advanced21 min studyLesson 860 in Physics

Inflationary background dynamics and reheating

Separate exact scalar-background dynamics, slow-roll estimates, and the formation of a thermal radiation bath.

Before you begin

On this page

The question for today

What you’ll understand

State the exact acceleration criterion for a canonical scalar background.
Estimate e-folds under the potential slow-roll approximation.
Distinguish the end of inflation, particle production, thermalization, and completed reheating.

An accelerated early expansion can extend the conformal history preceding the hot radiation era, but acceleration alone does not produce a thermal bath. Inflationary background models specify the field dynamics that drive expansion; reheating models specify how that energy becomes particles and eventually equilibrated radiation. We study a single canonical scalar minimally coupled to Einstein gravity in a flat FLRW universe, with natural units. The reduced Planck mass is $\bar M_{\rm Pl}=(8\pi G)^{-1/2}$ , distinct from the unreduced mass in the radiation-rate formula. A simple potential calculation will illustrate the method without asserting that its potential is an observationally preferred inflation model.

Negative pressure follows from a potential-dominated scalar

A homogeneous canonical scalar has positive kinetic energy and potential energy. Kinetic energy contributes positively to both density and pressure, whereas the potential contributes negatively to pressure. Potential domination can therefore support accelerated expansion. The scalar equation contains Hubble friction because work is done against expansion; that term alone does not mean the scalar has heated a surrounding radiation bath. Energy transfer into particles requires additional couplings and their dynamics.

Define the Hubble slow-roll parameter from the fractional change of the expansion rate. The acceleration equation makes its inequality below unity an exact criterion for acceleration in an expanding background. For the scalar-only model it also equals a kinetic-energy expression. This criterion does not require the potential slow-roll approximation. A transient accelerated solution can exist without the field having settled onto a long-lived slow-roll attractor, so acceleration and the approximation should be tested separately.

\rho_\phi=\frac12\dot\phi^2+V,\quad P_\phi=\frac12\dot\phi^2-V,\quad H^2=\frac{\rho_\phi}{3\bar M_{\rm Pl}^2},\quad \ddot\phi+3H\dot\phi+V'=0,\quad \epsilon_H=-\frac{\dot H}{H^2}=\frac{\dot\phi^2}{2\bar M_{\rm Pl}^2H^2}<1

The equalities assume the canonical scalar is the sole background source. Primes on the potential denote differentiation with respect to

\phi

. The final inequality selects accelerated expansion;

\phi

and

\bar M_{\rm Pl}

have energy units.↔ Scroll wide equations horizontally.

Slow roll converts a trajectory problem into a potential integral

If the potential is sufficiently flat and the solution has reached the appropriate attractor, neglect the scalar acceleration relative to Hubble friction and the potential force, and neglect kinetic energy in Friedmann's equation. Small potential slope and curvature parameters diagnose this regime. Both initial conditions and the range of validity matter; a locally flat potential does not ensure that a rapidly moving field has already slowed enough to obey these approximations.

Count expansion using e-folds, the logarithm of the scale-factor ratio. For a monotonically rolling field, the slow-roll equations turn the time integral into an integral of potential divided by its derivative. The endpoint defined by a potential parameter reaching unity is itself an approximate prescription near the breakdown of slow roll. Moreover, mapping a measured comoving scale to its exit time requires the subsequent reheating history, rather than a universal requirement that every observed mode exited exactly sixty e-folds before the end.

\epsilon_V=\frac{\bar M_{\rm Pl}^2}{2}\left(\frac{V'}V\right)^2,\qquad \eta_V=\bar M_{\rm Pl}^2\frac{V''}V,\qquad 3H\dot\phi\simeq-V',\qquad N\simeq\frac1{\bar M_{\rm Pl}^2}\int_{\phi_{\rm end}}^{\phi_{\rm start}}\frac{V}{V'}\,d\phi

Potential slow roll requires

\epsilon_V\ll1

and

|\eta_V|\ll1

along an appropriate trajectory. The displayed integral assumes the chosen field orientation rolls from the stated start toward the end without a turning point.↔ Scroll wide equations horizontally.

Reheating needs energy transfer and thermalization

After acceleration ends, oscillations near a quadratic potential minimum can average to pressureless-matter behavior when their frequency exceeds the expansion rate. Decays or nonperturbative particle production can then transfer energy, but the first produced distribution need not be thermal. Establishing both kinetic and chemical equilibrium is an additional process. Consequently the end-of-inflation energy scale, maximum bath temperature, and temperature at completed radiation domination can be different quantities.

In a sudden-decay estimate with rapid thermalization, equating a decay rate to the expansion rate yields $T_R\simeq[90/(\pi^2g_*)]^{1/4}\sqrt{\Gamma\bar M_{\rm Pl}}$ . This assumes the transferred energy dominates as thermal radiation and uses its energy count. Delayed thermalization or prolonged production invalidates an automatic identification with the highest temperature reached. Entropy production also means that the earlier adiabatic cooling law cannot be extrapolated through reheating unchanged. An energy-density fourth root supplies an instantaneous upper scale, not a complete reheating history.

Work it through

An explicitly approximate quadratic-potential e-fold count

For the pedagogical potential $V=\tfrac12m^2\phi^2$ , use the potential slow-roll approximation and define the approximate endpoint by $\epsilon_V=1$ . Start at $\phi_{\rm start}=\sqrt{242}\,\bar M_{\rm Pl}$ . Find the endpoint, the estimated e-fold count, and the starting slope parameter. No claim about this potential fitting current cosmic microwave background data is implied.

Differentiate the potential to obtain $V'/V=2/\phi$ . Therefore $\epsilon_V=2\bar M_{\rm Pl}^2/\phi^2$ , and the chosen endpoint prescription gives $\phi_{\rm end}=\sqrt2\,\bar M_{\rm Pl}$ on the positive-field branch.
Use $V/V'=\phi/2$ in the e-fold integral. It gives $N\simeq(\phi_{\rm start}^2-\phi_{\rm end}^2)/(4\bar M_{\rm Pl}^2)=(242-2)/4=60$ . This is an estimate under the prescribed endpoint, not a precision integration through the end of acceleration.
At the start, the slope parameter is $2/242=1/121\simeq0.00826$ . The curvature parameter is equal for this quadratic potential. Their smallness supports the starting slow-roll estimate if the trajectory has reached its attractor; it supplies no microscopic control of omitted large-field operators.

The result

The prescribed endpoint is $\sqrt2\,\bar M_{\rm Pl}$ , the estimated expansion is sixty e-folds, and the starting slope parameter is $1/121$ . The particle-production and thermalization history remains unspecified.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

A positive-curvature damping analogy after inflation

Reset and set both controls to damping ratio $\zeta=0$ , natural frequency $\omega_0=1$ , then Step once: $x=\cos(0.5)\simeq0.8776$ . Reset and set both controls to $\zeta=1$ , $\omega_0=1$ , then Step once: $x=1.5e^{-0.5}\simeq0.9098$ . Both start at unit displacement and zero velocity. This positive-stiffness oscillator illustrates damped motion near a minimum only; it does not model slow-roll inflation, time-dependent Hubble friction, energy transfer, or reheating temperature.

Displacement x(t)Current time

Displacement x(t): 1
Velocity v(t): 0
Regime: underdamped

Damping ratio ζ0.2Natural frequency ω₀1.4

t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA damped classical displacement can illustrate post-inflation motion but does not establish the creation of a thermal bath.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The estimate converts decay-era energy into a thermal radiation density. Delayed equilibration or prolonged energy transfer changes the thermal history and the relation to a maximum temperature.

Your turn

Try it yourself

At a fixed scale factor, a stipulated post-inflation energy density is $\rho=16(\pi^2/30)g_*(10^8\,\mathrm{GeV})^4$ . Hold $g_*$ constant. Find the temperature if all energy instantly becomes thermal radiation. If instead only one sixteenth is in an equilibrated bath at that instant, find its temperature and explain whether reheating is complete.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Temperature is the fourth root of thermal radiation energy density; distinguish bath energy from total energy.

Compare with the solution

Complete instantaneous conversion gives $T=16^{1/4}10^8=2\times10^8\,\mathrm{GeV}$ . If only one sixteenth of the energy is thermal, the bath has $T=10^8\,\mathrm{GeV}$ . The other fifteen sixteenths still lie outside that bath, so this second temperature does not describe completed radiation domination. Expansion during a real conversion process and changing degrees of freedom would require an evolving energy-transfer calculation.

Keep the idea

Take it with you

The acceleration condition is exact, while potential slow roll is an additional approximation.
An e-fold estimate depends on the trajectory, endpoint prescription, and later expansion history.
Particle production and thermalization are distinct steps in reheating.

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