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I know that in our particular case the inflaton field expanded the volume of the universe while simultaneously maintaining a mass-energy density close to the critical density all the while, thus the flatness.

But why that number? Why didn't the inflaton field maintain a density which would have caused some other kind of curvature instead? I thought inflation was supposed to solve the fine tuning problem but I still run into the problem.

Basically, what makes inflation "choose" which energy density to maintain? And if it always picks the critical density (for flat curvature), why is that?

Ocsis2
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2 Answers2

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Instead of positing that inflation selected a density of the universe, it is better to say that inflation stretched the size of the universe so that its geometry became nearly flat. In so doing, the inflation field adjusted the critical density needed for a flat universe to the actual energy density of the universe, rather than vice versa.

kleingordon
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  • How did it do that? I thought the critical density was a constant? – Ocsis2 Apr 09 '12 at 12:17
  • @Ocsis2 no, the critical density depends on the Hubble parameter – kleingordon Apr 09 '12 at 20:17
  • the hubble parameter is the evolution of the scale factor over time? how did the inflaton field adjust that? – Ocsis2 Apr 10 '12 at 20:01
  • @Ocsis2 The expansion of the universe is governed by the Friedmann equation, $H^2 = 8 \pi G \rho /3 - kc^2/a^2$. The closer $H^2$ gets to $8 \pi G \rho /3$, the flatter the universe becomes. – kleingordon Apr 11 '12 at 03:39
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Have a look at http://en.wikipedia.org/wiki/Flatness_problem#Inflation

The key point is the equation:

$$(\Omega^{-1} - 1)\rho a^2 = \frac {-3kc^2}{8 \pi G}$$

In inflation the scale factor, $a$, increases enormously, but at the same time the energy density, $\rho$, remains roughly constant because it's dominated by the inflaton field, so the product $\rho a^2$ increases enormously. However the right hand side of the equation is just constant, so this means $(\Omega^{-1} - 1)$ must get vanishingly small i.e. $\Omega$ is driven towards unity. This result is quite general and doesn't depend on the fine details of how inflation works.

John Rennie
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    Thanks, that clears it up!

    I still get confused though. Is there a known mechanism by which the inflaton field "adds" energy to maintain a constant energy density?

     – Ocsis2 Apr 09 '12 at 12:21
  • The inflaton field is generally taken to be a property of space, so it has a constant energy density by definition. I admit this sort of bypasses your question without answering it, but then we have no idea of the origin of the inflaton field. In fact we're similarly ignorant of it's younger brother, dark energy, and this behaves in exactly the same way. – John Rennie Apr 09 '12 at 13:37