The measurement of the WMAP satellite resulted a planar geometry of the universe with a 0.4% uncertainity (http://en.wikipedia.org/wiki/Shape_of_the_universe).

If there is a little deviation from the measured zero curvature, I think it could give a lower limit to the size of the universe (in case of positive curveture and spherical geometry). How big is it? How could it be calculated?


1 Answer 1


From the Friedmann equations, you can derive that $$ \dot{R}^2 - \frac{8\pi}{3}G\rho R^2 = -k c^2, $$ where $\rho$ is the total density of the universe and $k$ is a constant that determines the shape of the universe: $k=-1,0,1$ for an open, flat and closed universe, respectively. If the universe is a hypersphere ($k=1$), then $R$ can be thought of as its 'radius'.

Because the right-hand side is a constant, it is also equal to the present-day values $$ \dot{R}_0^2 - \frac{8\pi}{3}G\rho_0 R_0^2 = -k c^2, $$ or $$ \frac{\dot{R}_0^2}{R_0^2} - \frac{8\pi}{3}G\rho_0 = -\frac{kc^2}{R_0^2}, $$ and, introducing the Hubble constant $H_0=\dot{R}_0/R_0$, we get $$ H_0^2 - \frac{8\pi}{3}G\rho_0 = -\frac{kc^2}{R_0^2}. $$ If $k=0$, we have a flat universe, and the corresponding density equals the so-called critical density $$ \rho_{c,0} = \frac{3H_0^2}{8\pi G}. $$ The general case can thus be written in the form $$ H_0^2\left(1 - \frac{\rho_0}{\rho_{c,0}}\right) = -\frac{kc^2}{R_0^2}. $$ Finally, the factor between brackets is denoted as $\Omega_{K,0}$, so that $$ H_0^2\,\Omega_{K,0} = -\frac{kc^2}{R_0^2}. $$ In case of a universe with positive curvature, $k=1$ and $\Omega_{K,0}$ is negative, so that $$ R_0 = \frac{c}{H_0\sqrt{-\Omega_{K,0}}}. $$ The nine-year WMAP value for $\Omega_{K,0}$ is (see the last table on the wiki page) $$ \begin{align} \Omega_{K,0} &= -0.037^{+0.044}_{-0.042}\qquad&&\text{(WMAP only)},\\ &= - 0.0027^{+0.0039}_{-0.0038}\qquad&&\text{(WMAP + other obs.)}, \end{align} $$ and $H_0 = 70\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$. So we find $$ \begin{align} R_0 &\approx 22.3\;\text{Gpc}\approx 72.7\;\text{billion ly}\qquad&&(\text{for $\Omega_{K,0} = -0.037$}),\\ R_0&\approx 82.5\;\text{Gpc}\approx 269\;\text{billion ly}\qquad&&(\text{for $\Omega_{K,0} = -0.0027$}). \end{align} $$ This can be interpreted as the radius of the universe if it is a hypersphere, although the topology of the universe could be more complicated. The latest Plank results put even tighter constraints on the curvature of the universe (see page 40 in this paper).

  • $\begingroup$ If it is a hypersphere and it's finite, does that mean if i go far enough in 1 direction i will comw back to where i started? $\endgroup$ Dec 6, 2018 at 13:26
  • $\begingroup$ @Armend If the universe wasn't expanding, yes. But since it expands, its circumference keeps increasing. And it's increasing faster than you'd be able to go around. Even a beam of light isn't fast enough. $\endgroup$
    – Pulsar
    Dec 6, 2018 at 18:23
  • $\begingroup$ The question asked for a lower bound, but you used the central values from WMAP, so the values you calculated are too high. It would make more sense to use the lower end of a 95% CI or something, though I don't know how to calculate that from the reported values (because I doubt they're independent). $\endgroup$
    – benrg
    Sep 24, 2020 at 19:58
  • $\begingroup$ @Pulsar "The latest Plank results put even tighter constraints on the curvature of the universe (see page 40 in this paper)." This is a 2013 paper. In the eight years since then, has there been any further calculations for an approximate size of a hyper-spherical universe? $\endgroup$
    – Buzz
    Aug 23, 2021 at 15:43

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