It is my understanding that inflation solves the horizon problem associated with an FLRW description of the universe by introducing a period during which $\varrho + 3p = \varrho(1 + 3w) < 0$, so that the co-moving Hubble radius (for $w \neq -1$) $$ (aH)^{-1} = H_0^{-1}a^{(1+3w)/2} $$ shrinks, thus allowing the widely separated parts of the universe to once have been in thermal contact.

I also believe one usually considers all the energy content of the universe to be bound up in the inflaton field. During reheating, the field oscillates about the minimum of its potential, and decays into matter and radiation fields.

I have two questions, and I do hope my understanding is not too flawed:

  1. At the end of the inflationary epoch, the inflaton field remains in thermal equilibrium, from the period of causal contact. The decay ought to be a non-equilibrium process, however. I suppose that the argument is that the fields then themselves thermalize, and because the different regions were in equilibrium they remain so. Although this seems very plausible, I wonder if there is any quantitative argument that shows this?

  2. The decay of the inflaton must be described by an interaction Lagrangian that couples the inflaton to the fields it decays into. Which, if any, are the concrete models that succeed at this, yielding an acceptable quantitative description of the fields of the radiation era? Do different models differ in predictions in any fundamental way?


1 Answer 1


At the end of inflation the inflaton is not in thermal equilibrium. Because the universe expands by a huge amount ($\sim e^{60}$) any thermal bath of particles is diluted away, leaving only the $0$-mode of the inflaton. Historically, it was thought that inflation was preceded by another thermal era (this is why decay of the inflaton was dubbed reheating) but in actual fact we don't really know anything about the pre-inflationary universe. In any case, because the observable universe originates in the a causally connected region the value of the inflaton field is the same everywhere (up to $\sim 10^{-5}$ inhomogeneities generated by inflation) and once the inflaton decays this uniformity is inherited by the decay products, i.e., the temperature is the same everywhere (again up to $\sim 10^{-5}$).

The reheating stage is not very well constrained experimentally. All you need for successful Big Bang cosmology is a thermal soup of relativistic particles with temperature above a few MeV (in order to achieve successful Big Bang Nucleosynthesis). On the other hand, we know from the bound on primordial gravitational waves that the scale of inflation could be as high as $\rho^{1/4} \sim 10^{16}$ GeV ($\sim T_\mathrm{reh}$). The range of allowed temperatures is huge so it's not that difficult to find successful models of reheating.

As far as concrete models go, these are myriad and there are many phenomenological differences, with non-perturbative effects often being important. However, since reheating is poorly constrained most of them are "successful" in the sense of being capable of producing the initial conditions for the Hot Big Bang.

A simple example might be an interaction Lagrangian $\delta \mathcal L = - \sigma \phi \chi^2$ between the inflaton $\phi$ and a boson $\chi$ (this could be the Standard Model Higgs for example) or a Yukawa interaction $\delta\mathcal L = - y\phi\bar\psi\psi$ between the inflaton and a fermion. As the inflaton oscillates at the end of inflation it produces pairs of quanta of the fields it is coupled to. If the couplings are sufficiently small that non-perturbative effects are unimportant then the decay rate of the inflaton is (assuming harmonic inflaton potential)

\begin{equation} \Gamma(\phi\rightarrow\chi\chi) = \frac{\sigma^2}{8\pi m}, \qquad \Gamma(\phi\rightarrow \psi\psi) = \frac{y^2m}{8\pi} \end{equation} where $m$ is the inflaton mass. Since the inflaton loses energy also to the expansion of the universe, the decay due to this particle production becomes effective when the decay time becomes smaller than the Hubble time ($H\sim \Gamma$). Provided that interactions between decay products and SM are sufficiently strong these thermalize and provide the thermal initial condition for the subsequent Big Bang evolution.

  • $\begingroup$ Is there a reference concerning the change from the historical view to the one you present? It seems to me that you are saying that all thermalization occurs after reheating, but does not the horizon problem then return again? Specifically, once the inflaton field decays, the Hubble radius is once again monotonically increasing. $\endgroup$ Dec 18, 2017 at 19:39
  • $\begingroup$ Also, if I may, what are the reasons for reheating being poorly constrained? I would have thought one could consider primordial element abundances to get at least approximate values for quark abundancies? Is it that there are too many intermediate steps? Or have I, in my limited understanding, overlooked something more fundamental? $\endgroup$ Dec 18, 2017 at 19:40
  • $\begingroup$ @ErikJörgenfelt At the end of inflation you have a big patch of the universe filled with just the inflation field and across this patch the value of the field is pretty much constant. It then doesn't matter whether different subregions of this patch have enough time to interact with each other subsequently - as the inflaton decays it will produce the same temperature everywhere because the starting value of the field (and hence the energy stored in it is the same). $\endgroup$
    – Philo
    Dec 18, 2017 at 21:08
  • $\begingroup$ @ErikJörgenfelt As far as understand primordial element abundances are generally insensitive to initial conditions such as quark abundances or intial temperature. All unstable particles decay once the temperature decreases enough and so all heavy quarks eventually decay into lighter particles and below the energy of about $100$ MeV (pion, muon decay) only protons, neutrons, electrons, positrons, neutrinos and photons remain. The proton-to-neutron ratio is maintained while weak processes are in equilibrium but once neutrinos decouple neutrons start to decay into protons... $\endgroup$
    – Philo
    Dec 18, 2017 at 21:10
  • $\begingroup$ @ErikJörgenfelt ...continued: This happens until nucleosynthesis when protons and neutrons combine to form nuclei (neutrons are stable when in nuclei). This basically determines the final ratio of neutrons to protons and nuclear physics determines the eventual abundances of elements. The exact details are very complicated and I don't know much about them, but the point, I think, is that this is all determined by known, sub-GeV physics. So as long as initially the temperature was higher than this the predictions of BBN are unaffected. $\endgroup$
    – Philo
    Dec 18, 2017 at 21:12

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .