Age of a vaccum energy dominated universe

1. Feb 27, 2012

sri sharan

The other day, I was calculating the age of universe dominated by vacuum energy and it turned out to be infinity. What does age of the universe being infinite mean? On explanation I thought of is that may be this implies that such a universe has no beginning. Is it a proper explanation?

2. Feb 27, 2012

Nabeshin

No, it simply means the universe will never experience a big crunch, i.e. recollapse. Simply put, the scale factor never returns to zero. It is of course possible to have a universe which starts with a=0, but then persists indefinitely (as is the case with our own).

3. Feb 27, 2012

sri sharan

Hmm, isn't that more like the fate of the universe. What I was trying to calculate was what would be the present age of the universe in standard Friedman cosmology for a flat universe(sorry i didn't mention that before), as a function of the observed CMB redshift and Hubble. But what I got was that for vacuum dominated universe, the age would turn out to be infinity (irrespective of value of redshift and H), and the only meaningful explanation I could think of was universe with no beginning

4. Feb 27, 2012

cepheid

Staff Emeritus
Perhaps you were doing your integrals wrong? What values did you use for $\Omega_m$ and $\Omega_\Lambda$? Even if I plug in 0 for the former and 1 for the latter in here...

http://www.astro.ucla.edu/~wright/CosmoCalc.html

...I get about 37 Gyr, not ∞.

Also, what do you mean by, "as a function of the observed CMB redshift?" What does that have to do with anything? Isn't the only relevant value of z the value at which you want to compute the age of the universe (which would be z = 0 for the age at the present time)?

5. Feb 27, 2012

George Jones

Staff Emeritus
Yes.

6. Feb 27, 2012

cepheid

Staff Emeritus
Yeah, my bad. When I responded to the OP, I hadn't actually written out the equations (EDIT: and I'm assuming that this is a case for which the numerical calculator that I linked to simply breaks down). So tell me if I'm doing this right. With only dark energy (assuming it's in the form of a cosmological constant) the Friedmann equation is$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{\Lambda}{3}$$This assumes the universe is spatially flat. This becomes$$\frac{1}{a}\frac{da}{dt} = \left(\frac{\Lambda}{3}\right)^{1/2}$$which you can solve analytically to get $$a(t) = \exp\left[\left(\frac{\Lambda}{3}\right)^{1/2}(t-t_0)\right]$$where I arbitrarily chose t0 to be the time value when the scale factor is unity. The thing is, as you back in time, for t < t0, the scale factor asymptotically approaches 0, but never actually reaches it. So it would seem that indeed this type of cosmological model does not have a beginning.

I'm guessing that the OP tried to invert the differential equation and then integrate to solve for t(a), but obtained something proportional to $\int_0^1 \frac{1}{a}\,da$ which does not converge -- which is another way of showing the same result.

So I read that this is the de Sitter universe, and that it is also used as an approximation to inflationary models whose dynamics are similar. Is this idea of "no beginning" sort of the basis for "eternal inflation?"

Last edited: Feb 27, 2012
7. Feb 28, 2012

Chalnoth

The only issue here is that any amount of matter or radiation causes the universe to have a finite age. So it is not considered feasible for inflation to be past-eternal, because there will always be some matter or radiation, no matter how diffuse.

8. Feb 28, 2012

sri sharan

yeah, that's what I did. And thanks of the the De Sitter info . Didnt know about that before