# Einstein-de sitter universe

1. Aug 20, 2010

### AWA

Up until the 80's this was the model favored by the mainstream and it admittedly was in accordance with observations, could we recover it if some day DE and DM are shown to not exist (hypothetically) and are explained by some other effect? Or does it have too many problems on its own?

2. Aug 20, 2010

### cepheid

Staff Emeritus
If I recall correctly, the Einstein de Sitter Universe is one in which:

$$\Omega_m = 1,$$

$$\Omega_\Lambda = 0$$

The major problem with this model is that numerous different observations intended to estimate the total matter density (including both dark matter and baryons) have come up with the result that it is much less than the critical density. The "concordance" value (the one that is the most consistent with all of the observations) is around Ωm = 0.27 (or something). If you get rid of dark matter, the problem becomes much worse (EDIT: estimates that are strongly constrained by big bang nucleosynthesis models put Ωb ~ 0.05).

When you combine that with the fact that the observations strongly favour a non-zero cosmological constant, well...

Last edited: Aug 20, 2010
3. Aug 20, 2010

### AWA

But you are limiting yourself to the flat Einstein-de Sitter model, the way it was told at the time you had also the close and the open universes, in the case of the open solution, Ωm was less than 1, no need to reach the critical density.
I know the CMB seems to point to flatness but this result are very recent and could be subject of posterior revisions.

4. Aug 21, 2010

### Chalnoth

Hypothetically, of course, as an effective theory. But observationally this model has utterly failed.

Btw, looks like the Einstein de Sitter universe is a flat model:
http://www.britannica.com/EBchecked/topic/139301/cosmology/27596/The-Einstein-de-Sitter-universe

5. Aug 21, 2010

### AWA

But not as much as the L-CDM

Sure, in my second post I was refering also to the open Friedmann model which basically only defers in that it doesn't reach the critical density so cepheid was right, I wasn't debating him that but I guess it looked like that., but to be precise yes, the Einstein-de Sitter model is the spatially Euclidean one.

6. Aug 21, 2010

### Chalnoth

Uh, what? Are you actually claiming that L-CDM has failed observational tests? More so than the Einstein-de Sitter universe?

7. Aug 21, 2010

### cepheid

Staff Emeritus
Emphasis mine. I'm assuming you meant "differs" in which case it seems like you're advocating for an open universe with no cosmological constant. (I'm not trying to be a jerk and nitpick your grammar, I'm just trying to be very clear about how I'm interpreting your posts so that we can communicate effectively). Anyway, the model you propose definitely doesn't match observations as well as LCDM. In fact, such an open model is excluded with pretty high confidence, for example even just by SNe Type Ia observations alone. This is illustrated nicely in Figure 7 from this paper:

I also find the figure really nifty and generally useful.

Chalnoth's incredulity ---> seconded. If this is what you meant, where did you get this idea?

Last edited by a moderator: Apr 25, 2017
8. Aug 22, 2010

### AWA

Sorry about the word swapping, should have read it before post it.
I am not advocating for it, but in an open friedmann model maybe the SNIa observations could be interpreted geometrically rather than as cosmological constant, with k<1 the volume of a sphere is > than 4/3pir^3 and therefore the light of SN spreads over a surface bigger than 4pir^2 and is perceived as fainter than expected for a certain distance.

9. Aug 22, 2010

### Chalnoth

Well, yes, the supernova observations alone do not put much constraint on whether it is dark energy or curvature. However, in order to go all the way to having zero cosmological constant, you need to have either a Hubble rate or supernova intrinsic brightness that is inconsistent with other observations (and not just slightly inconsistent, but completely and utterly wrong).

What is done in cosmology is the degeneracy between various parameters from one particular sort of observation is removed by combining it with different observations. For instance, if you combine supernova, WMAP, and BAO data, you end up with very tight errors around $\Omega_k = 0$. Just to make sure we're not doing something else wrong, though, we can also use the distance ladder to get an estimate of the supernova intrinsic brightness, and nearby Hubble expansion measurements to get a handle on the Hubble expansion rate. Add these observations to the supernova data, and you get much the same result (albeit with larger error bars).

10. Aug 22, 2010

### AWA

Ok, I was leaving WMAP, and BAO data aside just for the moment (but notice that for instance BAO data alone permits a lambda=0-see: http://supernova.lbl.gov/Union/ ).
I am interested in the error bars with Ho, what kind of Ho would be necesary to fit the SNIa data? 90-100 km/s/Mpc? Bigger? smaller?

11. Aug 22, 2010

### Chalnoth

But again, in order to get BAO data alone to permit $\Lambda = 0$, you need other parameters to be absurd. You can do the same with WMAP alone (and get a different set of completely wrong parameters).

It's been quite a while since I looked at this in detail. But right now, $H_0$ is known to within about 5%.

Anyway, I looked a bit more, and I'd have to do the work over again to be sure, but according to the Supernova Legacy Survey first-year release (four years ago), it's actually not possible to fit $\Omega_k = 0$. You can see the paper here:
http://xxx.lanl.gov/abs/astro-ph/0510447

The relevant figure is figure 5, where even the 99.7% confidence contours nowhere intersect $\Omega_\Lambda = 0$ where $\Omega_m > 0$. I'd have to look into the situation in more detail to see if this can be modified by changing $H_0$, but I'm not sure any longer that supernovae can actually fit a universe with no dark energy.

12. Aug 22, 2010

### AWA

So how exactly do they get to the fit with Probability of $$\Omega_\Lambda$$>0=0,99?
Only from the magnitudes and the redshifts? or they have more factors constraining that confidence limit?

13. Aug 22, 2010

### Chalnoth

Well, the way it works is you perform the following calculation for each supernova:

$$D_L = (1+z) D_M$$
$$D_M = \frac{c}{H_0}\frac{1}{\sqrt{\Omega_k}}sinh\left(\sqrt{\Omega_k}\int_0^z\frac{dz}{E(z)}\right)$$
$$E(z) = \frac{H(z)}{H_0} = \sqrt{\Omega_m(1+z)^3 + \Omega_\Lambda + \Omega_k(1+z)^2}$$

Note that as $\Omega_k \to 0$, the hyperbolic sine and the factors of $\Omega_k$ just disappear. Also, if $\Omega_k < 0$, the hyperbolic sine becomes a sine (since $sin(x) = sinh(ix)/i$).

The luminosity distance, $D_L$, is what we measure when we look at the brightness of the supernova. So we sample this value at many different redshifts by looking at many different supernovae. The effect of the curvature, as you can see, is two fold. First, it modifies the Hubble expansion rate, which changes the value of the integral. Second, there is an additional geometric factor.

So, as you can see, the dependence of this observation on $\Omega_k$ is non-trivial, and in principle, sampling enough supernovae can potentially give constraints on $\Omega_k$. The key is getting a broad redshift range.

Last edited: Aug 22, 2010