Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why is inflation initial patch length ~ 1 Hubble length ?

  1. Jul 27, 2017 #1
    I'm asking many questions on inflation, I hope this is the last one...

    If inflation is correct, our observable Universe would have been a tiny homogeneous patch before inflation started (if it started at all). The length of that initial patch is estimated to be of the order of 1 Hubble length at the time before inflation (see for instance Guth's notes, and 1992 paper, or this video lecture). This turns out to be something like 10-26 cm or 10-28 cm, that kind of crazy size, depending on the sources.

    Why is the initial patch length estimated to be of the order of 1 Hubble length at that time ? I understand that a patch of that size is causally connected, providing homogeneity. But when we talk of the horizon distance problem of the standard Big Bang, we use the particle horizon as the criteria for causal connection, not the Hubble distance.
  2. jcsd
  3. Jul 27, 2017 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Section 2.2.2 "Hubble Radius vs. Particle Horizon" from Baumann's notes might (or might not) answer your questions about this.
  4. Jul 28, 2017 #3
    Thank you, that reading does address the issue. However I have the impression he just basically says that we use the Hubble radius simply because it's easier than using the particle horizon. Am I understanding it correctly ?

    Since the Hubble radius is easier to calculate than the particle horizon it is common to use the Hubble radius as a means of judging the horizon problem. If the entire observable universe was within the comoving Hubble radius at the beginning of inflation—i.e. (aIHI)−1 was larger than the comoving radius of the observable universe (a0H0)−1—then there is no horizon problem. Notice that this is more conservative than using the particle horizon since χph(t) is always bigger than (aH)−1(t). Moreover, using (aIHI)−1 as a measure of the horizon problem means that we don’t have to assume anything about earlier times t < tI .
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted