Lifetime of Universe: Limits & Expansion Explained

[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), won't the universe simply keep expanding? If I read right, our universe is mostly flat, right? Then what is driving the expansion? Dust?

 

[Orodruin]

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.

 

[unscientific]

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.

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 $a(t) \propto e^{mt}$.

 

[unscientific]

anyone?

 

[PeterDonis]

I suppose at late times the curvature dominates and universe starts contracting?

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.

Are they simply referring to $\int_0^{\infty} dt$ ?

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

Do you think this makes sense

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

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

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.