# Big Crunch time

For a universe that is flat, has matter and a cosmological constant, we can write the Friedmann equation in the following way:

$$\frac{H^{2}}{H^{2}_{0}} = \frac{\Omega_{m,0}}{a^{3}} + (1 - \Omega_{m,0})$$

I understand that if the second term is negative ($$\Omega_{m,0}>1$$) then the final fate of the universe is that it is going to collapse again in the Big Crunch!

I understand that I can calculate the maximum value of the scale factor doing $$H^{2}=0$$ and that I can rewrite the above equation as an ODE just with some algebra to have the following expression:

$$H_{0}t = \int_{0}^{a} \frac{da}{(\Omega_{m,0}/a + (1 - \Omega_{m,0})a^{2})^{1/2}}$$

that relates the cosmic time with the scale factor $$a$$. My question is then the following: How can I calculate the Big Crunch time, (e.g. the time the we will again have $$a=0$$?

I though about it being the twice the above integral with the upper limit being the $$a$$ to which we have $$H(t) = 0$$, but it does no make a lot of sense for me. Also, the Introduction to cosmology by Barbara Ryden says that the time I'm looking for is (eq. 5.98):

$$t_\text{crunch} = \frac{2\pi}{3H_{0}}\frac{1}{(\Omega_{m,0}-1)^{1/2}}$$

What I think that suggest that I'm trying the wrong approach. Can someone help me? How can I find the above equation?

• I understand that if the second term is negative ($\Omega_{m,0}>0$) ... $\Omega_{m,0}>0$ doesn’t make the second term negative. – G. Smith Oct 8 '20 at 0:08
• Oh, I'm sorry, it was a typo. – kplt Oct 8 '20 at 0:10

That integral has a closed-form solution, $$H_0 t = \frac{2 \sinh^{-1} \left(\sqrt{\frac{1-Ω_{m,0}}{Ω_{m,0}}}\,a^{3/2}\right)}{3 \sqrt{1-Ω_{m,0}}} = \frac{2 \sin^{-1} \left(\sqrt{\frac{Ω_{m,0}-1}{Ω_{m,0}}}\,a^{3/2}\right)}{3 \sqrt{Ω_{m,0}-1}}.$$
For a recollapsing universe this gives you only the expanding half of the evolution, which ends when the arcsine reaches its maximum value of $$π/2$$. The total age is therefore $$2 \frac{2 (π/2)}{3 \sqrt{Ω_{m,0}-1}}/H_0$$.