# FRW cosmology $Ω_{Λ}$ domination

1. Jul 8, 2014

### ChrisVer

Suppose you have a universe where $Ω_{Λ}=1$ and the rest are zero. Then the Friedmann equations are:
$(\frac{H}{H_{0}})^{2} + \frac{k}{(aH_{0})^2} = \Omega_{\Lambda}=1$

since $\Omega= \Omega_{\Lambda}=1$ we have a flat universe and so $k=0$
This leaves us with:
$(\frac{H}{H_{0}})^{2}=1$

or

$\frac{H}{H_{0}}=1 \rightarrow H=H_{0}$

If I try to write find the age of this universe today, it will give me:

$\frac{1}{a} \frac{da}{dt} = H = H_{0}$

If I integrate from $a=0$ to $a=1$ and so time from 0 to T (today):
$\int_{0}^{1} \frac{da}{a} = H_{0} T$

I am getting negative age T.... because the integral is logarithmically divergent at 0.

An additional problem also appears by solving for $a$ as a differential equation and get:

$a(t) = a_{0} e^{H_{0} t}$

from which you see that to solve $a(t)=0$ (so to find the time connecting $a_{0}$ to $a=0$) you need to get $H_{0} t= -∞$
Both these cases seem unphysical to me...? Any help for clarification?

2. Jul 8, 2014

### George Jones

Staff Emeritus
This in one (the k = 0; there are also k = 1 and k = -1 slicings of de Sitter) of the slicings of de Sitter spactime. de Sitter spacetime is not singular. The k = -1 slicing does have a(0) = 0, but this is just a coordinate singularity.

3. Jul 8, 2014

### ChrisVer

For a flat spacetime you have $k=0$... in my model $\Omega=\Omega_{\Lambda}=1$. This can also be seen by taking the Friedman equation again for today ($H=H_{0},~~ a=a_{0}$:

$1 + \frac{k}{(a_{0}H_{0})^{2}} =1$
$\frac{k}{(a_{0}H_{0})^{2}} =0$
$k=0$

4. Jul 8, 2014

### George Jones

Staff Emeritus
Sorry , trying to watch the World Cup (5-0 for Germany after 29 minutes!!). de Sitter spacetime is a particular 4-dimensional, non-singular spacetime manifold. There are 3 different ways that this one spacetime manifold can be foliated by 3-dimensional spacelike hypersurface, one for k = 1 closed space, one for k = 0 flat space, and one for k = -1 open space.

You are looking at the k = 1 slicing of de Sitter

Again it is the same non-singular spacetime manifold for each type of spatial slicing!

If you have access to it, the book "Einstein's General Relativity with Applications in Modern cosmology" by Gron and Hervik has a nice discussion that starts on page 309, with a very illustrative diagram on page 310.

Last edited: Jul 8, 2014
5. Jul 8, 2014

### ChrisVer

Ah then $a(0)=0$ is not an appropriate choice -due to a horizon. Thanks

6. Jul 8, 2014

### George Jones

Staff Emeritus
I made at least two mistakes in my previous post.

I meant k = 0.

Should be "starts on page 305".

You can see your situation (exponential scale factor for k = 0) in the middle portion of equation (12.10)

7. Jul 9, 2014

### ChrisVer

Yes I understand this, mathematically it works... It's what I deduced and I was able to check it in various sources...
eg in Kolb & Turner they deal with a flat matter+vacuum universe, and in the case ΩΛ=1 they also get infinite age. One can say that due to this, the model is unphysical and it doesn't really make sense.
However I'd like to ask what we mean by "Age"...the age of such a universe is going to start from infinity because:

$H_0 \int dt = \int_{0}^{x} d(\ln a)$
diverges for any $x \ne 0$

I can see the formulas, but I cannot "get" the meaning of the quantities... Is the age really defined as the time elapsed to go form $a=a_0=1$ to $a=0$?

8. Jul 9, 2014

### George Jones

Staff Emeritus
Right. Since de Sitter is a non-singular spacetime, it does not have a Big Bang Singularity, and thus has no "beginning".

Elapsed time in general relativity is calculated the same way it is in special relativity, integrate proper time along a worldline.