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What is math showing inflation solves horizon problem?

  1. Aug 10, 2015 #1
    One aspect of the Horizon Problem is that the observed equal CBR temperature from opposite directions of the sky is a mystery. If we choose two points, say A and B, at opposite ends of a arbitrary diameter of the observable universe, at the present time the distance between A and B would be D = D(t0) (= approx. 93 Gly) (see https://en.wikipedia.org/wiki/Observable_universe), where t0 is the current age of the universe (= approx 13.8 Gy)

    If a(t) is the scale factor at time t, and a(t0∫) = 1, then D= 2* c* ∫SUB][/SUB]t(dt/a(t)).
    (See https://www.physicsforums.com/threa...e-of-observable-universe.827068/#post-5194532)

    Since the maximum distance light can travel since t=0 and t0 is D/2 (= approx. 46.5 Gly), nothing originating from A could ever reach B (or vise versa) at any time in the history of the universe, unless cosmic inflation can produce a form for a(t) that allows this to happen.

    I have three questions:
    (1) What are the values of the various components of Ω (= 1 by definiton) during inflation, say between ts (start of inflation) and te (end of inflation)?
    (2) What is the form of a(t) that results from cosmic inflation?
    (3) How does the math show that with this a(t) the correponding value for D(t0) would be sufficiently small for light to travel between A and B during the time interval between t = 0 and t = t0?

    I looked the related threads listed below, and none appear to answer the questions I am asking.
    "Explanation on how inflation solves the horizon and flatness problem"
    "Horizon problem - why do we need inflation?"
    "How inflation solves the horizon problem?
    "Inflation and particle horizon"
    "How inflation solves the horizon problem"
    This thread shows an interpretation that would be an explanation is it were OK, but I think there is a problem about constant energy density during inflation that I can't post there because the thread is closed. From https://en.wikipedia.org/wiki/Inflation_(cosmology) :
    During inflation, the energy density in the inflation field is roughly constant. However, the energy density in everything else, including inhomogeneities, curvature, anisotropies, exotic particles, and standard-model particles is falling, and through sufficient inflation these all become negligible. This leaves the Universe flat and symmetric, and (apart from the homogeneous inflation field) mostly empty, at the moment inflation ends and reheating begins.​
    Last edited: Aug 10, 2015
  2. jcsd
  3. Aug 10, 2015 #2
    Personally I always liked Barbers Rydens solution in page 243 "Introductory to Cosmology". It's a simple and elegant solution.

    first consider inflation as a temporary cosmological constant with equation of state w=-1. As inflation is dominant we can ignore the other EoS. ( matter etc)

    [tex]a(t)\propto H_i t[/tex]


    E is e folds [tex]N=H_i(t_f-t_i)[/tex]

    Barbera sets N=100 so the scale factor grow by 10^43.

    If we compare Omega and unity before and after inflation.
    [tex]\mid1-\Omega(t_f)\mid=e^{-2N}\mid1-\Omega_{t_i}\mid\sim 10^{-87}\mid 1-\Omega(t_i)\mid[/tex] so even if you had strong curvature before inflation it would be flattened like a proverbial pancake.

    The horizon at the end of inflation is
    [tex]d_{hor}(t_f)=e^n c(2t_i+H_i^{-3})[/tex]

    So if the horizon distance is 10^-28 m prior to inflation, after inflation it is roughly 0.8 pc.

    Without inflation the horizon distance is roughly 0.4 Mpc after inflation its roughly 10^43 Mpc. At least according to Ryden who sets inflation starting at GUT 10^-36 seconds with 100 e folds. Number of e folds needs to be greater than 60. This is sufficient to place the surface of last scattering in causal contact.

    I'm positive others here may have better solutions, One reason I enjoyed Barbera Rydens book is she sticks to the FLRW metric.

    (I still consider her book one of the better introduction level textbooks, particularly her single and multicomponent toy universes)

    Also keep in mind I didn't cover slow roll and reheating.
    Last edited: Aug 11, 2015
  4. Aug 10, 2015 #3


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    Buzz, I think you meant to write:
    $$d_{\mathrm{horizon}} = c \int^{t_0}_0 \frac{dt}{a\left(t\right)},$$
    except with a factor of 2 so it could be the diameter D instead of the radius (of the observable region)
    and somehow it got mixed up and appeared as:
    a(t0∫) = 1, then D= 2* c* ∫SUB][/SUB]t(dt/a(t)).
    Last edited: Aug 10, 2015
  5. Aug 10, 2015 #4
    I forgot to add Barbera Ryden uses the following for Inflation.

    [tex]\rho\phi=\frac{1}{2}\frac{1}{\hbar c^3}\dot{\phi}^2+V(\phi)[/tex]

    [tex]p\phi=\frac{1}{2}\frac{1}{\hbar c}\dot{\phi}^2-V(\phi)[/tex]

    Which you will notice the connection to the equations of state for scalar modelling.


    One reason the above formula is used is at the time of inflation all particle species are in thermal equilibrium. So we can model the universe at this time based on kinetic vs gravitational potential energy.
    Last edited: Aug 10, 2015
  6. Aug 11, 2015 #5


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    The potential energy term in those equations is not gravitational. It is a field self-interaction term.
  7. Aug 11, 2015 #6
    oops yeah your right, not sure why I included gravitational, thanks for the correction
    Last edited: Aug 11, 2015
  8. Aug 11, 2015 #7
    Thanks Modred for your excellent answers to my questions, and for the citation of the Rydens book.

    Thanks also to Marcus for noting my typo. I tend to avoid using the equation writing tool I have becasue it is awkward. I tried to instead use just the editing tools provided at this site, and I don't know why my proofreading missed my omiting "0".

    Thanks also to Orodruin for your post.

    Regards to all,
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