# Hubble parameter

1. Jun 20, 2009

### keepitmoving

what was the value of the Hubble parameter when the universe was decelerating?

2. Jun 20, 2009

### Hal King

There is no solid evidence that the universe has ever 'decelerated' -- or 'accelerated' its expansion.
That is still in debate.

http://www.cfa.harvard.edu/~huchra/hubble/

This link -- one of many -- has some nice graphs comparing the various accepted values of the
Hubble parameter over time.

Our interpretation has a lot more 'deceleration' and 'acceleration' than the universe.

3. Jun 20, 2009

### keepitmoving

if the Hubble parameter is 1, is the universe accelerating?

4. Jun 20, 2009

### Hal King

Not directly related. Hubble parameter is more closely related to the 'age of the universe' (rather the time it took for light to reach us from its beginning) and not its 'dynamic' properties -- (i.e. acceleration/deceleration).

Sorry ... maybe someone else could describe it better.

5. Jun 21, 2009

### Chalnoth

Hmmm, that's not something that's usually computed. And, unfortunately, the precise value of the Hubble parameter when the universe stopped decelerating and started accelerating is not known to much precision (it's a rather noisy measurement). It is also worth mentioning that our universe was decelerating for most of its history, from the end of inflation to relatively recently (when dark energy took over). This covers an incredibly wide range of Hubble parameters.

So, I'll just see if I can answer what the Hubble parameter was when the deceleration ended, for the special case of cold dark matter with a cosmological constant, and taking the best-fit WMAP parameters. This should at least be in the rough ballpark of the real answer (say, +/- 20% or so).

Basically, whether or not a universe is accelerating depends upon its equation of state. This is the relationship between the pressure and energy density. For example, normal matter and dark matter have zero pressure on cosmological scales, and so have $$w = 0$$. A cosmological constant has negative pressure equal to its energy density, $$w = -1$$. The deceleration parameter q is written as:

$$q = \frac{1}{2}\left(1 + 3w\right)$$

If, for example, we have pure cosmological constant, the deceleration parameter $$q = -1$$, which indicates an accelerating universe. If, on the other hand, we have just normal matter, then $$q = \frac{1}{2}$$, which would be a decelerating universe. The transition from deceleration to acceleration, then, occurs when the $$w = \frac{-1}{3}$$ for the entire universe.

This parameter is defined as follows:

$$p = w\rho$$

We can now decompose the left and right sides into the two main components of interest, normal matter and a cosmological constant:

$$0 + -\rho_\Lambda = w \left(\rho_m + \rho_\Lambda\right)$$

So:

$$w = \frac{-\rho_\Lambda}{\rho_m + \rho_\Lambda}$$

A standard trick is to redefine the energy densities in terms of the critical density $$\rho_c$$ at the current time when $$a = 1$$:

$$\Omega_\Lambda = \frac{\rho_\Lambda(a=1)}{\rho_c(a=1)}$$
$$\Omega_m = \frac{\rho_m(a=1)}{\rho_c(a=1)}$$

Now, we also know how these two things scale with time: the cosmological constant is independent of expansion, while the normal matter dilutes as $$1/a^3$$. So our equation for $$w$$ becomes:

$$w = \frac{-\Omega_\Lambda}{\frac{\Omega_m}{a^3} + \Omega_\Lambda}$$

Now we just need to solve the above equation for $$a$$. I get:

$$a = \left(\frac{\Omega_m}{-\left(1 + \frac{1}{w}\right)\Omega_\Lambda}\right)^\frac{1}{3}$$

With $$w = -1/3$$, $$\Omega_m = 0.26$$ and $$\Omega_Lambda = 0.74$$, I get:

$$a = 0.56$$

So that means a Hubble parameter of:

$$H(a) = H(a=0)\left(\frac{\Omega_m}{a^3} + \Omega_\Lambda\right)$$

Plugging in $$a = 0.56$$, I get:

$$H(a) = 2.22H(a=0)$$

So that means that the Hubble parameter when the universe stopped decelerating was in the ballpark of twice its current value, or around 150km/s/Mpc.

6. Jun 21, 2009

### keepitmoving

thank you Chalmoth. The math was above me but i did expect the Hubble parameter to be more at the end of deceleration than now.
By the way, im not a real physicist but i did have dinner at a Holiday Inn last year (a joke that was on TV).
Anyway, but what would the Hubble parameter be when the universe was actually decelerating? If it was still expanding during deceleration wouldnt a given object be moving faster relative to a reference point than it was the previous Myr? Its hard for me to picture expanding without a given object moving away from a reference point at a speed that is less than it was moving away during the previous Myr.

7. Jun 21, 2009

### Chalnoth

Well, larger. It varied dramatically with time. The smallest would have been roughly twice the current value, as I calculated above. The largest would have been whatever it was at the end of inflation, which we don't know precisely yet. A rough ballpark would be somewhere around $$10^{22}$$ to $$10^{25}$$ km/sec/Mpc.

Edit: fixed the error from the square root noticed by Ich below.

Last edited: Jun 22, 2009
8. Jun 21, 2009

### keepitmoving

maybe im on the wrong track. Is the acceleration really the acceleration of acceleeration?

9. Jun 21, 2009

### marcus

To get a feel for what the Hubble rate has been in the past, there is a quick easy way.
You just go to one of the online calculators and have it tell you what H has been at various redshift eras. Like for example z = 1090, which is the redshift of the CMB---this will get you info about the time when the cosmic microwave background light was originally emitted.

For example google "cosmos calculator". That will get you this online calculator:
http://www.uni.edu/morgans/ajjar/Cosmology/cosmos.html

To start a session you have to type in 3 presentday parameters:
I would recommend typing in .25, .75, and 74
You will see the three boxes on the left: matter density (put in .25), cosm. const. (put in .75) and Hubble parameter (put in 74)

Then you can put in any redshift and it will tell you what the Hubble parameter was in the past when light was emitted that we receive today with that redshift.

If you put in z = 1090, it will say the Hubble parameter back then was 1.3 million
=============

Cosmos calculator rounds off ages to the nearest tenth of a billion years. The output, especially for redshifts over 1000, is a rough approximation, but it gives you idea of the magnitudes.

For a higher precision online calculator, google "wright calculator".
Both are based on the standard cosmology model, but Ned Wright's has more decimal places of precision.
==============

Note that when people talk about accel/deceleration they are talking about the scalefactor a(t). Accelerating expansion means that the time derivative a'(t) is increasing. It does not mean that H(t) is increasing. In fact according to standard cosmo model, H(t) is currently decreasing and scheduled to continue decreasing for the foreseeable future.

================

Chalnoth, thanks for estimating when deceleration stopped! I see your estimate of the H parameter is 2.22 times presentday. If I take the presentday value of 74 (which Adam Riess et al just came out with) that means that H was about 164 when the switch from decel to accel occurred. (I'll save rounding off for later. )

Again using Riess et al numbers, that would mean the switch occurred around z = 1.5. Does that seem about right to you?
This puts the age of expansion when the changeover occurred at 4.3 billion years, or 9 billion years ago. Again, does that seem right? I had the impression that changeover was more recent, not 9 billion years ago.

Last edited: Jun 21, 2009
10. Jun 21, 2009

### Chalnoth

The acceleration is the acceleration of the scale factor.

11. Jun 21, 2009

### keepitmoving

please forgive my calculus, it was 45 years ago.
Would i be correct to ask if the acceleration of the scale factor is an acceleration with respect to time or with respect to previous acceleration?

12. Jun 21, 2009

### marcus

You can think of acceleration as the second time-derivative of the scalefactor. The scalefactor a(t) is our handle on the "size" of the universe. It is normalized so at the present time a(present) = 1.

The first time derivative is a'(t) or if you like "d by dt" notation, say da/dt.
The fact that the U is expanding is simply that a'(t) > 0.

The second time derivative is a"(t) or you can write it with "d by dt" notation.
When people say U is accelerating what they basically mean is that a"(t) > 0.

Or in other words, a'(t), which is the rate of expansion, is increasing.

There are some other fancier notations, but a(t) the scalefactor is the mathematical bedrock they are built on.
For example H(t) the Hubble parameter is another name for a'(t)/a(t). It is a useful ratio.

There is also the q parameter, also defined in terms of a(t) and its derivatives. Another handy ratio. But the fundamental object is the scalefactor a(t) which gives the basic history of U expansion and is the basis on which the other stuff is defined.

So the easiest, most elementary way to say what acceleration means is in terms of what the scalefactor is doing. It is increasing at an increasing rate.

13. Jun 21, 2009

### keepitmoving

Marcus, thanks. Just to confirm that i understand - during the deceleration phase the scale factor was increasing at a decreasing rate? Would the Hubble parameter go up or down as you go from, say, 5 Byr`s ago to 10 Byrs ago?

14. Jun 21, 2009

### Chalnoth

The Hubble parameter is given by the Friedman equation:

$$H^2 = \frac{8 \pi G}{3} \rho$$

...where $$\rho$$ is the energy density of the universe. Since the energy density has been monitonically decreasing with time due to the expansion, the Hubble parameter has also been monotonically decreasing.

15. Jun 22, 2009

### Ich

Yes, that's a nice shortcut for the calculation of H at the end of deceleration:
$$H_{then}=H_{now} \sqrt{\frac{\Omega_{\Lambda now}}{\Omega_{\Lambda then}}}=H_{now} \sqrt{3\Omega_{\Lambda}}$$
It seems that you forgot the sqrt in your derivation, the factor being 1.5 rather than 2.22?

16. Jun 22, 2009

### Chalnoth

Eh, you're right. My mistake.

17. Jun 22, 2009

### marcus

Ich suggested the ratio is 1.5 instead of 2.22 (a missed square root). So let me revise the above. If the present H is 74, then the value at changeover would be 1.5 x 74, or about 110.
Just as a rough estimate (again using the new Riess et al numbers) this corresponds to z = 0.81, and an expansion age of 6.7 billion years. This is closer to what I remember.

Last edited: Jun 22, 2009
18. Jun 22, 2009

### marcus

The Hubble parameter goes up as you go further back into the past. Chalnoth gave a concise algebraic reason for this, so I'm just agreeing.
It looks like we now have a rough ballpark figure for what the Hubble parameter was when the U stopped decelerating. So that helps respond to your original questions.

Based on the discussion in this thread, I would estimate that the value of H was around 110 km/s per megaparsec when deceleration gradually halted and gave way to acceleration. That changeover was gradual so it doesn't define a moment in time very sharply. Estimates like this depend somewhat on what you take to be the right numbers to plug into the model. We are also assuming spatial flatness here, which is a common and very reasonable assumption--- reasonable because the evidence is that space is either flat or nearly so. It simplifies most calculations.

Making reasonable assumptions we can say that H has been approximately proportional to the square root of the energy density, as Chalnoth indicated, so in early times when the density was much higher then H must have been much greater. And it gradually declined until at the changeover H was around 110. And then acceleration gradually started up, and H continued declining until it is now around 74.

Last edited: Jun 22, 2009
19. Jun 22, 2009

### keepitmoving

thanks for those explanations.

20. Jun 23, 2009

### marcus

You are welcome! Everybody appreciates getting questions like that. I believe that Chalnoth and Ich would say the same. It's fun responding.

21. Jun 24, 2009

### Chalnoth

Indeed. Wouldn't do it otherwise :)

22. Jun 24, 2009

### zeebo17

Could someone explain why the hubble parameter was constant during inflation?

23. Jun 24, 2009

### Hal King

I'd like a simple clarification of the term 'hubble parameter'.

When using that term is the intent to refer too the parameter often used in the GR relations or is it intended to be a general term for an 'expansion rate'? How does it differ from 'hubble constant' in general concept? (I think this is causing confusion)

My review of various models would seem to indicate it could have different values simply due to the model being used.

24. Jun 24, 2009

### Chalnoth

Well, nearly constant. Basically the proposal of inflation is that during that time, the universe was dominated by some sort of field which had a large energy density that varied very slowly with time. By the Friedman equation, then, the Hubble parameter would have also been nearly constant.

The way that this is usually considered is that the physics of this field are such that it experiences some potential energy, but it is not at the minimum: it's rolling down towards the minimum. If the parameters of the field are a certain way, this roll will be slowed by the expansion, such that it rolls down this potential energy hill very, very slowly towards the bottom (and when it hits the bottom, inflation ends).

Hal King: The Hubble parameter is defined as follows:

$$H = \frac{\dot{a}}{a}$$

...where the dot denotes a derivative with respect to time. The Hubble constant has been used as a misleading name for the same thing, and also the value of the Hubble parameter at the current time.

25. Jun 24, 2009

### Hal King

Thank you.

As I thought model dependent when using the term 'hubble parameter' -- General Relativity.

I suspected earlier posters were confused by that.