# I Evolution of the Energy Density Parameters

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1. Jan 26, 2017

### Jorrie

I am busy with an effort to show how the energy density parameters evolve over time in an update of the LightCone7 calculator. See the posts on the thread Steps on the way to Lightcone cosmological calculator. As part of this effort, I ran into some difficulties with deciding how to find and present the evolution of the various Omega's over time for non-flat geometries.

$\Omega$ without subscript is usually defined as the ratio: present total energy density to the present critical density ($\rho_{crit} = 3Ho^2/(8 \pi G)$), with the latter the density required to make space flat, given the present expansion rate Ho.
$\Omega = \Omega_\Lambda + \Omega_m + \Omega_r = 1$ for a spatially flat universe.

The evolution of Omega over time (or redshifts) is a 'weighted' sum of the components of present matter density $\Omega_m$, radiation density $\Omega_r$ and the cosmological constant's energy equivalent, $\Omega_\Lambda$, i.e.
$\Omega(z) = \Omega_\Lambda + \Omega_m (z+1)^3 + \Omega_r (z+1)^4$
and the individual components are given by the terms in the function.

However, it seems not to be so simple when $\Omega <> 1$. I have tried an approach through the actual energy densities, i.e.
$\rho_{z=0} = \Omega \rho_{crit}$
$\rho_\Lambda =$ constant
$\rho_{m} = \rho_{z=0} - \rho_\Lambda$
$\rho_{r} = \rho_{m}/(z_{eq}+1)$ (z_eq indicates radiation-matter density equality)
These present densities can be translated into density evolution as functions of z:
$\rho(z) = \rho_\Lambda + \rho_{m}(z+1)^3 +\rho_{r}(z+1)^4$

My problem is to decide by which $\rho_{crit}$ should I divide the above individual components to get the corresponding Omega's for the $\Omega <> 1$ case. In the latest test version of LightCone7, I have used the $\rho_{crit}(z)$ values for the flat ($\Omega = 1$) case. But is this correct? Should I have not have used the "non-flat critical density" based on the "non-flat H(z)", that now has a different profile?

Does anyone know of a straightforward method that has been published?

PS. I've corrected the last ($\rho(z)$) equation. There is no (z) required for the $\rho_\Lambda$ term, because it is $(1+z)^0$.

Last edited: Jan 27, 2017
2. Jan 27, 2017

### Chronos

See https://arxiv.org/abs/1701.07261 for an example of Einstein's approach. To his unending chagrin, he struggled to make sense of the issue with infinite boundary conditions. This paper; https://arxiv.org/abs/1309.6590, The Mythical Snake which Swallows its Tail: Einstein's matter world, may also offer some insights. I suspect others have probed it in more detail, so this may be no more than a teaser. He was a pretty bright guy so chances are not many people have arrived at unambiguous solutions he failed to consider.

3. Jan 27, 2017

### Chalnoth

I'm not sure I understand the confusion. $\rho_{crit} = \rho_\Lambda + \rho_m + \rho_r$. This is true at any time, and whether or not the universe is flat.

4. Jan 27, 2017

### Jorrie

So does this mean $\rho_{crit}$ is not defined as the energy density for a flat-space universe?

5. Jan 27, 2017

### Chalnoth

You're right, sorry. You also have to include the curvature term if you want to calculate the critical density from the densities. An easier way to define it is just in terms of H:

$$\rho_c = {3 H^2 \over 8\pi G} = \rho_\Lambda + \rho_m + \rho_r - {8 \pi G k c^2\over 3 a^2}$$

6. Jan 27, 2017

### Jorrie

Yup, that's the way I have it, but then my issue was how to interpret the evolution of the individual density parameters for the $\Omega<>1$ case. E.g. should it be
$$\Omega_m(z) = \frac{\rho_m(z)_{\Omega<>1}}{\rho_c(z)_{\Omega<>1}}$$
or should it be a division by the critical density of the flat case for the same z? E.g.
$$\Omega_m(z) = \frac{\rho_m(z)_{\Omega<>1}}{\rho_c(z)_{\Omega=1}}$$
The $\rho_c(z)$ evolve differently for the two cases.

I need the $\Omega(z)$ for the calculator, because they are visually a lot more meaningful than plain old densities.

7. Jan 28, 2017

### Chalnoth

The first one. You're calculating the density fractions at different times, so it doesn't make sense to include an energy density at the current time when computing the fraction at past times.

8. Jan 28, 2017

### Jorrie

The reason for my doubt was that the second method does not take $\rho_c$ at the current time, but at the same z as the first one, just for a spatially flat case. But I now realize that the curvature term in the equation for $\rho_c$ correctly compensates for the curvature.

The second method does not, I think, for the reason that even at the same z, it is still not for the same cosmic time, as you implied.

I will change the LighCone7 trial version accordingly.

9. Jan 28, 2017

### Mordred

On the calc I have one question.

How does the calc handle the high redshift beyond Hubble horizon? To define what I mean I had previously been curious as to how to correct the redshift equations beyond Hubble horizon as the common redshift equation we are all familiar with is only accurate for low redshift value.

I can't recall where I got this formula from and its been difficult to find good papers covering this (I did come across one from Bunn) his equation was slightly different extremely close to this one

$$z=\frac {H_0l^2}{c+\frac {1}{2}(1+q_0)H^2_0l^2/c^2+O (H^3_0l^3)}$$

Which uses the evolution of matter/radiation and Lambda terms in the equation you use for H/H_0 column in your calc. I was curious how your calc handles the high redshift.

If I'm thinking correct in your older calc this modified formula to use stretch may be involved in the above question.
$$H = H_0 \sqrt{\Omega_\Lambda + (1-\Omega) S^2 + \Omega_m S^3 (1+S/S_{eq})}$$

As I recall the Bunn paper used the standard version. (using z insteas of stretch). So was curious how your calc handled high z corrections.

I know it was handling it as your numbers were always incredibly accurate on cross checks comparisions. Am I correct you have been applying the last formula?

Ps just a confirmation my understanding is correct. I need to practice the derivitaves so don't want the full derivitaves.

Last edited: Jan 28, 2017
10. Jan 28, 2017

### Jorrie

I'm not sure which common redshift equation you refer to, but the Lightcone7 calculator uses basically only the equation
$$H = H_0 \left(\Omega_\Lambda + (1-\Omega) (z+1)^2 + \Omega_m (z+1)^3 + \Omega_r (z+1)^4\right)^{0.5}$$
as you have given in S-form. It takes inputs as z-values and calculates the rest by integrating the equation for various limits and constants. It hence does not run into a problem or singularity at the Hubble radius.

It is obviously hard to extract z in terms of H, etc. from it (if one want to include radiation, matter and Lambda). I think various approximate solutions for z may run into problems beyond the horizon, but I'm not sure.

BTW, the method for the Omegas that I agreed upon with Chalnoth above, is in the LightCone 7, Cosmo-Calculator (2017-1) version that I asked you to comment on before this thread started (and later thought to be in error). If we find no other errors in it, it may become the linked version in our signatures.

11. Jan 28, 2017

### Mordred

kk thats what I thought, thanks for the clarification. I have seen numerous approximations for high z and agree some of the derivitaves do run into problems. The methodology you are using is the more recommended method AFIAK. One of reasons why your calculator is far more accurate than those that use $$z=\frac {\lambda_{obs}-\lambda_{emit}}{\lambda_{emit}}$$ which is only good for low recessive velocities signifantly less than c.

Last edited: Jan 28, 2017
12. Jan 28, 2017

### Mordred

One further question. Does your new temperature column also run off the individual density correlations as per above formulas. As each Matter/radiation/lLambda has different temperature contributions via their individual EoS. (equation of state) as a functional evolution over cosmic time? Or are we using the combined methodology of the three aforementioned components on essentially the Gibb's law relations to temp/scale factor averaging?

ie inverse scale factor/temp relation of each individual component as opposed to the average of the three?
edit: The detail on the calc used will be useful as I am helping beta test your new version when I have time, that and on private convo I offered to help update your support links (user manual etc). Not sure if you caught as on edit.

On simple readability all culumns added in format display x.xxx e (exp) I think would be better displayed on default in column options for degree of accuracy to single digit. ie x.x e exp. Those columns are less an initial eyesore if the column definition decimal setting is set to 1

Last edited: Jan 28, 2017
13. Jan 29, 2017

### Jorrie

The latter, since the calculator only goes back to z=20000. I have used the simple 1/(z+1) relationship only. I'm not sure if "background CMB temperature" in the tooltip description is enough, though.
OK, I'll look into a better initial or default display options. I ended up with some in exponent and others in fixed, because it is a little difficult to make them dynamically changing.
Thanks for the feedback. ;)

14. Jan 29, 2017

### Jorrie

No, I think this relationship always holds - it is simply z = 1/a - 1, which is exactly the definition of cosmological redshift.
I think the problem comes in when the approximation $z \approx v_{rec}/c$ is used for larger recession rates.

15. Jan 29, 2017

### Mordred

Yes I believe your correct on that and rereading what I wrote I see that I misimplied it was the redshift equation that was the issue. When I should have detailed the V=Hd formula departures. Good catch and thanks for clarifying the temp column

16. Jan 30, 2017

### Jorrie

Here is a trial version with consistent column number formats. I have tried 2 decimal places as default and I must say it looks less intimidating than the mixed bag of the previous versions. Temp-link: LightCone7-2017-01-30 Cosmo-Calculator.

I have added a column for the evolution of the total density parameter (OmegaT, just the sum of matter-, radiation- and Lambda-energy). It is uninteresting when Omega=1, but the evolution of Omega is very interesting in cases where the present Omega <> 1, e.g. for Omega = 0.75 (the minimum, that the calculator allows):
Note how OmegaT starts at ~1, evolves away from 1, all the way down to OmegaT~0.5, and then back up to approach unity again (as OmegaL dominates). As a sanity check, it passes through 0.75 at present time (z=0), which for Omega=0.75 is at T=17.4 Gy.

I hope I'm right with these calculations in the software - if anyone cares to check the results, it will be highly appreciated.

Last edited: Jan 30, 2017
17. Jan 31, 2017

### Mordred

Looks much better, I'll play around with it but looks good thus far

18. Jan 31, 2017

### Chronos

shouldn't the formula be assymptotic?

19. Jan 31, 2017

### Jorrie

If you are referring to the total density parameter $\Omega$, when $\Omega_\Lambda > 0$, it goes asymptotically to 1.0. I think if $\Omega_\Lambda =0$ identically, it depends on whether $\Omega_{now}$ is larger or smaller than 1, going asymptotically to either infinity or zero. Not sure of this.

The calculator seems to indicate that for $\Omega_\Lambda \approx 0$ and $\Omega_{now} > 1$, the eventual collapse will cause an exponential rise in $\Omega$ towards infinity. Which seems right for any 'big crunch'.

Last edited: Jan 31, 2017
20. Feb 3, 2017

### Jorrie

As a further check, does this correctly represents what the LCDM model predicts for the value of $\Omega(T)$, given a present (hypothetical) $\Omega$=1.5?

I.e. the total density had to start very close to unity, then evolved away from that during the radiation/matter dominated era. Dark energy domination then drives it asymptotically back to approach unity again.

At early times, $\Omega(T)$ appears like an abrupt deviation from unity, but a zoom to early times indicates that it is also an asymptotic deviation from near-unity.

At the earliest time that the calculator can handle (z=20000, t~1000 yrs), $\Omega(T)$ ~ 1.0005 and it evolves like this, according to the calculator under test. So, the 'abruptness' is just a matter of the scale.

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