# What happens for a "closed" universe without any content?

Let an universe with no content and positive curvature. Friedmann-Lemaître equation $$H^2=\frac{8\pi G}{3}\left(\rho_m+\rho_r+\rho_{\Lambda}\right)-\frac{k\, c^2}{a^2},$$ where $$a$$ corresponds to the scale factor of Friedmann-Lemaître-Robertson-Walker Metric, and the $$\rho$$ to the density of the contents, will become $$H^2=-\frac{\, c^2}{a^2},$$ as positive curvature (closed universe) corresponds to $$k=+1$$.

So, as $$H=\frac{\dot{a}}{a}$$ with $$\dot{a}=\frac{da}{dt}$$,

$$da=\pm i \, c \, dt$$

$$a(t)=\pm i \, c \, t + cst$$

what am I doing wrong ?

• FYI a native English speaker would say "a universe". Sep 28 '18 at 4:04

I see that you are using the convention that a has units of distance and is like the radius of curvature of the universe, and you are getting that either a or t would need to be imaginary. So you can interpret that in two ways, either you say that is impossible, and conclude that it is impossible for an empty universe to be closed, or you attribute some physical meaning to imaginary radius of curvature or imaginary time. I'm not sure what that significance would be, but what seems clear is there is something bizarre, or even bogus, about claiming that an empty universe can have positive curvature.

Another way to see this is to interpret your equation as an equation for k via H=Kc/a, where K^2 = -k. If you further demand that nothing be imaginary, this requires that k be either 0 or -1. Empty universes are flat or open, or else they are imaginary in some way.

You are not doing anything wrong. Positively curved empty universes are forbidden. Since as you figure it out the scale factor would be imaginary.