# Hubble's Constant

1. Aug 24, 2007

### K.J.Healey

And the rate of expansion/ reduction in density of the universe.

I'm wondering if anyone has links to the physical research that has been done to calculate the rate at which distant, stars, galaxies, etc are moving away from us/each other.
Theres a million and one theories on how the universe is expanding, but I'm wondering if anyone has links/information of any actual empirical results, perhaps due to the redshift of distant galaxies.

I've searched the internet and its mainly results rather than data. Such as saying the HC is 70 km/s/megaparsec (10% error), but not actually giving the data on the distant galaxies they used to get this value.

Any help is appreciated! Thank you!

-KJH

2. Aug 24, 2007

### mgb_phys

Most of the results will be in published journals
I can't think of classic paper references of hand but anything by W.L. Freedman (http://www.ociw.edu/research/wfreedman/) should give you the answers

3. Aug 24, 2007

### George Jones

Staff Emeritus
One of Ned Wright's pages gives links to papers on a number methods for measuring the Hubble constant.

I'm not sure this is all you want, though. The value of the Hubble constant is not constant in time, it's constant in space at any instant of time. The value of the Hubble constant that you quoted is its value now, which is different from its value in the past and future.

A common misconception is that, since the expansion of the universe is at present accelerating, the value of the Hubble constant is presently increasing. Actually, it is presently decreasing.

To figure all this out, one needs the scale factor as a function of time, i.e, one needs to solve the Friedmann equation, so I think you want to know how observations lead to a particular best-fit solution to the Friedmann equation. But I could be wrong.

4. Aug 25, 2007

### K.J.Healey

5. Aug 25, 2007

### K.J.Healey

So saying that the Acceleration of the universe's expansion is increasing, while Ho is decreasing would mean what? The rate of acceleration of the universe (is that called Jerk?) is changing negatively/slowing down?

6. Aug 26, 2007

### hellfire

Increasing acceleration means that the second time derivative of the scale factor $a$ is increasing $\ddot a > 0$.

The scale factor measures the relation between a past cosmological distance and today for example a = 0 for t = 0 and a = 1 for today.

On the other hand, the decrease of the Hubble parameter means $\dot H < 0$. If you apply the definition of the Hubble parameter $H = \dot a / a$ you get:

$$\frac{d}{dt} \left( \frac{\dot a}{a} \right) = \frac{a \ddot a - \dot a^2}{a^2} = \frac{\ddot a}{a} - H^2$$

As you see these are two different things and $\ddot a > 0$ does not necessarily imply $\dot H > 0$.

If you insert the Friedmann equations without cosmological constant in the last relation, at least for a flat space k = 0 you can easily show that an increasing acceleration with decreasing Hubble parameter implies an equation of state with state parameter w > -1.

I.e., if the universe contains some dark energy with w = -1, then there still exists a non negligible amount of matter so that the expansion still does not get exponential (with the definition of the Hubble parameter I gave you above you can show that H = constant means a -very fast- exponential expansion a ~ exp Ht).

Last edited: Aug 26, 2007
7. Aug 27, 2007

### Jorrie

I have done an integration of the Lambda_CDM Friedmann equation with present best values and was surprised to learn that the Hubble constant H will decrease by about 10 km/s/Mpc for a doubling of the expansion factor. After that H remains more or less constant, since the cosmological constant then dominates the expansion curve completely.

Does this sound more or less right?

8. Aug 27, 2007

### hellfire

You result is correct. For expansion factor a = 2 (with a = 1) today, the Hubble parameter is H = 0.87 Ho (Ho today = 71 Km/s Mpc).

Last edited: Aug 27, 2007