# Thread: Does this count as an accelerating universe?

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## Does this count as an accelerating universe?

For a model universe with the scalefactor 'a', if a(dot)/a=constant, is that called an accelerating universe?
(where the dot is a time derivative)

Could it not be that our universe has always had a(dot)/a = constant, everywhere and at all times?

In such a universe would the Hubble parameter be H(z)=constant or H(z)=(1+z)*constant?

Thankyou.

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The first question: It depends. Acceleration would be determined by a(double dot). See the second Friedmann equation, sometimes called the Friedmann acceleration equation.

The middle question: No because the Hubble parameter H(z) = a(dot)/a has been measured in our universe and is not constant.

In a hypothetical universe with a(dot)/a = constant, the Hubble parameter is constant because it is defined as a(dot)/a.

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On the first question: if a(dot)/a = constant (e.g. K) then a=exp(Kt), (where t is time going forward). Doesn't this determine a(double dot) to be K^2exp(Kt) or K^2a...is this called an accelerating universe?

middle, ok thanks

On third question: In a hypothetical universe with a(dot)/a constant at all times, everywhere...if all the physical objects in that universe (and every length dimension, including all distances, atoms, scientists etc..) were part of the 'expansion', presumably locally it would appear static If attempts were made to measure H(z) using distant sources in such a universe...would H(z) now be measured as constant or (1+z)*H(z) or other?

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If we assume that d(a)/dt = aH , which is the same as a(dot)/a = constant. Here we have a Hubble constant which is truly constant.

Then by simply trying a = eHt, we find that the scale factor is exponential in t.

It's pretty easy to calculate the second derivative and find that a(double dot) = H2eHt. Since a(double dot) > 0, then yes, it is accelerating. So you had it right the first time.

This cosmology is called a DeSitter universe (it's basically a universe of pure dark energy):
https://en.wikipedia.org/wiki/De_Sitter_universe
Last edited by ShinAce; 2017-Aug-10 at 02:12 PM.

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Originally Posted by john hunter
On third question:...
The third question is answered: a(dot)/a = constant means H which is defined as = a(dot)/a = constant.
The 4th question starts with ignoring gravity in a thread about GR! It is gravity that means that solar systems, galaxies and groups of galaxies in an expanding universe do not expand. Also remember that we measured that our universe was expanding before we found that that the expansion was accelerating.
Last edited by Reality Check; 2017-Aug-11 at 04:40 AM.

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ShinAce and Reality Check,

The reason for these questions originated many years ago by considering the cause of gravity.

If we allowed all length dimensions…all lengths, physical constants, atoms etc… to take part in the ‘expansion’, where a quantity Q with n length dimensions changes as Q=Q(0)exp(nHt). Then we have an apparently static universe with a redshift, due to conservation of energy of photons, as E=hf(emission), becomes E=h*exp(2Ht)*f(arrival), so a redshift is needed to conserve energy.

Also for a mass with total energy (mc^2-GMm/R),…(M=mass of universe, R=radius of universe)… to conserve energy, which is later (mc^2-GMm/R)*exp(2Ht), then mc^2-GMm/R=0, and the flatness problem isn’t a problem anymore…i.e the cause of gravity is to conserve energy as the universe ‘expands’.

G=Rc^2/M (Newton) or omega(m)=1( Einstein)

This would be a De-Sitter universe but purely matter.

Gravity would be caused by the 'expansion', but have no effect on the rate of 'expansion'.

Hence the question 4 of what would be observed for H(z) in such a universe by the observers who themselves are ‘expanding’…it’s to help test whether our universe could be of this type.
Last edited by john hunter; 2017-Aug-11 at 08:00 PM.

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A deSitter universe isn't flat. Our observable universe is flat. That sounds like a flatness problem to me.

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There is a nice simple solution, with k=0, lambda=0, H constant, p=-c^2(rho), rho = 3H^2/(8pi*G)...or G=3H^2/(8pi*rho)

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Originally Posted by john hunter
There is a nice simple solution, with k=0, lambda=0, H constant, p=-c^2(rho), rho = 3H^2/(8pi*G)...or G=3H^2/(8pi*rho)
k=0, means it is flat. So it is not deSitter. At least we can get that out of the way.
lambda=0, implies no dark energy.
pressure proportional to -rho implies it is entirely dark energy.
rho = 3H^2/(8pi*G), I don't have pen and paper, but I presume this is simply the critical density for k=0.

Your solution is not consistent. Is there dark energy or not? If there is, why isn't k=1? If there is no dark energy, why is the pressure negative?

With all due respect, you appear well versed in introductory cosmology yet you seem to ignore introductory cosmology. I can't figure out if you've studied cosmology or not.
Last edited by ShinAce; 2017-Aug-12 at 12:15 AM.

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Yes, a flat universe with k=0 (By De-sitter, it just meant a constant H type of solution so a=exp(Ht)).

Lambda = 0, no dark energy, so your point of why the pressure is negative is a good one.

This type of universe would appear static, so what is to prevent all matter collapsing in a kind of 'big crunch'? It would be a kind of outward pressure which, on the large scale, must balance gravity.

E.g matter falls inwards towards the centres of galaxies, but for regions where m/r reaches c^2/G gravity reduces (in a way not yet understood) and AGNs result, firing the matter back out and providing the 'pressure'.

It could happen on larger scales too for galactic clusters which collapse and 'bounce', giving the foam-like large scale structure...and also on the largest scale of all to give our 'big bang' which has been such a successful model.

------------------------------

'Ignoring introductory cosmology' comes about as there seems to be a definite problem with it, so some parts should be questioned and probably changed...yet some parts are almost impossible to question and change.

Problem: The coincidence problem. Why are omega(lambda) and omega(matter) similar sizes at the moment? Solution: we should try and find a way that omega(lambda) can be zero.
'Solid' parts of cosmology: Big Bang. Very successful, although possibly could be changed from a single bang to multiple 'bangs'.
General Relativity: Very successful, just a possibility of change for regions where m/r is near c^2/G.
Perhaps room to reinterpret 'expansion' in cosmological models.

So does anyone have an answer for question 4, (from post 3, named question 4 by Reality Check) what would we observe for H(z) in the universe described above? Thanks.
Last edited by john hunter; 2017-Aug-12 at 10:25 AM.

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I would expect any solution to be consistent with three things:
1) The Friedmann equations, which you have done.
2) The equation of state parameter used to solve the second Friedmann equation, which is the one were arguing about.
3) Energy conservation given k and the proposed solution to 2).

If k=0, then the total energy of the universe is zero. The sources of energy are matter, dark energy, pressure, and gravitational potential. If you set k, p, and rho, then you should be able to calculate the total energy to see if it's consistent.

Lots of people are searching for solutions to the cosmological constant problem, but that isn't yet mainstream and will surely find itself in ATM. One example is Thanu Padmanabhan CosMIn proposal. While it seems cool, there's no framework(quantum gravity) to work on it properly.

Trying to fudge the numbers in a standard cosmology just seems like it won't work. LambdaCDM is here to stay a while longer. The vast array of observations(CMB, BAO, Supernovae data, elemental abundances, gravitational waves) today won't be explained with a universe of pure dark energy.

I would recommend reading "Modern Cosmology" by Dodelson. If you've taken undergrad cosmology you should be able to follow it, but it's really intended as a graduate level textbook.

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Thanks for the book recommendation,

1) OK

2)
Originally Posted by ShinAce
pressure proportional to -rho implies it is entirely dark energy.

Couldn’t it be that neither gravity or the pressure term have any effect on the rate of ‘expansion’. …which just exists as a constant H and continues ‘expanding’ all physical distances and constants with length dimensions, regardless of gravity or pressure. The universe would appear static, but with a redshift as described above.

On top of the ‘expansion’ we would add on motion of stars etc…which sometimes fall together in dense regions, e.g. galaxies, until the mass/radius reaches c^2/G …then explode as AGNs, (little bangs). The 'falling together' could occur on a much larger scale too, and probably did so 14 Billion years in our part of the cosmos, i.e the 'Big Bang'.

The interpretation of the pressure term p=-c^2*rho (the equation of state) could be for the matter, caused by gravity itself, which keeps the universe under a kind of 'tension', i.e. negative pressure. The average matter density would be constant, as in this form of the FLRW equations, https://en.wikipedia.org/wiki/Friedm...Interpretation
(interpretation of first equation, is constant matter density because if p=-c^2*rho, rho(dot)=0)

for 3) Since universe is apparently static but with continuous interchange of matter from dense regions, firing back out to less dense regions etc...i.e. kinetic energy to gravitational energy, total energy should be conserved.
Last edited by john hunter; 2017-Aug-13 at 12:54 PM.

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Originally Posted by john hunter
To my knowledge, dark energy is the only thing in all of physics which has negative pressure. Admittedly, dark energy is a name for an effect without a cause and hence not a physical object until proven otherwise. I doubt it will be proven to exist because it does have negative pressure implying it can do work for free in a closed system. Nonetheless, if you want to invoke negative pressure, please show me how you plan to do it without dark energy.

If lambda is zero, then you have no way of explaining negative pressure.
Last edited by ShinAce; 2017-Aug-14 at 01:12 AM.

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Just to recap the question to anyone interested: What would H(z) appear to be for this type of universe?...(so it can be tested)

Model universe.

a(dot)/a = constant (H), everywhere and at all times, so a=a(0)expHt

'expansion' applies to every length, physical quantity and constant with length dimension 'n', all distances, sizes of people, atoms etc... according to Q=Q(0)exp(nHt).
i.e. Q=Q(0)*a(t)^n, time going forward.
Redshift then because E=hf conserved and h=h(0)exp(2Ht), so H half of usual expansion rate, and a=1/sqrt(1+z)

Solution of the Friedmann equations k=0, lambda=0, G=3H^2/(8*pi*rho)...(omega(m)=1, omega(lambda)=0)...avoiding coincidence problem.
p=-c^2*rho...negative pressure, but with a lambda=0
----------

1) Coincidence problem.
2) Measurements of omega(m) approx. 0.25 e.g. this recent paper https://arxiv.org/abs/1708.01530, naturally explained as omega(m) would appear to be 0.25 from definition involving H^2 on denominator.
3) Match to supernovae data (orange line, 2(sqrt(1+z)-1)) with only h as a variable parameter https://www.desmos.com/calculator/hkawxpmtcj the Betoule data ...prediction that LCDM would give omega(m) just under 1/3, by considering binomial expansion for this dl = z-1/4*z^ 2 and LCDM int (E(z)^-0.5) dz which gives z-3/4*m*z^2...i.e match at low z if m=1/3, (higher z needs under 1/3)

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Interpretation:
In post 12. Briefly, gravity and the value of the value of G is caused by the 'expansion'...
...to conserve energy for each mass m (mc^2-GmM/R)exp(2Ht)=0...G=Rc^2/M ,Newton, M= mass of universe...G=3H^2/(8*pi*rho), Einstein.

Neither gravity or the negative pressure influence the rate of expansion, which is independent and constant H.

The negative pressure being interpreted as an effect of gravity (ShinAce) which provides a kind of 'tension', in an infinite universe, average rho(dot)=0, but with lambda=0...resulting in a flat universe, with a redshift. Matter being in a continuous dynamic state in which there are 'little bangs' and 'big bangs'.
Last edited by john hunter; 2017-Aug-14 at 09:26 AM.

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Originally Posted by john hunter
The negative pressure being interpreted as an effect of gravity (ShinAce) which provides a kind of 'tension', in an infinite universe, average rho(dot)=0, but with lambda=0...resulting in a flat universe, with a redshift. Matter being in a continuous dynamic state in which there are 'little bangs' and 'big bangs'.
I'm not buying your toy model. You're using the Friedmann equations, so clearly it satisfies general relativity.

You've set lambda to 0 so where is this negative pressure coming from? You say it is so but it violates general relativity and therefore the Friedmann equations as well.

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It has some advantages. Anyone got an answer to the H(z) question?

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Originally Posted by john hunter
It has some advantages. Anyone got an answer to the H(z) question?
How can you assert that a static universe with no dark energy but with negative pressure has advantages without proving such? You bear the burden of proof. You can't appeal to general relativity and violate it in the same sentence and expect others to do your homework.

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Originally Posted by ShinAce
You're using the Friedmann equations, so clearly it satisfies general relativity.
but maybe you're right about the burden of proof, investigations will continue...if anyone wants to join in with the H(z) answer or other contributions, that's fine.

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Originally Posted by john hunter
Just to recap the question to anyone interested: What would H(z) appear to be for this type of universe?...(so it can be tested)
There is not enough information. Look up H(z) in the literature and see that the result has matter, dark matter and dark energy densities in it. Where are they in your model?
ETA: Utility of observational Hubble parameter data on dark energy evolution equation 3 is H(z) for a ΛCDM model with no curvature term ( a flat universe).
This universe is expanding and has measured matter, dark matter and dark energy densities. That model is not not our universe thus
• It cannot be tested in our universe.
• It is not a solution to any of the issues in our universe you list as advantages.

Nothing material in this model universe exists since it throws away all physics except GR. It has atoms, astronomers, planet, stars, galaxies not being bound systems and expanding.
Last edited by Reality Check; 2017-Aug-15 at 01:21 AM.

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Originally Posted by Reality Check
Nothing material in this model universe exists since it throws away all physics except GR. It has atoms, astronomers, planet, stars, galaxies not being bound systems and expanding.
You seem to be thinking that gravity changes the rate of 'expansion', but with the new model the 'expansion' causes gravity, so that energy is conserved (mc^2-GmM/R )exp(2Ht)=0
gravity does not effect this 'expansion' (in quotes, the one which effects all length scales, physical constants etc...)

Where in GR does it explain that 'expansion' only effects distance between galaxies and not other distances? How does it say that gravity changes the rate of 'expansion' which is not a physical object and has no mass?

However, if you are worried about observations that show expansion (without quotes) they are included in the model. The matter is in a continuous dynamic state in which there are 'bangs' of various sizes including the 'Big Bang'. So Big Bang is included in this model, (gravity effects the rate of expansion, without quotes).

The 'expansion' in quotes gives rise to the redshift.

It is proposed that the model is our universe.

It obeys the Friedmann equations and conservation of energy. Since 'expansion' cannot be measured locally it can't be ruled out on those grounds. It matches supernovae observations https://www.desmos.com/calculator/hkawxpmtcj and measurements of omega(m)h^2 from CMB, BAO which conclude that omega(m) is approx. 0.25. (as 2H is used in omega(m)h^2, instead of H, so really omega(m)=1 and omega(lambda)=0)...and it's nice and simple...just a reinterpretation of the formulae.

Hence the question. This new model has not been convincingly ruled out so perhaps H(z) would do that.

It is not scientific to exclude all other models if they don't match the mainstream one and then say they don't represent our universe and so cannot be tested.
Last edited by john hunter; 2017-Aug-15 at 11:09 AM.