# Limiting value of Hubble constant

1. Feb 8, 2010

### Ranku

As the universe approaches perfect exponential expansion the Hubble constant approaches a limiting value. What is the limiting value and how long from now will it be reached?

2. Feb 8, 2010

### marcus

Take the current value, approximately 71 and multiply by the square root of the dark energy fraction.
If you like 0.73 for the dark energy fraction (these are the default values for those parameters that you find, for instance, in Ned Wright's online cosmo calculator, and they are roughly what you get from the confidence intervals in the WMAP7 report that came out this month)
then what you want is 71*sqrt(.73) = 60.66
which rounds to either 60 or 61, whichever you like.

It would be gauche to overstate the accuracy. I would say the asymptotic value of H is "around 60" according to the standard LambdaCDM model parameters.

You can see the reasoning for this immediately from the Friedman equations, if you want to check it out. Google Friedman equations.

Since it is an asymptotic value, where the declining H kind of levels out but which is never quite attained, one can't say when it will be reached. But certainly by the time the scalefactor is 10 times what it is today we will be very close.

How close? Instead of .73+.27 the situation will be .73+.027 = .757 and the value will be
71*sqrt(.757) = 61.77 which rounds to 62. So when Uni has expanded ten-fold we will be pretty close to the asymptotic Hubble.

People are sometimes puzzled by this because they have the misconception that "acceleration" means that the Hubble rate is increasing, but it is not, and I think you understand that it is the scalefactor a(t) that is increasing and acceleration means that a(t) increase is speeding up, the second time-derivative a"(t) is positive.

Last edited: Feb 8, 2010
3. Feb 9, 2010

4. Feb 9, 2010

### Ich

For a constant H (a good approximation), you get the solution $a \propto e^{Ht}$. That's an e-fold expansion in 1/H, and tenfold in 2.3/H ~ 31 Gy.
Yes.

5. Feb 11, 2010

### Ranku

Thank you both.