Why was it cold during inflation?

Question

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?

Answer 1

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.

Answer 2

"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.

Answer 3

The ‘re’ in ‘reheating’ comes from assuming the universe was hot before inflation, so reheating restores the initial temperature $T$ (or at least, $T=T_{max}$) the temperature of the Hot Big Bang. During inflation, all the energy is in the inflaton.

So, a textbook kind of answer - say you decided to use the Planck 2018 slow roll upper constraint of $\sim6.6 \times 10^{15} \ GeV$ as $T$ (which is basically the hyperphysics timeline) , as well as say $\sim 60$ e-folds of inflation. The universe grows by $e^{60}\approx 10^{26}$ and the temperature reduces to $T/10^{26} \sim 0.06 eV$. Compared to $T, 0.06 ev$ 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 having a lower $T$).

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.