- #1

- 3

- 0

- Thread starter iidartzii
- Start date

- #1

- 3

- 0

- #2

Nabeshin

Science Advisor

- 2,205

- 16

We say light from certain galaxies will never reach us because they are currently receding from us at faster than the speed of light! So no matter how long we wait, the signals will never arrive. If this "recession faster than the speed of light" idea bothers you, it's actually not a violation of relativity, and there are numerous threads here which discuss why.

- #3

marcus

Science Advisor

Gold Member

Dearly Missed

- 24,738

- 785

The standard cosmology model depends on several key parameters that have to be estimated and what we expect the Hubble rate to do in the future depends on what values you plug into the model

A common cosmology calculator that people use a lot is Ned Wright's, here is the basic version:

http://www.astro.ucla.edu/~wright/CosmoCalc.html

It assumes the Hubble rate is 71, and the dark energy fraction is 0.73. It makes the standard assumptions about the character of dark energy (constant energy density etc.)

With those values, the Hubble rate will continue decreasing indefinitely and approach an asymptotic value of

71sqrt(.73). That equals 60.66, call it 61

In that scenario H keeps decreasing forever but as it nears 61 it decreases slower and slower so that it never reaches 61. It kind of declines but levels out at 61.

You can find explanations of what the numbers represent and what the units are, by looking around in Ned Wright's cosmology tutorial website, where the calculator is.

You can calculate the Hubble radius c/H yourself. At any given time it is the distance which is increasing at rate c. If a galaxy is at that distance, c/H, then its distance from us is increasing at the speed of light. Eventually this means that when H = 61, anything farther than c/H will not be able to get light to us.

If you want help calculating c/H, please ask.

If further more accurate measurements cause astronomers to change the parameters slightly, we might for example get a situation where the present value of H is estimated to be 72, and the dark energy fraction is estimated to be 0.74.

That might happen. The current numbers are supported by millions of datapoints, a huge body of data which the standard model fits remarkably well. But there is still some uncertainty.

If the estimates change like that, you can see that the asymptotic value of H, that it gradually declines to, will be

72 sqrt(.74), which is 61.94, call it 62.

So that number 61 I mentioned is subject to change a little, as more data is gathered, but roughly speaking it is probably pretty good already.

Last edited:

- #4

- 3

- 0

The standard cosmology model depends on several key parameters that have to be estimated and what we expect the Hubble rate to do in the future depends on what values you plug into the model

A common cosmology calculator that people use a lot is Ned Wright's, here is the basic version:

http://www.astro.ucla.edu/~wright/CosmoCalc.html

It assumes the Hubble rate is 71, and the dark energy fraction is 0.73. It makes the standard assumptions about the character of dark energy (constant energy density etc.)

With those values, the Hubble rate will continue decreasing indefinitely and approach an asymptotic value of

71sqrt(.73). That equals 60.66, call it 61

In that scenario H keeps decreasing forever but as it nears 61 it decreases slower and slower so that it never reaches 61. It kind of declines but levels out at 61.

You can find explanations of what the numbers represent and what the units are, by looking around in Ned Wright's cosmology tutorial website, where the calculator is.

You can calculate the Hubble radius c/H yourself. At any given time it is the distance which is increasing at rate c. If a galaxy is at that distance, c/H, then its distance from us is increasing at the speed of light. Eventually this means that when H = 61, anything farther than c/H will not be able to get light to us.

If you want help calculating c/H, please ask.

If further more accurate measurements cause astronomers to change the parameters slightly, we might for example get a situation where the present value of H is estimated to be 72, and the dark energy fraction is estimated to be 0.74.

That might happen. The current numbers are supported by millions of datapoints, a huge body of data which the standard model fits remarkably well. But there is still some uncertainty.

If the estimates change like that, you can see that the asymptotic value of H, that it gradually declines to, will be

72 sqrt(.74), which is 61.94, call it 62.

So that number 61 I mentioned is subject to change a little, as more data is gathered, but roughly speaking it is probably pretty good already.

Is it just a co-incidence that Hubble's radius is the almost the same as the age of the universe?

- #5

marcus

Science Advisor

Gold Member

Dearly Missed

- 24,738

- 785

Some other people may have opinions about that and want to discuss it. What I can say is that the standard cosmo model is well supported by observations and it gives a simple coherent picture of the history and future of the broad outlines of the U.Is it just a co-incidence that Hubble's radius is the almost the same as the age of the universe?

And according to that model it was NOT true in the past that age approximately equaled 1/H,

Nor will it be true in the future.

You can see for yourself using another calculator, the Morgan "cosmos calculator".

It gives the value of the H parameter at past times in the universe's history.

So you can look back, say to when the U expansion age was 1 billion years and see what the H was then.

It won't be approximately 13 times what it is today. You can check.

Google "morgan cosmos calculator"

You will get:

http://www.uni.edu/morgans/ajjar/Cosmology/cosmos.html (I keep that link in my sig.)

You have to put in parameters .27, and .73, and 71

for matter density, dark energy or Lambda, and Hubble

Then you put in a redshift z, like 6.

If you put in 6 and press calculate it will say what Hubble H was back when the U was about 1 billion years old. (actually 0.94 billion but you can play around--change z a little--and get it to be exactly a billion or whatever you want.)

It will say that H was 686 instead of 71, at that time.

But if the age was always going to be 1/H, then since 1 billion is about 1/13 of the present age, we ought to have the H then being 13 times the 71 we have now, and that is more than 900. So since it was only 686, that is not right.

If you get some hands-on, and learn how to use tools like this, then you can answer questions on your own and check what other people tell you. Plus it's fun.

Last edited:

- #6

sylas

Science Advisor

- 1,646

- 7

That's a good question. My guess is ... yes it is.Is it just a co-incidence that Hubble's radius is the almost the same as the age of the universe?

I think this is the implication of Marcus' answer as well; and he posted while I was writing, so I can quote him instead:

In other words, we happen to living in the epoch where age is close to 1/H, and that seems to be just a co-incidence.And according to that model it was NOT true in the past that age approximately equaled 1/H,

Nor will it be true in the future.

There are some non-standard notions as well which try to give a reason for this co-incidence, and which have been published; but they are speculative and not as good as the conventional ΛCDM model for explaining observations.

In the ΛCDM model, you have the age of the universe equal to 1/H at around about Ω

Current estimates tend to be around Ωm ~ 0.27, but 0.2624 falls within the confidence limits. This value decreases over time as the universe expands. So in the conventional model, the point of equality for 1/H and age was most likely very recently, in cosmological terms. Co-incidence? Yes, most likely; but this is not completely definite.

Cheers -- sylas

- #7

sylas

Science Advisor

- 1,646

- 7

Urk. Correcting myself. I mean the point of actual equality is likely to be in the near future. So if the usual ΛCDM models are correct, then it is a co-incidence now and will be an even stronger co-incidence soon (on cosmological time scales, that is, of the order of 10 million years). After that the co-incidence will weaken again.Current estimates tend to be around Ωm ~ 0.27, but 0.2624 falls within the confidence limits. This value decreases over time as the universe expands. So in the conventional model, the point of equality for 1/H and age was most likely very recently, in cosmological terms. Co-incidence? Yes, most likely; but this is not completely definite.

Cheers -- sylas

- #8

- 83

- 0

I'm sorry but something is very wrong with my conception about expansion. The fact that expansion is accelerated doesn't mean that value of H is increasing with time? Why do you say that will gradually decline?If the estimates change like that, you can see that the asymptotic value of H, that it gradually declines to, will be 72 sqrt(.74), which is 61.94, call it 62.

- #9

sylas

Science Advisor

- 1,646

- 7

Imagine a simple universe in which expands at a constant rate. This means that everything is moving away from everything else, and there are no accelerations of this movement.I'm sorry but something is very wrong with my conception about expansion. The fact that expansion is accelerated doesn't mean that value of H is increasing with time? Why do you say that will gradually decline?

Consider a galaxy that is is a distance d away, and moving away at velocity v. The value of H is the relation between d and v. Now wait for a long time. The galaxy becomes further away, but because there's no acceleration, it is still moving away at v. The value of H is now smaller, because d is larger but v is unchanged.

H always decreases as the universe expands, as long as the expansion is anything less than exponential.

If the expansion is accelerating exponentially, then the velocity of any particular galaxy is also increasing exponentially. Suppose you have a galaxy where the distance from us to the galaxy is an exponential function. d = d

The recession velocity of this galaxy is obtained by differentiating. v = k.d

Cheers -- sylas

- #10

marcus

Science Advisor

Gold Member

Dearly Missed

- 24,738

- 785

I see Sylas already answered. I'll see if I can give an alternative response that says much the same thing in a different way. It might help to have two.I'm sorry but something is very wrong with my conception about expansion. The fact that expansion is accelerated doesn't mean that value of H is increasing with time? Why do you say that will gradually decline?

Above all to read or talk cosmology you have to be familiar with the scale factor a(t).

When they plot the history of expansion they don't plot H(t). They plot a(t) the scale factor as a function of time.

It is sometimes called "the average distance between galaxies", or the "size" of the universe. But it is just a handle on the size---we don't know the actual size, it might be infinite and therefore undefined.

The rigorous meaning of a(t) is that it is the scale factor that appears in the standard cosmo model metric, the Friedman metric, or FRW metric, that everybody uses.

There are various conventions. Commonly a(t) is normalized so a(present time) = 1.

So when distances have expanded to double, then a will equal 2. And when distances were only 1/100 what they are today, then a was equal 0.01.

a(t) is the important thing.

The mathematical definition of H(t) is simply a'(t)/a(t) where a'(t) is the time-derivative of a(t).

The current H(t) means that a(t) increases by about 1/130 of one percent every million years.

When people say that the universe is expandig (or that distances between CMB stationary points are increasing) they mean that a(t) is increasing, IOW that a'(t) is positive.

This says nothing about H(t). H(t) is a fractional rate of change. What it does depends on both its numerator and its denominator. It will do whatever. Since it is defined to be

a'(t)/a(t), if it happens that a is growing faster than a' then H will decrease! If the denominator is growing faster then the fraction will decrease regardless, even though both numerator and denominator are increasing.

Please tell me if that was clear. If it was, and if you want, I will show a simple way to calculate what H(t) will do in the future. And I will explain why H(t) has to continue to decline down to some asymptotic limit that is about sqrt(0.74) of what it is now. Or Sylas or Twofish can jump in and explain that. It is kind of nice. Uses an elegant little equation called the Friedman equation, plus a tiny bit of 9th grade algebra.

Skolon, it was good you asked that I was hoping someone would. Keep asking stuff!

Last edited:

- #11

- 83

- 0

And yes again, I will be pleased if you will explain how and why H(t) will continue to decline.

- Replies
- 8

- Views
- 8K

- Last Post

- Replies
- 19

- Views
- 13K

- Last Post

- Replies
- 4

- Views
- 2K

- Replies
- 8

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 4K

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 8

- Views
- 3K

- Last Post

- Replies
- 20

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 540