Suppose I'm modelling a Universe with non-zero curvature, filled with matter, radiation and dark energy (further described by quintessence). The appropriate Friedmann equation would be of the form:

$$ H^{2}=\Big(\frac{\dot{a}}{a}\Big)^{2}=H_{0}^{2}\big(\Omega_{m,0}a^{-3}+\Omega_{r,0}a^{-4}+\Omega_{\Lambda,0}a^{-3(1+w)}+\Omega_{k,0}a^{-2}\big),$$

where $\Omega_{k}=-\frac{k}{a^{2}H^{2}}$. Now suppose I'm interested in the turning point when the Universe goes from an accelerating to a decelerating one (or vice versa) - I would achieve such a thing by setting the deceleration parameter $q=-\frac{\ddot{a}(t)a(t)}{\dot{a}^{2}(t)}$ to zero (which is in fact nothing more than setting $\ddot{a}(t)$ to zero). Taking the acceleration equation:

$$ \frac{\ddot{a}(t)}{a(t)}=-\frac{4\pi G}{3}\sum_{i}\rho_{i}\big(1+3w_{i}\big), $$

I see no way in which the curvature might manifest itself. Is this to be expected? Or did I make some sort of banal mistake in the process?


If you re-arrange the Friedmann equation (multiplying by $a^2$ and moving terms around) into $$ \dot{a}^{2} - H_{0}^{2}a^2\big(\Omega_{m,0}a^{-3}+\Omega_{r,0}a^{-4}+\Omega_{\Lambda,0}a^{-3(1+w)}\big) = H_{0}^{2}\Omega_{k,0}, $$ it resembles a classical mechanic equation where the Hubble term $$ \dot{a}^{2} $$ is the kinetic energy. The matter + radiation + CC terms $$ - H_{0}^{2}a^2\big(\Omega_{m,0}a^{-3}+\Omega_{r,0}a^{-4}+\Omega_{\Lambda,0}a^{-3(1+w)}\big) $$ constitute the potential energy. The punch line is that the curvature term $$ H_{0}^{2}\Omega_{k,0} $$ amounts to the total (conserved) energy, which of course does NOT dictate the acceleration/deceleration of the Universe (manifested as no impact on the acceleration equation, since $w_{k}=-1/3$ hence $1+3w_{k}=0$). Acceleration/deceleration is instead controlled by the shape of the aforementioned potential energy.

The curious fact is that the total energy (curvature) $\Omega_{k,0}$ is observed to be very close to zero, which prompted some ad hoc explanations, such as the inflationary scenario.


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