Age of a dark energy dominated universe

Question

In a flat universe that is dominated by dark energy (or cosmological constant), the Friedmann equation can be written as:

$H^2 = (\frac{\dot a}{a})^2 = \frac{8\pi G\varepsilon_{\Lambda}}{3c^2}$

Where $H$ is the Hubble parameter, $a$ is scale factor, $\dot a$ is scale factor's time derivative, $\varepsilon_\Lambda$ is the energy density of cosmological constant. Currently, our belief is that $\varepsilon_\Lambda$ is a constant, so the right side can be replaced by Hubble constant $H_0$

Rearrange Friedmann equation and integrating each side:

$a(t) = e^{H_0(t-t_0)}$

$t_0$ is the time right now.

From here, my textbook just says that a flat universe containing nothing but a cosmological constant is infinitely old, but I certainly do not see why this is the case, it is because cosmological constant will drive the universe to expand forever?

Answer 1

In non-dark-energy-dominated phases, $a\propto (t-t_1)^\gamma$ for some $\gamma>0$ and some reference time $t_1$. This means that $a=0$ when $t=t_1$, so that time $t_1$ is the beginning of the Universe. (Normally we take $t_1=0$.)

However, for dark energy domination, you have

$a(t) = e^{H_0(t-t_0)}$

which tells you that there is no finite time at which $a=0$. Instead, $a=0$ is only reached in the limit that $t\to-\infty$. That's the sense in which the Universe would be infinitely old (assuming it was always dark energy-dominated).

It's also worth noting that an eternally dark energy-dominated Universe is static. Its density is constant in time, and its metric can be written in a manifestly static form. Since there is no evolution, the Universe can have no beginning.