When numerically solving the Friedmann equations for varying cases of an open universe (i.e. $\Omega_0 < 1$) I get the following evolution plots,

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where the left plot is for an open universe containing only matter and the right plot is for an open universe containing 30% matter and 70% dark energy. As expected, in the dark energy case, the scale factor exponentially increases with time (accelerated expansion of the universe).

However, when looking at the current age of the universe (the time at which the scale factor $a=1$) it can be seen that for each age in the left plot, the corresponding line in the right plot has a significantly greater age. I would have thought that the inclusion of dark energy would cause the age of the universe to be smaller than the corresponding matter only case due to the negative pressure anti-gravity properties of dark energy pushing everything apart.

If anyone can explain the reason for this, I'd appreciate it.


In both plots the blue lines correspond to $\Omega_o = 0.1$, however the left plot is $\Omega_{m,0}=\Omega_0=0.1$, whereas the right plot is $\Omega_{\Lambda,0}+\Omega_{m,0}=\Omega_0=0.1$ (with $\Omega_{\Lambda,0}=0.07$ and $\Omega_{m,0}=0.03$), where $\Lambda$ and m denoted dark energy and matter respectively.

The time axes aren't very clear, apologies, the left plot axis starts at negative because I've set t=0 as the current age of the universe ($t(a=1)$) to more easily compare the shapes of the plots - I have taken these plots from a paper I've written so they may seem strange out of context.

  • $\begingroup$ Are the plots on the right flat universes with the same $\Omega_0$ as on the left? And why does $t$ go negative on the left? $\endgroup$
    – Pulsar
    Commented May 1, 2018 at 9:42
  • $\begingroup$ @Pulsar No the plots on the right are still open universes, with each colour corresponding to the same $\Omega_0$ as the left plots, however the left is matter only and the right contains dark matter - I've put an edit at the end of the post to explain this and the time axes more clearly. $\endgroup$
    – Brudalaxe
    Commented May 1, 2018 at 10:34

1 Answer 1


If I understood your question correctly, you're asking why a universe with dark energy is older than a corresponding universe without dark energy but with the same matter density and Hubble constant.

The current age of a FRW universe is $$ t = \int_0^1\frac{\text{d}a}{\dot{a}} = \int_0^1\frac{\text{d}a}{aH(a)}, $$ where, ignoring the radiation density, $$ H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_Ma + (1-\Omega_M -\Omega_\Lambda)a^2 + \Omega_\Lambda a^4}. $$ since $$ \Omega_\Lambda (a^4 - a^2) \leqslant 0 $$ for $\Omega_\Lambda\geqslant 0$ and $a\leqslant 1$, we get $$ \begin{align} H_\text{nde}^2(a) &= H_0^2[\Omega_Ma + (1-\Omega_M)a^2]\\ &\geqslant H_0^2[\Omega_Ma + (1-\Omega_M -\Omega_\Lambda)a^2 + \Omega_\Lambda a^4]\\ &= H_\text{wde}^2(a) \end{align} $$ for $a\leqslant 1$, where '$\text{nde}$' and '$\text{wde}$'' stand for 'no dark energy' and 'with dark energy', respectively. Consequently, $t_\text{nde}\leqslant t_\text{wde}$.

The key point here is that we kept the present-day value $H_0$ constant. By definition $H_\text{nde}(1) = H_\text{wde}(1) = H_0$. But the expansion rate of a universe with dark energy eventually starts to accelerate, which means that $\dot{a}_\text{wde} = aH_\text{wde}$ must increase more rapidly than $\dot{a}_\text{nde} = aH_\text{nde}$ at a given $a$. We can only reconcile these two facts if $\dot{a}_\text{wde} \leqslant \dot{a}_\text{nde}$ for $a\leqslant 1$, and that $\dot{a}_\text{wde}$ is 'catching up' with $\dot{a}_\text{nde}$ until they are equal at $a=1$.


The reasoning remains the same if you're comparing a universe where $\Omega_M = \Omega_0$ with a universe where $\Omega_M + \Omega_\Lambda = \Omega_0$ for some fixed value of $\Omega_0$. In this case $$ \Omega_\Lambda (a^4 - a) \leqslant 0 $$ and $$ \begin{align} H_\text{nde}^2(a) &= H_0^2[\Omega_0a + (1-\Omega_0)a^2]\\ &\geqslant H_0^2[(\Omega_0 - \Omega_\Lambda)a + (1-\Omega_0)a^2 + \Omega_\Lambda a^4]\\ &= H_\text{wde}^2(a). \end{align} $$


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