0
$\begingroup$

Recently I have watched videos on how the inflation theory solves some problems that arise from the Big Bang Theory. In particular, I have a few questions regarding the flatness problem and how the inflation theory manages to resolve it. But first I would like to explain my understanding of the flatness problem.

The flatness problem occurs where the current universe has a density so very close to the critical density, but that is weird because that would require the universe to have omega almost = 1 from the VERY beginning (and this probability is extremely small). It is said that if omega 'much' smaller/bigger than 1, then it would drive the universe to expand/contract exponentially.

My questions:

  1. How is this process of exponentially expansion/contraction different from the evolution of closed/open universe? Does inflation theory straight-up rejects any possibility of open/closed universe (which have omega more/less than 1)?

  2. How is inflation different from expansion? In the sense that if there was no inflation but only expansion, then the density would deviate from critical density exponentially, but somehow if there was inflation, then the inflation would drive whatever initial density towards critical density, why's that?

I would really appreciate any clarification on this matter!

$\endgroup$
1
$\begingroup$

Inflationary expansion is characterized by the fact that the expansion rate, quantified in terms of the scale factor, $a(t)$, is accelerating. That is, $\ddot{a} > 0$.

1) Inflation imposes no requirements on the global geometry of the universe. If the universe is closed, inflation just makes a much bigger closed universe. Locally, this has the effect of making the universe look more flat. And similarly for an open universe.

2) Inflation is special kind of expansion, namely, expansion with $\ddot{a} > 0$. From the Friedmann equation which governs the evolution of the homogeneous and isotropic universe, we can write the density parameter in terms of the scale factor as $$\Omega(a) = \left[1-3 ka^{1+3w}\right]^{-1},$$ where we've set the reduced Planck mass to unity. Here $k$ is the curvature parameter and $w$ determines the equation of state of the cosmological fluid, via $p = w\rho$ where $p$ is its pressure and $\rho$ is the density. For non-relativistic matter, $w = 0$; this is an example of "normal" un-accelerated expansion. You can see that if, say, $\Omega < 1$ initially, with $w=0$ the density parameter will get driven to zero as the universe grows under matter-dominated expansion.

If, however, the universe is inflating, we have $w < -1/3$. Of particular interest is the case $w=-1$, for which the inflationary energy density is constant (so-called de Sitter expansion). Here, you can see that $\Omega$ gets driven to 1. Here is a simple plot showing this behavior with $\Omega = 0.5$ initially, with $a$ along the x-axis and $\Omega(a)$ along y,

enter image description here

These same arguments carry over to closed models as well.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! Your explanation was clear, even though the extent of my knowledge did not include the density parameter equation and everything afterwards. I'd like to ask if you know of any source that is good for beginners to understand the latter parts? $\endgroup$ – Clara May 2 at 10:03
  • $\begingroup$ If you know a little calculus, this could be a useful reference: ned.ipac.caltech.edu/level5/Peacock/Peacock3_2.html. $\endgroup$ – bapowell May 2 at 12:44
  • $\begingroup$ One more question--according to the equation Ω(a)=[1−3ka(1+3w)]^−1, if w=0(in the case of normal un-accelerated expansion), then Ω(a)=[1−3ka]^−1. It would seem that no matter k>0 or k<0, as 'a' approaches infinity, Ω(a) will approach 0. How then can big crunch happen? $\endgroup$ – Clara May 3 at 8:54
  • $\begingroup$ The initial value of $a$ is only implicit in that expression. Really, $k \rightarrow k/a_i$. Try again with, say, $\Omega = 1.5$ at $a_i = 1$. When do you get your Big Crunch? $\endgroup$ – bapowell May 3 at 14:37
  • $\begingroup$ Do I sub a as 1, therefore 1.5=[1-3k]^-1, hence big crunch happens when initial k=1/9 ? $\endgroup$ – Clara May 5 at 3:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.