# Lifetime of the universe - what are the limits?

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1. Apr 25, 2015

### unscientific

I am studying general relativity from Hobson and came across the term 'lifetime' of a closed (k>0) universe, $t_{lifetime}$.

I suppose at late times the curvature dominates and universe starts contracting? Are they simply referring to $\int_0^{\infty} dt$? If so, would the bottom expression be right?

$$\int_0^{\infty}dt = \int_1^{0} \frac{1}{a(t) H(t)} dt$$

Do you think this makes sense since $a(t)$ is decreasing and will eventually reach $0$?

Then for an open universe (k<0), wont the universe simply keep expanding? If I read right, our universe is mostly flat, right? Then what is driving the expansion? Dust?

2. Apr 25, 2015

### Orodruin

Staff Emeritus
In the closed universe (without dark energy), there will be a past singularity as well as a future singularity, i.e., there will be a maximal value for the cosmological time t.

This is purely academic as we seem to be living in a universe with a dark energy component that has already started to dominate, meaning the universe will forever undergo accelerated expansion.

3. Apr 25, 2015

### unscientific

Ok, so it would be
$$\int_{t_0}^{t_{life}}dt = \int_1^{0} \frac{1}{a(t) H(t)} dt$$

So there is a 'cosmological constant' after all? Meaning since $\rho_{\Lambda} = constant$, then solving the friedmann equation gives $a(t) \propto e^{mt}$.

4. Apr 30, 2015

### unscientific

anyone?

5. Apr 30, 2015

### Staff: Mentor

No. What happens is that the density of ordinary matter in the universe is sufficient to cause it to recollapse. The second Friedmann equation, for $\ddot{a} / a$, is the key to the dynamics; note that there is no curvature term in this equation.

No. This just gives $\infty - 0 = \infty$.

No. I don't understand what you think this integral represents.

There is a cosmological constant according to our best current model, yes. But I don't see what the integral you wrote down has to do with it.