Simple no-pressure cosmic model gives meaning to Lambda

[marcus]

My PoV is not necessarily decisive in this thread, of course, but I'll tell you my impression about matter density. I don't think we can estimate it at all accurately, what with dark matter clouds and gas and all kinds of stuff besides stars. Even the estimates of luminous matter in galaxies are rather uncertain. So I think the matter density estimate just comes from observing the lack of overall curvature and calculating the (matter+radiation only) CRITICAL.

As far as concerns me, Lambda is not an energy and does not contribute to flatness the way matter density does. In any case not for the model discussed here.
Friedmann equation , for our purposes in this thread, has Lambda on the left hand side as reciprocal square time.
$$H^2 - \frac{\Lambda}{3} = \frac{8\pi G}{3c^2}\rho^*$$
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:
$$H^2 - H_\infty^2 = \frac{8\pi G}{3c^2}\rho^*$$

EDIT: I deleted a reference to "ρ_crit" when it was pointed out in a helpful comment that this might be confusing. As additional guard against confusion I put an asterisk on the density as a reminder that, as the density of matter and radiation, it doesn't involve a Lambda component. The equation must be satisfied for spatial flatness and so, in that sense, ρ* is critical for spatial flatness once the two expansion rates H and H_∞ have been determined.

A few posts back, Wabbit pointed out a useful version of the Friedmann equation (for matter-era on into indefinite future, since there is no "Lambda era") that saves a fair amount of bother, writing density, and constants like π and G etc. I'll write it using the wavestretch factor that Jorrie introduced in the Lightcone calculator. S=1 denotes the present.

$$H(s)^2 - H_\infty^2 = (H(1)^2 - H_\infty^2)s^3$$

For me, in this thread, the main topic is this model in which Λ, or more precisely T_∞ serves as a time scale. So to proceed we should evaluate the terms in the equation. Obviously the present value of the Hubble constant is 173/144 = 1.201... and its square is 1.443. Obviously, in the timescale we are using, H_∞ = 1 and its square is 1.
The RHS of the Friedmann equation evaluates to:
$$H(s)^2 - H_\infty^2 = (H(1)^2 - H_\infty^2)s^3 = 0.443s^3$$
And in our time scale the Friedmann simplifies to:
$$H(s)^2 - 1 = (H(1)^2 - 1)s^3 = 0.443s^3$$

 

[Jorrie]

But it does rely on the FRW model. I must retract my incorrect suggestion above that this might be a model free formula. It think it might be possible to do that in principle but I don't know how.

Interesting equation, thanks.
Matter density can be obtained from other independent observations, perhaps most importantly from grav. lensing. If that's accurate enough, Lambda is indirectly available for the flat space case. Still never quite model-free, I guess...

 

[Jorrie]

Since we observe spatial flatness, near enough to it anyway, that defines the (matter-radiation) critical density ρ_crit
$$H^2 - H_\infty^2 = \frac{8\pi G}{3c^2}\rho_{crit}$$

One must be careful not to confuse here, because isn't the present (matter-radiation) critical density only 30% of the 'standard' (quoted) critical density?

If so, shouldn't you give it a different subscript?

 

[marcus]

It is the combined density of all known forms of energy that is required for spatial flatness. Do you have any suggestions?
How about "rho_flatness" or "rho flat"?
See how you think these would work:
$$H^2 - H_\infty^2 = \frac{8\pi G}{3c^2}\rho_{flat}$$
$$H^2 - H_\infty^2 = \frac{8\pi G}{3c^2}\rho_\flat$$

Not using LaTex involves using the word as in ρ_flat
or having the symbol available to paste (since I don't know how to type it, with a Mac)
ρ_♭

 

[marcus]

Referring back to post #61:
In our time scale the Friedmann simplifies to:
$$H(s)^2 - 1 = (H(1)^2 - 1)s^3 = 0.443s^3$$ Referring also to a post or two on previous pages:

So adapting your equation in post#35
$$ D(a)=\int_a^1\frac{da}{a^2H(a)}=\int_0^z\frac{dz}{H(z)} $$we can write:
$$ D(S) = \int_1^S T(s)ds = \int_1^S (( (\frac{17.3}{14.4})^2 - 1) s^3 +1)^{-1/2}ds$$

This is where we got the equation (with help of Wabbit's post #35) for the present distance to a source we are now receiving light from that is stretched by factor S
$$ D(S) = \int_1^S T(s)ds = \int_1^S (0.443 s^3 +1)^{-1/2}ds$$
This formula is the basic tool that allows astronomers to directly tell the cosmological time-scale constant Λ from wavestretch-distance data. To consider an example such data could for instance consist of pairs of numbers (s, D) each giving the stretch factor of some light received and the standard candle estimate of current distance to its source.

The procedure basically relies on assuming near spatial flatness, which is supported by a variety of evidence. Given that, and that you have independently determine the present Hubble time 14.4 billion years, you choose some alternative time-scales to try out: 16.3 billion years, 17.3 billion years, 18.3 billion years. Each one will change the 0.443 number somewhat.
Then for each observed wave-stretch factor in your sample you compute the D(S) distance that light should have covered (don't forget to multiply by the distance scale). And you see if that matches the "standard candle" distance that was also part of the data.

It turns out that the expansion time and distance scale 17.3 billion years gives the best fit to the wavestretch-distance data, at least so far. The point I'm emphasizing is the sense in which it is directly observable without having to know the value of the matter density or assuming any model specifics. Sure it depends on General Relativity (from which the Friedmann equation is derived) and on the assumption of near spatial flatness, but those are widely accepted general assumptions.

 

[marcus]

So if we want to compare two assumptions H_∞ = 1/17.3 per billion years, or = 1/20 per billion years, using (s, D) data, we calculate
(20/14.4)² - 1 = 0.929
(17.3/14.4)² - 1 = 0.443
And we evaluate these integrals which give the distances in billions of lightyears:
$$ D(S) = 17.3 \int_1^S T(s)ds = \int_1^S (0.443 s^3 +1)^{-1/2}ds$$
$$ D(S) = 20 \int_1^S T(s)ds = \int_1^S (0.929 s^3 +1)^{-1/2}ds$$

I've tried it using an online definite integrator and the latter (the "20" one) gives noticeably smaller distances, especially in the higher wavestretch range such as S > 1.5 and even more so for S > 2.

The H_∞ = 1/20 per billion years is after all closer to zero than 1/17.3 per billion years.
So the "20" case is more like having zero cosmological constant. What woke people up to the fact of a positive cosmological constant was that measured distances to standard candle supernovae were distinctly larger than theoretically predicted assuming zero Lambda.
H_\infty = \sqrt{\Lambda/3} is the operative form of the cosmological curvature constant here.
And I find its reciprocal, the longterm Hubble time T_∞ = 1/H_∞ ≈ 17.3 billion years, is its most useful, easiest-to-remember quantitative expression.

 

[Jorrie]

It is the combined density of all known forms of energy that is required for spatial flatness. Do you have any suggestions?
How about "rho_flatness" or "rho flat"?
See how you think these would work:
$$H^2 - H_\infty^2 = \frac{8\pi G}{3c^2}\rho_{flat}$$

This may still confuse readers, because "flat" is normally associated with total energy density being critical. I noticed that The Perlmutter et al. paper of 1998 (http://arxiv.org/abs/astro-ph/9812133) made use of a super- and subscript to indicate the "matter density component of a spatially flat cosmos", i.e. \Omega^{flat}_M, so \rho^{flat}_M may be a good solution for clarity here.

 

[wabbit]

This may still confuse readers, because "flat" is normally associated with total energy density being critical. I noticed that The Perlmutter et al. paper of 1998 (http://arxiv.org/abs/astro-ph/9812133) made use of a super- and subscript to indicate the "matter density component of a spatially flat cosmos", i.e. \Omega^{flat}_M, so \rho^{flat}_M may be a good solution for clarity here.

I think the discussion of flatness and its impact on interpreting things can be confusing here - would it not be better to separate this as successive independent steps such as :
(a) we know that to a good approximation the universe is spatially flat and was so during the period concerned,
(b) with that in mind (and also perhaps an observational argument for stating that radiation is negligible over that period), we can think of the universe as made of matter, with a CC, and nothing else
(c) on this basis, follow with marcus' presentation of how we can measure the CC/matter proportion etc.

These steps may not be really that independent, but as a first introduction it still seems a fair simplification to me. Maybe not though, not quite sure here.

 

[Jorrie]

And we evaluate these integrals which give the distances in billions of lightyears:
$$ D(S) = 17.3 \int_1^S T(s)ds = \int_1^S (0.443 s^3 +1)^{-1/2}ds$$
$$ D(S) = 20 \int_1^S T(s)ds = \int_1^S (0.929 s^3 +1)^{-1/2}ds$$

I've tried it using an online definite integrator and the latter (the "20" one) gives noticeably smaller distances, especially in the higher wavestretch range such as S > 1.5 and even more so for S > 2.

While the derivation of this approximation is interesting and educational, the result is more easily (and probably more accurately) obtained by Lightcone 7. To simulate a "no Lambda" flat universe, just copy and paste the max allowable R_\infty (999999) into the box, set S_upper to 2 (or whatever). Calculate and look at the value of D_now at (say) S=2 or lower. If you make R_\infty just marginally larger than R₀ (say 14.41), you get a near-Lambda-only, 0.1% matter flat universe. The calculator is not designed for matter closer to zero than that.

 

[marcus]

This may still confuse readers, because "flat" is normally associated with total energy density being critical. I noticed that The Perlmutter et al. paper of 1998 (http://arxiv.org/abs/astro-ph/9812133) made use of a super- and subscript to indicate the "matter density component of a spatially flat cosmos", i.e. \Omega^{flat}_M, so \rho^{flat}_M may be a good solution for clarity here.

Thanks Jorrie,
I went back and edited post #61, eliminating the notation ρ_crit. I shall use the notation ρ* and put in frequent reminders that the cosmological constant term is here on the left side, so there is no Lambda contribution to the energy density.

 

[marcus]

While the derivation of this approximation is interesting and educational, the result is more easily (and probably more accurately) obtained by Lightcone 7. To simulate a "no Lambda" flat universe, just copy and paste the max allowable R_\infty (999999) into the box, set S_upper to 2 (or whatever). ...

Thanks. I'm glad you approve of the exercise. Interest and hands-on learning are the main aims, in fact. The simple model here (I think George Jones first suggested it to us, it may be well-known) seems to approximate Lightcone numbers fairly well as long as one stays away from the radiation dominated era ( e.g. don't go earlier than, say, year million). Basically AFAICS Lightcone is the online gold standard. I'm constantly checking number calculated with this simple "by hand" method to see how well they agree. I see the two things working together in the same learning experience.

I hadn't planned on including a "no Lambda" calculation. The "20" calculation was just a sample of a substantially smaller Lambda. But I see one could get the required distances as you describe---using a large Hubble radius.

 

[Jorrie]

I shall use the notation ρ* and put in frequent reminders that the cosmological constant term is here on the left side, so there is no Lambda contribution to the energy density.

I'm comfortable with this as a valid interpretation of the standard model, assuming that Lambda is really constant. If proved otherwise, then "simple, no-pressure" may have to be abandoned, but that's surely not for beginners. I think this is a very useful introductory approach...

 

[George Jones]

A related pedagogical paper: "A new standard model of the universe" by Gron,

http://arxiv.org/abs/0801.0552

 

[marcus]

Øyvind Grøn of Oslo University. Neat! I see he has some of the same-shaped curves! coth for H(t) and sinh^2/3 for the scale of a sample distance growing over time. Crucial differences--I wouldn't want to mix discussion of the two models--but interesting. Thanks GJ, I'm glad to know of Grøn's paper

 

[marcus]

I'm exploring ways to present the flat LambdaCDM as it works any time after the early years when radiation was the dominant content. As soon as matter takes over from radiation as the main component of their combined energy density ρ* , the standard cosmic model equations simplify and there is a reasonable facsimile of LambdaCDM that we can work by hand.
Here we use the same two basic inputs used in the Lightcone calculator: present and future Hubble times 14.4 and 17.3 billion years, and the same independent variable S = z+1=1/a, the wave and distance stretch factor. It is the reciprocal of the normalized scale factor, so S=1 denotes the present and its definition extends into the future.

I assume that the two Hubble growth rates H_now and H_∞ have been determined to be 1/14.4 and 1/17.3 per billion years.
The essential point here is that if we make 17.3 billion years and 17.3 billion light years our units of time and distance then the Friedmann equation (the main cosmic model equation) takes a remarkably simple form.

First notice that the present Hubble growth rate is simply 17.3/14.4 = 1.2014 per unit time. If you call our new unit of time a "Uday" then the present rate is 1.2 per Uday, and the eventual Hubble growth rate is exactly one. H_∞ = 1 per Uday.
An important term that appears in the Friedmann equation involves the square growth rates:
H_now² - H_∞²
And that turns out to be 1.443 - 1 = 0.443

To make a short story even shorter, when we use these units of time and distance the Friedmann equation basic to cosmology simplifies to:

H(s)² - 1 = 0.443s³

I've described earlier in the thread how this is derived from the more recognizable form of Friedmann:
H² - H_∞² = [const.] ρ*
I'll review that derivation later, and it's not hard so you may see for yourself how it goes. But now let's just look at the extremely simple version of the cosmic equation.
Something to notice is how close this equation is to observation---observation of actual wavelengths of light.
If you are visiting an observatory at night and the telescope is trained on a galaxy and you are told the stretch factor is 2, the wavelengths coming in are TWICE what they were when the light was emitted and headed our way,
then you can almost do in your HEAD the calculation of what the Hubble expansion rate was back then when the light was emitted. 2 cubed is 8.

And 8 times .443 is about 3.5.
Plus 1 is 4.5, whose square root is 2.1.

So you can remark to your astronomer friends: "I know what the Hubble growth rate was back then when the light was emitted! It was a little over twice the longterm rate."
If you wanted to spell it out in terms of years instead of the longterm rate unit, it is 2.1 times the longterm rate of 1/17.3 per billion years, so 2.1/17.3, or about 1/8.1, per billion years.
So you could add: "The Hubble time must have been around 8.1 billion years back then!"

 

[marcus]

To take another example, suppose your friend is observing a galaxy and says the incoming light has stretch S = 3.
3 cubed is 27, times 0.443 is about twelve.
Add one and take the square root. The square root of 13 is about 3.6.
So the Hubble growth rate back then, when the light was emitted, was 3.6 times the longterm rate.

The longterm expansion rate H_∞ is our natural UNIT for picturing growth rates, and I am trying to think in terms of that. So I won't bother to work out what that would be in "per-billion-Earth-year" terms. But you could, if you want, divide 3.6 by 17.3.

To review a bit. We look at the standard cosmic model, ΛCDM, focusing on the spatial flat matter-dominated case. and we let an important feature of that model, namely Λ, determine our scale of time and distance. The square root of Λ/3 is the longterm growth rate H_∞, in the standard model. And the corresponding Hubble time, 1/H_∞ is generally estimated to be around 17.3 billion years. We treat that as one "universe day", or Uday.

so the longterm growth rate H_∞ is now our unit growth rate. As it happens the current rate of distance growth H_now is about 20 percent larger. It is 17.3/14.4 = 1.2014. That number and its square, 1.443, help to characterize the present for us.
In terms of our new time scale, H_now = 1.201 per Uday and H_now² is 1.443 per Uday²

The Friedmann equation, the basic cosmic model equation, simplifies on our time-scale to:
$$H(s)^2 -1 = 0.443s^3$$
where s is the stretch factor, s = z+1. In the Lightcone calculator the stretch factor is capital S. I may have to switch over to that, or alternate s and S.
S = 2 denotes a time in the past when distances were 1/S = 1/2 their present size.

The wave stretch of light received today tells us the distance it has traveled---how far it is now from its source matter. This includes the effect of distance expansion while the light was in transit.
$$D(S) = \int_1^S \frac{ds}{\sqrt{0.443s^3 + 1}}$$

 

[Jorrie]

The Friedmann equation, the basic cosmic model equation, simplifies on our time-scale to:
$$H(s) -1 = 0.443s^3$$
where s is the stretch factor, s = z+1.

You seem to have missed a squaring in the above: should be \small H(s)^2 -1 = 0.443s^3 (?)

 

[marcus]

Corrected. Thanks for catching that.

 

[Jorrie]

Corrected. Thanks for catching that.

Not wanting to be pedantic, but why not just stick the constant 17.3 in front of the proper distance integral and let the answer come out in Gly, like we are used to on this forum? You are using the constant 0.443, which comes from it anyway...

 

[marcus]

I know what you mean, Jorrie. Tried that earlier. This is exploratory. I want to see what the approach looks like when you go all the way over to the new units. Time scale 17.3 Gy and distance scale 17.3 Gly. If I was doing a practical calculation where I wanted to compare results with your calculator, I'd certainly stick a 17.3 out in front and save a step in the calculation as was done here.

...
And we evaluate these integrals which give the distances in billions of lightyears:
$$ D(S) = 17.3 \int_1^S T(s)ds = \int_1^S (0.443 s^3 +1)^{-1/2}ds$$
...

BTW if you have the time and want to help explore this and post some equations/derivations/explanations here in thread, it would be very welcome.

 

[marcus]

To review a bit. We look at the standard cosmic model, ΛCDM, focusing on the spatial flat matter-dominated case. and we let an important feature of that model, namely Λ, determine our scale of time and distance. The square root of Λ/3 is the longterm growth rate H_∞, in the standard model. And the corresponding Hubble time, 1/H_∞ is generally estimated to be around 17.3 billion years. We treat that as one "universe day", or Uday.

so the longterm growth rate H_∞ is now our unit growth rate. As it happens the current rate of distance growth H_now is about 20 percent larger. It is 17.3/14.4 = 1.2014. That number and its square, 1.443, help to characterize the present for us.
In terms of our new time scale, H_now = 1.201 per Uday and H_now² is 1.443 per Uday²

In this scale of time and distance the Friedmann equation simplifies to $$H(s)^2 -1 = 0.443s^3$$
and a flash of light's distance from its source simplifies to $$D(S) = \int_1^S \frac{ds}{\sqrt{0.443s^3 + 1}}$$

I guess my observation about the number 0.443 is that it doesn't depend on Earth years or any particular human units. Rather it depends on this present moment in universe history. Imagine people on a different planet with a different year, perhaps not even using years to measure time but some "second" defined using their own atomic clock.

They could still measure H_{now} and H_\infty

and they would still find that
$$\frac{H_{now}}{H_\infty} = 1.201...$$ $$(\frac{H_{now}}{H_\infty})^2 = 1.443...$$

 

[Jorrie]

In this scale of time and distance the Friedmann equation simplifies to $$H(s)^2 -1 = 0.443s^3$$
and a flash of light's distance from its source simplifies to $$D(S) = \int_1^S \frac{ds}{\sqrt{0.443s^3 + 1}}$$

I guess my observation about the number 0.443 is that it doesn't depend on Earth years or any particular human units. Rather it depends on this present moment in universe history.

I agree, but am still unsure about the wisdom of scaling the equations to the so-called "long term Hubble time (or radius)" 1/H_∞. The conventional method is to scale to 1/H₀, which is an LCDM-model-independent observable (AFAIK). I don't think 1/H_∞ is.

The penalty for doing so is (perhaps) a slightly messier equation, but then it correlates directly with what everyone reads in the textbooks.
Starting with your $$H(S)^2 - H_{\infty}^2 = (H_0^2 - H_{\infty}^2)S^3 $$ (which is 'cool') and normalizing for H₀=1 instead of (your) H_∞, we get $$H(S)^2 - \Omega_\Lambda = (1-\Omega_\Lambda)S^3$$ or $$H(S)^2 - \Omega_\Lambda = \Omega_m S^3$$
Any student will recognize this as the matter + cosmological constant 'first Friedmann' for the spatially flat LCDM model.

 

[marcus]

Hi Jorrie, thanks for pointing out that alternative. I think they are both interesting ways to present the standard LambdaCDM cosmic model in a simple post-radiation-era spatial flat version.
I think you could argue that they each have advantages and it's somewhat a matter of taste. Maybe both should be developed in tandem and both should be available for teacher and student inspection.

It's clear that in our lifetimes the quantity 14.4 billion years is not going to change (except to be determined more precisely) so even if in principle it is an accident special to the present moment in time one can and does use it as a a time-scale! I think that's the essence of what you suggest here.
The current Hubble growth rate 1/14.4 per billion years is not in a strict sense a "constant" but it is practically speaking constant enough to provide a convenient timescale. So why not use it? At least it is less arbitrary than "years" the orbit period of this one particular planet in this one little solar system! I can see the point. It also makes the equations nice. Probably we should explore both.

Also we don't KNOW that Einstein's curvature constant Lambda is actually a constant of nature's geometry. We are admittedly in the dark about that. I think what I am doing is testing something out. We don't know that the speed of light is constant, but so far evidence supports the idea. We don't know that Planck's hbar is a constant but so far it seems to be. And so far it looks like there's this curvature constant Lambda in nature. Let's suppose it really is a constant and tailor a bit of expository writing around that---treat it as such IOW.

 

[marcus]

So, Jorrie, the premise here (at least within the confines of this thread) is that there is this growth rate H_∞ whose square is Λ/3, a third of Einstein's curvature constant. And that this growth rate H_∞ is a universal constant of nature which is not going to change "as long as the rivers run to the sea" (as the European colonists and the Native Americans used to write forever in their treaties)

I don't know if you take much interest QG threads in BtSM forum---you may or may not have noticed this development: Several approaches to formulating quantum geometry/gravity use simplexes to represent quanta of geometry (triangles, tetrahedra, pentachora).
It's been found interesting and possibly useful, instead of using flat ones, cut out of flat space, to use simplexes with a constant curvature Lambda. It's conceivable that nature's simplexes are actually not flat, but very slightly curved.

This is one motivation for wanting to try treating Lambda as a geometric constant of nature, which, as such, defines a natural time scale.

 

[Jorrie]

This is one motivation for wanting to try treating Lambda as a geometric constant of nature, which, as such, defines a natural time scale.

I have no problem with Lambda as a constant - both approaches assume that. My concern is more that we may confuse students with this unconventional 'natural time scale', while the Hubble time is well known and documented. AFAIK, it is also the more directly observable one; the value of 1/H_∞ is still more uncertain.

'Pushing the boundaries' may not be appropriate for beginners courses, but more advanced students will certainly find this stimulating. So the 'tandem approach' that you mentioned may be a good option.

 

[marcus]

Just have time to begin this post, company just arrived. Will edit these quotes from earlier posts down to essentials as soon as I have time
==revised excerpts==
I think the matter density estimate just comes from observing the lack of overall curvature and calculating the (matter+radiation only) CRITICAL. As far as concerns me, Lambda is not an energy and does not gravitate the way matter density does. It's just the original Einstein curvature constant. In the model discussed here, the Friedmann equation , for our purposes in this thread, has Lambda on the left hand side as reciprocal square time (the square of a a distance growth rate).
$$H^2 - \frac{\Lambda}{3} = \frac{8\pi G}{3c^2}\rho^*$$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:
$$H^2 - H_\infty^2 = \frac{8\pi G}{3c^2}\rho^*$$This shows that as the matter&radiation density thins out the growth rate H must approach a longterm limit growth rate H_∞. I put an asterisk on the density as a reminder that, as the density of matter and radiation, it doesn't involve a Lambda component. The equation must be satisfied for spatial flatness and so, in that sense, once the current and longterm expansion rates H and H_∞ have been determined, ρ* is critical for spatial flatness.
A few posts back, Wabbit pointed out a useful version of the Friedmann equation (for matter-era on into indefinite future, since there is no "Lambda era") that saves a fair amount of bother, writing density, and constants like π and G etc. I'll write it using the wavestretch factor that Jorrie introduced in the Lightcone calculator. s=1 denotes the present.$$H(s)^2 - H_\infty^2 = (H(1)^2 - H_\infty^2)s^3$$The main novelty in this thread is the way in which Λ, or more precisely T_∞ = 1/H_∞ serves as a time scale. So to proceed we should evaluate the terms in the equation. Obviously the present value of the Hubble constant is 173/144 = 1.201... and its square is 1.443. Obviously, in the timescale we are using, H_∞ = 1 and its square is 1.
The RHS of the Friedmann equation evaluates to:
$$H(s)^2 - H_\infty^2 = (H(1)^2 - H_\infty^2)s^3 = 0.443s^3$$
And in our time scale the Friedmann simplifies to:
$$H(s)^2 - 1 = (H(1)^2 - 1)s^3 = 0.443s^3$$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.
My observation about the number 0.443 is that it doesn't depend on Earth years or any particular human units. Rather it depends on this present moment in universe history. Imagine people on a different planet with a different year, perhaps not even using their planet's years to measure time but some other natural cycle. They could still measure H_{now} and H_\infty and as long as they are our contemporaries they would still find that
$$\frac{H_{now}}{H_\infty} = 1.201...$$ $$(\frac{H_{now}}{H_\infty})^2 = 1.443...$$==end of revised quotes==

 

[marcus]

The distance growth rate (Hubble parameter) as a function of time is given by an unusually simple equation. I'll use x to denote time in units of
1/H_∞ = T_∞ = 17.3 billion years. Exploring the use of that scale is basically what we are doing in this thread. On that scale the present epoch is 0.8 , more precisely 0.797 but 0.8 is close enough. It is just the usual age in billions of years, 13.8, divided by 17.3.
Nature seems to like that scale because in those terms the growth rate at any given time in history is simply the hyperbolic cotangent$$H(x) = \coth(\frac{3}{2} x) $$The scale factor, the size of a generic cosmological distance as it grows over time, is also give by a hyper-trig function, the hyperbolic sine$$u(x) = \sinh^\frac{2}{3}(\frac{3}{2} x) $$For some purposes it's nice to have the scale factor NORMALIZED to equal one at the present epoch. To normalize this we just find out its value at x = 0.8 and divide by that so that the new scale factor a(x) is forced to take the value 1.$$a(x) = u(x)/u(0.8) = u(x)/1.311$$

 

[marcus]

There was an earlier curve that also belongs to this model and that I want to recall. It is the INVERSE of the normalized scale factor a(x) as a function of time x. Remember we are measuring time in "Universe days" or "Udays" of 17.3 billion years so the present is x=0.8.
What the normalized scale factor does , given a time x, is tell you the size of distances back at time x, compared with now. The formula is
$$a(x) = \sinh^\frac{2}{3}(\frac{3}{2}x)/1.311$$The 1.311 is just a normalization factor put into make sure that a(x) = 1 at the present epoch.
We might want to solve that in reverse. For instance some light comes in at the observatory and we put it through a grating to spread out the various colors/wavelengths and we recognize a hot sodium line but the wavelength is THREE TIMES that of the yellow light that hot sodium makes in the lab.
So while that light was traveling to us, wavelengths and distances were stretched three-fold. Distances were 1/3 their present size. So a(x_then)=1/3.
And we wonder WHEN WAS THAT? When was it when that galaxy emitted the light I just put through the grating? What were things like? What was the distance expansion rate back then? How far away is that galaxy now?
So we want to find the time x, for which a(x) = 1/3. Want the reverse or inverse function of the normalized scale factor a(x)
Call it x(a) if you like, the time x that corresponds to a scale factor a.
$$x = \frac{2}{3}\ln(\sqrt{(1.311a)^3} + \sqrt{(1.311a)^3+1})$$It's basically just the inverse of the hyperbolic sine, sinh. Here's the plot. It should look sort of like one half of that earlier ("antelope horns") a(x) picture turned over on its side:

Notice that it gives the right age of the universe. Light with scale factor a = 1 is emitted and received in the present, without any stretch. For a=1 the curve says the time x = 0.8. That is 0.8 Udays or 80% of our time unit 17.3 billion years---namely 13.8. Here's some earlier discussion which gives some other examples.

To recap, in this thread we are examining the standard LambdaCDM cosmic model with some scaling of the time variable that simplifies the formulas, and makes it easy to draw curves: curves showing how the expansion rate (percent per billion years) changes over time, how a sample distance grows over time, and so on.
Time is measured in units of 17.3 billion years. If it makes it easier to remember think of that as a day in the life of the universe: a Uday.
I've been using the variable x to denote time measured in Udays,
Our unnormalized scale factor u(x) = sinh^2/3(1.5x) shows the way a generic distance grows.
Normalizing it to equal 1 at present involves dividing by u(x_now) = 1.311.
The normalized scale factor is denoted a(x) = u(x)/1.311.
The fractional distance growth rate H(x) = u'(x)/u(x) = a'(x)/a(x) = coth(1.5x)

Note that the normalized scale factor a is something we observe. When some light comes in and we see that characteristic wavelengths are twice as long as when they were emitted, we know the light began its journey when a = 0.5, when distances were half the size they are today.

So by the same token the unnormalized u = 1.311a is observed. We just have to measure a and multiply it by 1.311.

Knowing the wavelength enlargement we can immediately calculate the TIME x (measured in Udays) when that light was emitted and started on its way to us. Here is x as a function of the number u.
$$x = \frac{2}{3}\ln(\sqrt{u^3} + \sqrt{u^3+1})$$Since u = 1.311a, we can also write time as a function of the more readily observed number a.$$x = \frac{2}{3}\ln(\sqrt{(1.311a)^3} + \sqrt{(1.311a)^3+1})$$Here's a plot of the time x(a) when the light was emitted as a function of how much smaller the wavelength was at emission compared with the wavelength on arrival.
View attachment 81741
You can see, for example, that if the wavelength at emission was 0.2 of its present size, then that light was emitted at time x = 0.1. since time unit is 17.3 billion years, that means emitted in year 1.73 billion.
On the other hand if the wavelength was originally 0.8 of its present size, which mean it has only been enlarged by a factor of 1.25, then the light was emitted more recently: at time x = 0.6.
Multiplying by our 17.3 billion year time unit we learn it was emitted around year 10.4 billion.

You might think the curve is useless to the right of a = 1 (which denotes the present). But it shows the shape and, come to think of it, tells something about the future. If today we send a flash of light which later is picked up in a distant galaxy by which time its wavelengths are 40% larger---1.4 of their present size---what year will that be?
You can see from the curve that it will be time x=1.1
which means the year will be 17.3+1.73=19.03---about year 19 billion.

And the curve also tells how long the universe has been expanding, just evaluate the time at a=1.

 

[marcus]

So just for fun (or for hands-on practice) let's do an example. I realize I didn't put in the curve for a(x) yet, I only put the curve for the Unnormalized scale factor u(x), where we didn't divide by 1.311 to make it equal one at the present day. So here is the normalized scale factor a(x). This is the most basic curve of the model. It shows how the size of the universe (or a sample distance) changes over time:

You can see how a(x) equals one at the present time (0.8 Uday).
I think of this as the "antelope horns" picture. : ^)
So imagine we are at an observatory and some light comes in that has been wave-stretched 5-fold. What does it tell us? It comes from a time when distances were 1/5 their present size, so 0.2 on the vertical scale.
When was that? We can see from the red curve that the time was x = 0.1, one TENTH of a Uday, one tenth of 17.3 billion years. But we are using the Uday time scale so we don't bother to convert to billions of Earth years, we just think of it as time x = 0.1

What was the expansion rate H(x) back then? We just put x=0.1 into our tanh formula (that gives the Hubble time, one over that gives the Hubble rate)
tanh(1.5*0.1)
Google says:
tanh(1.5 * 0.1) = 0.14888503362
IN OTHER WORDS GROWTH BACK THEN WAS ABOUT SEVEN TIMES THE LONGTERM GROWTH RATE
The eventual longterm growth rate is a good unit to think in terms of. We are used to it by now in many guises: one per 17.3 billion years, 1/173 percent per million years. 1/17.3 per billion years, about 0.06 per billion years. It is the eventual growth rate the universe is heading towards. and 1/0.1488 is about 7.
So when that light was emitted which we just analyzed, the growth rate was 7 times eventual.

If you want Hubble time in billions of Earth years, just multiply the 0.1488.. by 17.3 and you get 2.5757...billion years. I would read the Hubble growth rate from that as 1/26 of a percent per million years. It's somewhat a matter of taste and habit what you find the most comfortable way to express and think about the Hubble distance growth parameter.

 

[marcus]

How far is that galaxy from us now? We were thinking of an example were some light came in with stretch factor 5, from a galaxy back in the time when distances were 1/5 what they are today. And we used our model to say WHEN that was (time x = 0.1, or year 1.73 billion)
and also used our model to say what the growth rate H(x) was, back then (about 7 times the universe's longterm rate).

So suppose now we ask how far that galaxy is now? How far has this flash of light traveled from its source? (on its own and aided by expansion).

This is a job for the "number empire" integrator. We integrate out to the incoming stretch factor s=5.
$$\int_1^5 \frac{ds}{\sqrt{.443s^3 + 1}}$$
You google "number empire definite" and you get
http://www.numberempire.com/definiteintegralcalculator.php
and you just type in the integrand and the limits 1 and 5. It's easy.
The integrand is (0.443*s^3+1)^(-1/2). You press "compute"
It gives 1.38, that is the distance to the galaxy in light Udays our distance unit when we use this time scale. So if you like, mulitiply the 1.38 by 17.3 billion light years to get it in those Earth year terms.

And we can tell right away what the distance to the galaxy was THEN, when the light was emitted and started on its way to us. Distances were 1/5 their present size back then so divide 1.38 by 5.

Basically the light you receive tells you all these things by showing you how much it was stretched.