# Open Big Bang-less universe?

This came up in discussion around a class I'm taking. For a Universe with $\Lambda$ and matter contributions to energy density (and implicitly curvature, but no radiation), can you have a universe with open geometry ($\Omega_\Lambda + \Omega_m < 1$) that fits the description of a "big bounce" universe?

All the possible descriptions/behaviours of such universe models are summarized in this diagram: An equivalent way of asking this question is: Does the line separating Big Bang/No Big Bang models approach the $\Omega_m=0$ line asymptotically, or does it meet it at $\Omega_\Lambda=1$? We had a go at sorting this out but couldn't come to any agreement...

My suspicion is that "if open geometry then Big Bang". Follow up question assuming this is the case: Is there an intuitive interpretation as to why open geometries MUST have a Big Bang (given the restrictions of these models of course, e.g. positive matter energy density, no relativistic species)?

## 1 Answer

The dividing line meets at $(\Omega_m,\Omega_\Lambda)=(0,1)$. From the Friedmann equations, it follows that the scale factor $a(t)$ satisfies the relation $$\frac{\dot{a}^2}{H_0^2} = \Omega_m a^{-1} + (1 - \Omega_m - \Omega_\Lambda) + \Omega_\Lambda a^2.$$ The universe has no big bang singularity if the above expression is negative (or zero) for some (small) values of $a$. So we have to examine under which conditions $$\frac{\dot{a}^2}{H_0^2} \leqslant 0.$$ Let us first assume $\Omega_m = 0$. Then the condition is $$(1 - \Omega_\Lambda) + \Omega_\Lambda a^2 \leqslant 0,$$ which implies $\Omega_\Lambda\geqslant 1$. The value $(\Omega_m,\Omega_\Lambda)=(0,1)$ is in fact a special case, because then we have $$\frac{\dot{a}^2}{H_0^2} = \Omega_\Lambda a^2,$$ with solution $a(t) \sim \exp(tH_0\sqrt{\Omega_\Lambda})$, which has no big bang, since $a\rightarrow 0$ if $t\rightarrow-\infty$, i.e. the singularity lies in the infinite past.

If $\Omega_m > 0$, then we require even higher values of $\Omega_\Lambda$ to get a universe without a big bang. This follows from the fact that, for small values of $a$, we have $$\begin{multline} \Omega_m a^{-1} + (1 - \Omega_m - \Omega_\Lambda) + \Omega_\Lambda a^2 >\\ \Omega_m + (1 - \Omega_m - \Omega_\Lambda) + \Omega_\Lambda a^2 = (1 - \Omega_\Lambda) + \Omega_\Lambda a^2, \end{multline}$$ so for a given $a$, the values of $\dot{a}^2/H_0^2$ in the general case are larger than the $(\Omega_m = 0)$ case.