# The future of our universe

1. Mar 5, 2015

### Quarlep

I learned friedmann eq and there I learned that k is equal zero in our universe it means that universe will grown infinite. I am asking to you am I correct.Universe is infinite and every second hubble constant is changing isnt it according to 2/3t.

2. Mar 5, 2015

### ChrisVer

This is not true because k=0 ... The fact that k=0 means that the universe is spatially flat and infinite.
In order to check what will happen with the Universe, you have to try and solve the Friedmann equations for the cosmological parameters that fit best the data... If you do so, and because our universe in this phase is dominated by dark energy, you will find that the Universe will expand forever.
http://map.gsfc.nasa.gov/universe/uni_fate.html

3. Mar 5, 2015

### Quarlep

You mean that normally universe should grow like a p equals zero but in our observation p is lower than zero cause of dark energy.Like a negative pressure I understand it from nasa.so low pressure causes low energy and it causes more accelaration.If this is true than hubble constant must be higher than 2/3t cause in this idea we got p=0 but we undertand that (cause of dark energy) p<0

4. Mar 6, 2015

### ChrisVer

What do you mean "normally should grow like p=0"?

Well you can take two cases... first take the case of a matter dominated universe: so the solution for the scale factor $a(t)$ is
$a(t) = a_0 t^{2/3}$ and $H= \frac{\dot{a}}{a} = \frac{2}{3t}$
So your result for the Hubble constant takes into account that you have only a matter dominated universe.

On the other hand, a cosmological constant implying a constant density gives from the Friedmann equation:

$H(t) = H_0 \Omega_\Lambda^{1/2}=const$ (so the Hubble parameter is a constant, at least if the cosmological constant is a constant-there are searches about that) and so $\frac{\dot{a}}{a} = \frac{d \ln a}{dt} = H_0 \Omega_\Lambda^{1/2}$

The solution for $a(t)$ is $a(t)=a_0 e^{H_0 \sqrt{\Omega} t}$ and the Universe so grows exponentially.

To get a better result, you have to solve the Friedmann equation for $\Omega_m \approx 0.3$ and $\Omega_\Lambda \approx 0.7$. The result will be the red-line in the figure in the link above. In that case, the $a(t)$ doesn't have a very nice form (but nevertheless the exponential will dominate).

5. Mar 6, 2015

### Chronos

It is currently [and may forever be] uncertain if the universe is infinite, ginormous, but, finite; or finite, but, unbounded. It is worth noting that FLRW, the current accepted cosmological model, is not necessarily well behaved in an infinite universe. The Friedmann equations assume a subtle symmetry that may not apply in an infinite universe. For any interested to see how Friedmann equations were derived, see http://www.cesura17.net/~will/Ephemera/Nerdliness/Relativity/flrw.html

Last edited: Mar 7, 2015
6. Mar 7, 2015

### Staff: Mentor

Were you intending the article you linked to as a reference for this? I don't see anything to this effect in the article.

7. Mar 7, 2015

### Chronos

No. It was a throw in for a little background on the Friedmann equation.

8. Mar 7, 2015

### Staff: Mentor

I ask because I'm confused by the statement that there is a "subtle symmetry that may not apply in an infinite universe". What does that refer to?