Value of Hubble Parameter in Decelerating Universe

In summary, the value of the Hubble parameter when the universe was decelerating is still a topic of debate and there is no precise measurement for it. However, based on the accepted values of the Hubble parameter over time, it is believed that the universe was decelerating for most of its history until relatively recently when dark energy took over. The transition from deceleration to acceleration depends on the equation of state, and for a universe with a cosmological constant, it is estimated to have been around twice its current value. The Hubble parameter has varied dramatically with time and was at its smallest during deceleration. The largest value would have been at the end of inflation, estimated to be around 10^{22} to
  • #1
keepitmoving
97
0
what was the value of the Hubble parameter when the universe was decelerating?
 
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  • #2
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
if the Hubble parameter is 1, is the universe accelerating?
 
  • #4
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
keepitmoving said:
what was the value of the Hubble parameter when the universe was decelerating?
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 [tex]w = 0[/tex]. A cosmological constant has negative pressure equal to its energy density, [tex]w = -1[/tex]. The deceleration parameter q is written as:

[tex]q = \frac{1}{2}\left(1 + 3w\right)[/tex]

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

This parameter is defined as follows:

[tex]p = w\rho[/tex]

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

[tex]0 + -\rho_\Lambda = w \left(\rho_m + \rho_\Lambda\right)[/tex]

So:

[tex]w = \frac{-\rho_\Lambda}{\rho_m + \rho_\Lambda}[/tex]

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

[tex]\Omega_\Lambda = \frac{\rho_\Lambda(a=1)}{\rho_c(a=1)}[/tex]
[tex]\Omega_m = \frac{\rho_m(a=1)}{\rho_c(a=1)}[/tex]

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

[tex]w = \frac{-\Omega_\Lambda}{\frac{\Omega_m}{a^3} + \Omega_\Lambda}[/tex]

Now we just need to solve the above equation for [tex]a[/tex]. I get:

[tex]a = \left(\frac{\Omega_m}{-\left(1 + \frac{1}{w}\right)\Omega_\Lambda}\right)^\frac{1}{3}[/tex]

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

[tex]a = 0.56[/tex]

So that means a Hubble parameter of:

[tex]H(a) = H(a=0)\left(\frac{\Omega_m}{a^3} + \Omega_\Lambda\right)[/tex]

Plugging in [tex]a = 0.56[/tex], I get:

[tex]H(a) = 2.22H(a=0)[/tex]

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
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, i`m 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 wouldn`t a given object be moving faster relative to a reference point than it was the previous Myr? It`s 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
keepitmoving said:
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, i`m 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 wouldn`t a given object be moving faster relative to a reference point than it was the previous Myr? It`s 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.
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 [tex]10^{22}[/tex] to [tex]10^{25}[/tex] km/sec/Mpc.

Edit: fixed the error from the square root noticed by Ich below.
 
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  • #8
maybe i`m on the wrong track. Is the acceleration really the acceleration of acceleeration?
 
  • #9
keepitmoving said:
what was the value of the Hubble parameter when the universe was decelerating?

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. :smile:)

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.
 
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  • #10
keepitmoving said:
maybe i`m on the wrong track. Is the acceleration really the acceleration of acceleeration?
The acceleration is the acceleration of the scale factor.
 
  • #11
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
keepitmoving said:
maybe i`m on the wrong track. Is the acceleration really the acceleration of acceleeration?

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
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
keepitmoving said:
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?
The Hubble parameter is given by the Friedman equation:

[tex]H^2 = \frac{8 \pi G}{3} \rho[/tex]

...where [tex]\rho[/tex] 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
The Hubble parameter is given by the Friedman equation
Yes, that's a nice shortcut for the calculation of H at the end of deceleration:
[tex]H_{then}=H_{now} \sqrt{\frac{\Omega_{\Lambda now}}{\Omega_{\Lambda then}}}=H_{now} \sqrt{3\Omega_{\Lambda}}[/tex]
It seems that you forgot the sqrt in your derivation, the factor being 1.5 rather than 2.22?
 
  • #16
Ich said:
Yes, that's a nice shortcut for the calculation of H at the end of deceleration:
[tex]H_{then}=H_{now} \sqrt{\frac{\Omega_{\Lambda now}}{\Omega_{\Lambda then}}}=H_{now} \sqrt{3\Omega_{\Lambda}}[/tex]
It seems that you forgot the sqrt in your derivation, the factor being 1.5 rather than 2.22?
Eh, you're right. My mistake.
 
  • #17
marcus said:
...
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. :smile:)

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.

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.
 
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  • #18
keepitmoving said:
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?

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.

keepitmoving said:
what was the value of the Hubble parameter when the universe was decelerating?

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.
 
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  • #19
thanks for those explanations.
 
  • #20
You are welcome! Everybody appreciates getting questions like that. I believe that Chalnoth and Ich would say the same. It's fun responding.
 
  • #21
marcus said:
You are welcome! Everybody appreciates getting questions like that. I believe that Chalnoth and Ich would say the same. It's fun responding.
Indeed. Wouldn't do it otherwise :)
 
  • #22
Could someone explain why the Hubble parameter was constant during inflation?
 
  • #23
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
zeebo17 said:
Could someone explain why the Hubble parameter was constant during inflation?
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:

[tex]H = \frac{\dot{a}}{a}[/tex]

...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
Thank you.

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

I suspected earlier posters were confused by that.
 
  • #26
Hal King said:
Thank you.

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

I suspected earlier posters were confused by that.
The only way in which the Hubble parameter is model dependent is that it depends upon the description of our universe as being approximately homogeneous and isotropic on large scales being accurate. Other than that, it is completely model independent.
 
  • #27
As a 'term' and approximate value I agree.

As the relation cited is another thing. The relation only has meaning in context of the definition of 'a'.
That can vary.

What is better is a description of the concept. This is why sometimes a conceptual description is better than a formula. A formula is sometimes 'too precise' -- in that it carries with it the assumptions that allowed the formula to be written.

Of course, that is often looked at as the 'goal' in physics -- the 'formula'.
 
  • #28
Chalnoth said:
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.

Is this energy density that of dark energy? I know that after inflation matter and radiation dominated the total energy density and then the universe transitioned into a state where dark energy is now dominating the total energy density. But what were these densities like before and during inflation?
 
  • #29
zeebo17 said:
Is this energy density that of dark energy?
No. It would have had to have been vastly, vastly higher. We'll know more about precisely what the energy density during inflation was if we can get a positive detection of what is known as the B-mode polarization of the CMB (which is expected to be a very small, difficult to extract signal in the most optimistic scenarios...so far nobody has yet confirmed a detection, though, so we don't know at what level this sort of polarization actually is).

zeebo17 said:
I know that after inflation matter and radiation dominated the total energy density and then the universe transitioned into a state where dark energy is now dominating the total energy density. But what were these densities like before and during inflation?
Bear in mind that the density of the inflaton field right as inflation ended would have been identically equal to the total energy density of the particles it decayed into immediately afterwards. And we know the energy density of the very early universe had to be obscenely high to produce dark matter and the asymmetry between matter and anti-matter (as we haven't seen the physics for either of these in particle accelerators yet).
 
  • #30
During inflation the scale factor behaves as [tex]a \propto e^{Ht}[/tex] and during a radiation dominated era it behaves as [tex]a \propto t^{1/2}[/tex] and then during a matter dominated era behaves as [tex]a \propto t^{2/3}[/tex].

So then is it correct to say that during inflation: [tex]H= \dot{a}/a \propto constant[/tex]
radiation dominated era: [tex]H= \dot{a}/a \propto 1/t[/tex]
matter dominated era: [tex]H= \dot{a}/a \propto 1/t[/tex]
from simply taking the derivative and dividing by a?

Does the Hubble parameter behave as 1/t during these eras? How would you find how it behaves during an era with mixed conditions, such as the present?

Thanks!
 
  • #31
zeebo17 said:
During inflation the scale factor behaves as [tex]a \propto e^{Ht}[/tex] and during a radiation dominated era it behaves as [tex]a \propto t^{1/2}[/tex] and then during a matter dominated era behaves as [tex]a \propto t^{2/3}[/tex].

So then is it correct to say that during inflation: [tex]H= \dot{a}/a \propto constant[/tex]
radiation dominated era: [tex]H= \dot{a}/a \propto 1/t[/tex]
matter dominated era: [tex]H= \dot{a}/a \propto 1/t[/tex]
from simply taking the derivative and dividing by a?

Does the Hubble parameter behave as 1/t during these eras? How would you find how it behaves during an era with mixed conditions, such as the present?

Thanks!
Not easily. Basically you'd have to go back to evaluate the integrals numerically (they can't be solved by hand).
 

FAQ: Value of Hubble Parameter in Decelerating Universe

What is the Hubble parameter and how is it related to the expansion of the universe?

The Hubble parameter, denoted as H0, is a measure of the rate at which the universe is expanding. It is named after Edwin Hubble, who first discovered the expansion of the universe. The Hubble parameter is related to the expansion of the universe through the Hubble's law, which states that the velocity of recession of a galaxy is directly proportional to its distance from us.

How is the Hubble parameter calculated and what are its units?

The Hubble parameter is calculated by dividing the velocity of recession of a galaxy by its distance. It is usually expressed in units of kilometers per second per megaparsec (km/s/Mpc). This means that for every megaparsec (3.26 million light years) of distance, the velocity of recession increases by a certain amount.

What is the significance of the Hubble parameter in a decelerating universe?

In a decelerating universe, the Hubble parameter is used to measure the rate at which the expansion of the universe is slowing down. This is because the Hubble parameter is inversely proportional to the age of the universe. In a decelerating universe, the Hubble parameter decreases as the universe ages, indicating a slower expansion rate.

What is the current value of the Hubble parameter and how has it changed over time?

The current accepted value of the Hubble parameter is approximately 70 km/s/Mpc. However, this value has been a subject of debate and has changed over time as new measurements and observations are made. In the past, the value of the Hubble parameter was thought to be higher, indicating a faster expansion rate. However, recent observations have shown that the expansion rate of the universe is actually slowing down.

How does the value of the Hubble parameter affect our understanding of the universe?

The value of the Hubble parameter is crucial in understanding the past, present, and future of the universe. It helps us determine the age of the universe, the rate at which it is expanding, and the composition of the universe. By studying the changes in the value of the Hubble parameter, scientists can gain insights into the evolution and fate of the universe.

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