Were There Any Proposed Non-Expanding Cosmological Models Without Lambda?

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Discussion Overview

The discussion centers on the historical context of non-expanding cosmological models without the cosmological constant (lambda) in the framework of general relativity (GR). Participants explore whether such models have ever been proposed, particularly in light of assumptions of isotropy and homogeneity.

Discussion Character

  • Debate/contested
  • Historical
  • Technical explanation

Main Points Raised

  • Some participants note that historically, the only non-expanding cosmological models with lambda are the Einstein 1917 model and the de Sitter model, which is considered to be expanding with the right choice of coordinates.
  • One participant mentions that Einstein initially believed in a static universe without lambda in his 1915 publication of GR, but critics pointed out that this would lead to gravitational collapse, prompting him to introduce lambda.
  • Another participant questions how the original 1915 equations without lambda could lead to gravitational collapse, suggesting that there was no model to critique at that time.
  • Some argue that the boundary conditions in GR do not mandate inward acceleration or deceleration in an open universe, leading to discussions about the implications of these conditions on static solutions.
  • Participants reference the Friedmann equations, noting that they are solutions for expanding models and questioning their relevance to the discussion of static models without lambda.
  • There is mention of the possibility of a static equilibrium with lambda, but it is characterized as unstable, where any perturbation would lead to expansion or contraction.
  • Some participants express uncertainty about whether any non-expanding models without lambda exist in the history of cosmology, emphasizing the historical aspect of the inquiry.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the existence of non-expanding cosmological models without lambda. There are competing views regarding the implications of the original GR equations and the role of boundary conditions, as well as the relevance of the Friedmann equations to the discussion.

Contextual Notes

Limitations in the discussion include unresolved assumptions about boundary conditions and the historical context of the models being discussed. The relationship between the original field equations and the Friedmann equations is also a point of contention.

  • #31
Chalnoth said:
If you prefer, you could also compute the derivative of the energy density and show that it increases with time. To do this you'd need more specific information about the matter content than simply p \ge 0 (or really, p > -\rho/3), but it could definitely be done.
Let's go back for a minute to the second Friedmann equation, you said that starting from a static setting it led to collapse because the left side is <0, wouldn't that mean that in the expanding setting with a positive energy density it leads to a decelerating expansion? But we observe accelerating expansion, you would need to change the sign of some parameter, is that physically plausble?
 
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  • #32
AWA said:
Let's go back for a minute to the second Friedmann equation, you said that starting from a static setting it led to collapse because the left side is <0, wouldn't that mean that in the expanding setting with a positive energy density it leads to a decelerating expansion? But we observe accelerating expansion, you would need to change the sign of some parameter, is that physically plausble?
The accelerating expansion comes about because we have a cosmological constant (or, potentially, something else that behaves a lot like one). Looking at the second Friedmann equation, we have:

{\ddot{a} \over a} = -{4\pi G \over 3}\left(\rho + {3p \over c^2}\right) + {\Lambda c^2 \over 3}

What has been happening here is that the pressure term dropped to essentially zero in the very early universe, so it no longer is relevant, and as the expansion continued, the energy density term \rho has been dropping as well. As this occurred, the cosmological constant \Lambda remained the same. So as the universe expanded, the deceleration of the universe got slower and slower. But it never stopped, so by the time the energy density dropped below the cosmological constant, the acceleration became positive, and has been becoming more and more positive ever since.

Now the energy density is just going down faster, and the acceleration of the expansion is approaching a constant (given by the cosmological constant).
 
  • #33
Ich said:
1.

Or, if you like it better,
\sqrt{\frac{c^2}{4\pi G \rho}}
Right, and I naively reasoned that a derivative of any order of 1 is zero and therefore the left side of the second Friedmann equation was zero, but Chalnoth said it couldn't be done, that that is not a degree of freedom allowed by the theory, and the left side is determined by the right side of the equation, not by a priori setting the scale factor to 1. If I undestood correctly.

This kind of leaves me where I was, If it is not permitted to independently set the left side of the equation to reflect an static scenario, I would say that the equation is not compatible with any static solution, but I think is kind of obvious, given the fact that it is derived by solving the FRW line element equation with the Einstein field equations.
 
  • #34
Right, because there is no stable static solution to a homogeneous, isotropic universe in General Relativity.
 
  • #35
If I undestood correctly.
You didn't.

Chalnoth said:
Right, because there is no stable static solution to a homogeneous, isotropic universe in General Relativity.
Why stable?
Just set rho+3p=0 (including lambda), and you're done. A perfect unstable static solution to a homogeneous, isotropic universe.
 
  • #36
Ich said:
Why stable?
Just set rho+3p=0 (including lambda), and you're done. A perfect unstable static solution to a homogeneous, isotropic universe.
Well, there is obviously an unstable static solution. But there is no stable static solution. You need a stable static solution for it to be at least somewhat reasonable for a universe to be in that configuration.
 
  • #37
Ich said:
Why stable?
Just set rho+3p=0 (including lambda), and you're done. A perfect unstable static solution to a homogeneous, isotropic universe.

That equation is what I had in mind from the first moment but Chalnoth said it couldn't be equated to zero, that it had to be <0 so I'm not sure who is misleading me here.

Now you need to explain me what is unstable about it:

\rho =-3p

This looks like a physically plausible solution(but wrong of course since our universe is expanding), given the fact that currently a negative pressure is not considered unphysical but likely in an accelerating expansion universe.
 
  • #38
That equation is what I had in mind from the first moment but Chalnoth said it couldn't be equated to zero
Chalnoth said that it can't be zero with normal matter only. You need Lambda or, generally, negative pressure to make it zero.
Now you need to explain me what is unstable about it:
Any perturbation will grow without bound. If there is a region with Lambda winning by a 10^-50 percentage, that region will grow to a de Sitter universe. If there is an overdense region, it will collapse. If the whole balance is off by a minuscule amount, you get either de Sitter or Big Crunch.
 
  • #39
AWA said:
Now you need to explain me what is unstable about it:

\rho =-3p

What happens if there is a small perturbation \epsilon to this equation of state (with c=1), so that\rho + \epsilon =-3p?
 
  • #40
Ich said:
Any perturbation will grow without bound. If there is a region with Lambda winning by a 10^-50 percentage, that region will grow to a de Sitter universe. If there is an overdense region, it will collapse. If the whole balance is off by a minuscule amount, you get either de Sitter or Big Crunch.

There is no lambda in this equation. It was derived from the second Friedmann equation with Lambda=0
 
  • #41
AWA said:
There is no lambda in this equation. It was derived from the second Friedmann equation with Lambda=0
True, but all normal matter, dark matter, and radiation have p \ge 0. To get p \le -\rho/3, you need some form of dark energy (well, cosmic strings have p = -\rho/3, but obviously you'd need nothing but cosmic strings to have a static universe of cosmic strings, and even then they tend to annihilate with one another to produce standard model particles, which would cause a static cosmic string universe to start collapsing).
 
  • #42
George Jones said:
What happens if there is a small perturbation \epsilon to this equation of state (with c=1), so that


\rho + \epsilon =-3p?

I thought both parameters were considered constant in cosmological homogenous and isotropic perfect fluid. Wasn't that assumption considered valid in the cosmological level regardless local inhomogeneities, I guess if you admit that inhomogeneity in the cosmological realm you are not talking anymore about an isotropic and homogenous universe, I've heard there are also theoretical solutions for isotropic inhomogenous perfect fluids but that is a different story.
 
  • #43
AWA said:
I thought both parameters were considered constant in cosmological homogenous and isotropic perfect fluid.
Not for any real fluid. For normal matter at low densities, for instance, the pressure is dependent upon its temperature. As the temperature becomes relativistic, the pressure approaches p = \rho/3. As the temperature tends to zero (again at low densities), the pressure becomes p = 0.

Similarly, if normal matter interacts, it will produce photons under a variety of circumstances, and photons have p = \rho/3. So if you start with a balanced universe made up of nothing but gas, for instance, then once that universe starts to form stars, that little bit of extra pressure from the photons will unbalance the equations and cause the universe to collapse.
 
  • #44
Chalnoth said:
Not for any real fluid. For normal matter at low densities, for instance, the pressure is dependent upon its temperature. As the temperature becomes relativistic, the pressure approaches p = \rho/3. As the temperature tends to zero (again at low densities), the pressure becomes p = 0.
All modern cosmologies work under the assumption of an idealized perfect fluid, that is not obviously the real fluid we experiment in normal life, but it is considered a valid assumption nevertheless. Actually this perfect fluid is adiabatic so it doesn't get inhomogenous due to local temperature

Chalnoth said:
Similarly, if normal matter interacts, it will produce photons under a variety of circumstances, and photons have p = \rho/3. So if you start with a balanced universe made up of nothing but gas, for instance, then once that universe starts to form stars, that little bit of extra pressure from the photons will unbalance the equations and cause the universe to collapse.
Remenber we are talking here about a static universe,if you start with a balanced universe, that's how it's going to remain, IOW it doesn't really start, it has no beguinning and no end, so the pressure and density at the cosmological scale remain constant. Perhaps this is difficult to imagine because we have our mind set on our evolving, dynamical universe.
 
  • #45
Suppose a spherical cloud of dust with diameter D and uniform density \rho were suddenly created in empty space at time t=0. Suppose D and \rho are such that the escape velocity is <<c. We can now get a good GR approximation of what would happen inside just by using Newtonian gravity, although we must add that gravitons travel at c.

Two test masses inside separated by distance r would initially experience zero acceleration toward each other. Only after time t=D/c would all internal test masses experience acceleration toward each other at 4\pi\rhorG/3. This is because each test mass only experiences the gravitational signal within a sphere of radius ct centered on that test mass. Only after that test mass' expanding sphere passes the cloud boundary does the test mass begin to accelerate with respect to the cloud.

Although one cannot instantaneously create such a cloud, it can be assembled rapidly so that at t=0, a given test mass inside the cloud would not yet "know" the initial spherical distribution of matter, or more precisely, the test mass would experience the gravitational signal from something other than the initial shape of the cloud. Given that, the initial acceleration would be something other than 4\pi\rhorG/3, and it would not become 4\pi\rhorG/3 until time D/c.

If D is the diameter of an infinite universe, that would be a long time.
 
  • #46
There is no lambda in this equation. It was derived from the second Friedmann Equation with Lambda=0
It was derived from the second Friedmann equation with Lambda factored into the energy and pressure. I said so explicitly, look up the details at http://en.wikipedia.org/wiki/Friedmann_equations#The_equations".
I thought both parameters were considered constant in cosmological homogenous and isotropic perfect fluid. Wasn't that assumption considered valid in the cosmological level [etc. etc.]
Do you understand the difference between a valid solution and a stable solution? The eternal universe is a valid unstable solution. To check for stability, you have to check the response to small perturbations. It seems you are not aware what http://en.wikipedia.org/wiki/Stability_theory" means.

BTW, our universe is unstable, too. The resulting process of local collapse is called structure formation.
 
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  • #47
AWA said:
All modern cosmologies work under the assumption of an idealized perfect fluid, that is not obviously the real fluid we experiment in normal life, but it is considered a valid assumption nevertheless. Actually this perfect fluid is adiabatic so it doesn't get inhomogenous due to local temperature
This isn't true, as the early universe starts off with very high temperatures and cools, resulting in lots of conversion of energy between radiation and matter. If you want to get the right answers for BBN and the CMB, you have to take these factors into account.

AWA said:
Remenber we are talking here about a static universe,if you start with a balanced universe, that's how it's going to remain, IOW it doesn't really start, it has no beguinning and no end, so the pressure and density at the cosmological scale remain constant. Perhaps this is difficult to imagine because we have our mind set on our evolving, dynamical universe.
If it has a non-zero temperature, however, there will always be fluctuations around the mean density, which means that it won't stay static for long.
 
  • #48
BillSaltLake said:
Suppose a spherical cloud of dust with diameter D and uniform density \rho were suddenly created in empty space at time t=0. Suppose D and \rho are such that the escape velocity is <<c. We can now get a good GR approximation of what would happen inside just by using Newtonian gravity, although we must add that gravitons travel at c.

Two test masses inside separated by distance r would initially experience zero acceleration toward each other. Only after time t=D/c would all internal test masses experience acceleration toward each other at 4\pi\rhorG/3. This is because each test mass only experiences the gravitational signal within a sphere of radius ct centered on that test mass. Only after that test mass' expanding sphere passes the cloud boundary does the test mass begin to accelerate with respect to the cloud.

Although one cannot instantaneously create such a cloud, it can be assembled rapidly so that at t=0, a given test mass inside the cloud would not yet "know" the initial spherical distribution of matter, or more precisely, the test mass would experience the gravitational signal from something other than the initial shape of the cloud. Given that, the initial acceleration would be something other than 4\pi\rhorG/3, and it would not become 4\pi\rhorG/3 until time D/c.

If D is the diameter of an infinite universe, that would be a long time.
The speed of gravity can't be thought of in such a naive manner for real systems. This may be true if you instantaneously create a large mass, but it won't be true for any real assemblage of matter. See, for example, this paper by Steve Carlip:
http://arxiv.org/abs/gr-qc/9909087

The way this basically works in General Relativity is that when there is a change in the configuration of the system that requires information propagation, that information is propagated through gravitational waves. Therefore, if your change to the system doesn't emit gravitational waves, then it is felt instantaneously everywhere. This may seem a bit paradoxical, but it is largely explained by the gravitational field of a particle moving along with that particle. That's a bit overly-simplistic, and Carlip does a good job of going through it in detail, but there you have it.
 
  • #49
Ich said:
It was derived from the second Friedmann equation with Lambda factored into the energy and pressure. I said so explicitly, look up the details at http://en.wikipedia.org/wiki/Friedmann_equations#The_equations".
Rather than wikipedia, look up any cosmology textbook, you'll find that it can be derived from the original field equations without Lambda.

Ich said:
Do you understand the difference between a valid solution and a stable solution? The eternal universe is a valid unstable solution. To check for stability, you have to check the response to small perturbations. It seems you are not aware what http://en.wikipedia.org/wiki/Stability_theory" means.

BTW, our universe is unstable, too. The resulting process of local collapse is called structure formation.
I don't know why you deliver this kind of cryptic, puzzling comments, that verge on ATM, do you want to clarify or confuse? The Einstein model was rejected among other things (like absence of redshift) because it was unstable, if you say that our universe is unstable too that is going to confuse some people, and I think PF is meant to be an educational site. I am here to learn too, and your cryptic, condescending replies don't help much.
 
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  • #50
AWA said:
I don't know why you deliver this kind of cryptic, puzzling comments,
That comment was perfectly sensible to me. Why did it confuse you? A homogeneous universe is inherently unstable, in that it becomes more and more inhomogeneous with time.
 
  • #51
Chalnoth said:
This isn't true, as the early universe starts off with very high temperatures and cools, resulting in lots of conversion of energy between radiation and matter. If you want to get the right answers for BBN and the CMB, you have to take these factors into account.
What is not true?. I have in fron of me a couple of cosmology books (Schutz and Ryder) that say in cosmology and GR the perfect fluid assumption is used and considered valid, and further that perfect fluids are considered adiabatic, now if textbooks are also wrong...

Chalnoth said:
If it has a non-zero temperature, however, there will always be fluctuations around the mean density, which means that it won't stay static for long.
Well of course there will be local fluctuations, but we are talking about idealized cosmological solutions that globally consider the universe as isotropic and homogenous and therefore those fluctuations wouldn't be a problem in order to remain static since they would cancel out. Otherwise you'd be talking about an inhomogenous solution, there you would be right that it wouldn't be static for long.
 
  • #52
Chalnoth said:
That comment was perfectly sensible to me. Why did it confuse you? A homogeneous universe is inherently unstable, in that it becomes more and more inhomogeneous with time.
Then why physicists rejected Einstein model due to instability? In the end it was an empirical rejection because of the redshift and other things, but at first it was rejected because it was unstable, or so they say in many books.

If our universe is becoming more and more inhomogenous, logic says that eventually it'll become inhomogenous, that is against the main assumption of our standard model.
 
  • #53
AWA said:
What is not true?. I have in fron of me a couple of cosmology books (Schutz and Ryder) that say in cosmology and GR the perfect fluid assumption is used and considered valid, and further that perfect fluids are considered adiabatic, now if textbooks are also wrong...
What isn't true is that these fluids don't interact and change with temperature. They do interact, and their behavior changes quite a lot with temperature. A perfect fluid, after all, is merely defined as a fluid which can be fully described by its energy density and pressure, and dealing with these interactions doesn't break that assumption.

AWA said:
Well of course there will be local fluctuations, but we are talking about idealized cosmological solutions that globally consider the universe as isotropic and homogenous and therefore those fluctuations wouldn't be a problem in order to remain static since they would cancel out. Otherwise you'd be talking about an inhomogenous solution, there you would be right that it wouldn't be static for long.
No, temperature fluctuations don't cancel out. As long as you have non-zero temperature, the fluctuations will be random, and you will get some deviation from the average density on all scales.

Of course, the deviations from the average density will be smaller on larger scales, but you only need the deviation to be non-zero for the system to start collapsing/expanding.

I think you're thinking in terms of classical thermodynamics, where a gas remains essentially uniform after small perturbations. But this is only the case because it's in a stable configuration, one that tends to go back to the equilibrium configuration after being perturbed. This isn't the case for gravitational systems.
 
  • #54
Chalnoth said:
I think you're thinking in terms of classical thermodynamics, where a gas remains essentially uniform after small perturbations. But this is only the case because it's in a stable configuration, one that tends to go back to the equilibrium configuration after being perturbed. This isn't the case for gravitational systems.
Be that as it may, I've got the feeling that here we face some arbitrariness towards assumptions like homogeneity and the ideal gas or fluid, so that whenever it's convenient they hold with all the consequences and when it's not convenient they don't hold (and I don't mean the local case as you remark this happens at all scales) and it seems doubtful that this has anything to do with gravitation or GR.
This is great as an argumentative tool but I'm not sure it makes up for the apparent lack of intellectual honesty.
 
  • #55
AWA said:
Be that as it may, I've got the feeling that here we face some arbitrariness towards assumptions like homogeneity and the ideal gas or fluid, so that whenever it's convenient they hold with all the consequences and when it's not convenient they don't hold (and I don't mean the local case as you remark this happens at all scales) and it seems doubtful that this has anything to do with gravitation or GR.
This is great as an argumentative tool but I'm not sure it makes up for the apparent lack of intellectual honesty.
It's merely a matter of being aware of the realistic limits of our assumptions. Homogeneity, for instance, is never expected to be exact, but only approximate. Since it is only expected to be approximate, it is important to recognize when even tiny deviations from the precise assumption lead to dramatically different behavior.

With a homogeneous, static gas balanced by a cosmological constant, for instance, you can prepare it in a homogeneous state just fine. But it won't stay homogeneous if there is any non-zero temperature. Because it won't stay homogeneous, it is necessary consider what happens when the assumption of homogeneity breaks down, and in this case you'll first get structure formation, but in addition to that you'll end up with the universe as a whole either collapsing or expanding (I think collapse is the more likely, because some of the energy density of the gas will become photons, which have positive pressure).
 
  • #56
AWA said:
I was wondering if historically there was ever proposed some non-expanding cosmological model without the lambda term, I've read somewhere it's not possible with the assumption of isotropy and homogeneity so I am only interested in terms of the history of cosmology.

As to your original question, Harrison says: "The last static steady state universe was conceived at the university of Chicago in 1918 an elaborated in the 1920s by Astronomer William MacMillan"

And you are right about Eddington - although he received a copy of the 1915 publication early on - the War would most likely have prevented much in the way of communication - so I can't find where I read who influenced Einstein with regard to problem of potential G collapse - but there was almost a 2 year gap between the publication of the first version and the 1917 version - so plenty of time for inputs -
 
  • #57
yogi said:
As to your original question, Harrison says: "The last static steady state universe was conceived at the university of Chicago in 1918 an elaborated in the 1920s by Astronomer William MacMillan"

And you are right about Eddington - although he received a copy of the 1915 publication early on - the War would most likely have prevented much in the way of communication - so I can't find where I read who influenced Einstein with regard to problem of potential G collapse - but there was almost a 2 year gap between the publication of the first version and the 1917 version - so plenty of time for inputs -
Thanks yogi, I've read in an article of 1930 by de sitter that there was even another static solution in the famous book by mathematician Levi-Civita: The absolute differential calculus from 1924 but I haven't found this book yet.
 

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