A few more comments:

(1) Right, for all and, for the modes of interest, . The large effect on the power spectrum isn’t because we have a lot of particles, it’s because of the IR cascade. We’re not at all claiming that the production of iso-inflaton modes is particularly violent. The effect is all coming from what happens after the iso-inflatons are produced. The same can produce lots of IR by multiple rescatterings in less than an e-folding. Because of multiple rescatterings, you can have even though . (Recall that the large contribution to the power spectrum is from the cascading inflaton modes, not from the iso-inflaton particles.)

(2) If you agree with formula (8) in the text then the naive estimation is trivial. Set to make life simpler. Now set and (because we’re talking about modes near the horizon after about an e-folding). Then you trivially find which is very large. Putting the factors of (which make a difference) you find which is consistent with fig 3 up to factors order unity, as it should be.

(3) For the issue of generating super-horizon at the moment we may have to agree to disagree. I have already given the intuitive argument for why nothing interesting happens. Also, I have explained that we already checked by computing the Bogoliubov coefficients numerically.

NOTE: Of course we did that check in de Sitter space. You don’t need a computer to solve for the Bogoliubov coefficients with , it’s a text-book case. We didn’t put this in the paper because it’s not at all interesting.

However, I want to impress upon you that even if you did somehow generate super-horizon fluctuations in the tiny fraction of an e-folding that was light, it still wouldn’t change our final answer. There are several reasons:

FIRST, note that all the contribution to the IR inflation power comes from modes near (see equation 6). This is what I mean when I say that all the interesting modes are near . The fact that the power peaks near is precisely the reason that you get the factor of in equation (8).

SECOND, remember than was MASSIVE for some number of e-foldings before . I really want to stress this: heavy fields in dS have exponentially damped fluctuations on large scales! So any non-trivial effect would have to be exponentially large to overcome this damping. Heavy fields in dS behave very differently from light fields on large scales. Heavy fields do not acquire super-horizon correlations.

THIRD, even if you somehow did generate a lot of superhorizon fluctuations (you don’t, but let’s pretend) it would make our effect even stronger; you would get get a much larger effect on the large scale inflaton power spectrum.

(4) Thanks again for the opportunity to post and for all the discussion. Jolly good fun!

Best,

Neil

First off, the Schwinger result is not exponentially suppressed in the interesting regime .

No, but is of the order 1 for the modes you are interested in. There is no kinetic amplification of rescattering familar to us from the physics of preheating, just rigid gas of particles. But as I said above – the effect is what it is and it is definitely there.

Next, onto the issue of rescattering. What?s important is the power in produced IR inflaton modes. That this is larger than one might naively expect is precisely the central claim of the paper.

What is naive estimation? The particle production rate is proportional to , to initial occupation number of particles, and to the phase volume (just write a collision integral)/tex. That is exactly your formula (8).

But 1/10 of an e-folding is not enough to generate any significant effect outside the horizon. We?ve already verified this by computing the Bogoilubov coefficients numerically.

As I said above, effect is certainly small but your effect is also small, so one needs to honestly compare them. Regarding the Bogolyubov coefficients – did you calculate them with fixed ? If yes, I thought I already convinced you above that this is not very good idea since is of the order .

Well, thanks for the discussion anyway, it was really great. Without exaggeration, I enjoyed it a lot.

Cheers,
Dmitry.

For the record, the expansion is consistently included in the numerics, we just left it out of the analytical calculation in appendix A for simplicity.

Thanks a lot, Neil! Everything is much clearer to me now.

Cheers,
Dmitry.

Next, on scales there were no super-horizon fluctuations of for because it was massive. Recall that the mode function for a massive field in dS damps exponentially on large scales. Of course the mass is decreasing and the field is effectively light very near to . But of an e-folding is not enough to generate any significant effect outside the horizon. We’ve already verified this by computing the Bogoilubov coefficients numerically.

Next, onto the issue of rescattering. What’s important is the power in produced IR inflaton modes. That this is larger than one might naively expect is precisely the central claim of the paper. Intuitively you can understand this by noting (1) that production of very IR inflaton modes is very cheap, and, (2) that the same iso-inflaton can produce many inflaton modes. The time scale for this cascading is which is fast compared to the Hubble time. So there is lots of interesting dynamics before the dilution becomes important.

Obviously this intuitive argument is not completely convincing. That’s why in the paper we backed that argument up in two ways: (1) analytical computation of the correlators using QFT tools, and, (2) fully nonlinear lattice field theory simulations. Both agree with each other (see the power spectrum above, the dots are numerics and the solid lines are analytics). Moreover, both agree with the intuitive argument I made above.

Such a term dominates the effective frequency of the large scale inflaton modes but it doesn?t lead to uncontrolled super-horizon growth.

Ok, sure. What I want to say is that therу was some interesting near-horizon effect (as for any light field in dS background), but for a very short time – as long as the field was effectively massless. The effect was extremely weak, but on the other hand you also talk about exponentially small effect (Schwinger rate is exponentially suppressed). So, one has first to compare two small particle production effects and see which one is more important.

Suppose that your effect is more important. You say that it is anyway irrelevant, since what affects dynamics of the inflaton is rescattering. Let me now understand this statement, if you did not get bored too much by the discussion.

Particle production rate of from the diagram you draw is proportional to the particle number density of generated from the Schwinger process. This particle number density is a) small from the very beginning (according to (4) for interesting modes), b) number density is actually proportional to . Why is there some nontrivial kinetics before the particles of get exponentially diluted?

Cheers,
Dmitry.

It’s very misleading to refer to that term as “tachyonic”. Such a term dominates the effective frequency of the large scale inflaton modes but it doesn’t lead to uncontrolled super-horizon growth.

Best,

Neil

Let us work in conformal coordinates.

is equal to , that is, it is of the order near .

In principle, I could drop it out as you say since

,

while

,

so the curvature term is suppressed by the power for modes peaked near .

It is however not suppressed for modes with , moreover, it introduces tachyonic mass for them, which leads to their growth. So, I disagree that nothing interesting happens for modes with .

Cheers,
Dmitry.

Think about comparing the and terms in the equation of motion. If the field varies appreciably over a time interval then, to first approx, you can ignore the friction term.

The term gives the in my formula. For all but a tiny interval, this term is subdominant to the logarithm (from the mass). Even at the massless point, for the interesting modes, , this term is subdominant. At nothing interesting is going on. Remember that was heavy for some number of e-foldings before so it has no superhorizon fluctuations anyway.

you need to estimate a derivative, not – that is, you need to divide small by another small quantity – . The result is not so small quantity which has the order of magnitude of , i.e., square root of de Sitter curvature ( is also not small, especially compared to effective mass, it is of the order of dS curvature).

obviously there will be some small corrections to the Schwinger formula near

That was exactly my question from the beginning – why they are small and what is the small parameter that suppresses them. It seems to me that you neglect the curvature term exactly in the regime where it is important. Numerics is a good argument, I just don’t see physically why the curvature term can be neglected.

Cheers,
Dmitry.

Explicitly: setting you find that the logarithm is subdominant over an interval
. During that interval the scale factor changes by an amount for .

Obviously there will be some small corrections to the Schwinger formula near but those are irrelevant for our calculation because the power in produced peaks near which is inside the horizon.

One could re-compute the Bogiliubov coefficients including the Hubble friction numerically and compare to eqn (4). We already did that and, as expected, the effect is small.

although is large, logarithm is small in the regime your are interested in (and exactly equal to zero at ).

On the other hand, the term is of the order near and cannot be neglected during the conformal time interval

.

as follows from your formula for frequency.

Since is not terribly larger than , this seems to be more or less the interval where you produce particles due to non-adiabaticity breakdown.

Cheers,
Dmitry.

P.S. Sorry for being buggy.

No, the inflaton is light () but for the particle the situation is opposite. The iso-inflaton is not a light field.

The particle has mass

Because this is less than only over a time interval (less than an e-folding). So the scale factor is almost constant during that brief massless interval.

In you prefer to work in conformal time, the effective frequency is:

Up to slow roll parameters this gives:

(I’ve set the origin of time so that is the moment when .) Because the term coming from is subdominant.

Cheers,

Neil