<p>What exactly did the book say? It all depends on what you mean/define as "curvature". What you describe appears to be a description of the behaviour of $\Omega$. Inflation does indeed drive $\Omega$ towards unity and simultaneously flattens space because the radius of curvature grows exponentially bigger. </p>

<p>If $\Omega &lt; 1$ at some early epoch then it should decrease quickly with time such that $\Omega &lt;&lt; 1$ in the present day - this means that the universe has negative curvature, but does not mean it is becoming more curved.</p>

<p>In the Friedmann equation, the curvature parameter $k$ is a constant $(1,0,-1)$
$$H^2 = \frac{8\pi G\rho}{3} - \frac{kc^2}{a^2}$$</p>

<p>Here, the spatial curvature is $k/a^2$ and the radius of curvature is $a$ if $k=+1$. Thus as the universe expands and $a$ gets larger, any curvature becomes smaller.</p>

<p>In a little more detail - one can write the above equation in terms of the density parameter $\Omega$, the ratio of density to the critical density $3H^2/8\pi G$:
$$(\Omega ^{-1}-1)\rho a^{2}={\frac {-3kc^{2}}{8\pi G}}$$</p>

<p>During inflation, the energy density $\rho c^2$ remains constant as $a$ grows exponentially. In order to keep the left hand side equal to the right hand side (which is just a collection of constants), then $\Omega$ must be driven very close to unity, while $k/a^2$ will tend towards zero.</p>

<p>After inflation then $\rho$ will vary with $a$ depending on whether the expansion is dominated by matter ($a^{-3}$) or radiation $(a^{-4})$. In both these cases $\rho a^2$ will decrease as the universe expands, such that if $k \ne 0$, then $(\Omega^{-1} -1)$ must increase, which means that $\Omega$ must either grow or shrink away from unity. But $k/a^2$ continues to get smaller as the universe expands. </p>

CC BY-SA 3.0
2016-08-07T15:08:38.777Z
The book says "... if the Universe is close to critical now, it must have been extremelyextremely close to critical in the past...". Am I wrong by saying that it means curvature grows with time?
velut luna
CC BY-SA 3.0
2016-08-07T14:41:37.790Z
2016-08-07T14:52:25.097Z
2016-08-07T14:52:25.097Z
ProfRob
<p>Why it is said that in standard Big Bang model space curvature grows with time when the Universe expands? My intuition is that it should be the opposite.</p>

<p>I saw this in a book talking about the problem of standard Big Bang model and why we need inflation theory. It says that the Universe is observed to the nearly flat now. And because the space curvature grows with time, it should be even flatter in the past, which cannot be explained. But the inflation theory solves the problem because even if space is initially far from flat, inflation will smear out the curvature and make it flat.</p>

<big-bang-theory><cosmological-inflation>
Standard Big Bang model and space curvature
velut luna
# Standard Big Bang model and space curvature

(Asked by: username_0 on 2016-08-07)

Why it is said that in standard Big Bang model space curvature grows with time when the Universe expands? My intuition is that it should be the opposite.

I saw this in a book talking about the problem of standard Big Bang model and why we need inflation theory. It says that the Universe is observed to the nearly flat now. And because the space curvature grows with time, it should be even flatter in the past, which cannot be explained. But the inflation theory solves the problem because even if space is initially far from flat, inflation will smear out the curvature and make it flat.

**Tags:** big-bang-theory, cosmological-inflation


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## 1 answers

### Answer by username_1 on 2016-08-07. Score: 3 | Accepted answer

What exactly did the book say? It all depends on what you mean/define as "curvature". What you describe appears to be a description of the behaviour of $\Omega$. Inflation does indeed drive $\Omega$ towards unity and simultaneously flattens space because the radius of curvature grows exponentially bigger.

If $\Omega < 1$ at some early epoch then it should decrease quickly with time such that $\Omega << 1$ in the present day - this means that the universe has negative curvature, but does not mean it is becoming more curved.

In the Friedmann equation, the curvature parameter $k$ is a constant $(1,0,-1)$
$$H^2 = \frac{8\pi G\rho}{3} - \frac{kc^2}{a^2}$$

Here, the spatial curvature is $k/a^2$ and the radius of curvature is $a$ if $k=+1$. Thus as the universe expands and $a$ gets larger, any curvature becomes smaller.

In a little more detail - one can write the above equation in terms of the density parameter $\Omega$, the ratio of density to the critical density $3H^2/8\pi G$:
$$(\Omega ^{-1}-1)\rho a^{2}={\frac {-3kc^{2}}{8\pi G}}$$

During inflation, the energy density $\rho c^2$ remains constant as $a$ grows exponentially. In order to keep the left hand side equal to the right hand side (which is just a collection of constants), then $\Omega$ must be driven very close to unity, while $k/a^2$ will tend towards zero.

After inflation then $\rho$ will vary with $a$ depending on whether the expansion is dominated by matter ($a^{-3}$) or radiation $(a^{-4})$. In both these cases $\rho a^2$ will decrease as the universe expands, such that if $k \ne 0$, then $(\Omega^{-1} -1)$ must increase, which means that $\Omega$ must either grow or shrink away from unity. But $k/a^2$ continues to get smaller as the universe expands.

> **Comments:**
> **username_0**: The book says "... if the Universe is close to critical now, it must have been extremelyextremely close to critical in the past...". Am I wrong by saying that it means curvature grows with time?  

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