How does inflation explain why the curvature in the early universe was flattened out and now results in a flat universe?
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3 Answers
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This is really simple once you get it:
The usual analogy here is to look at the surface of a balloon (which is a surface in 3D space, rather than 3D space itself, but the analogy still works)
A partially inflated balloon has tight curvature, but as it is inflated, the curve decreases - the surface at any point becomes 'flatter'
From an article on this topic on scienceblogs.com:
The universe has expanded a lot since the big bang - any curves, ripples or inconsitencies have largely been flattened, so it is difficult to detect them now. This article on space.com quotes:
The Harvard-Smithsonian study spotted gravitational waves as ripples in space-time possible left over from the rapid expansion of the universe (called inflation) right after the Big Bang nearly 13.8 billion years ago.
answered May 13, 2014 at 19:57
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$\begingroup$ Could it be that because the universe is so vast, we don't really see the curve, but there could still be one? How do we know for sure that every curve, ripple, or inconsistency was completely flattened out? $\endgroup$– user1542Commented May 13, 2014 at 21:01
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$\begingroup$ Your comment makes no sense. We know that the ripples have not been entirely flattened out, as I stated in my answer. There is a curve, but it is exceedingly flat - as I stated in my answer. $\endgroup$ Commented May 13, 2014 at 21:02
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$\begingroup$ Updated to link to one of many articles and papers on the subject $\endgroup$ Commented May 13, 2014 at 21:14
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$\begingroup$ @RoryAlsop I think gravitational waves doesn't provide any evidence for the flatness of the universe, as you point out towards the end of your answer. What I mean is, the existence of those gravitational waves do provide some evidence for the inflationary process the universe overcame in the very beginning but it tells us nothing about it being curved or flat. $\endgroup$ Commented May 13, 2014 at 23:30
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$\begingroup$ @harogaston You don't get gravitational (or any other) waves without the change in local geometry. $\endgroup$ Commented May 14, 2014 at 1:10
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Curvature of spacetime is predicted by General Relativity and is a function of the density of the universe. Because inflation rapidly expands the volume of the universe it also rapidly decreases the density, so flattening space time.
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$\begingroup$ -1 this explanation might make some sense if critical density were zero, but that is not the case. $\endgroup$ Commented May 15, 2014 at 4:12
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One of the Friedmann equations describing the cosmological evolution of the universe is: $$H^2 \equiv \frac{\dot{a}^2}{a^2} = \frac{8\pi G}{3}\rho - \frac{k}{a^2}\text{,}$$ where $a$ is the scale factor describing by how much the space expands (related to the Hubble parameter by $H = \dot{a}/a$), and $k\in\{-1,0,+1\}$ is a parameter corresponding to whether the spatial curvature is negative, zero, or positive, respectively. Note that this parameter doesn't tell us the magnitude of the spatial curvature, only its sign, e.g., a very slightly positively curved universe would still have $k=+1$. Also, the cosmological constant is considered to be part of the energy density $\rho$.
If we want a spatially flat universe, we can put it $k = 0$ and infer a critical density that the universe would have to have in order to be flat: $$\rho_\text{c} = \frac{3H^2}{8\pi G}\text{.}$$ Using this value, we can re-write the first Friedmann equation as: $$\rho_\text{c} = \rho - \frac{3k}{8\pi G a^2}\text{,}$$ or in terms of the density parameter $\Omega = \rho/\rho_\text{c}$: $$(\Omega-1)\rho_\text{c}a^2 = \frac{3k}{8\pi G}\text{.}$$ What inflation does is exponentially increase scale factor $a$, which keeps $H$ and therefore $\rho_\text{c}$ constant. Since the right-hand side is always constant, that means $(\Omega-1)\rightarrow 0$, i.e., the universe is driven towards the density that makes it flat.
