Based off of the study that found the universe has a positive curvature, how big would the whole universe be?

Question

This study found that the universe has a positive curvature https://www.nature.com/articles/s41550-019-0906-9. I didn't really want to buy it and see if it says how big the universe might be. I know it didn't have a low enough p-value to be considered confirmed. Is there a big margin of error for the whole size of the universe?

Answer 1

Surprisingly small!

(To me, at least.)

The paper referred can be found on the arXiv as Di Valentino et al. (2019).

As is customary with Planck results, the exact values of the cosmological parameters depend on how much confidence you put in auxiliary data, such as baryonic acoustic oscillation data (from surveys such as the 6dF Galaxy Survey, SDSS, and BOSS), supernova data, and Big Bang nucleosynthesis models. In this answer, I'm going to assume that the inferred value of the curvature parameter is given by their "$\Lambda\mathrm{CDM}$$+$$\Omega_K\!$" model, with their 99% confidence limits of $-0.007>\Omega_K>-0.095$, i.e. I'll use $$ \Omega_K = -0.0438^{+0.0368}_{-0.0512}. $$ The dynamics of the Universe is given by the Friedmann equations, which can be rearranged like this: $$ \Omega_K = -\frac{kc^2}{R_0^2\,a(t)^2 H(t)^2}, $$ where $k=+1$ for a closed universe, $a\equiv1$ today, $H=H_0$ is the Hubble constant today, and $R_0$ is the radius of curvature, and $c$ is the speed of light. The value of $H_0$ depends somewhat on the assumed value of $\Omega_K$; here I'll use $H_0 \sim 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.

I this case, we get that the radius of the Universe is roughly $$ R_0 \sim 67_{-21}^{+100}\,\mathrm{billion\,light\text{-}years}.\qquad (99\%\,\mathrm{C.L.}) $$

Accepting the "mainstream" cosmological parameter the observable Universe has a radius of $46.3\,\mathrm{Glyr}$. This result will change somewhat in the case of a closed Universe, but I can't seem to identify the author's preferred set of cosmological parameters. If we just use this value nonetheless, it means that we are currently able to observe a volume fraction of $$ f = \frac{V_\mathrm{obs}}{V_\mathrm{tot}} \simeq \left(\frac{46.3}{67_{-21}^{+100}}\right)^3 \sim 33_{-31}^{+67}\% $$ of the total Universe, i.e. we see somewhere between a few percent and everything, but most likely "a third".