# Variation of Density Parameters with Redshift

1. Dec 25, 2009

### cepheid

Staff Emeritus
I was looking at section 7.6 of Longair's Galaxy Formation, and in it, he is talking about the flatness problem, or fine-tuning problem. In it, he shows that if you define a general (i.e. varies with cosmic time and hence with redshift) density parameter $\Omega_m$ for matter by analogy with $\Omega_{m,0}$ (the matter density parameter at the present epoch), it would be defined as follows:

$$\Omega_m = \frac{8 \pi G \rho_m}{3H^2}$$ ​

Where H(z) is the Hubble parameter (so that H0 = H(0))

Since H = $\dot{a}/a$, and a = 1/(1+z), we can get directly from the Friedman equation:

$$H = H_0 [(1+z)^2(\Omega_{m,0} z +1) - \Omega_{\Lambda ,0}z(z+2)]^{1/2}$$ ​

I've used a slightly non-standard notation of including a zero in the subscript of the dark energy density parameter to explicitly indicate that I'm talking about the value of this parameter at the present epoch. The reason for doing so is that, later on, I'll be talking about the value of this parameter not at the present epoch. Anyway, from this, Longair shows that

$$\Omega_m = \frac{\Omega_{m,0}}{\left(\frac{\Omega_{m,0} z + 1}{1+z}\right) - \Omega_{\Lambda ,0} \left(\frac{1}{1+z} - \frac{1}{(1+z)^3}\right)}$$ ​

From this, it can be shown that if $\Omega_m$ had a value that differed even slightly from 1 initially, then the current value $\Omega_{m,0}$ would be drastically different from 1 (it would be either huge or miniscule...small variations initially would cause it to diverge rapidly as z approached 0). Another way of looking at this is that there is a very wide range of value of the parameter $\Omega_{m,0}$ that all basically converge to 1 at high redshifts. It's certainly true for any value between 0 and 1. This is the flatness, or "fine-tuning" problem. Why is the universe seemingly required to have been exceedingly close to flat initially in order to accord with present-day observations?

This result got me thinking. It is shown (elsewhere in the book) that the present condition for flatness is that the total density parameter is 1:

$$\Omega_{\textrm{tot},0} = \Omega_{m,0} + \Omega_{\Lambda ,0} =1$$ ​

We think that this is probably true (e.g. they are ~0.3 and ~0.7, speaking very loosely). But, I was bothered by the fact that if the universe is flat now, then it must always have been flat, suggesting that:

$$\Omega_{\textrm{tot}} = \Omega_{m} + \Omega_{\Lambda} =1$$ ​

is true generally (where these are now functions of z). But I have just shown that $\Omega_m \rightarrow 1$ all by itself as z approaches infinity! Therefore, the only way that this condition can be satisfied is if $\Omega_{\Lambda} \rightarrow 0$ as z approaches infinity. Longair makes no mention of this, so I set about to figure out if this was true on my own. I came up with what I thought was a reasonable defnition for the total density parameter (again, by analogy with the existing ones):

$$\Omega_{\textrm{tot}} = \frac{8 \pi G \rho_{\textrm{tot}}}{3H^2}$$

$$\Omega_{\textrm{tot}} = \frac{8 \pi G (\rho_m + \rho_v)}{3H^2}$$ ​

where the 'm' and 'v' stand for 'matter' and 'vacuum energy' respectively:

$$\Omega_{\textrm{tot}} = \frac{8 \pi G \rho_m }{3H^2} + \frac{8 \pi G \rho_v }{3H^2}$$ ​

The first term is just the matter density parameter from before, therefore the second term must be the dark energy density parameter:

$$\Omega_{\textrm{tot}} = \Omega_m + \frac{8 \pi G \rho_v }{3H^2}$$

$$\Rightarrow \frac{8 \pi G \rho_v }{3H^2} \equiv \Omega_{\Lambda}$$ ​

The question is, does it go to zero in the limit of infinite z? Using the definition of H from before:

$$\Omega_{\Lambda} = \frac{8 \pi G \rho_v }{3H_0^2 [(1+z)^2(\Omega_{m,0} z +1) - \Omega_{\Lambda ,0}z(z+2)]}$$ ​

Everything that is not in square brackets is just, by definition, $\Omega_{\Lambda, 0}$

$$\Omega_{\Lambda} = \frac{\Omega_{\Lambda ,0}}{[(1+z)^2(\Omega_{m,0} z +1) - \Omega_{\Lambda ,0}z(z+2)]}$$ ​

Now, the nice thing about this is that IF $\Omega_{m,0} + \Omega_{\Lambda ,0} =1$, then you can show using a little bit of algebra that the denominator (the part in square brackets) becomes:

$$\Omega_{\Lambda} = \frac{\Omega_{\Lambda ,0}}{[\Omega_{m,0}(1+z)^3 +\Omega_{\Lambda ,0}]}$$ ​

I'm confident that that does indeed approach zero as z approaches infinity. My question is simple: have I attempted to reconcile the apparent problem in the proper way, or is my attempt to define a dark energy density parameter that varies with cosmic epoch totally out to lunch?

2. Dec 26, 2009

### Chalnoth

For a while, most theorists just assumed that $$\Omega_\Lambda = 0$$. The correct statement is that $$\Omega_{tot}$$ must be absurdly close to one in the distant past for it to be anywhere near one now.

3. Dec 29, 2009

### cepheid

Staff Emeritus

Yes, I understand that the observational evidence for a non-zero cosmological constant is relatively recent. However, when most people talk about $\Omega_\Lambda$ or any density parameter for that matter, they are usually talking about that parameter at the present epoch. My post and my question deals with how these parameters evolve with redshift. I was struck by the fact that the matter density parameter itself tends towards unity as z tends to infinity for almost any reasonable value of said parameter at the present epoch. For the flatness condition to be true, this must mean that the dark energy density parameter tends to zero as z tends to infinity. I set about trying to show that, and I managed to, but I am wondering whether my analysis is correct.

4. Dec 29, 2009

### Chronos

Fine tuning is definitely a problem in cosmology, and it is probably true Omega had to be almost exactly 1 over the entire history of the universe for reasons you noted. Why is the mystery, expansion dilutes the matter density of the universe and dark energy compensates almost exactly. Weird.

5. Dec 30, 2009

### Chalnoth

In a spatially-flat universe, $$\Omega$$ is just the fraction of the total density of any given matter or energy type. So whatever type of matter/energy dilutes the slowest will always tend toward $$\Omega = 1$$.

If you have no dark energy, and the universe is perfectly flat, then that is matter. But if you have dark energy, which is required to dilute more slowly than normal matter in order for it to explain the current acceleration, then it is the dark energy instead that will approach $$\Omega = 1$$, not the matter. Dark energy has the added feature that it also dilutes more slowly than the effect of the curvature, such that you no longer need a perfectly-flat universe for this to be true.

(In a matter-dominated universe with some curvature, either $$\Omega_m \rightarrow 0$$ and $$\Omega_k \rightarrow 1$$ (open case), or $$\Omega_m \rightarrow \infty$$ and $$\Omega_k \rightarrow -\infty$$ such that $$\Omega_m + \Omega_k = 1$$ (closed case).)