# Cosmological Coincidences

1. Mar 15, 2007

### Garth

One problem in the standard $\Lambda$CDM model is the near equality to an OOM of the densities of baryonic matter (4%), non-baryonic Dark Matter (23%) and Dark Energy (73%). The coincidence is more striking if it is realised that the proportion of DE, if due to the cosmological constant, will grow with the volume of the universe and the universe has expanded by something of the order of 1060 since the Planck era.

If DE is not a cosmological constant but some form of quintessence with $\omega$ <,>, -1, then its density proportion will also grow or possibly shrink with cosmic expansion.

In either case the question is why should this density be in the same 'ball-park' as that of the matter content?

A second coincidence is related to this.

If the presence of DE with negative pressure is accepted it allows the universe to accelerate. A purely accelerating universe can have any age from just over Hubble time (the inverse of Hubble's constant) upwards to infinity (Such as in the Steady State model).

Without any DE acceleration, but with the presence of ordinary matter and energy with a positive pressure, the universe purely decelerates. A decelerating universe has an age less than Hubble time.

The standard model first decelerates, then massively accelerates (inflation), then decelerates through the BBN era until the recent epoch (~1 < z < 0) when it accelerates again.

So what has been the result of this deceleration/acceleration process on the age of the universe?

The present best accepted values of cosmological parameters
(using the table at http://lambda.gsfc.nasa.gov/product/map/dr2/params/lcdm_all.cfm [Broken])
H0 = 70.4 km/sec/Mpsc
$Omega_{\Lambda}$ = 0.732
$Omega_{matter}$ = 0.268

Feeding these values into Ned Wright's Cosmology Calculator:
The age of the universe is = 13.81 Gyrs.
But with h100 = 0.704,
Hubble Time = 13.89 Gyrs.

Strange that the age of the universe should be equal to Hubble Time to within an error of 0.6%, almost as if the universe had been expanding linearly at the same rate all the way along!

Just food for thought.

Garth

Last edited by a moderator: Apr 22, 2017 at 4:19 PM
2. Mar 15, 2007

### sylas

Yes! I noted this myself previously, and started up a thread about it on "Bad Astronomy" some time ago. I also corresponded with Ned Wright directly. Ned considers it is probably just a co-incidence; but it is remarkable.

The thread I wrote at Bad Astronomy is Cosmic Coincidences (April 2005). I love it that we picked almost the same title!

Cheers -- Sylas

3. Mar 16, 2007

### Garth

I raised the question of the Hubble time/Universe Age coincidence at last Friday's meeting of the RAS after a lecture on DE. Michael Rowan Robinson offered the explanation that the Age used to be always near Hubble time (TH) and so a coincidence should not be too significant.

However that was in the days of purely decelerating universes, when the age lay somewhere between TH and 2/3 TH depending on $\Omega$.

If we allow DE and acceleration into our model then the story changes.

Now it could be anything from 2/3 TH to infinity depending on the DE/Matter (DM and baryonic) ratio.

So why should the observed composition give an age that is so close to TH?

Garth

Last edited: Mar 16, 2007
4. Mar 16, 2007

### hellfire

The expression for the age of the universe in a general cosmological model is:

$$T = \frac{1}{H_0} \int_0^1 \frac{da}{\sqrt{ \Omega_{k, 0} + \displaystyle \frac{\Omega_{m,0} }{a} + \displaystyle \frac{\Omega_{r,0} }{a^2}+ \Omega_{\Lambda,0} a^2 \right)}}}$$

Neglecting the radiation density, for the age to be equal to $T = 1/H_0$, it must hold that:

$$\mathcal{I}(\Omega_{m,0}, \Omega_{\Lambda,0}) = \int_0^1 \frac{da}{\sqrt{ \Omega_{k, 0} + \displaystyle \frac{\Omega_{m,0} }{a} + \Omega_{\Lambda,0} a^2 \right)}}} = 1$$

With $\Omega_{k, 0} = 1 - \Omega_{m,0} - \Omega_{\Lambda,0}$.

It would be nice to see graphically how the surface $$\mathcal{I}(\Omega_{m,0}, \Omega_{\Lambda,0})$$ behaves depending on different values of $\Omega_{m,0}$ and $\Omega_{\Lambda,0}$ (for example, between [0, 1]). Unfortunately I do not have the tools to do such graphics.

Last edited: Mar 16, 2007
5. Mar 17, 2007

### Garth

As $$\int_0^1 da = 1$$

Then one possible solution would be

$$\Omega_{k, 0} + \displaystyle \frac{\Omega_{m,0} }{a} + \Omega_{\Lambda,0} a^2 \right) = 1$$

As $\Omega_{k, 0} = 1 - \Omega_{m,0} - \Omega_{\Lambda,0}$ then this would require

$$\Omega_{m,0}\left(1- \frac{1}{a}\right) = \Omega_{\Lambda,0} \left(a^2 -1\right)$$

or $$\Omega_{m,0}= a\left(a + 1\right)\Omega_{\Lambda,0}$$

Garth

6. Mar 18, 2007

### Chronos

Nice exchange Garth, I perceive you did not take the bait. It proves nothing, but the exercise is refreshing. A key variable is still missing, IMO. I am reluctant to call it a hidden variable, but something is clearly missing.

7. Mar 18, 2007

### Garth

What I did not point out before was that in a (non-standard) static universe model, where cosmological red shift would not be explained by expansion but instead by a mass field effect, then

$$a(t) = a_0 = 1$$

and if

$$\Omega_{m,0}= a\left(a + 1\right)\Omega_{\Lambda,0}$$

then

$$\Omega_{m,0}= 2 \Omega_{\Lambda,0}$$

And it just so happens that in http://en.wikipedia.org/wiki/Self_creation_cosmology [Broken] we find $\Omega_{m,0} = 0.22$ and $\Omega_{\Lambda,0} = 0.11$

Of course in my post #5 if a = 1 the solution becomes degenerate so you would have to say:

$$\Omega_{k, 0} + \displaystyle \frac{\Omega_{m,0} }{a} + \Omega_{\Lambda,0} a^2 \right) = X$$

if a = 1 then

$$\Omega_{k, 0} + \Omega_{m,0} + \Omega_{\Lambda,0} = X$$

but as $\Omega_{k, 0} = 1 - \Omega_{m,0} - \Omega_{\Lambda,0}$ then

X = 1 as required.

Of course in the static model case the concept of Hubble time and an 'Age of the universe' has to be reinterpreted....

But just food for more thought.

Garth

Last edited by a moderator: Apr 22, 2017 at 4:22 PM
8. Oct 19, 2007

### Garth

For those interested in other coincidences.....

Eventually we have published on the ArXiv Brandon Carter's previously unpublished paper The significance of numerical coincidences in nature
Better late than never!!

Garth

9. Oct 20, 2007

### SpaceTiger

Staff Emeritus
I'm not a big fan of cosmology by coincidence. It worked well for the flatness and horizon problems because they were so extreme and required a great deal of fine-tuning in the absence of inflation.

These "coincidences", however, are less striking. The trouble is that, because of randomness in nature, there will be apparently unusual things in any astronomical data set even if nothing is unusual going on. Unless the coincidence is overwhelming, there's no way to know whether it's just an artifact of selective attention or a real indication of new physics.

10. Oct 20, 2007

### Wallace

The 'age' co-incidence is of little consequence. The reason for this is that the things that matter, e.g. what can be observed, generally do not depend on the total age of the universe. This includes formation and evolution of structure, distance measures and observed CMB fluctuations. The age of the Universe is a quantity that you get by integrating a model that has been determined by fitting its parameters to what has been observed.

What you claim is that there is significance in the fact that two models, one of which is well fit by observations and the other at odds with observational data have a similar value for a quantity that is intrinsically unobservable.

What matters is what we can see, not numerology!

11. Oct 20, 2007

### Chronos

I concur, Wallace. The issue must be decided by observational evidence. WMAP3 is a powerful tool, but not inscrutable. The 'axis of evil hypothesis' will eventually be proven to be a selection effect, IMO.

12. Oct 20, 2007

### Garth

Agreed on the last point Wallace, but do you think that the near equality (to an OOM) of the densities of baryonic matter (4%), non-baryonic Dark Matter (23%) and Dark Energy (73%) is also an insignificant coincidence?

As I said, this coincidence is more striking if it is realised that the proportion of DE, if due to the cosmological constant, will grow with the volume of the universe and the universe has expanded by something of the order of 1060 since the Planck era.

The two coincidences are linked because it is the mix of DE/Matter that determines the age of the universe in terms of Hubble's constant.

Garth

Last edited: Oct 20, 2007
13. Oct 20, 2007

### Wallace

The energy density co-incidence is interesting, as is the related relisation that 'dark energy' comes to dominate at the point at which the universe becomes void dominated.

Both of these are interesting because a) they are the result of things that can be measured and b) the quantities in the co-incidence (energy densities, the state of structure growth) have a bearing on the evolution of the universe. The age of the universe in and of itself does not which is why the age co-incidence is pure numerology and nothing more.

14. Oct 21, 2007

### Garth

We should add: "in the standard GR $\Lambda$CDM model"....

Actually the age of the universe is the result of

$$T = \frac{1}{H_0} \int_0^1 \frac{da}{\sqrt{ \Omega_{k, 0} + \displaystyle \frac{\Omega_{k, 0} }{a} + \displaystyle \frac{\Omega_{r,0} }{a^2}+ \Omega_{\Lambda,0} a^2 \right)}}}.$$

I still find it interesting that, ignoring radiation density, our present day values for $\Omega_{k, 0}$, $\Omega_{k, 0}$, $\Omega_{\Lambda,0}$ yield
TH0 = 0.994 and wonder whether that $\pm$0.6% near equality with unity might be telling us something about that standard GR $\Lambda$CDM model....

Garth

Last edited: Oct 21, 2007
15. Oct 21, 2007

### Wallace

Right, the age of the universe is the result of the integral, but the age doesn't appear in any calculations in cosmology, it is the result of other calculations and observations.

What is it do you think this co-incidence 'is telling us'?

16. Oct 21, 2007

### oldman

Ah yes. Or we might be concocting new physics out of the fact that we can just observe total eclipses, because of the near-equality of the angles subtended here by the sun and moon.

That way lies anthropic reasoning and much else ....

17. Oct 21, 2007

### Garth

Well, either it is just a coincidence or that for some as yet unknown reason

$$1 = \int_0^1 \frac{da}{\sqrt{ \Omega_{k, 0} + \displaystyle \frac{\Omega_{k, 0} }{a} + \displaystyle \frac{\Omega_{r,0} }{a^2}+ \Omega_{\Lambda,0} a^2 \right)}}}.$$

This would place a constraint on the DE equation of state so that, for example, it resulted in
$$\Omega_{\Lambda,0}= \frac{1}{a\left(a + 1\right)}\Omega_{m,0}$$,

(ignoring radiation density) as discussed above in post#5.

Garth

Last edited: Oct 22, 2007
18. Oct 21, 2007

### Wallace

Placing a constraint means that a parameter of a model is examined by seeing the effects of different values of that parameter on observable and confronting that with observation data.

What you are suggesting it that we first constrain the parameters by looking at available data, then given that model the data prefers calculate the unmeasurable age of the Universe that the model implies. Once we have that value of the age, you suggest that we then ignore the data and claim that the region of allowed models is constrained to those which give the same age. This is an incredibly silly way to do cosmology, since most of the region of allowed models under your suggesting will be ruled out by the very data that arrived at the age of unity in the first place!

To be clear about this, the process you suggest is somewhat analogous to the following: We have an unknown function describing the universe. We can measure this function at some points to some accuracy. Based on these measurements and known physics we come up with a plausible physical model with some free parameters that we use the measured points to constrain. Having arrive at our model, which describes the unknown function we then can integrate the function from time zero (or close to it) to today. Having done that integral you suggest therefore that any set of values of the parameters which give you the same integrated value between the end points is therefore allowed and suggest that this is a 'constraint'. It is not. The constraint comes from whether the curve goes through the measured points, not the total integrated value since that is what we can observe!

Last edited: Oct 22, 2007
19. Oct 22, 2007

### Garth

All I am pointing out is that in the present epoch, with the observed values determined by the 'precision cosmology' mainstream model,

$$\int_0^1 \frac{da}{\sqrt{ \Omega_{k, 0} + \displaystyle \frac{\Omega_{m, 0} }{a} + \displaystyle \frac{\Omega_{r,0} }{a^2}+ \Omega_{\Lambda,0} a^2 \right)}}}\approx 1 .$$

We note that according to that mainstream $\Lambda$CDM model this near identity with unity ($\pm$0.6%) did not hold at various earlier epochs and indeed it need not hold in the present epoch. Depending on the mix of DE and matter, and the equation of state of DE, the integral could have a value between 0.66 and $\infty$. It therefore seems a suspicious coincidence that the present value of the integral should be so near to unity.

Furthermore there is a related coincidence in the near equality of density of the components of that DE/DM/Matter mix in the present epoch.

It is not "an incredibly silly way to do cosmology", but a complementary observation that may indicate we are missing something in that standard model.

Garth

Last edited: Oct 23, 2007
20. Oct 22, 2007

### Garth

Another example in today's physics arXiv of somebody (Zhao) using an apparent coincidence as a hint that the standard model may need modification, the coincidence may or may not be significant: Coincidences of Dark Energy with Dark Matter - Clues for a Simple Alternative?

Garth

Last edited: Oct 22, 2007