Is it correct that the radius of curvature scales with the scale factor $R(t) = a(t) R_o$? If so, in an expanding universe, the radius of curvature gets larger and larger, does that make the curvature smaller and smaller, i.e. towards being flat?

Yet, in the motivation for the inflation scenario, $$1 -\Omega(t) = - \frac{\kappa c^2}{R^2_o a^2(t) H^2(t)},$$ in the radiation or matter dominated era, the deviation from flatness $1 -\Omega(t) $ grows with time, which means the universe becomes more and more curved with expansion.

How do I reconcile these two seemingly contradicting intuition.


1 Answer 1


The radius of curvature $R(t)$ and the dimensionless energy density in curvature $1-\Omega(t)$ are two different measures of curvature, that need not (and do not) behave in the same way.

The reason that $R(t)$ increases with time is that every length scale will grow with time in an expanding Universe.

The reason that $1-\Omega(t)$ grows with time (in a Universe with matter+radiation+curvature) is that in the Friedman equation \begin{equation} \frac{3}{8\pi G}H^2 = \rho_{r,0} a^{-4} + \rho_{m,0} a^{-3} - \frac{3k}{8\pi G} a^{-2} \end{equation} the last term, representing the curvature, dominates at large $a$. This means that at late times, the contribution of curvature dominates the energy density budget of the Universe. Even though the energy density of curvature is decreasing, the energy density of matter and radiation is decreasing faster.

  • $\begingroup$ Well done Andrew. I like your clarity. $\endgroup$
    – Buzz
    Dec 13, 2022 at 16:12

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