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I have assumed, for reasons a bit too detailed to go into here, that if $\Omega_{k}$, the curvature term in the FLRW metric, is equal to 1, then the radius of curvature is equal to 13.8 billion light years. Is anyone able to confirm (or otherwise) this?

If it helps, for simplicity, assume $\Omega_{k}$ is the only term.

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The Gaussian curvature is $$\frac{k}{a^2} = \frac83 πGρ - H^2 = \left(\frac{ρ}{ρ_c} - 1\right) H^2 = -Ω_k H^2$$

so the radius of curvature is $\displaystyle \frac{1}{H\sqrt{|Ω_k|}}$.

If $Ω_k=1$ then that's equal to the Hubble distance, and in our universe at the present time, the Hubble distance is roughly 14 billion light years. But in our universe, $Ω_k$ is nowhere near $1$.

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  • $\begingroup$ Could you please define H? It must have units of reciprocal length. Do you mean K = 1/(H Omega(k)^2)? So if Omega(k) =0.5, the radius of curvature would be 4 H? $\endgroup$
    – John Hobson
    Commented Jul 25, 2023 at 11:02
  • $\begingroup$ @JohnHobson $H$ is the Hubble parameter, which has units of reciprocal time. I'm using units with $c=1$. With explicit factors of $c$, the Gaussian curvature is $-Ω_kH^2/c^2$ and the radius of curvature is $c/(H\sqrt{|Ω_k|})$. $\endgroup$
    – benrg
    Commented Jul 25, 2023 at 15:38

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