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

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?

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".

• So the circumference of the whole universe is 4.52 times larger than the diameter of the observable universe? Nov 12 '19 at 17:21
• How big the would the observable universe be compared to the total universe? Nov 12 '19 at 17:30
• Geometry for a closed universe is a little different than for a flat one, but if I've calculated correctly, the difference is on the ~1% level, so roughly, with a diameter of the obs. Universe of ~2×46 Glyr, and a circumference of the whole Universe of $C\simeq2\pi R_0\simeq434$ Glyr, you're right. The obs. Universe then comprises roughly $(43/67)^3 \sim 25\%$ of the total Universe.
– pela
Nov 12 '19 at 21:45
• @user1781498 Observable universe is not ~14 Glyr - the light we see from distant objects is more like 46 Glyr away now - the light has been traveling for nearly 14 glyr but the distant object has been moving away from us all that time and is now very much further away. See Davis and Lineweaver for a nice discussion. arxiv.org/pdf/astro-ph/0310808.pdf Nov 14 '19 at 8:40
• @Reign Hmm… no, that's not correct: The radius of curvature $R_0$ is a scalar with the dimension of distance. For a positively curved universe, R0 can be identified as the actual radius of the Universe. There's a good introduction to this in Ryden (2003).
– pela
Nov 21 '19 at 9:53