During inflation, spacetime is approximately de Sitter and there is a cosmological horizon; in analogy with Hawking radiation, there should be a de Sitter temperature $T \sim H$. This temperature has nontrivial effects, for it is responsible for the fluctuations measured in the CMB, as covered in almost all cosmology textbooks.

In most inflationary models $H$ is taken to be very high, so it seems that during inflation, the temperature should be high. However, in every treatment I have seen, the default assumption seems to be that it's cold during inflation. Because some comments and answers have disagreed on this point:

  1. The process by which a thermal plasma is created at the end of inflation is called reheating, and the reheating temperature is at most of order $H$. If we already had a thermal plasma of temperature $H$ at this point, there would be no need to talk about reheating; the universe would already be hotter than the maximum possible reheat temperature.
  2. The scenario where there is a thermal plasma during inflation is called warm inflation. In this case, the temperature has to be put in separately, and papers about warm inflation emphasize that it is not the default; usually there is no temperature.

If there isn't a temperature $T \sim H$ during inflation, how is that consistent with the de Sitter temperature? If there is a temperature $T \sim H$ during inflation, why do we bother talking about reheating and warm inflation?

  • $\begingroup$ cold is a relative term. If the energy density of de Sitter thermal radiation is tiny compared with vacuum energy, it could be ignored as a first approximation, even if it is large by the scales of later epochs. $\endgroup$
    – A.V.S.
    Jan 21, 2019 at 18:36
  • $\begingroup$ @A.V.S. I was thinking that could be the case, that there is a thermal plasma that just isn't very important, but then what is "warm inflation" supposed to be about? Proponents of warm inflation say there is a thermal plasma, and place themselves in contrast to everybody else, who says by default there isn't. $\endgroup$
    – knzhou
    Jan 21, 2019 at 18:44
  • $\begingroup$ For warm inflation thermal quanta are produced by another mechanism (not by horizon), as a consequence the role of thermal plasma could be larger, smoothly interpolating from cold to radiation domination. $\endgroup$
    – A.V.S.
    Jan 21, 2019 at 18:53

3 Answers 3


If the inflaton is expected to be coupled to the thermal particle production associated with the de Sitter horizon, then indeed these degrees of freedom should be accounted for, e.g. via the thermal inflation scenario.

  • $\begingroup$ I'm not sure what this answer is saying -- can you give a bit more detail? Are you saying that the de Sitter thermal particle production must create a thermal plasma of temperature $H$? $\endgroup$
    – knzhou
    Jan 22, 2019 at 16:28
  • $\begingroup$ Yes. But how that plasma affects the inflaton is model-dependent. $\endgroup$
    – bapowell
    Jan 22, 2019 at 18:51
  • $\begingroup$ But if the plasma is already there, why bother talking about reheating? Isn't the universe automatically heated? $\endgroup$
    – knzhou
    Jan 22, 2019 at 19:02
  • $\begingroup$ My understanding is that the de Sitter plasma could well reheat the universe. This is an old idea I think originally advanced by Ford (see journals.aps.org/prd/pdf/10.1103/PhysRevD.35.2955). Gravitational particle creation of this type is used to reheat the universe in quintessential inflation where the inflaton does not decay. But, in models where there is decay, it's possible that the decay products dominate the density of the de Sitter bath and so the latter isn't relevant. I'm not sure about that point, though. $\endgroup$
    – bapowell
    Jan 22, 2019 at 19:30

"Cold" and "hot" is model dependent.

There are models where the effect of the composition in particles and the inflaton field is taken into account for the temperature.

In this analysis , the universe starts with infinite temperature which falls.

Also in this timeline it starts very hot and it cools continuously due to the expansion and fall of density.


So it is cooler, due to the falling density of energy but not "cold"

Here is a model that uses Hawking radiation:

We consider the effect of the Gibbons-Hawking radiation on the inflaton in the situation where it is coupled to a large number of spectator fields. We argue that this will lead to two important effects - a thermal contribution to the potential and a gradual change in parameters in the Lagrangian which results from thermodynamic and energy conservation arguments.

If one is really interested one should spend the time to look at the mathematical details of the models.


The ‘re’ in ‘reheating’ comes from assuming the universe was hot before inflation, so reheating restores the initial temperature $T$. During inflation, all the energy is in the inflaton.

So, a textbook kind of answer - say you decided to use the MSSM prediction of $10^{16}GeV$ as $T$ (which is basically the hyperphysics timeline) , as well as say $\sim 53$ e-folds of inflation. The universe grows by $e^{53}\approx 10^{23}$ and the temperature reduces to $T/10^{23} \sim 100 eV$. Compared to $T, 100 ev$ is essentially zero, thus ‘cold’. However, this is still an active research area. For one, as pointed out by Matt Strassler, assuming the universe was hot before inflation is just a guess, and also, the ‘cold’ temperature should be even colder, a tiny fraction above absolute zero (one way to achieve that is by increasing the number of e-folds).

Finally, at the end of reheating, one can estimate that essentially all the energy density of the universe is in radiation, so the standard equation (5.57) i.e. how the OP’s idea that $H \sim T$ comes into it, is: \begin{equation} \notag H(T)^2 = g_{*}(T) \frac{\pi^2}{90} \frac{T^4}{M_{P}^2} \end{equation}

where $ g_{*}(T)$ is the effective number of degrees of freedom and has the value g∗(T) = 106.75 in the Standard Model at high temperatures. $M_{P}$ is the reduced Planck mass.


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