Inflationary Kinematics and the Comoving Horizon

Inflation is an interval of accelerated expansion. Its decisive local effect is that the comoving Hubble radius shrinks: a fixed Fourier scale can begin well inside the Hubble radius, cross it during inflation, and re-enter during a later decelerating era. This statement is kinematical. It does not by itself identify the matter driving inflation, select a quantum state, or assert the existence of a global event horizon.

Required background. The Einstein Equation supplies Einstein Equations for the FLRW acceleration and curvature relations. Classical Field Modes and Canonical Variables supplies Flat-Spacetime Quantum Field Theory for the Fourier-mode convention and the distinction between comoving and physical wavenumber.

Helpful background. Relativistic Matter and Compact Objects supplies stress–energy language, while Relativistic Cosmology supplies the FLRW background and Friedmann equations used below.

Acceleration and the shrinking Hubble radius

Let $t$ be cosmic time in an expanding FLRW spacetime, with

$$H\equiv\frac{\dot a}{a}\gt 0, \qquad \epsilon_H\equiv-\frac{\dot H}{H^2}.$$

The physical Hubble radius is $H^{-1}$ and its comoving counterpart is $(aH)^{-1}$. Since $aH=\dot a$,

$$\frac{\ddot a}{a}=H^2(1-\epsilon_H), \qquad \frac{d}{dt}(aH)^{-1} =-\frac{\ddot a}{\dot a^2} =-\frac{1-\epsilon_H}{a}.$$

Therefore, for an expanding background,

$$\boxed{ \ddot a\gt 0 \quad\Longleftrightarrow\quad \epsilon_H\lt 1 \quad\Longleftrightarrow\quad \frac{d}{dt}(aH)^{-1}\lt 0 }.$$

Negative pressure is required if the background obeys Einstein gravity with the cosmological constant, if any, absorbed into the total source: the acceleration equation gives $\ddot a/a=-4\pi G(\rho+3p)/3$, so $\rho+3p\lt 0$. The equivalence above is more primitive—it follows from the definition of $H$ without specifying a matter model. The basic inflationary kinematics and their relation to the flatness and horizon problems are developed in Ellis, Vennin, and Wands 2025/2026, §23.1, printed pp. 1–3, eqs. (23.1)–(23.6) (Official PDF) and Baumann 2012, §§4.2–5.2, printed pp. 24–28, eqs. (36)–(55) (Open PDF).

Constant-ε expansion

For constant $\epsilon_H=\epsilon$, the definition $\dot H=-\epsilon H^2$ can be integrated exactly. With values $a_i,H_i$ at $t_i$,

$$H(t)=\frac{H_i}{1+\epsilon H_i(t-t_i)},$$

and, on the branch where the denominator is positive,

$$a(t)= \begin{cases} a_i\left[1+\epsilon H_i(t-t_i)\right]^{1/\epsilon}, & \epsilon\ne0,\\[4pt] a_i e^{H_i(t-t_i)},&\epsilon=0. \end{cases}$$

Thus $0\lt\epsilon\lt1$ gives accelerated power-law expansion, while $\epsilon=0$ gives exponential expansion, the flat-slicing de Sitter history when $K=0$. Values $\epsilon\lt0$ also satisfy the kinematical inflation condition but have $\dot H\gt 0$; in spatially flat Einstein gravity they require a source that violates the usual null-energy condition.

Write $N\equiv\ln(a/a_i)$. A fixed comoving Fourier mode $k$ obeys

$$\frac{d}{dN}\ln\!\left(\frac{k}{aH}\right) =-(1-\epsilon).$$

For constant $\epsilon\lt 1$,

$$\frac{k}{aH} =\left(\frac{k}{a_iH_i}\right)e^{-(1-\epsilon)N}.$$

Its physical reduced wavelength is $a/k$. A mode with $k/(aH)\gg1$ is initially shorter than $H^{-1}$; it crosses at $k=aH$ and subsequently has $k/(aH)\ll1$. During a later decelerating epoch, $\epsilon_H\gt 1$ and the trend reverses, permitting re-entry. “Horizon exit” is common shorthand for this Hubble-radius crossing, not proof that the mode crosses a causal boundary.

For constant $\epsilon\lt 1$, conformal time may be shifted so that the power-law future lies at $\eta=0^-$. Then

$$aH=-\frac{1}{(1-\epsilon)\eta}, \qquad a\propto(-\eta)^{-1/(1-\epsilon)}.$$

This form makes the shrinking comoving Hubble scale especially transparent: $\lvert(aH)^{-1}\rvert=(1-\epsilon)\lvert\eta\rvert$.

Why curvature is diluted

Use $K$ for the constant spatial-curvature parameter and reserve $k$ for a Fourier wavenumber. Define

$$\Omega_K\equiv-\frac{K}{a^2H^2}, \qquad \Omega_{\rm tot}-1=\frac{K}{a^2H^2}=-\Omega_K.$$

Whenever $K\ne0$,

$$\frac{d\ln\lvert\Omega_K\rvert}{dN}=-2(1-\epsilon_H).$$

Inflation therefore drives the fractional curvature contribution toward zero. For constant $\epsilon\lt 1$ over $N$ e-folds,

$$\frac{\lvert\Omega_{K,f}\rvert}{\lvert\Omega_{K,i}\rvert} =e^{-2(1-\epsilon)N}.$$

This does not set $K$ itself to zero; it makes $K/a^2$ small relative to $H^2$. A later decelerating era makes $\lvert\Omega_K\rvert$ grow again, so a complete flatness estimate must include the entire post-inflationary expansion history.

Three different causal scales

The comoving particle horizon from an initial boundary $t_i$ is

$$\chi_{\rm p}(t)=\int_{t_i}^{t}\frac{dt'}{a(t')}.$$

It is the greatest comoving distance a light signal could have travelled since that boundary and is nondecreasing. Inflation does not make this integral shrink. Instead, it can take a region that was initially within one causal patch and stretch it to encompass scales much larger than the Hubble radius at the end of inflation.

The comoving event-horizon radius, when it exists, is

$$\chi_{\rm e}(t)=\int_t^{t_{\max}}\frac{dt'}{a(t')}.$$

Its convergence depends on the entire future spacetime. A finite inflationary stage followed by sufficiently long decelerating expansion need not produce a global event horizon. Conversely, an eternally accelerating future may do so. Neither conclusion follows from the instantaneous crossing condition $k=aH$. These distinctions are essential whenever “horizon” is used without a qualifier.

The particle-horizon, event-horizon, and Hubble-radius comparison develops the null-ray interpretation of these integrals. A useful test of any inflationary history is to ask two separate questions: did the relevant region have time to exchange signals before inflation stretched it, and what happened to $(aH)^{-1}$ during the stretch? The second calculation cannot replace the first or specify the initial state of that region.

Exercises

An inflationary stage has constant $\epsilon=0.02$ and lasts $N=60$ e-folds. Find the change in (a) the comoving Hubble radius and (b) the magnitude of the curvature fraction. Does either result establish a global event horizon?

Solution

The comoving Hubble radius changes by

$$\frac{(aH)^{-1}_f}{(aH)^{-1}_i} =e^{-(1-\epsilon)N} =e^{-58.8} =2.91\times10^{-26}.$$

The curvature fraction changes by the square of this factor,

$$\frac{\lvert\Omega_{K,f}\rvert}{\lvert\Omega_{K,i}\rvert} =e^{-117.6} =8.45\times10^{-52}.$$

Both are local-to-the-inflationary-interval statements. An event horizon requires convergence of $\int_t^{t_{\max}}dt'/a(t')$ and therefore cannot be decided without specifying the future expansion after inflation.

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.