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

[AWA]

In most cosmology books it says that historically there is only two(some include here also the Minkowski spacetime but I'd rather leave it aside since it is a flat spacetime) non-expanding cosmological models with lambda, the Einstein 1917 model and the de Sitter model, this last was afterwards shown to be actually expanding too with the right choice of coordinates.
I think it was Tolman who showed these are the only possible static models with Einstein equations containing the cosmological constant lambda.

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.

 

[yogi]

In the 1915 publication of GR, there was no Lambda - Einstein believed the universe to be static - but critics were quick to point out that the equations would lead to a gravitational collapse - so to preserve the static status, Einstein introduced a term that exactly balanced gravity - the problem then became that of instability - any slight variance would upset the precarious static state and lead to run away expansion or gravitational collapse - but neither was observed - so things were more or less unresolved - as the years passed more and more data was being collected by Hubble, Slipher and others - and you know the rest of the story - Einstein eventually considered the addition of Lambda as his worst mistake as it mislead him away from the possibility of being the first person to predict expansion

To his credit, he structured the universe to fit what appeared to be good data at the time - I have read some accounts that claim Einstein even went so far as to inquire of astronomers whether there was any evidence of large scale motion - but of course he asked the wrong astronomers

IMO Lambda was actually a great foresight but misinterpreted for many years - viewed as we now know the universe to be accelerating, it explains both gravity and the present state of the universe as a de Sitter expansion

 

[AWA]

In the 1915 publication of GR, there was no Lambda - Einstein believed the universe to be static - but critics were quick to point out that the equations would lead to a gravitational collapse

I fail to see how exactly the original 1915 equations without lambda lead to gravitational collapse.

Also,I think until 1917 Einstein didn't publish any cosmological model and this first already contained the lambda term, how could critics point out collapse of any cosmological model if there wasn't any model yet, and, do you happen to know who were those critics?
I know he had correspondance with de sitter in those early years, was he one of them?

 

[BillSaltLake]

GR by itself does not mandate that the overall mass-energy density in an open (zero or negative curvature) Universe will cause inward acceleration or deceleration of the expansion. It depends on what boundary conditions are chosen. In essence, if the gravitational "signal" from distant mass-energy has not had time to travel here, there will be no inward acelleration from the overall mass-energy density. We have to assume that this "signal" pre-existed everywhere since the beginning. The deSitter model does assume this inward acceleration. The results apparently don't match observation unless inward acceleration is assumed.
An eternal Universe would probably be interpreted as having inward acceleration, so that before mid-century, there would have been criticism that the GR Universe would collapse.

 

[Ich]

I fail to see how exactly the original 1915 equations without lambda lead to gravitational collapse.

Then please go and have a look at Friedmann equation #2. Or, if you want to be closer to the "original equations", have a look at http://arxiv.org/abs/gr-qc/0103044" .
Starting with a homogeneous, isotropic static state, collapse is a rather direct consequence.

 

[AWA]

GR by itself does not mandate that the overall mass-energy density in an open (zero or negative curvature) Universe will cause inward acceleration or deceleration of the expansion. It depends on what boundary conditions are chosen.

So I guess there is a problem with this boundary conditions that forbids static solutions with the original GR equations,

An eternal Universe would probably be interpreted as having inward acceleration, so that before mid-century, there would have been criticism that the GR Universe would collapse.

This seems to contradict your first paragraph, if GR does not mandate it, why suppose that before mid-century there would have been such criticism?

 

[AWA]

Then please go and have a look at Friedmann equation #2. Or, if you want to be closer to the "original equations", have a look at http://arxiv.org/abs/gr-qc/0103044" .

Yeah, well, the Friedmann equations came later than the period we are centering on right now and they are solutions to expanding models (with or without lambda), we are referring to the original field equations previous to the discovery of expansion either theoretically by Friedmann or empirically by Hubble.

Starting with a homogeneous, isotropic static state, collapse is a rather direct consequence.

That is what seems to be the case, but nothing in the references you give explains it, can you?

 

[Chalnoth]

Yeah, well, the Friedmann equations came later than the period we are centering on right now and they are solutions to expanding models (with or without lambda),

The Friedmann equations are a general solution for a homogeneous, isotropic universe. They can be thought of as a reduction of the Einstein Field Equations for this special case.

 

[Ich]

As Chalnoth said, the Friedmann equations are the field equations, applied to cosmology. "Expansion" is not an assumption needed to derive them, it's a result of their application.

That is what seems to be the case, but nothing in the references you give explains it, can you?

Look at the beginning of chapter 3 in Baez's article. You'll find that a volume filled with static dust will start contracting. End of story. For the detailed math how you get from the original EFE to Baez's slimmed-down version, see chapter 5.

 

[Chalnoth]

I'd also like to add that if you add a cosmological constant to the situation, you can manufacture a situation where a static cloud of homogeneous, isotropic dust will neither expand nor contract. However, this is an unstable equilibrium, and any tiny perturbation will cause runaway expansion/contraction.

 

[AWA]

The Friedmann equations are a general solution for a homogeneous, isotropic universe. They can be thought of as a reduction of the Einstein Field Equations for this special case.

I'd also like to add that if you add a cosmological constant to the situation, you can manufacture a situation where a static cloud of homogeneous, isotropic dust will neither expand nor contract. However, this is an unstable equilibrium, and any tiny perturbation will cause runaway expansion/contraction.

Exactly, as you admit in the second quote the friedmann equations are not the only solutions to a homogenous and isotropic universe, as you point out there is the static unstable Einstein model with the lambda added. What I asked in the OP is , are there any other solutions but without lambda for an isotropic , homogenous universe in the history of cosmology? Regardless of its viability, this is more of a history of cosmology kind of question.

As Chalnoth said, the Friedmann equations are the field equations, applied to cosmology. "Expansion" is not an assumption needed to derive them, it's a result of their application.

As Chalnoth and I said said above this is not completely accurate, as shown by the Einstein model, or do you argue that this model is not derived from the field equations plus the assumption of homogeneity and isotropy?

Look at the beginning of chapter 3 in Baez's article. You'll find that a volume filled with static dust will start contracting. End of story.

That gravity attracts is a conclusion you don't need GR to reach. Newton is fine at that level. When you say collapse is a direct consequence of gravity that is a trivial truth for the local case. We are talking about cosmology where such simplistic statements are misleading, i.e, even though gravity attracts our univese is not collapsing but expanding. Even in a static setting there could be unknown variables capable of keeping the universe from global collapse. Like spatial infinity for instance.

 

[JDoolin]

As Chalnoth said, the Friedmann equations are the field equations, applied to cosmology. "Expansion" is not an assumption needed to derive them, it's a result of their application.

Uniform density is the assumption.

 

[Chalnoth]

That gravity attracts is a conclusion you don't need GR to reach. Newton is fine at that level. When you say collapse is a direct consequence of gravity that is a trivial truth for the local case. We are talking about cosmology where such simplistic statements are misleading, i.e, even though gravity attracts our univese is not collapsing but expanding. Even in a static setting there could be unknown variables capable of keeping the universe from global collapse. Like spatial infinity for instance.

It's not really that difficult. Just take the second Friedmann equation:

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

For now, let's consider normal matter, where p >= 0 and \Lambda = 0. We can see trivially that in this case, \ddot{a}/a < 0, which indicates collapse if we start from a static configuration.

Making the universe spatially infinite won't change anything, as the Friedmann equations already assume this.

 

[AWA]

It's not really that difficult. Just take the second Friedmann equation:

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

For now, let's consider normal matter, where p >= 0 and \Lambda = 0. We can see trivially that in this case, \ddot{a}/a < 0, which indicates collapse if we start from a static configuration.

Making the universe spatially infinite won't change anything, as the Friedmann equations already assume this.

This is not really related to my OP but anyway..
In a static configuration the left side of the equation is already zero to begin with (the second derivative of a constant is zero) so the whole equation is of no use.

 

[BillSaltLake]

A uniform cloud of heavy dust will accelerate inward assuming equilibrium as a starting condition. However, if a sphere of heavy dust is suddenly brought into place, there is initially little or no inward acceleration near the center. Only after the gravitational signal from the edge of the sphere has reached all points in the sphere will there be inward acceleration equal to the Friedmann equation. Although it's not actually possible to bring a heavy dust cloud into place "instantly", if the cloud was just assembled from distant particles being brought in at say half the speed of light, the initial inward acceleration (at the moment that the sphere is completed) would be less than the Friedmann equation predicts.

The problem here is that at the beginning of time, do we assume that all the energy had pre-existed (so the gravitational signal is in equilibrium) or do we assume that the energy was suddenly placed where is was. If the latter, we might conclude that locally, test particles are not yet influenced by the gravity of remote energy. Therefore there would be no inward acceleration from the overall average density, and there would in fact never be such acceleration.

 

[Chalnoth]

This is not really related to my OP but anyway..
In a static configuration the left side of the equation is already zero to begin with (the second derivative of a constant is zero) so the whole equation is of no use.

No. The first derivative of the scale factor can be set to zero. You can't set the second derivative to zero arbitrarily, as that is set by gravity.

 

[Chalnoth]

The problem here is that at the beginning of time, do we assume that all the energy had pre-existed (so the gravitational signal is in equilibrium) or do we assume that the energy was suddenly placed where is was.

We don't bother with any such assumptions in cosmology. For nearly all cosmological studies, we just take a set of conditions at a specific time and evolve those forward (or, equivalently, take the current conditions and evolve them backward in time to a certain point). For the most part we don't worry about what set up those "initial" conditions (initial is in quotes because it's not really initial per se, just that we don't know what happened before).

Now, there are cosmologists that are very interested in what happened earlier, but most of working cosmology takes place after that point and is completely independent of what set up the initial conditions.

If the latter, we might conclude that locally, test particles are not yet influenced by the gravity of remote energy. Therefore there would be no inward acceleration from the overall average density, and there would in fact never be such acceleration.

This would presume that our universe has regions which are and always were causally disconnected. Isotropy would be impossible under such conditions. Since we observe a nearly isotropic universe, all of the observable universe must have been in causal contact at some time. This is one of the primary motivations for cosmic inflation.

 

[AWA]

No. The first derivative of the scale factor can be set to zero. You can't set the second derivative to zero arbitrarily, as that is set by gravity.

But it's not arbitrary, you set the condition of staticity, which makes the scale factor constant.
I think any order derivative of a constant wrt to any variable is 0.

 

[AWA]

We don't bother with any such assumptions in cosmology. For nearly all cosmological studies, we just take a set of conditions at a specific time and evolve those forward (or, equivalently, take the current conditions and evolve them backward in time to a certain point). For the most part we don't worry about what set up those "initial" conditions (initial is in quotes because it's not really initial per se, just that we don't know what happened before).

Who is we? I think it's advisable for you to try and speak for yourself.

This would presume that our universe has regions which are and always were causally disconnected. Isotropy would be impossible under such conditions. Since we observe a nearly isotropic universe, all of the observable universe must have been in causal contact at some time. This is one of the primary motivations for cosmic inflation.

That pressumption is supported by the finiteness of the speed of light. Isotropy only implies rotational invariance or no preferred direction, nothing to do with causality. you are conflating it with Temperature homogeneity of the CMB which is what demands inflation.

 

[Chalnoth]

But it's not arbitrary, you set the condition of staticity, which makes the scale factor constant.
I think any order derivative of a constant wrt to any variable is 0.

But you don't have those degrees of freedom available if you also set the matter content of the universe. Once you set the matter content (and the cosmological constant), the second derivative of the scale factor determined by gravity. Declaring differently would require a different law of gravity.

Who is we? I think it's advisable for you to try and speak for yourself.

I'm speaking of the majority of working cosmologists. We accept that there are limits to our theories, and while we may speculate about what may lie beyond current observational limits, we don't usually assume any particular model. This is my impression of living and working with people like this for some years now.

That pressumption is supported by the finiteness of the speed of light. Isotropy only implies rotational invariance or no preferred direction, nothing to do with causality. you are conflating it with Temperature homogeneity of the CMB which is what demands inflation.

Isotropy has everything to do with causality. While the connection to homogeneity is more obvious, I specifically mentioned isotropy because that is on firmer observational footing. Basically, if we see a patch of the CMB, and in another direction a different patch of the same CMB that has nearly the same temperature, then those two different patches must have, at some point in time, been in causal contact. Thus observed isotropy implies such causal contact.

Inflation solves this issue, at least at this level, because everything in the observable universe was in causal contact at some time during inflation. This still leaves open the question of how inflation got started to begin with, but it still solves the issue for the behavior of our observable universe.

 

[BillSaltLake]

The question of self-acceleration vs. no self-acceleration would have been a valid question to ponder when GR was first proposed, and I would guess that it was initially considered by some investigators. Therefore a static Universe might have made sense then. This was in answer to the initial question on this topic.
As for the causal contact problem, before there was any theory that predicted an expected statistical temperature distribution, one could not declare the observed distribution to be too uniform unless there was causal contact. What is "too uniform" relative to? We must remember that physical cosmology deals with quantifiables.

 

[Chalnoth]

As for the causal contact problem, before there was any theory that predicted an expected statistical temperature distribution, one could not declare the observed distribution to be too uniform unless there was causal contact. What is "too uniform" relative to? We must remember that physical cosmology deals with quantifiables.

If we are completely ignorant as to the fundamental physics, we would expect deviations for causally-disconnected regions to be of order one in dimensionless numbers, meaning roughly Planck temperature deviations.

Looking into it in a bit more detail, we might tend to expect that universes might tend towards lower temperatures in time, from which we'd largely come to the conclusion that the temperature almost everywhere should be right at absolute zero, so we'd only expect to see a small causally-connected region instead of an entire observable universe (that is, we'd expect to see everything in our causally-connected region, with everything outside of that at zero temperature and thus invisible).

In any case, we definitely do not expect causally-disconnected regions to, for instance, stop undergoing inflation at exactly the same time.

 

[Ich]

As Chalnoth said, the Friedmann equations are the field equations, applied to cosmology. "Expansion" is not an assumption needed to derive them, it's a result of their application.

As Chalnoth and I said said above this is not completely accurate, as shown by the Einstein model, or do you argue that this model is not derived from the field equations plus the assumption of homogeneity and isotropy?

Do you argue that the Einstein model is not a solution of the Friedmann equations?

That gravity attracts is a conclusion you don't need GR to reach.

So it shouldn't be surprising that it still holds in GR.

 

[yogi]

I fail to see how exactly the original 1915 equations without lambda lead to gravitational collapse.

Also,I think until 1917 Einstein didn't publish any cosmological model and this first already contained the lambda term, how could critics point out collapse of any cosmological model if there wasn't any model yet, and, do you happen to know who were those critics?
I know he had correspondance with de sitter in those early years, was he one of them?

If I remember correctly, Eddington was one of the first to point out the collapse problem,

 

[AWA]

But you don't have those degrees of freedom available if you also set the matter content of the universe. Once you set the matter content (and the cosmological constant), the second derivative of the scale factor determined by gravity. Declaring differently would require a different law of gravity.

Right, that is precisely what makes that equation not usable in a static setting.

Isotropy has everything to do with causality. While the connection to homogeneity is more obvious, I specifically mentioned isotropy because that is on firmer observational footing. Basically, if we see a patch of the CMB, and in another direction a different patch of the same CMB that has nearly the same temperature, then those two different patches must have, at some point in time, been in causal contact. Thus observed isotropy implies such causal contact.

Did you read what I wrote? I specifically said causality has everything to do with isotropy in the CMBR case.

Do you argue that the Einstein model is not a solution of the Friedmann equations?

Explain how, what is the scale factor a in the Einstein model? There is no time-dependent rescaling of space in this model.

 

[AWA]

If I remember correctly, Eddington was one of the first to point out the collapse problem,

He pointed out the instability problem of the Einstein model, with lambda added to the original equations.

 

[Chalnoth]

Right, that is precisely what makes that equation not usable in a static setting.

But the Friedmann equations are the solution to the Einstein field equations in the homogeneous, isotropic case. The only degrees of freedom available are:

1. The energy density of each type of matter/energy.
2. The relationship between energy density and pressure for each type of matter/energy.
3. The instantaneous expansion rate at one specific time.

Once you have set these three degrees of freedom, General Relativity describes everything else. If we set, for instance, only normal matter with some energy density and zero expansion rate, we necessarily get a universe that subsequently collapses.

To get out of this you have to propose something different from General Relativity, but that is going to be very difficult to do given the successes of General Relativity in solar system experiments and the successes of the expanding universe model. A static universe with positive matter density and no dark energy is simply not a solution to Einstein's equations.

 

[AWA]

But the Friedmann equations are the solution to the Einstein field equations in the homogeneous, isotropic case. The only degrees of freedom available are:

1. The energy density of each type of matter/energy.
2. The relationship between energy density and pressure for each type of matter/energy.
3. The instantaneous expansion rate at one specific time.

Once you have set these three degrees of freedom, General Relativity describes everything else.

Fine, with the exception you mentioned of the unstable Einstein model with cosmological constant.

If we set, for instance, only normal matter with some energy density and zero expansion rate, we necessarily get a universe that subsequently collapses.

So according to you in this case the second Friedmann equation has the left side <0 and that means gravitational collapse because it implies the second derivative of a is <0 and this is set by gravity? A negative gravity means collapse?

To get out of this you have to propose something different from General Relativity, but that is going to be very difficult to do given the successes of General Relativity in solar system experiments and the successes of the expanding universe model. A static universe with positive matter density and no dark energy is simply not a solution to Einstein's equations.

Sure, I'm not proposing anything. I was asking if such a solution had been ever proposed.

 

[Chalnoth]

Fine, with the exception you mentioned of the unstable Einstein model with cosmological constant.

It's not an exception. It's just one particular solution to the equations.

So according to you in this case the second Friedmann equation has the left side <0 and that means gravitational collapse because it implies the second derivative of a is <0 and this is set by gravity? A negative gravity means collapse?

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 &gt; -\rho/3), but it could definitely be done.

Sure, I'm not proposing anything. I was asking if such a solution had been ever proposed.

Well, yes, I'm pretty sure this sort of discussion is exactly the sort of discussion that led to Einstein inserting the cosmological constant in the first place.

 

[Ich]

Explain how, what is the scale factor a in the Einstein model?

1.

Or, if you like it better,
\sqrt{\frac{c^2}{4\pi G \rho}}

 

[AWA]

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 &gt; -\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?

 

[Chalnoth]

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).

 

[AWA]

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.

 

[Chalnoth]

Right, because there is no stable static solution to a homogeneous, isotropic universe in General Relativity.

 

[Ich]

If I undestood correctly.

You didn't.

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.

 

[Chalnoth]

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.

 

[AWA]

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.

 

[Ich]

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.

 

[George Jones]

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?

 

[AWA]

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

 

[Chalnoth]

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).

 

[AWA]

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.

 

[Chalnoth]

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.

 

[AWA]

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

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.

 

[BillSaltLake]

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.

 

[Ich]

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.

 

[Chalnoth]

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.

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.

 

[Chalnoth]

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.

 

[AWA]

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.

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.

 

[Chalnoth]

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.