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?


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? $\endgroup$ – Armend Veseli Dec 6 '18 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 '18 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.