# Cosmologies without lambda

1. Sep 19, 2010

### 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.

2. Sep 19, 2010

### 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

3. Sep 19, 2010

### AWA

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?

4. Sep 19, 2010

### 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.

5. Sep 19, 2010

### Ich

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" [Broken].
Starting with a homogeneous, isotropic static state, collapse is a rather direct consequence.

Last edited by a moderator: May 4, 2017
6. Sep 19, 2010

### AWA

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

7. Sep 19, 2010

### AWA

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

Last edited by a moderator: May 4, 2017
8. Sep 20, 2010

### Chalnoth

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.

9. Sep 20, 2010

### 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.
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.

10. Sep 20, 2010

### 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.

11. Sep 20, 2010

### AWA

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 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?

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.

12. Sep 20, 2010

### JDoolin

Uniform density is the assumption.

13. Sep 20, 2010

### Chalnoth

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.

14. Sep 20, 2010

### AWA

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.

15. Sep 20, 2010

### 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.

16. Sep 20, 2010

### Chalnoth

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.

17. Sep 20, 2010

### Chalnoth

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.

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.

18. Sep 20, 2010

### AWA

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.

19. Sep 20, 2010

### AWA

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

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 wich is what demands inflation.

20. Sep 20, 2010

### Chalnoth

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.

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.

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.