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Inflation, comoving Hubble radius and particle horizon

  1. Dec 1, 2015 #1
    I have a question regarding the exact formulation of the mechanism of Inflation.

    In the cosmology lecture notes by Baumann he uses ##\frac{d}{dt} \frac{1}{aH} < 0## as an definition of inflation. I see that it yields ## \ddot a > 0##, but my confusion lies in the interplay between the particle horizon and the comoving Hubble radius.
    Why do we require ## \frac{1}{aH} \ll \chi_p## in the early universe? He says
    But i dont see how that goes about.

    I am especially confused by figure 2.3 on page 33 (i cropped and attached it for your convenience),
    Screen Shot 2015-12-01 at 17.30.41.png

    I see that the horizon problem gets solved because the points p and q now have overlapping particle horizons, but what does the comoving hubble sphere have to do with it?
    They would still have overlapping particle horizons if i did not draw the Hubble sphere or if i drew it differently. Or is it not possible to have both things at the same time? (a different Hubble sphere AND overlapping particle horizons of p and q).
    To me it just seems that "adding more conformal time before the initial singularity and shift it to -inf. or less" s.t. p and q have overlapping particle horizons would do the job just fine, not worrying about the comoving Hubble radius.

    So, i am obviously missing something, but what is it?

  2. jcsd
  3. Dec 1, 2015 #2


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    The shrinking comoving Hubble radius is a consequence of the accelerated rate of expansion, and nothing more. Indeed, simply "adding more conformal time before the initial singularity..." is all you need to ensure that the particle horizons overlap, but there is no room to simply "add more" in the standard cosmology. We must postulate a period of inflation to give us the extra room, and the shrinking comoving Hubble scale is a result of this requirement.
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