In a closed universe without dark energy, it departs rapidly from flatness and become more curved over time. The expansion of the universe eventually stops and starts to collapses into a big crunch.

Will a closed universe with dark energy still collapse into a big crunch or will it expand forever?


3 Answers 3


The question whether or not a closed universe will collapse depends on the roots of the Friedmann equations. For $\Lambda$CDM models, these are $$\begin{align} \dot{a}^2 &= H_0^2\left(\Omega_{M,0}\,a^{-1} + \Omega_{K,0} + \Omega_{\Lambda,0}\, a^2\right),\tag{1}\\ \ddot{a} &= H_0^2\left(-\frac{1}{2}\Omega_{M,0}\,a^{-2} + \Omega_{\Lambda,0}\, a\right),\tag{2} \end{align} $$ where $\Omega_{M,0}$ and $\Omega_{\Lambda,0}$ are the present-day matter and dark energy parameters, we ignore the (small) contribution of radiation, and $\Omega_{K,0} = 1 - \Omega_{M,0} - \Omega_{\Lambda,0}$. We can rewrite $(1)$ as $$ f(a) = \frac{a\dot{a}^2}{H_0^2} = \Omega_{M,0} + \Omega_{K,0}\, a + \Omega_{\Lambda,0}\, a^3,\tag{3} $$ along with its derivative in $a$ $$ f'(a) = \Omega_{K,0} + 3\,\Omega_{\Lambda,0}\, a^2.\tag{4} $$ Consider the following example: enter image description here

This plot shows $f(a)$ for three models with $\Omega_{M,0}=2.5$. The green model, with $\Omega_{\Lambda,0} = 0.15$, expands forever. The blue model, with $\Omega_{\Lambda,0} = 0.05$, has a root at $a_0 = 1.8015$. Since $\ddot{a}<0$ at this root, $\dot{a}$ changes from positive to negative, so this model will collapse. The red model is a boundary case: here, both $\dot{a}$ and $\ddot{a}$ are zero at the same point, $a_0 = 2.3490$, so the expansion comes to a temporary halt, but then continues. To find these boundary models, we need to obtain an expression for $\Omega_{\Lambda,0}$ for a given value $\Omega_{M,0}$, such that $$ f(a_0) = f'(a_0) = 0, $$ where $a_0 > 1$. Instead of solving for $\Omega_{\Lambda,0}$ directly, we will solve for $\Omega_{K,0}$ first. By plugging $$ f'(a_0) = \Omega_{K,0} + 3\,\Omega_{\Lambda,0}\, a_0^2 = 0 $$ into $f(a_0) = 0$, we can eliminate $\Omega_{\Lambda,0}$ and obtain $$ 3\,\Omega_{M,0} + 2\,\Omega_{K,0}\,a_0 = 0.\tag{5} $$ We plug this back into $f'(a_0) = 0$ to eliminate $a_0$:

$$ 4\,\Omega_{K,0}^3 + 12\,\Omega_{\Lambda,0}\,\Omega_{K,0}^2\, a_0^2 = 4\,\Omega_{K,0}^3 + 27(1 - \Omega_{K,0} - \Omega_{M,0})\,\Omega_{M,0}^2 = 0, $$ or $$ \Omega_{K,0}^3 - \frac{27}{4}\,\Omega_{M,0}^2\,\Omega_{K,0} + \frac{27}{4}(1 - \Omega_{M,0})\,\Omega_{M,0}^2 = 0. $$ This is a cubic equation in $\Omega_{K,0}$ of Cardano form $t^3 + pt + q = 0$. Its three roots are

$$ \Omega_{K,0}^{(k)} = -\frac{3}{2}\Omega_{M,0}^{2/3}\left[e^{4\pi ik/3} \left((1 - \Omega_{M,0}) + \sqrt{1 - 2\,\Omega_{M,0}}\right)^{1/3} +\right. \\ \left. e^{-4\pi ik/3} \left((1 - \Omega_{M,0}) - \sqrt{1 - 2\,\Omega_{M,0}}\right)^{1/3}\right], $$ with $k=0,1,2$. If $\Omega_{M,0}\geqslant 1/2$, these three roots are real, and we can write

$$ (1 - \Omega_{M,0}) + \sqrt{1 - 2\,\Omega_{M,0}} = (1 - \Omega_{M,0}) + i\sqrt{2\,\Omega_{M,0}-1} = re^{i\theta}, $$ with

$$\begin{align} r &= \sqrt{(1 - \Omega_{M,0})^2 + 2\,\Omega_{M,0}-1} = \Omega_{M,0},\\ \theta &= \arccos\left(\frac{1 - \Omega_{M,0}}{\Omega_{M,0}}\right), \end{align} $$ so that $$ \Omega_{K,0}^{(k)} = -3\,\Omega_{M,0}\cos\left(\frac{\theta + 4\pi k}{3}\right). $$ If $\Omega_{M,0}\geqslant 1$, the $k=1$ root defines the collapse boundary. Indeed, $\pi/2\leqslant\theta < \pi$, so that $-3/2\,\Omega_{M,0} < \Omega_{K,0}^{(1)} \leqslant 0,$ and from $(5)$ we get $a_0 > 1$. One can further verify that the $k=2$ root is unphysical ($a_0 < 0$), while the $k=0$ root defines the boundary of models with no Big Bang ($a_0 < 1$).

Therefore, $$ \begin{align} \Omega_{\Lambda,0}^{(\text{collapse})} &= 1 + \Omega_{M,0}\left[ 3\cos\left(\frac{\theta + 4\pi }{3}\right) - 1\right] = 4\,\Omega_{M,0}\cos^3\left(\frac{\theta + 4\pi}{3}\right), \end{align} $$ where we used the identity $3\cos x = 4\cos^3 x - \cos 3x$. The plot below shows this boundary, between the red and the yellow area. The red dot corresponds with the red model in the first plot. Note that the $\Lambda$CDM model corresponding with our universe (black dot) will not collapse.

enter image description here

  • 1
    $\begingroup$ Seeing such complex notation correspond to my own conclusion (basically a guess) is kind of reassuring: I can't see why this answer hasn't been either accepted by, or commented on by, the OP. $\endgroup$
    – Edouard
    Commented Mar 15, 2021 at 4:42

A spatially closed universe can expand forever if the vacuum energy density is not zero.

Yes, a universe without dark energy will expand decelerated and collapse into a big crunch. This is still true if small amounts of vacuum energy, respectively $\Omega_\Lambda$ is added. The big crunch is avoided if the density parameter $\Omega_\Lambda$ exceeds a critical value. This value corresponds to a closed universe which expands forever. The formula hereto is given in Peacock's "Cosmological Physics" page 82. To answer your question with respect to dark energy is not as strict because its nature is unknown. Up to now the data are consistent with the assumption that the observed accelerated expansion of the universe is due to the cosmological constant $\Lambda$.


I believe you might be confusing the curvature of the space-time manifold with the spatial curvature, once you differentiate the two, one would also need to supply some reasonable initial conditions to make your question a bit more precise. In any case I will try answering your question as best as possible.

To be on the same page let us assume the $\Lambda$CDM-model of cosmology. You will see in the article that the basis for it is the FLRW-metric which contains a variable $k$ which can only take three values a priori, in your case for a closed universe the value of $k$ corresponds to $+1$. Now consider the Friedmann equation which comes out of Einstein's field equations and the FLRW metric: $$H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}{\rho} - \frac{kc^2}{a^2}+\frac{\Lambda c^2}{3}$$ So to answer your question exactly one would need to specify the matter content, that is specify $\rho$ or at least its scaling with $a$ (the scale factor). If it was the case, as it is now, that matter density scales as $a^{-3}$, you can say that eventually the Dark energy term $\Lambda$ will dominate the expansion. However you can ask whether we could reach the current state of the universe within a closed-universe scenario, but for that you will have to specify the content for different epochs. The only way in which you can contract as you can see from the equation is that the middle term of the right hand side dominates, and that would only happen for very specific stages (small $a$ but not small enough that so that the $\rho$ term dominates).

  • $\begingroup$ In your equation k needs to be in units of 1/m² (same as Λ) in order for the units to match, if you set it to 1 you get 1/sec²=m²/sec². $\endgroup$
    – Yukterez
    Commented Aug 9, 2023 at 2:23

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