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Consider a universe with a nonzero curvature and matter.

One can write the Friedmann Equation in this universe as such:

$$\frac{H(t)^2}{H_0^2} = \frac{\Omega_0}{a^3}+\frac{1-\Omega_0}{a^2}$$

Where $H(t)$ is the Hubble parameter, $\Omega_0$ is the density parameter for matter (since it is the only component in the universe aside curvature, I will ignore the subscript) and $a$ is scale factor.

Now consider a universe where $\Omega_0$ is greater than 1 (positive curvature), leading in a negative term of $\frac{1-\Omega_0}{a^2}$. It is obvious that the universe will expand first, then come to a halt when $H(t) = 0$.

However, when $H(t)$ turns negative, LHS of the equation is again positive, which is no difference from the expanding scenario. Thus I conclude this universe will expand forever, however this universe will meet its end in the Big Crunch. Which means the universe will first expand until it reach a maximum of scale factor, then contract until everything crash together.

Where did I made a mistake?

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  • $\begingroup$ @paulina, I conclude that the universe will expand forever, not contract. That is why I am asking where did I made a mistake. $\endgroup$
    – Polaris5744
    Commented Jul 2 at 0:24

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The first Friedmann equation says nothing about the sign of $H$. During expansion, it says how fast the universe is expanding. During contraction, it says how rapidly the universe is contracting. Think of it like a conservation-of-energy expression in classical mechanics. If you know a particle's conserved energy (analogous to $1-\Omega_0$) and position (analogous to $a$), that tells you the particle's speed (analogous to $|aH|$) but not the direction of its velocity (analogous to $aH$).

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  • $\begingroup$ So how did cosmologist find that the universe is contracting, by using the fluid equation? $\endgroup$
    – Polaris5744
    Commented Jul 2 at 0:31
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    $\begingroup$ At the point where $H=0$ (so $\dot a=0$), you can use the second Friedmann equation to show that $\ddot a<0$. Then you know the sign of $\dot a$ (and hence $H$) is changing from positive to negative, and so it must be negative thereafter. $\endgroup$
    – Sten
    Commented Jul 2 at 0:40
  • $\begingroup$ Alternatively, you can simply integrate the second Friedmann equation from the outset. The analogy with classical particle mechanics is that the first Friedmann equation is like $E=\frac{1}{2}m\dot{\vec{x}}^2+m\phi(\vec x)$, while the second Friedmann equation is like $\ddot{\vec{x}} = -\nabla\phi(\vec x)$. The first is convenient, but all you need in principle is the second. $\endgroup$
    – Sten
    Commented Jul 2 at 0:40

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