# Why are the theoretical and measured sizes/ages of the universe so similar?

1. Feb 24, 2005

### warpsmith

Since space expands everywhere evenly, objects 2x as far apart recede twice as fast. It has been calculated that at a distance of 14 billion light-years (the Hubble distance), objects recede faster than light. This is OK since it is space itself is expanding.

This dictates a theoretical limit to the size of the observable universe (perhaps up to 150 B LY due to the expanding Hubble radius catching up with old light rays). This size of the observable universe is entirely due to the rate of expansion.

However, isn't it a HUGE coincidence that the rate of expansion matches almost exactly the best estimates for the size of the universe (also 14B LY) as determined by the ages of stars and other rate-independent measurements?

In other words (ignoring expansion acceleration and other wrinkles), won't the universe always seem to be about 14B years old, based solely on the Hubble constant? Couldn't the universe be MUCH older than 14B years?

2. Feb 24, 2005

### Aether

Hubble's "constant" is not constant over time.

3. Feb 24, 2005

### warpsmith

Yes, I'm aware of that

But it seems that the Hubble radius is 14B LY or so. This is due to the current rate of expansion. If for some reason the universe were 100B years old (not that I believe this, just making a point), it seems that at the same rate of expansion, we would still have a similar Hubble radius (since objects are expanded away FTL beyond that point).

I recognize that the observable universe is tied to but not equal to the Hubble radius, but these values seem independent of the age of the universe.

4. Feb 24, 2005

### chronon

If you ignore such things as gravity and the cosmological constant, then the rate of expansion would be constant. So when the universe is twice its current age, a galaxy will be twice as far away as it is now, but will have the same speed of recession. Hence the hubble constant will be half what it is now.

Note that the limit of the observable universe is not due to objects travelling away faster than light, it is when we see back to the big bang (although the CMBR means that we can't actually see that far back). I've written more about this in http://www.chronon.org/Articles/cosmichorzns.html

5. Feb 24, 2005

### warpsmith

chronon,

very nice site!

I am still not understanding this. I am not trying to be obtuse or trying to propose some weird alternative theory - I just cannot get this straight in my head!

As I understand it, if you have dots one foot apart along an infinite elastic string, and the string is expanding at the same rate as the universe, then dots that are 14B LY apart are mathematically receding from each other at the speed of light. This is irrespective of how 'old' the string is or for how long it has been expanding. If the string had been expanding for 100B years, and you were to hop onto it and look down along it, then the dots 14B LY away would be receding FTL (you could still see a lot further, due to light leaving earlier in time, which makes the observable radius much further).

But if you didn't know how old the string was, wouldn't you just assume it was 14B years old based on the rate of expansion, when in fact it could be much older?

I'm not trying to argue static or regenerative models - I am just trying to grasp this one mathematical point.

Thanks for the help!

6. Feb 24, 2005

### Aether

Actually, $$R(t) = \int_0^t \dot R(t) dt$$ rather than $$R(t) = \dot R(t) \cdot t$$ as is commonly assumed. This is a crucial distinction when $$\ddot R \neq 0$$.

If you assume the same rate of expansion, then when the universe is 100Gyr old the Hubble radius should span a distance several times greater than it is currently observed to.

If the two ends of the string had been moving apart at the speed of light for 100Gyr, then the dots 14Gly away would only be receding at 0.14 times the speed of light, would they not?

Last edited: Feb 24, 2005
7. Feb 24, 2005

### chronon

You seem to have a hidden assumption of some sort of acceleration in the expansion of the universe. If you assume this (i.e. a cosmological constant) then the universe would be older than 1/(Hubble constant). This was why the cosmological constant was taken seriously when 1/(Hubble constant) was thought to be about 2 billion years, that is younger than the Earth.

However, if you don't have any acceleration then Hubble's constant will decrease over time, since it is equal to speed/distance and the speed is constant while the distance is increasing. Hence your estimate of the age of the universe, which is based on 1/(Hubble constant) will increase.

8. Feb 24, 2005

### warpsmith

I feel unusually obtuse in the presence of two patient and kind tutors.

No, I am assuming an infinite string with constant rate of expansion such that dots 14B LY apart are receding FTL (which models our universe).

Almost got a flicker of comprehension from this. But I do not understand the part about the speed being constant, since from a fixed point the perceived recession rate increases with distance. Maybe I am off by one derivative of speed somehow...

Let me put forth my (a)ssumptions and (c)onclusions for you to demolish:

1a - assume no acceleration, gravity, or cosmological constant
2a - assume an infinite and eternally old string with dots at regular intervals
3a - assume the string between each dot expands at a fixed and constant rate
3c - this means that dots recede from all other dots at geometrically increasing rates as a function of distance
4a - assume the rate of expansion is equal to today's rate of expansion of the universe
4c - mathematically, dots 14B LY apart recede FTL
- (aside) this gives observable universes of up to 156B LY depending on whose math you believe in other posts
5a - you arrive at this string and attempt to determine the age of it
5c - you look in either direction and can see or calculate that the dot at 14B LY must be moving away at light speed
5c - you figure that dot must have been at your position 14B years ago
5c - you determine the string is 14B years old

I am completely aware that I am missing something painfully obvious. I just don't know what!

9. Feb 24, 2005

### Nereid

Staff Emeritus
You might like to read some of the papers by Lineweaver and Davis (several are referenced in the sticky here), or their recent article in Scientific American.

It may be that what you are interested in is beyond the scope of what addressed by Lineweaver and Davis, but at least it should rule out many of the common misunderstandings.

10. Feb 24, 2005

### warpsmith

Actually, it was Lineweaver's current article in SciAm (Misconceptions about the Big Bang) that sent me here! I thought I had it all straight until that article...

Thank you for the references. I will go off and do some reading.

11. Feb 24, 2005

### hellfire

Yes, I belive this is a coincidence.

In the current model with H0 = 71 Km /s Mpc, $\Omega_m = 0.27$ and $\Omega_{\Lambda} = 0.73$ the "Hubble age" 1/H0 is nearly equal to the age of the universe. In this model 1/H0 = 13.7 Gyr. This implies that the comoving distance which enters the Hubble law to give a recession velocity c (this distance is called Hubble radius) is DcH = c / H0, equal to 13.7 GLyr. As an aside, the particle horizon is located at 46 GLyr (comoving distance).

This coincidence does not happen for every model. For example, an Einstein de-Sitter universe (flat space with $\Omega_m = 1$), has an age of 9 Gyr at the time at which H0 = 71 Km /s Mpc. At this time t0, the Hubble age is obviously different to the age of the universe. Of course, at t0, the comoving distance at which the Hubble sphere is located is equal to 13.7 GLyr. The particle horizon is located at 27 GLyr.

This coincidence does also happen for other models, such as the Milne universe, with H0 = 71 Km /s Mpc, $\Omega_m = \Omega_{\Lambda} = 0$, but there are lots of universes for which this coincidence does not occur. I cannot tell you whether there is a deeper reason behind this.

Last edited: Feb 25, 2005
12. Feb 24, 2005

### hellfire

I have been thinking about this and I came to the conclusion that the fact that this coincidence arises in the Milne universe (linearly expanding universe) is natural (actually it is not a coincidence in that model). In that model t = 1/H, for every t, since $\dot a = k$ (the first time derivative of the scale factor is a constant; note that $H = \dot a / a$). I wonder now whether there is an explanation for the the similarity between the Milne universe and the concordance model today at H0 = 71 Km/s Mpc.

Last edited: Feb 24, 2005
13. Feb 24, 2005

### Garth

Some think the Milne universe i.e. here ought to be the concordance model!

14. Feb 25, 2005

### Chronos

I object. Nucleosynthesis does not work in the Milne model. It could, however, work in other reference frames.. like BD.

15. Feb 25, 2005

### Garth

Others would disagree, for example Kolb Nucleosynthesis in a Simmering Universe and the Indian team (Daksh Lohiya, Annu Batra, Shobhit Mahajan, Amitabha Mukherjee, Department of Physics & Astrophysics, University of Delhi) A coasting cosmology

Of course the GR Milne model is empty and so has no nucleosynthesis, however the freely coasting model is modified GR theory which is based on a Milne universe (R ~ t, k = -1)

Note the SCC Jordan conformal frame model is (R ~ t, k = +1) but the k = +1 and k = -1 models converge in the early universe, and so the freely coasting nucleosynthesis applies to SCC as well.

Garth

Last edited: Feb 25, 2005