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I have just started studying Cosmology and we have been asked to prove that in an expanding FRW Universe which obeys the strong energy condition: $$\rho + 3P >0$$ Then there must exist a Big Bang singularity. I can see that this condition implies $$\ddot{a}/a \leq 0$$ which when you plot against t gives an always decreasing rate of expansion over time. This when plotted then shows that if you extend the curve far back enough will cross the t axis at some finite t-value representing a Big Bang singularity. My question is why can't the universe begin with some finite non-zero a value? is there some physical reason that we expect the Universe to begin with a=0?

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  • $\begingroup$ Can an FRW universe just have an arbitrary starting point at some particular t value? $\endgroup$
    – Steve Linton
    Commented Oct 25, 2020 at 11:48

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It does not have to start from the big bang. There are also different universe models such as the big bounce. Where the universe has an infinite past, such that it expands and then contract.

See

https://www.quantamagazine.org/big-bounce-models-reignite-big-bang-debate-20180131/

https://www.wired.com/story/what-if-the-big-bang-was-actually-a-big-bounce/

For more detailed explanations you can look some cosmology textbooks

enter image description here

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  • $\begingroup$ Ok I’m not sure this quite answers the question- the idea is more that why does a=0 at t=0 necessarily in a Big Bang singularity model? $\endgroup$
    – Astroguy1234
    Commented Oct 25, 2020 at 11:58
  • $\begingroup$ @Astroguy1234 Clearly you can set the origin of t whenever you want, but the point is a=0 somewhere in the past. GR doesn't really cover the case of the universe just spontaneously snapping into being. $\endgroup$
    – Steve Linton
    Commented Oct 25, 2020 at 12:39
  • $\begingroup$ @Astroguy1234 how can you start the universe by setting $t=0$, $a \ne 0$ ? The only way is create cosmological models such as big bounce. But in that case the universe has infinite age. Otherwise we need a singularity as Steve Linton pointed out. $\endgroup$
    – seVenVo1d
    Commented Oct 25, 2020 at 14:05

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