Accelerating universe, decreasing hubble constant

  • Thread starter Ranku
  • Start date
  • #1
260
2

Main Question or Discussion Point

I wanted to know a bit more about the fact that in the presently accelerating expansion of the universe the Hubble constant is still decreasing. When the universe was decelerating the Hubble constant was decreasing. It is still decreasing in an accelerating universe. Does that mean the Hubble constant is decreasing in an accelerating universe at a lower rate than it was when the universe was decelerating? In other words, when we say the universe is accelerating do we really mean that it is decelerating less than before?
 

Answers and Replies

  • #2
Ich
Science Advisor
1,931
1
The Hubble constant is defined as (da/dt) / a. Pick any function for a(t), where d²a/dt² > 0 means "accelerating", and have a look at the Hubble constant.
You'll see that even if a(t) increases exponentially with time, H merely stays constant.
 
  • #3
marcus
Science Advisor
Gold Member
Dearly Missed
24,738
785
... In other words, when we say the universe is accelerating do we really mean that it is decelerating less than before?
Ich answered perfectly. I will just repeat for emphasis. No, we do not mean what you said. When we say the universe is accelerating, we mean that the expansion is accelerating (an increase in the time rate of increase in the scalefactor)

Our handle on the size of the universe is the scalefactor a(t). What plugs into the Friedman metric and makes distances expand.
The scalefactor is the key thing in the Friedman model. The two Friedman equations govern the evolution of the scalefactor.

(Look up Friedman equations on Wiki)

Do not worry about the Hubble rate. It is just a conventional symbol for the ratio a'/a.
It relates in a convenient way to observations, and is extremely useful, but is not fundamental.
The real thing you must think about is a(t). It is the backbone and the guts of the model.

When people say "the universe is expanding" that means nothing else than that
a'(t) > 0

When people say "the expansion is accelerating" that means nothing else than
a"(t) > 0.

There is no reason to be surprised that H(t) is decreasing. H(t) is merely something defined to be equal a'(t)/a(t) and it is quite normal for a function which is increasing at an increasing rate to have that particular ratio be decreasing. Here is an example:

Let f(t) = t3 on the positive real axis.

Then f'(t) = 3t2 which is positive so f is increasing

and f"(t) = 6t which is positive so the increase is accelerating.

But what would be analogous to H, namely f'(t)/f(t) = 3t2/t3 = 3/t,

is obviously decreasing.

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

Ranku, thanks for posting this question on the forum. It is constructive. Many people do not understand what "accelerating expansion" means and get confused in exactly the way you did. If there is something about this that you now do not understand, please ask more questions. Keep asking until it is clear. :smile:
 
Last edited:
  • #4
260
2
Hi Marcus, thank you for your detailed explanation; (and to Ich's help too). I do not have much of a math background, but I get the idea that taking the time derivative of a(t) gives the required results.
What I would like to know is how does it relate to the astronomical observation of the value of the Hubble constant, and so I repeat that part of the question: is the Hubble constant decreasing at a lower rate in an accelerating universe than the rate it was decreasing when the universe was decelerating?
Regarding using the scale factor a(t), what is its unit, how do we get the values for it, how is it measured?
 
  • #5
Ich
Science Advisor
1,931
1
What I would like to know is how does it relate to the astronomical observation of the value of the Hubble constant, and so I repeat that part of the question: is the Hubble constant decreasing at a lower rate in an accelerating universe than the rate it was decreasing when the universe was decelerating?
We can measure the current value of the Hubble constant directly, not its time derivatives. So there is no relation to the direct measurement of H0.
H is decreasing wth time, but that has nothing to do with acceleration/deceleration. If a(t) were any power funtion, H~1/t would be always true.
Regarding using the scale factor a(t), what is its unit, how do we get the values for it, how is it measured?
It's usualy taken as a dimensionless variable and defined to be 1 in the present epoch. The exact definition does not matter, all you can measure is ratios like a(t1)/a(t2) or a'(t)/a(t) and such.
a(t1)/a(t2) is direcly related to the measured redshift of comoving objects: a(now)/a(time of light emission) = 1+z.
 
  • #6
260
2
Thanks for the response, Ich. But I still haven't got the exact answer I am looking for. I understood that the Hubble constant is decreasing whether the universe is accelerating or decelerating. What I want to know is about the rate of decrease of the Hubble constant. Is the rate of decrease lower in an accelerating universe than in a decelerating universe?
 
  • #7
Ich
Science Advisor
1,931
1
Is the rate of decrease lower in an accelerating universe than in a decelerating universe?
Ok, if you could measure H and H', you can construct the deceleration parameter from the ratio H'/H². That means, if H'/H²>-1, the expansion is accelerating.
So, yes, if you define "rate of decrease" as -H'/H², your statement is true.
 
  • #8
260
2
Thank you Ich for your answer. Until the next time.
 
  • #9
Chronos
Science Advisor
Gold Member
11,408
738
Modern cosmology, based on observational evidence [as Ich noted], provides one answer. Im not aware of any viable alternatives to date.
 

Related Threads for: Accelerating universe, decreasing hubble constant

Replies
4
Views
3K
  • Last Post
Replies
10
Views
3K
Replies
14
Views
5K
Replies
18
Views
4K
  • Last Post
Replies
4
Views
2K
Top