TITLE: Why was it cold during inflation?
QUESTION [2 upvotes]: 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:

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.
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?

REPLY [0 votes]: 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.
{"set_name": "stack_exchange", "score": 2, "question_id": 455707}