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A End of Inflation

  1. Apr 3, 2017 #1
    From cosmology, the end of inflation is usually defined as when the Hubble slow roll parameter is equal to one,

    ##\epsilon_{H} = -\frac{\dot H}{H^2} = -\frac{d\ln H}{dN} = 1##

    Is this really the definition of the end of inflation? Or is it for the standard cold inflation only? How about the eternal inflation model? Also, what should be the physical meaning of the Hubble slow roll parameter being equal to one?
     
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  3. Apr 3, 2017 #2

    bapowell

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    Yes, this is the definition for the end of inflation at the spacetime point at which it is evaluated. Eternal inflation refers to the condition that inflation is always going on somewhere in the initial inflationary patch (though certainly some regions within the initial patch of ceased inflating). It is generally characterized by [itex]P(k) \approx 1[/itex], i.e. the power spectrum amplitude is of order 1, indicating that inflaton field fluctuations are sufficiently large to keep driving the inflaton back up the potential somewhere, continuing inflation in that region.

    The condition [itex]\epsilon = 1[/itex] physically corresponds to an equation of state that separates accelerated from decelerated expansion.
     
  4. Apr 3, 2017 #3

    Chalnoth

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    The calculations used to produce inflation make use of an approximation, where a couple of the terms in the equations of motion can be ignored because they're tiny. One of them is that slow roll parameter, ##\epsilon_{H}##. Once this parameter gets large, the slow roll approximation no longer holds, so the universe can't be described as following the inflationary equations. The choice of ##\epsilon_{H} = 1## as the precise cutoff is, I think, arbitrary. But it's as good as any, since apparently the slow roll parameter grows rapidly once inflation ends. For example, if my calculations are correct, a radiation-dominated universe (which is what immediately follows inflation) has approximately ##\epsilon_H = 2##. Later, during matter domination, ##\epsilon_H = 1.5##.
     
  5. Apr 3, 2017 #4

    bapowell

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    No, it's not arbitrary. It relates directly to the threshold equation of state separating inflationary from non-inflationary expansion.
     
  6. Apr 3, 2017 #5
    I don't think that it is arbitrary, the condition for the comoving horizon for inflation to take place is given by,

    ##\frac{d}{dt}(aH)^{-1} < 1##

    we can differentiate the above equation and taking note that ##\dot H = \frac{\ddot a}{a} -H^2## so that,

    ##\frac{d}{dt}(aH)^{-1} = -(aH)^{-2} ( a \dot H + \dot a H) = -(aH)^{-2} (\ddot a - aH^2 + \dot a H) = -(aH)^{-2} ( \ddot a - aH^2 +aH^2) = -\ddot a (aH)^{-2}##

    we can express ##\ddot a## in terms of ##\dot H## by equating the second and last equation above,

    ##\ddot a = a \dot H + \dot a H \quad \Rightarrow \quad \frac{\ddot a}{a} = \dot H + H^2 = H^2(1 + \frac{\dot H}{H^2}) = H^2(1 - \epsilon_H)##

    where we define ##\epsilon_H = -\frac{\dot H}{H^2}##, and in order for there to have an accelerated expansion (##\ddot a >0##), we need ##\epsilon_H <1## otherwise accelerated expansion halts. So I think ##\epsilon_H## is not arbitrary but as what bapowell said, it relates directly to the threshold equation of state separating inflationary from non-inflationary expansion.


    Another question arose to me, why define ##\epsilon_H = -\frac{\dot H}{H^2}##, there can be other ways of defining a certain equation.
     
  7. Apr 4, 2017 #6

    Chalnoth

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    I think the more accurate statement that it separates accelerated expansion from decelerated expansion. So I guess it makes sense to use this as the cutoff. Inflation itself is a bit more than just accelerated expansion: it's nearly-exponential expansion. This is required to produce the nearly scale-invariant power spectrum that is observed.

    Certainly. Usually the answer is history. You can get at very close to the same exact equation using the deceleration parameter instead:
    https://en.wikipedia.org/wiki/Deceleration_parameter

    Note that ##\epsilon_H = 1+q##.
     
  8. Apr 4, 2017 #7
    So how does this condition justify say, eternal inflation? That inflation is always going on somewhere in the initial inflationary patch?
     
  9. Apr 4, 2017 #8

    bapowell

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    Did you see my reply to your first post? I answer this. Does it make sense?
     
  10. Apr 4, 2017 #9

    Chalnoth

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    I think you're asking about how we know when a particular patch is still inflating vs. no longer inflating?

    I think the details of that will depend upon the specific dynamics of the inflaton field in question. Presumably at some point if the locally-measured ##\epsilon_H## is too low, then the expansion will not be close enough to exponential to allow eternal inflation. My naive guess is that this happens while ##\epsilon_H## is still substantially less than 1, but when this threshold is reached (which will be inflaton model-dependent), then the entire region is guaranteed to pass ##\epsilon_H## after a short time, progressing inevitably towards reheating. It's still perfectly fine to call ##\epsilon_H## the "end of inflation", and the earlier threshold is just the point where that end becomes inevitable.
     
  11. Apr 4, 2017 #10

    bapowell

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    As I state in my initial response, epsilon tells you only about local expansion: less than one, inflating; greater than one, not. The closer epsilon is to zero the closer one is to true exponential (de Sitter) expansion. However, eternal inflation is a global phenomenon. The condition for eternal inflation is determined by the size of the field fluctuation: if it's large, then there will be regions that are kept inflating on account of large fluctuations driving the field back up the potential. This condition is generally taken to be [itex]P(k) \sim 1[/itex].
     
    Last edited: Apr 5, 2017
  12. Apr 4, 2017 #11
    So we can't define an "end" of inflation globally? Only to some specific patches that stopped inflating? Say, after 60~70 e-folds. Basically, my question is that can we use the epsilon = 1 as a definition of the end of inflation globally, but as you've said it is constrained only locally. I want to know if there is any on-going accepted definition of terminating inflation.
     
  13. Apr 5, 2017 #12

    bapowell

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    No, we cannot define an end to inflation globally. In fact, inflationary perturbations owe their existence to this fact: they are the result of inflation ending at slightly different times across the initial inflationary patch. So there actually is no notion of "terminating" inflation that holds globally -- only across regions of space in which [itex]\epsilon = 1[/itex] holds.
     
  14. Apr 5, 2017 #13

    Chalnoth

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    One problem with this discussion so far is that we're talking about a variable, ##\epsilon_H##, which is defined in terms of global variables, variables which assume a perfectly homogeneous, isotropic universe so that they can be defined globally.

    But the actual universe isn't like that: the density varies from place to place, and in the context of inflation these variations can become so extreme that it doesn't make sense to talk about global variables like the expansion rate at all. Instead, we have to consider local variables.

    The problem with that is it's vastly more complicated, and not really feasible to get into on a forum like this. Just figure that we're sort of waving our hands and saying that these things defined as global, such as the expansion rate, are instead going to be interpreted as local variables that only work well for a single Hubble volume (in an inflating universe, the Hubble volume is tiny compared to the overall size).
     
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