5
$\begingroup$

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:

enter image description here

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)?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

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.