Why is the current cosmological constant insufficient to solve the flatness problem?

The flatness problem can be very succinctly summarized by the fact that the density parameter $$\Omega(t)$$ has an unstable equilibrium point at $$\Omega = 1$$ (which corresponds to a Euclidean universe). If it is slightly higher than $$1$$ (resulting in a closed universe), $$\Omega(t)$$ will be driven higher and higher. If it is slightly lower than $$1$$ (resulting in an open universe), $$\Omega(t)$$ will be driven lower and lower.

Of course, accelerated expansion is able to drive the density parameter $$\Omega(t)$$ back to $$1$$ (I can add more detail if anyone asks). This is one of the motivations for inflation, which makes total sense.

Now we know that dark energy today drives an accelerated expansion, leading to a positive cosmological constant $$\Lambda = 1.1056\times 10^{-52}\text{ m}^{-2}$$ (taken from this wikipedia page as of 4/21/22).

My question is, if we assume a constant cosmological constant (yes you read that right) throughout the history of the universe, why is this not enough to explain the flatness problem?

• This question looks related. May 3, 2022 at 2:33

For a small cosmological constant the energy density of the early universe is NOT dominated by vacuum energy but rather radiation. The total energy density of the universe as a function of the scale factor $$a$$ is given by: $$\rho (a) = \rho_r/a^4 + \rho_m/a^3 + \rho_{\Lambda}$$ where the second term is matter density and $$\rho_{\Lambda} \sim \Lambda$$ is the constant vacuum energy density. As $$a \to 0$$ the first term dominates. Therefore, $$\rho_{\Lambda}$$ cannot heal the flatness problem (if $$\Lambda$$ is small).