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After the BICEP2 results, we now know that $n_s = 0.96$ and $r = 0.2$.

From what I understand, this fits extremely well with the basic chaotic inflation model given by $V(\Phi) = \lambda \Phi^4$.

We also know that amplitude of density fluctuations is $\approx 10^{-5}$ and Energy scale of inflation is around $10^{16}$ GeV.

My question: Given also this information, can we now make an educated guess for how many e-foldings happened during inflation ? Or at least a theoretical lower (upper ?) bound ?

PS: I am only referring to inflation of the patch of space that now contains our observable universe, not inflation of Universe as a whole.

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There is a lower bound of $\sim 60$ e-folds in order for inflation to solve the problems it was supposed to. – Danu Mar 18 '14 at 10:26
Yes, but that's kind of a lower bound based on homogenity etc. Will all these extra details in hand, can we make a better calculation ? – user42761 Mar 18 '14 at 10:31
@Danu isn't 60 cited as an upper bound? The flatness problem is solved as long as N>37 according to equation 19 in the reference cited in my answer below. – DavePhD Mar 19 '14 at 18:16
@DavePhD I know from research experience in the field that this is what theoreticians use as a lower bound for what a model must produce in order to be realistically viable. Actually, it's more like a minimum of $\sim 50-55$, but $60$ is considered the standard. – Danu Mar 19 '14 at 18:28
@Danu OK, I was seeing 60 as the upper bound for observable e-folds (observable in the CMB), rather than total e-folds. – DavePhD Apr 19 '14 at 22:02

1 Answer 1

$r = 8(1-n_s)-\frac{8}{N_*}$ for monomial potential inflation models, equation 206 of this reference:

where $N_*$ is "the number of e-folds between horizon crossing for observables scales and the end of inflation" (basically the observable number of e-folds).

so for $r=0.16$ (the dust-corrected value of the BICEP2 paper) and $n_s = 0.96$

$N_* = 50$

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