<p>This is really simple once you get it:</p>

<p>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)</p>

<p>A partially inflated balloon has tight curvature, but as it is inflated, the curve decreases - the surface at any point becomes 'flatter'</p>

<p>From an article on this topic on <a href="http://scienceblogs.com/startswithabang/2009/07/31/the-size-of-the-universe-a-har/" rel="nofollow noreferrer">scienceblogs.com:</a></p>

<p><img src="https://i.sstatic.net/Mp40Y.jpg" alt="enter image description here"></p>

<p>The universe has expanded <strong>a lot</strong> since the big bang - any curves, ripples or inconsitencies have largely been flattened, so it is difficult to detect them now. <a href="http://www.space.com/25088-gravitational-waves.html" rel="nofollow noreferrer">This article</a> on space.com quotes:</p>

<blockquote>
  <p>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.</p>
</blockquote>

CC BY-SA 3.0
2014-05-13T21:01:50.523Z
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?
user1542
CC BY-SA 3.0
2014-05-13T21:02:47.077Z
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.
Rory Alsop
CC BY-SA 3.0
2014-05-13T21:14:20.043Z
Updated to link to one of many articles and papers on the subject
Rory Alsop
CC BY-SA 3.0
2014-05-13T23:30:32.143Z
@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.
harogaston
CC BY-SA 3.0
2014-05-14T01:10:25.593Z
@harogaston You don't get gravitational (or any other) waves without the change in local geometry.
TildalWave
CC BY-SA 3.0
2014-05-13T19:57:04.190Z
2014-05-13T21:14:01.990Z
2014-05-13T21:14:01.990Z
Rory Alsop
<p>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.</p>

CC BY-SA 3.0
2014-05-15T04:12:27.603Z
-1 this explanation might make some sense if critical density were zero, but that is not the case.
Stan Liou
CC BY-SA 3.0
2014-05-14T12:18:26.670Z
2014-05-14T12:18:26.670Z
adrianmcmenamin
<p>One of the <a href="http://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow">Friedmann equations</a> 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 $\rho$.</p>

<p>If we want a spatially flat universe, we can put it $k = 0$ and infer a <em>critical density</em> 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.</p>

CC BY-SA 3.0
2014-05-15T04:08:08.423Z
2014-05-15T04:08:08.423Z
Stan Liou
<p>How does inflation explain why the curvature in the early universe was flattened out and now results in a flat universe?</p>

<early-universe>
Flatness of the universe?
user1542
# Flatness of the universe?

(Asked by: username_0 on 2014-05-13)

How does inflation explain why the curvature in the early universe was flattened out and now results in a flat universe?

**Tags:** early-universe


---

## 3 answers

### Answer by username_1 on 2014-05-13. Score: 8

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:](http://scienceblogs.com/startswithabang/2009/07/31/the-size-of-the-universe-a-har/)

![enter image description here](https://i.sstatic.net/Mp40Y.jpg)

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](http://www.space.com/25088-gravitational-waves.html) 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.

> **Comments:**
> **username_0**: 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?  
> **username_1**: 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.  
> **username_1**: Updated to link to one of many articles and papers on the subject  
> **username_2**: @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.  
> **username_3**: @username_2 You don't get gravitational (or any other) waves without the change in local geometry.  

---

### Answer by username_4 on 2014-05-14. Score: 1

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.

> **Comments:**
> **username_5**: -1 this explanation might make some sense if critical density were zero, but that is not the case.  

---

### Answer by username_5 on 2014-05-15. Score: 0

One of the [Friedmann equations](http://en.wikipedia.org/wiki/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.


---

