# Why is a negatively curved universe older than a flat one?

And a positively curved universe younger than a flat one?

I get that negatively curved means an open universe, and positively curved is a closed one. But how can this affect their age?

Is it because the geometry affects the +/- signs on the right hand side of the Friedmann equation? If so how?

Your question isn't quite right - not all closed universes are younger than all flat universes, and all open universes are not older than all flat universes.

As a reference Universe, pick your favourite flat Universe - the density may be in several components (matter, radiation, $\Lambda$), but it must be exactly critical.

Suppose we want an open universe - this means we need sub-critical energy density. Suppose we steal a little bit of energy density from $\Lambda$. This component drives expansion, so the universe had a slightly harder time expanding, and took a little longer to reach its present size: it is older than the reference flat universe. Conversely, if we steal a little bit of energy density from the matter component (which inhibits expansion), the universe had a slightly easier time expanding and reached its present size faster: it is younger than the reference flat universe. Likewise for the closed universe - we need super-critical energy density, and get the opposite effect of removing a bit of energy density from different components.

More generally, yes, you just need to look to the Friedmann equation. The age is:

$$t = \int_0^1\frac{{\rm d}a}{\dot{a}}$$

And you get $\dot{a}$ from:

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}\left[\rho_{m,0}\left(\frac{a_0}{a}\right)^3+\rho_{r,0}\left(\frac{a_0}{a}\right)^4+\rho_{\Lambda,0}\right]-\frac{Kc^2}{a^2}$$

But you can't just look at the curvature term ($K=0$ is flat, $K=+1$ is closed, $K=-1$ is open) and draw a conclusion about the relative ages of various universes, because $K$ is not independent of the density parameters.

• awesome that's really helpful – Tom Apr 29 '15 at 19:20