at a potential maximum; it needs only a sufficiently flat stretch of \(V\). "Exponential expansion" is an approximation — \(H\) drifts slowly, which is essential, because a truly constant \(H\) (\(\varepsilon=0\)) would never end. And slow-roll does not mean the field is static: it rolls, just slowly compared with the Hubble rate.

Worked examples

Chaotic inflation, \(V(\phi)=\tfrac{1}{2}m^2\phi^2\): find \(\varepsilon,\eta\), the field value 60 e-folds before the end, and \(n_s\).

Compute the slow-roll parameters symbolically, then insert numbers. Use \(M_{\mathrm{Pl}}=2.4\times10^{18}\,\)GeV. B

\[ V'=m^2\phi,\quad V''=m^2\ \Rightarrow\ \varepsilon=\frac{M_{\mathrm{Pl}}^2}{2}\left(\frac{m^2\phi}{\tfrac12 m^2\phi^2}\right)^2=\frac{2M_{\mathrm{Pl}}^2}{\phi^2},\quad \eta=M_{\mathrm{Pl}}^2\frac{m^2}{\tfrac12 m^2\phi^2}=\frac{2M_{\mathrm{Pl}}^2}{\phi^2}=\varepsilon \]

Slope and curvature of a quadratic; both parameters equal \(2M_{\mathrm{Pl}}^2/\phi^2\). Slow-roll needs \(\phi\gg M_{\mathrm{Pl}}\). A

\[ N=\frac{1}{M_{\mathrm{Pl}}^2}\int_{\phi_{\mathrm{end}}}^{\phi_N}\frac{V}{V'}\,d\phi=\frac{1}{M_{\mathrm{Pl}}^2}\int_{\phi_{\mathrm{end}}}^{\phi_N}\frac{\phi}{2}\,d\phi=\frac{\phi_N^2-\phi_{\mathrm{end}}^2}{4M_{\mathrm{Pl}}^2} \]

Evaluate the e-fold integral for the quadratic. End of inflation \(\varepsilon=1\Rightarrow\phi_{\mathrm{end}}=\sqrt2\,M_{\mathrm{Pl}}\), negligible against \(\phi_N\). B

\[ N\approx\frac{\phi_N^2}{4M_{\mathrm{Pl}}^2}\ \Rightarrow\ \phi_N=2\sqrt{N}\,M_{\mathrm{Pl}}=2\sqrt{60}\,M_{\mathrm{Pl}}\approx15.5\,M_{\mathrm{Pl}}\approx3.7\times10^{19}\,\text{GeV} \]

Solve for the field value 60 e-folds before the end and put in numbers. A

\[ \varepsilon=\eta=\frac{2M_{\mathrm{Pl}}^2}{\phi_N^2}=\frac{1}{2N}=\frac{1}{120}\approx8.3\times10^{-3} \]

Evaluate both parameters at \(\phi_N\); for the quadratic \(\varepsilon=\eta=1/2N\). B

\[ n_s=1+2\eta-6\varepsilon=1-\frac{2}{N}=1-\frac{2}{60}\approx0.967,\qquad r=16\varepsilon=\frac{8}{N}\approx0.13 \]

Reading. The \(m^2\phi^2\) model needs super-Planckian field excursion (\(\phi\approx15\,M_{\mathrm{Pl}}\)) and predicts a red tilt \(n_s\approx0.967\) — pleasingly close to Planck's \(0.965\) — but a tensor ratio \(r\approx0.13\) now excluded by data, which is why pure \(m^2\phi^2\) is observationally dead. Units. \(\phi,M_{\mathrm{Pl}}\) in GeV; \(\varepsilon,\eta,n_s,r\) dimensionless; \(N\) dimensionless.

Starobinsky / plateau model, \(V(\phi)=V_0\left(1-e^{-\sqrt{2/3}\,\phi/M_{\mathrm{Pl}}}\right)^2\): find \(n_s\) and \(r\) at \(N=55\).

Compute \(\varepsilon,\eta\) on the plateau (\(\phi\gg M_{\mathrm{Pl}}\)), do the e-fold integral, evaluate observables. Let \(x\equiv e^{-\sqrt{2/3}\,\phi/M_{\mathrm{Pl}}}\ll1\). C

\[ \frac{V'}{V}=\frac{2\sqrt{2/3}\,x/M_{\mathrm{Pl}}}{1-x}\ \Rightarrow\ \varepsilon=\frac{M_{\mathrm{Pl}}^2}{2}\left(\frac{V'}{V}\right)^2=\frac{4}{3}\frac{x^2}{(1-x)^2}\approx\frac{4}{3}x^2 \]

Differentiate; on the plateau \(x\ll1\) so \((1-x)\approx1\). \(\varepsilon\) is exponentially small — the flat plateau. C

\[ N=\frac{1}{M_{\mathrm{Pl}}^2}\int\frac{V}{V'}\,d\phi\approx\frac{3}{4}\,e^{\sqrt{2/3}\,\phi_N/M_{\mathrm{Pl}}}=\frac{3}{4x}\ \Rightarrow\ x\approx\frac{3}{4N} \]

Leading-order e-fold integral on the plateau; invert for \(x\) at horizon exit. C

\[ \eta=-\frac{4}{3}\frac{x(1-2x)}{(1-x)^2}\approx-\frac{4}{3}x,\qquad \varepsilon\approx\frac{4}{3}x^2\approx\frac{3}{4N^2},\quad \eta\approx-\frac{4}{3}\cdot\frac{3}{4N}=-\frac{1}{N} \]

Curvature \(V''/V\) gives \(\eta\); insert \(x=3/4N\). Note \(\varepsilon\sim N^{-2}\) is much smaller than \(|\eta|\sim N^{-1}\) — curvature, not slope, limits inflation here. C

\[ n_s\approx1-\frac{2}{N}=1-\frac{2}{55}\approx0.964,\qquad r=16\varepsilon\approx\frac{12}{N^2}\approx\frac{12}{3025}\approx4.0\times10^{-3} \]

Reading. The plateau predicts the same red tilt \(n_s\approx0.964\) as the quadratic but a tiny \(r\approx0.004\), two orders of magnitude below \(m^2\phi^2\). This is the Planck-favoured "sweet spot": tensors are suppressed because \(\varepsilon\sim N^{-2}\ll|\eta|\sim N^{-1}\). Units. \(x,\varepsilon,\eta,n_s,r,N\) all dimensionless; \(\phi\) measured in units of \(M_{\mathrm{Pl}}\).

Problems

(A) For \(V=\tfrac12 m^2\phi^2\), show \(\varepsilon=\eta\) and find the field value where inflation ends.
Solution
\(V'=m^2\phi\), \(V''=m^2\). \(\varepsilon=\tfrac{M_{\mathrm{Pl}}^2}{2}(V'/V)^2=\tfrac{M_{\mathrm{Pl}}^2}{2}(2/\phi)^2=2M_{\mathrm{Pl}}^2/\phi^2\). \(\eta=M_{\mathrm{Pl}}^2 V''/V=M_{\mathrm{Pl}}^2 m^2/(\tfrac12 m^2\phi^2)=2M_{\mathrm{Pl}}^2/\phi^2=\varepsilon\). Inflation ends at \(\varepsilon=1\Rightarrow\phi_{\mathrm{end}}=\sqrt2\,M_{\mathrm{Pl}}\approx3.4\times10^{18}\,\)GeV.
(A) A model has \(H\approx10^{13}\,\)GeV nearly constant. How many e-folds accumulate in a time interval \(\Delta t=10^{-36}\,\)s? (\(1\,\text{GeV}^{-1}=6.58\times10^{-25}\,\)s.)
Solution
\(N=\int H\,dt\approx H\,\Delta t\). Convert \(\Delta t\) to natural units: \(\Delta t=10^{-36}\,\text{s}/(6.58\times10^{-25}\,\text{s GeV}^{-1})=1.52\times10^{-12}\,\text{GeV}^{-1}\). Then \(N=(10^{13}\,\text{GeV})(1.52\times10^{-12}\,\text{GeV}^{-1})\approx15\) e-folds. (So \(\sim4\times10^{-36}\,\)s of such expansion gives the \(\sim60\) e-folds needed.)
(B) Derive \(\varepsilon_H\equiv-\dot H/H^2\) in terms of \(\dot\phi\) and show it equals \(\varepsilon\) at leading order in slow-roll.
Solution
From \(H^2=\rho/3M_{\mathrm{Pl}}^2\), differentiate: \(2H\dot H=\dot\rho/3M_{\mathrm{Pl}}^2\). Conservation gives \(\dot\rho=-3H(\rho+p)=-3H\dot\phi^2\), so \(\dot H=-\dot\phi^2/2M_{\mathrm{Pl}}^2\). Thus \(\varepsilon_H=-\dot H/H^2=\dot\phi^2/(2M_{\mathrm{Pl}}^2 H^2)\). Using slow-roll \(\dot\phi=-V'/3H\) and \(H^2=V/3M_{\mathrm{Pl}}^2\): \(\varepsilon_H=\tfrac{(V'/3H)^2}{2M_{\mathrm{Pl}}^2 H^2}=\tfrac{(V')^2}{18M_{\mathrm{Pl}}^2 H^4}=\tfrac{(V')^2}{18M_{\mathrm{Pl}}^2}\cdot\tfrac{9M_{\mathrm{Pl}}^4}{V^2}=\tfrac{M_{\mathrm{Pl}}^2}{2}(V'/V)^2=\varepsilon\). Exact \(\varepsilon_H\) and potential \(\varepsilon\) agree to leading order.
(B) For quartic inflation \(V=\lambda\phi^4\), compute \(\varepsilon\), \(\eta\), the e-fold relation, and \(n_s\) at \(N=60\).
Solution
\(V'=4\lambda\phi^3\), \(V''=12\lambda\phi^2\). \(\varepsilon=\tfrac{M_{\mathrm{Pl}}^2}{2}(4/\phi)^2=8M_{\mathrm{Pl}}^2/\phi^2\); \(\eta=M_{\mathrm{Pl}}^2(12\lambda\phi^2)/(\lambda\phi^4)=12M_{\mathrm{Pl}}^2/\phi^2\). \(N=\tfrac{1}{M_{\mathrm{Pl}}^2}\int_{\phi_{\mathrm{end}}}^{\phi_N}\tfrac{\phi}{4}d\phi=\tfrac{\phi_N^2-\phi_{\mathrm{end}}^2}{8M_{\mathrm{Pl}}^2}\approx\phi_N^2/8M_{\mathrm{Pl}}^2\), so \(\phi_N^2=8N M_{\mathrm{Pl}}^2\) and \(\varepsilon=8M_{\mathrm{Pl}}^2/\phi_N^2=1/N\), \(\eta=12M_{\mathrm{Pl}}^2/\phi_N^2=3/2N\). Then \(n_s=1+2\eta-6\varepsilon=1+3/N-6/N=1-3/N=1-3/60=0.95\), and \(r=16\varepsilon=16/60\approx0.27\). Both \(n_s\) too low and \(r\) far too large — quartic inflation is robustly ruled out.
(C) Estimate the mass \(m\) in \(V=\tfrac12 m^2\phi^2\) required to match the observed scalar amplitude \(A_s\approx2.1\times10^{-9}\), given \(A_s=\dfrac{V}{24\pi^2 M_{\mathrm{Pl}}^4\,\varepsilon}\) evaluated at \(N=60\).
Solution
At \(N=60\): \(\phi_N^2=4NM_{\mathrm{Pl}}^2=240M_{\mathrm{Pl}}^2\), \(\varepsilon=1/2N=1/120\), \(V=\tfrac12 m^2\phi_N^2=120\,m^2 M_{\mathrm{Pl}}^2\). Then \(A_s=\dfrac{120 m^2 M_{\mathrm{Pl}}^2}{24\pi^2 M_{\mathrm{Pl}}^4(1/120)}=\dfrac{120\cdot120\,m^2}{24\pi^2 M_{\mathrm{Pl}}^2}=\dfrac{14400\,m^2}{24\pi^2 M_{\mathrm{Pl}}^2}=\dfrac{600\,m^2}{\pi^2 M_{\mathrm{Pl}}^2}\). Set equal to \(2.1\times10^{-9}\): \(m^2=\dfrac{2.1\times10^{-9}\,\pi^2}{600}M_{\mathrm{Pl}}^2=3.45\times10^{-11}M_{\mathrm{Pl}}^2\), so \(m=5.9\times10^{-6}M_{\mathrm{Pl}}\approx5.9\times10^{-6}\times2.4\times10^{18}\,\text{GeV}\approx1.4\times10^{13}\,\text{GeV}\). The inflaton mass is around \(10^{13}\,\)GeV — a GUT-scale mass, which is why inflation probes energies far beyond colliders.