# Constant Hubble parameter -> accelerating Universe?

1. Aug 10, 2011

### johne1618

Hi,

Let us assume that Hubble's Constant H is really constant. Therefore:

a' / a = H

where a is the scale factor.

The solution to this equation is:

a(t) = exp(H t)

This equation describes an accelerating universe with deceleration parameter q given by:

q = - a'' a / a'^2 = -1

This value of q is in agreement with current observations.

By the way, a constant value of H implies the Hubble radius which is a measure of the size of the observable Universe, R = c / H, is constant. If the above equation is true then in some sense the Universe is now static.

2. Aug 10, 2011

### marcus

According to the usual picture, the Hubble parameter is not constant in time. As far as I know it has never, in the history of modern cosmology, been considered to remain constant over time.
In standard expansion cosmology the change in H(t) is governed by the Friedmann equation. This is the basic eqn that all cosmo'ists use.

In the early universe the H was thousands of times larger than it is at this moment.

H has been declining for a long time and is still declining. But more and more slowly. Look at the Friedmann eqn to understand why. You can google it.

The Hubble radius is not equal to the size of the observable universe. Most of the galaxies we currently observe are more distant than that. There are various horizon distances.

But there is a grain of truth in what you say! Because of it's accelerated expansion, the U is in fact approaching a kind of "static" condition in which there will be a fixed radius to what we can observe. The cosmic event horizon will be stuck at a fixed distance. Because beyond that things will be receding too rapidly.

The simple model of exponentially expanding universe with no matter in it is the "de Sitter".
It has a fixed cosmic event horizon. So the observable region has a fixed size. We are gradually approaching that de Sitter picture because matter is thinining out. It is still some 27%. When the U has expanded a lot more, and matter is down near 0%, then we will be effectively de Sitter. Then what you say will actually be about right!
And then the Hubble parameter will be approximately constant.
And then the scale factor (distances) will be growing nearly exponentially.
And then the radius of the observable will be approximately constant.

Larry Krauss has an article on the arxiv describing what it will be like (billions of years from now) called "The Return of the Static Universe".

You might like the article. I'll get a link.
http://arxiv.org/abs/0704.0221

Fun article! Written for wide audience. Won a prize in an essay contest in 2007, as I recall. Like the world you describe, with constant Hubble parameter, but billion years in future, and described in much more detail.

3. Aug 10, 2011

### johne1618

Hi,

I think the equation

a(t) = exp(H t)

where H is constant

is a solution to the Friedmann equations provided that one assumes that the Universe is spatially flat and that the equation of state is:

p = - rho c^2

which is the equation of state of dark energy.

John

4. Aug 10, 2011

### marcus

I think you are right!
That is the de Sitter I was talking about. No matter (we have 27%) and only "drk enrgy" another name for cosmo constant if you assume that -1.

There is no evidence that we have that NOW. But our real universe can asympt.ly approach that condition over next few billions years, due to thinning. I'm jst repeating what I said earlier.

5. Aug 10, 2011

### bcrowell

Staff Emeritus
It's not a totally unreasonable model of the present-day universe. If you have to model the present universe using only a single component with a single equation of state, then a cosmological constant is the best approximation. But if you're an observational cosmologist, you're spending a lot of time looking at data from the earlier universe, in which this was not as good an approximation.