Marcus's 'no-pressure' model in different jacket

[Jorrie]

In this thread, I would like to discuss: "which is more natural for a simple cosmological model: taking the long term Hubble time as a timescale (as Marcus has done), or taking the Hubble time as a timescale". I will start by paraphrasing Marcus' OP and replacing his '$\Lambda$-timescale' with the '$H_0$-timescale'.

Here is the 'paraphrased' post:

The size scale of our universe (any time after year 1 million) is accurately tracked by the function
$u(y) = \sinh^{\frac{2}{3}}(\frac{3}{2}y)$
where $y$ is the cosmic time in years, scaled to a fraction of the Hubble time, $\frac{1}{H_0}$, which is presently 14.4 billion years.

That's it. That's the model. Just that one equation. What makes it work is scaling times (and corresponding distances) down by the cosmological constant. "Dark energy" (as Lambda is sometimes excitingly called) is here treated simply as a time scale. [see note 1]

So to take an example, suppose your figure for the present is year 13.79 billion. In the scaled units, present time is

$y_{now} = \frac{13.79\ billion\ years}{14.4\ billion\ years}=0.957$

When the model gives you times and distances in terms of similar small numbers, you multiply them by 14.4 billion years, or by 14.4 billion light years, to get the answers back into familiar terms. Times and distances are here measured on the same scale so that essentially c = 1.

If this seems to be a valid approach, we can jump to the next level of equations developed in the original thread.

[Notes]
1. If we use negligible matter in this model, i.e. essentially just $\Lambda$, the long term Hubble time would have been around 14.4 billion years. The actual long term Hubble time depends on the ratio of spacetime curvature caused by matter to the spacetime curvature caused by $\Lambda$.

2. The basic equation is from George Jones in this post, where he wrote it more specifically for the cosmological case:

For a spatially flat universe that consists of matter and dark energy (w = -1), but no radiation, the scale factor is given exactly by

$a\left(t\right) = A \sinh^{\frac{2}{3}} \left(Bt\right),$

where

$A = \left( \frac{1 - \Omega_{\Lambda 0}}{\Omega_{\Lambda 0}} \right)^{\frac{1}{3}}$

and

$B = \frac{3}{2} H_0 \sqrt{\Omega_{\Lambda 0}} .$

[3] My concern as expressed in Marcus' thread is that beginners may be confused by the 'new' natural timescale. I understand that in the end it might be a case of personal preference, but maybe there are more opinions about it.

 

[marcus]

Hi Jorrie, I can only offer encouragement. I think it's useful to explore different time-scales in the presentation of cosmology. Even if it never gets to the point of a textbook for beginners or a tutorial, it's good to try these things out. I won't try to understand or criticize this approach because it would just get me confused to try to develop and make sense of two new time-scales at the same time.
It seems to me you are suggesting using the present-day Hubble time of 14.4 billion years as a unit. That has some advantages. I wouldn't be surprised to learn that some professional cosmological modeling uses that very unit! Rather than the more familiar Earth year, or billions of Earth years. I just don't happen to know of a study that I could give a link to. But I shouldn't get in the way of your thread. I'll see how it goes.

 

[Jorrie]

It seems to me you are suggesting using the present-day Hubble time of 14.4 billion years as a unit. That has some advantages. I wouldn't be surprised to learn that some professional cosmological modeling uses that very unit! Rather than the more familiar Earth year, or billions of Earth years. I just don't happen to know of a study that I could give a link to. But I shouldn't get in the way of your thread. I'll see how it goes.

Just about every cosmo-model equation that relates to size and time has H_0 in them, either directly or indirectly. That's why I try to show it could be called the conventional approach. The only place I have ever seen H_\infty used is between a few of us around here - I think we started using it during the development of the Lightcone Cosmo-calculator.

There is nothing wrong with trying out a new choice of timescale, but the question is, how do students relate it to the conventional approach. In this thread I want to show the relationship with the approach based on the conventional Hubble time and then see which eventual benefits each can bring on.

 

[Jorrie]

To cut directly to the chase and add frills later, here is the part of Marcus' recent summary post, which requires no change when switching the time base.

$$H^2 - \frac{\Lambda}{3} = \frac{8\pi G}{3c^2}\rho^*$$Friedmann equation inherits that Lambda, on the LHS, directly from 1917 Einstein GR equation.
*Reminder: as I just said, ρ* is a matter&radiation density. It does not contain any "dark energy" component. The curvature constant Λ is explicitly on the left side. This equation must be satisfied for there to be overall spatial flatness.
By definition
$$H_\infty^2 = \frac{\Lambda}{3}$$Therefore the Friedmann can be written this way:
$$H^2 - H_\infty^2 = \frac{8\pi G}{3c^2}\rho^*$$

From here on there is the 'minor difference' in that I normalize for $\small H_0$ to be 1 (i.e. 14.4 Gy becomes the timescale) and hence $\small H_\infty^2 = (14.4/17.3)^2 = 0.693$

Marcus' $\small H(s)^2 - H_\infty^2 = (H(1)^2 - H_\infty^2)s^3 = 0.443s^3$ becomes:

$\small H(s)^2 - 0.693 = (1 - 0.693)s^3 =0.307s^3$
or
$\small H(s)^2 = 0.693 + 0.307s^3$

And the present distance to a galaxy at stretch factor S is given by
$$\small D(S) = \int_1^S \frac{ds}{\sqrt{0.693 + 0.307s^3}}$$

Since the Friedmann equation simplifies to $$H(s)^2 -1 = 0.443s^3$$a flash of light's distance wave stretch factor s = z+1 which we can read directly from it tells us how far it has gotten from its source, how far away its source now is:
$$D(S) = \int_1^S \frac{ds}{\sqrt{0.443s^3 + 1}}$$This can be used to DETERMINE H_∞ from standard candle distance redshift data as discussed earlier. So in that sense this is a universal form of the Friedmann equation at this point in the history of the cosmos.

I agree, but I would rather say that the equation is used to determine the combination of matter content and the cosmological constant, as is more apparent in the expression $$ \small D(S) = \int_1^S \frac{ds}{\sqrt{0.693 + 0.307s^3}}$$ It actually reflects the conventional $\small \Omega_\Lambda$ and $\small \Omega_m$ directly.

 

[Jorrie]

So far, it seemed like the 'jacket' fits rather well, i.e. the more standard approach of using the present Hubble time (or distance) as a unit seems to work well, if not better, than Marcus' "new" timescale, based on the long term Hubble time.

However, going further, it seems that the present Hubble time (as a scale) makes things much more messy - parts of the 'new jacket' do not fit so well anymore. Normalized to the present Hubble time (H_0 = 1), the Hubble parameter becomes:

$H(y) = \sqrt{1-\Omega_m} \coth(\sqrt{1-\Omega_m}\frac{3}{2} y)$,

compared to Marcus' much cleaner $H(x) = \coth(\frac{3}{2} x)$.

The same problem seems to plaque an expression for the scale factor u(y). I think the reason is that the present Hubble time is essentially just a point on the curve of the coth or tanh functions, both of which flatten out asymptotically as the argument grows large. In order to specify the curve from the present Hubble time, you must provide extra information about the present, e.g. the present 'critical' density of matter. This is essentially stating the value of the cosmological constant for a spatially flat universe.

Hence, using the long term Hubble time as a scale seems to be preferable above the present Hubble scale. I have to agree with Marcus that this could be a valuable pedagogical tool that he is developing here. :)