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To calculate the Big Crunch time (time since an expanding universe starts expanding until it collapses), starting from the first Friedmann equation:

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \frac{{\rho_{m,0}}}{a^3} - \frac{K}{a^2}$$

The Universe is initially only matter-dominated with no cosmological constant, and $\rho_{m,0}$ denotes matter density now. I have set up the following integral:

$$\pm \sqrt{\frac{3}{8 \pi G \rho_{m,0}}} \int_0^a \frac{da'}{\left(1/a' -\frac{3K}{8 \pi G \rho_{m,0}}\right)^{1/2}} = t.$$

It has the following solution from Wolfram Alpha: enter image description here

where

$$x = a', C = \frac{3K}{8 \pi G \rho_{m,0}}.$$

Please only indicate if I am going in the right direction.

Samama Fahim
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  • If you are integrating from $0$ to $a$ you will run into a problem with the $\frac 1x$ part. – joseph h Mar 29 '24 at 06:01
  • But the limit of it exists, doesn't it? – Samama Fahim Mar 29 '24 at 06:08
  • Then take the limit and see what answer you get. – joseph h Mar 29 '24 at 06:10
  • It can be calculated from here: https://www.integral-calculator.com. But if in this procedure I have made a mistake, please indicate that. – Samama Fahim Mar 29 '24 at 06:20
  • Do we have to assume a zero cosmological constant in the Big crunch scenario? I think we must because if it were there in this case, it would eventually dominate. – Samama Fahim Mar 29 '24 at 06:21
  • Yes, you must assume $\Lambda=0$ here (and of course, you need to multiply the RHS with $+\sqrt{C}$ and remove the integration constant). What answer did you get when you evaluated this? – joseph h Mar 29 '24 at 06:26
  • This is what I get: $$ \frac{\arctan(\frac{1-Ca}{\sqrt{Ca}})}{C^{3/2}} + \frac{\sqrt{a-Ca^2}{C} - \frac{\pi}{2C^{3/2}} = \int_0^a \frac{1}{\sqrt{1/a' - C}} da' $$ – Samama Fahim Mar 29 '24 at 06:35
  • Also, in the Big Rip scenerio, the only difference will be $K = -|K|$ in the Friedmann equation. Is that right? – Samama Fahim Mar 29 '24 at 06:42
  • Please consider solving as much as you can then posting questions (all you have done that far) as you go along. – joseph h Mar 29 '24 at 06:58

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