What can you learn about cosmology from Google calculator?

[wabbit]

"divide by the square root"...
Yes I have "multiplied" here, obvioulsy I made a silly mistake at some step
So, 200 bn ly, not 1. Seems more like it.

 

[wabbit]

The (hopefully correct now) formula I got for the curvature radius K is $$K=\frac{R_H}{\sqrt{-\Omega_k}}=\frac{R_H}{\sqrt{\Omega-1}}$$
This gives a way to derive K from published LCDM parameters, but it is perverse in a way as it hides the dependency of $\Omega$ on time, which could be explicited in an epoch where the matter/energy balance is stable, as is the case in this thread. For K, this is simply$$K(T)=K_0 a(T)$$and for $\Omega$ $$\Omega_k(T)=\frac{\Omega_{k,0}}{a(T)^2},\quad \Omega(T)=1-\frac{1-\Omega_{,0}}{a(T)^2}$$
As noted in another thread, I am using http://casa.colorado.edu/~ajsh/phys5770_08/frw.pdf , which covers non-zero K and provides a host of formulas while remaining easy to read. All errors here are strictly my own contribution however. And of course when we get an imaginary radius (unlikely from $\Omega=1.005\pm0.006$), this means we have a hyperbolic space, not a sphere.
The central value gives a 3-sphere of radius ~200 bn ly, and the upper value yields a minimum radius of ~130 bn ly.

 

[wabbit]

Oops I surreptitiously introduced an error in the a(T) expression (missing a 3/2). Corrected below, using your last formulation

Thanks Wabbit. It gets even nicer

$$H(T)=H_\infty \coth(\frac{3}{2} H_\infty T)$$
$$R(T)=R_\infty \tanh(\frac{3}{2}\frac{ c T}{R_\infty})$$
$$a(T)=\left(\frac{\sinh(\frac{3}{2}H_\infty T)}{\sinh(\frac{3}{2}H_\infty T_0)}\right)^{2/3}\simeq0.7636\cdot\sinh(\frac{3}{2}H_\infty T)^{2/3}$$

Silly question here: $\frac{2}{3H_\infty}\simeq11.5 bn yr$. What is that time? Everything depends on a ratio of T to this unit which is yet another formulation of the CC, as a time scale...

 

[marcus]

...
the 2015 Planck report on cosmology parameters has a confidence interval like that, from which one could estimate the smallest the universe could be, with 95% confidence. Assuming an S³ hypersphere, I mean. smallest radius of curvature, smallest circumference. Of course might be much larger but a lower bound is nice to have.

I looked it up:
http://arxiv.org/abs/1502.01589
Planck 2015 results. XIII. Cosmological parameters

==from the long (PDF) version of the abstract==
The spatial curvature of our Universe is found to be very close to zero with |Ω_K | < 0.005.
====
There is more detail in Table 5 on page 31
Planck by itself (column labeled TT) actually has a central value of Ω_K = -0.052
and the 95% confidence interval is all on the negative side(!) suggesting positive curvature and S³
Ω_K = -0.052^0.049_-0.055 translates to [-0.107, -0.003]

But then they have a column labeled TT+lensing+external
where external data is from other studies tagged BAO, JLA, and H₀
With correction for lensing of ancient light by intervening clusters of matter, and inclusion of data from the other studies, the central value is very close to zero. Ω_K = -0.0001^+0.0054_-0.0052
======

Then there is DISCUSSION of their results, and what they get by merging data with other studies, starting on page 37, in section 6.2.4 specifically devoted to curvature. They end up confirming what they said in the abstract which is basically [-0.005, +0.005].

I admit to being emotionally and philosophically (?) biased in favor of slight positive curvature and S³ so I'm tempted to take note of the pure Planck central value Ω_K = -0.052 .
14.4/(0.052)^(1/2) = 63
2pi*14.4/(0.052)^(1/2) = 397

 

[wabbit]

Thanks, I had tried to look there but that is one dense paper... And they provide so many different estimates too, it's hard to tell which one if any is "the latest Planck estimate". But using that value of $-0.0001\pm0.005$, this is now saying $$|K|\geq 200 \text{ Gly, sign unknown}$$It does seem to be homing in towards 0, and as discussed in another thread not much sign of muliply connected topology either... the good news I suppose, is that no observation will ever prove "exactly 0" (you think I'm stubborn?)

 

[marcus]

About your question in post #63, I think the 3/2 is only there because we are working in the matter-era and ignoring radiation part of energy density, which does OK after year 1 million or so. I think in the radiation-era the number would be 4/2. So by rights there should not be any special significance to 2/3 of the eventual Hubble time.
or to 3/2 of H_∞ the eventual Hubble rate. What matters is H_∞ and the 3/2 just reflects a convenient assumption about the physical makeup.
But I don't know . BTW I think Andy Hamilton is one of the best expositors of this material. Astro/cosmo GR Friedmann black holes and all. He's gifted and he seems to care. You found that University of Colorado class handout essay of his. The link should be posted in plain sight somewhere.
http://casa.colorado.edu/~ajsh/phys5770_08/frw.pdf

 

[wabbit]

I didn't know about him, but that was the best doc I found from googling. I can add it to the astro etc. bibilo thread, would that be a good place ? Also, to not hijack this thread too much I'll post my questions and comments about flatness in the thread I started about that.

And I think you're right, it's only $1/H_\infty$ that is meaningful, the 3/2 is a special case. Just checked post#46, it is 2 in a radiation-dominated epoch, as you suggested. Both appear to be of the form $a(T)\sim \sinh(\frac{1}{p}H_\infty T)^p$

 

[marcus]

...
$$H(T)=H_\infty \coth(\frac{3}{2} H_\infty T)$$
$$R(T)=R_\infty \tanh(\frac{3}{2}\frac{ c T}{R_\infty})$$
$$a(T)=\left(\frac{\sinh(\frac{3}{2}H_\infty T)}{\sinh(\frac{3}{2}H_\infty T_0)}\right)^{2/3}\simeq0.7636\cdot\sinh(\frac{3}{2}H_\infty T)^{2/3}$$

...

I wanted to check these using google calculator (the thread is partly a hat-tip to that scientific calculator for everybody)
and it does not have "coth" so we can use 1/tanh and write it this way:
H(T) = 1.83 attohertz/tanh(3/2*1.83 attohertz*T)
Try T = 13.787 billion years.
1.83 attohertz/tanh(3/2*1.83 attohertz*13.787 billion years)
gives the right thing. Basically it says present-day Hubble rate is 2.20 attohertz.

And the second equation we can write:
R(T) = 17.3 billion light years*tanh(3/2*1.83 attohertz*T)
again try T = 13.787 billion years.
17.3 billion light years*tanh(3/2*1.83 attohertz*13.787 billion years)
as one expects, the right thing again. Basically 14.4 billion light years.

And the third:
a(T) = (sinh(3/2*1.83 attohertz*T)/sinh(3/2*1.83 attohertz*T₀))^(2/3)
This time let's try T = 5.864 billion years (it's clearly normalized to have a=1 at present)
(sinh(3/2*1.83 attohertz*5.864 billion years)/sinh(3/2*1.83 attohertz*13.787 billion years))^(2/3)
Beautiful. 0.5000...

 

[wabbit]

the thread is partly a hat-tip to that scientific calculator for everybody

And a deserved one at that, this thing is quite impressive.

 

[marcus]

Agreed, part of the appeal is the "for everybody". For a given application there must surely be better-adapted online scientific calculators but everybody knows google, you don't have to find a link and go to any special place and adapt to slightly different conventions.

Here's a sample Jorrie table that could be used to check our (matter-era) googlable formulas.
{\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) &amp; R_{\infty} (Gly) &amp; S_{eq} &amp; H_{0} &amp; \Omega_\Lambda &amp; \Omega_m\\ \hline 14.4&amp;17.3&amp;3400&amp;67.9&amp;0.693&amp;0.307\\ \hline \end{array}} {\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&amp;S&amp;T (Gy)&amp;R (Gly) \\ \hline 0.167&amp;6.000&amp;1.1738&amp;1.76\\ \hline 0.200&amp;5.000&amp;1.5417&amp;2.30\\ \hline 0.250&amp;4.000&amp;2.1494&amp;3.19\\ \hline 0.333&amp;3.000&amp;3.2851&amp;4.80\\ \hline 0.500&amp;2.000&amp;5.8636&amp;8.11\\ \hline 1.000&amp;1.000&amp;13.7872&amp;14.40\\ \hline \end{array}}
Lightcone calculator (I'm also a fan of that) has this box you can check to get linear steps in S.

17.3 billion light years*tanh(3/2*1.83 attohertz*1.1738 billion years) ⇒ 1.75
17.3 billion light years*tanh(3/2*1.83 attohertz*1.5417 billion years) ⇒ 2.30
17.3 billion light years*tanh(3/2*1.83 attohertz*2.1494 billion years) ⇒ 3.18
17.3 billion light years*tanh(3/2*1.83 attohertz*3.2851 billion years) ⇒ 4.79
17.3 billion light years*tanh(3/2*1.83 attohertz*5.8636 billion years) ⇒ 8.10
17.3 billion light years*tanh(3/2*1.83 attohertz*13.787 billion years) ⇒ 14.39(sinh(3/2*1.83 attohertz*1.1738 billion years)/sinh(3/2*1.83 attohertz*13.787 billion years))^(2/3) ⇒ 0.167
(sinh(3/2*1.83 attohertz*1.5417 billion years)/sinh(3/2*1.83 attohertz*13.787 billion years))^(2/3) ⇒ 0.200
(sinh(3/2*1.83 attohertz*2.1494 billion years)/sinh(3/2*1.83 attohertz*13.787 billion years))^(2/3) ⇒
(sinh(3/2*1.83 attohertz*3.2851 billion years)/sinh(3/2*1.83 attohertz*13.787 billion years))^(2/3) ⇒
(sinh(3/2*1.83 attohertz*5.8636 billion years)/sinh(3/2*1.83 attohertz*13.787 billion years))^(2/3) ⇒
(sinh(3/2*1.83 attohertz*13.787 billion years)/sinh(3/2*1.83 attohertz*13.787 billion years))^(2/3) ⇒

 

[wabbit]

You were wondering earlier if we might extend that before the matter era. Perhaps this might work in two steps, if the energy-matter transition is short (it is, isn't it? Something like we reach a critical temperature and there's phase transition? Or is it just p=2/3 gradually winning over p=1/2?):
- find the matter-era parameters at the transition time;
- extrapolate backward using the p=1/2 formulas instead of p=2/3.
But it would still mean we just have two sets of formulas, one for each era, glued together at the transition.

Edit: nah it seems to be more the competing power thing. Might still work though, with good fits outside a time interval around the transition. 1/2 and 2/3 aren't that far apart so that should help too.
Also the transition time (equal energy matter balance) seems to be 47000 years only. (wikipedia) so the 2/3 model might work well before the ~My mentionned before.

Edit: google says a(47000y)=0.00019 and H(47000y)=0.45 picohertz or 450,000 attohertz

 

[marcus]

1/(17.3 billion years) ⇒ 1.8317... attohertz
The Lightcone calculator uses 17.3 billion light years as the Lambda representative default. If we want to get closer to Lightcone numbers maybe we should always use a closer H_∞ like 1.832

17.3 billion light years*tanh(3/2*1.832 attohertz*1.1738 billion years) ⇒ 1.75
17.3 billion light years*tanh(3/2*1.832 attohertz*1.5417 billion years) ⇒ 2.30
17.3 billion light years*tanh(3/2*1.832 attohertz*2.1494 billion years) ⇒ 3.19
17.3 billion light years*tanh(3/2*1.832 attohertz*3.2851 billion years) ⇒ 4.80
17.3 billion light years*tanh(3/2*1.832 attohertz*5.8636 billion years) ⇒ 8.11
17.3 billion light years*tanh(3/2*1.832 attohertz*13.787 billion years) ⇒ 14.40

Rounded to two decimal places it looks OK. Better with 1.832 though. And notice that at the earlier time of year 1.17 billion it is off in the second decimal place. Should say 1.76.

I think there is no possibilities of a satisfactory closed formula, taking account of the changing makeup of the energy density. We could ask Jorrie, I think he like everybody else resorts to numerical integration.
Number crunching in small S steps, or time steps.

Maybe Cai&W-E will follow up with more numerical work. I think they don't anything detailed from LQC.
Just the general idea of the bounce based on Ashtekar's modification of Friedmann equation.
This is also equation (1) of the Cai&W-E paper:
H(T)² = [Friedmann constant]ρ(1 - ρ/ρ_c)
Where the critical ρ_c is an extreme density comparable to the Planck density. Ashtekar et al often use 0.4ρ_Planck.
The Cai&W-E analysis is basically classical both contracting and expanding, with an extremely brief intervening interval when (1 - ρ/ρ_c) matters. You can see how the classical passages could be symmetric because the lefthand side is an H². The Friedmann equation can give very rapid contraction (negative H) just as it can give expansion (positive H), at high density.

At very high density, around bounce, one might suppose Lambda (or H_∞) to be negligible. One might suppose H(T) to be on the order of the Planck frequency, one over Planck time. But there is the brief interval when it goes from very negative to very positive. so it has to cross zero. Should Λ be included?
Should one use this form?

H(T)² - Λ/3 = [Friedmann constant]ρ(1 - ρ/ρ_c)
or equivalently
H(T)² - H_∞² = [Friedmann constant]ρ(1 - ρ/ρ_c)

 

[wabbit]

I think there is no possibilities of a satisfactory closed formula, taking account of the changing makeup of the energy density.

You are right. I just checked this, the solution is of the form $a=F^{-1}(T)$, where F is an elliptic integral.
This simplifies to sine or arcsine only when there the two dominant terms are CC and one density (CC+matter or radiation or curvature). I don't think it even simplifies this way for the radiation era (the dominant terms being radiation + matter), one has to drop to one term only then and we get just a power law for early time.

Also, the approximation I was proposing doesn't work for many reasons, the first one being there is no arcsine law in the radiation era as far as I can tell.

 

[marcus]

Here are Wabbit's formulas from a few posts back. These are approximate because strictly speaking they only apply to the matter-era where the density consists primarily of matter. They are solutions of the Friedmann equation which relates distance growth rate H(T) to energy density ρ(T).
a(T) keeps track of the "size" of a generic distance as it changes over time (normalized so a(present) = 1). The reciprocal of H is a time and that time multiplied by c is a distance, the Hubble radius, denoted here by R. It's a convenient handle on the growth rate H.
Google calculator doesn't have coth, so I rewrote the first formula using the equivalent 1/tanh:

$$H(T)=H_\infty \coth(\frac{3}{2} H_\infty T)$$
$$H(T)=H_\infty / \tanh(\frac{3}{2} H_\infty T)$$
$$R(T)=R_\infty \tanh(\frac{3}{2}\frac{ c T}{R_\infty})$$
$$a(T)=\left(\frac{\sinh(\frac{3}{2}H_\infty T)}{\sinh(\frac{3}{2}H_\infty T_0)}\right)^{2/3}\simeq0.7636\cdot\sinh(\frac{3}{2}H_\infty T)^{2/3}$$

 

[marcus]

... Only when there are only the cosmological constant and one other density contribution (or perhaps no CC and two other terms) does it reduce to sine or arcsine depending on sign (hey this rhymes )·
...

But the hyperbolic cotangent formula for H(T) is very evocative anyway, and the formulas are useful in the matter era. You mentioned a natural time-scale. That is a nice idea. Right now H_∞T = about 0.8
Let us temporarily call that number 0.8 the absolute time, or absolute age of universe expansion.
Then the formula says to multiply by 3/2 and take the 1/tanh, or coth
Multiply by3/2 and you get 1.2
And it just happens that coth(1.2) ≈ 1.2

Now your formula says to multiply that by H_∞ which is 1.832 attohz.
And when you do that you get the present-day H(now) = 1.832x1.2 = 2.20
======================
Now look at the graph of coth the hyperbolic cotangent. It describes a bounce universe with the origin of absolute time, X = 0, at the bounce. depicted as a minus-to-plus ∞ jump in the distance growth rate H(T).
Coming in on negative X it plunges down. H(T) becomes very negative. Faster and faster negative growth. that is the collapse to extreme density.

Then something, the quantum effects that kick in at extreme density, avoids a discontinuity and starts H(T) off at a very high value on the T positive side.
From which it starts to decline swiftly and then, around "absolute time" of order unity, it gradually levels out at value 1, so that the eventual H is H_∞

It's basically the picture that Cai and Wilson-Ewing are studying in their recent paper:

http://arxiv.org/abs/1412.2914
A ΛCDM bounce scenario
Yi-Fu Cai, Edward Wilson-Ewing

but they look in detail at what happens close in the origin that might replace the minus-to-plus discontinuity. Here is one of the figures from their paper. They also have plots of the scale factor (in conformal time which linearizes the scale factor) and of the densities.

 

[wabbit]

What you're saying about cross-bounce reminds me of a question I had about BH and Planck stars... I'll ask it in another thread later so as not to hijack this one:)

 

[wabbit]

I had another look at the elliptic integrals involved in solving the FLRW equations, and it turns out that in fact, for flat FLRW, there are indeed two formulas, one for early times and the other for late times, and with an overlap where they can be glued together. These are much nicer using a as parameter than T though.

I am not too sure about the exact results - units and other errors may have cropped in, but this is what I got for the radiation and early matter era (up to a ~ 0.1). $c=1$ here. $$\rho=\rho_{\lambda}+\frac{\rho_m}{a^3}+ \frac{\rho_r}{a^4};\quad H(a)= \sqrt{\frac{8\pi G}{3}\rho}$$
For $a\ll(\frac{\rho_m}{ \rho_{\lambda}})^{1/3},$
$$ \quad \sqrt{\frac{3}{8\pi G}}T(a)=\frac{4}{3} \frac{\rho_r^{3/2}}{\rho_m^2}\left(1- \sqrt{1+a\frac{\rho_m}{\rho_r}} \cdot\left(1-\frac{a}{2} \frac{\rho_m}{\rho_r}\right)\right)$$
$\simeq \frac{1}{2}\frac{a^2} {\sqrt{\rho_r}}$ when $a\ll\frac{\rho_r}{\rho_m}$, and $\simeq\frac{2}{3}\frac{a^{3/2}} {\sqrt{\rho_m}}$ when $a\gg\frac{\rho_r}{\rho_m}$

 

[marcus]

Recalling a couple of equations, to have them handy:
$$H(T)=H_\infty / \tanh(\frac{3}{2} H_\infty T)$$
$$a(T)=\left(\frac{\sinh(\frac{3}{2}H_\infty T)}{\sinh(\frac{3}{2}H_\infty T_0)}\right)^{2/3}\simeq0.7636\cdot\sinh(\frac{3}{2}H_\infty T)^{2/3}$$

... You mentioned a natural time-scale. That is a nice idea. Right now H_∞T = about 0.8
Let us temporarily call that number 0.8 the absolute time, or absolute age of universe expansion.
Then the formula says to multiply by 3/2 and take the 1/tanh, or coth
...

I like the idea of provisionally defining H_∞T_now = 0.8 as expansion age in absolute terms.
then I can plot our a(T) formula with a raw sinh(1.5x) without struggling to relabel the x axis. And I won't even bother to normalize it to equal one at present. Forget the "0.7636" out front. If the universe thinks it is 1.3 then let it be 1.3 at present, I just want the shape..
$$a(T) = |\sinh(\frac{3}{2}H_\infty T)|^{2/3}$$

For the moment the time is x = 0.8 and the unnormalized scale factor is y = 1.3. I want to see the bounce. The bounce is inherent in our formula. This is the raw |(sinh(1.5x))|^2/3
Btw, I think I see the inflection point in a(T), when acceleration starts. around x = 0.45
That would be about right because in our usual years measure it starts around year 8 billion
0.45 is to 0.8 (the present on this absolute scale) as 8 billion is to 13.787 more or less

This is supposed to go with the Cai&W-E modified cotangent figure a couple of posts back that shows H(T)
How it is negative (i.e. contraction ) before the bounce and goes VERY negative right before zero,
and then is positive after the bounce and is VERY positive immediately after zero. This corresponds to the slope of the a(T) curve here.

https://www.desmos.com/calculator does the graph and then "command-shift-4" makes a screen-shot

 

[wabbit]

Nice. I was going to nitpick about the behaviour near 0, which is in power 1/2 rather than 2/3, but I checked and it seems the lambda-matter model works well (within ~1 % ) down to a ~ 0.03 so even if you used numerical integration or a two-step formula, the difference would be completely invisible on the chart.

I think the inflection in a(T) should be at about the time of matter-CC balance, a~0.75,
Now I believe that time scale we were wondering about is closely related to this matter-lambda balance time, and is given by (some constant of order unity times) $a=(\frac{\rho_m}{\rho_{\lambda}})^{1/3}$

 

[marcus]

Wabbit, I think you may already have noticed this. We really only need the one equation, say for the unnormalized scale factor a(x) as a function of x = H_∞T. Making a change of time variable suggested by something you said earlier.
a(x) = sinh(1.5 x)^2/3

now let's differentiate and find H(x) = a'/a

a'(x) = (2/3)sinh(1.5 x)^-1/31.5 cosh(1.5x) by two simple applications of the chain rule.
2/3*1.5=1

now dividing by a(x) we get
a'(x)/a(x) = cosh(1.5x)/(sinh(1.5x)^1/3sinh(1.5x)^2/3 ) =cosh(1.5x)/sinh(1.5x) = coth(1.5x)
So the formula for H(x) = coth (1.5x) is an easy consequence of the a(x) formula.

 

[marcus]

So we seem to have boiled the universe down to a single equation for the (unnormalized) scale factor
$$a(T)^3 = \sinh^2(\frac{3}{2}H_\infty T)$$

Writing it that way takes fewer parentheses and has a reminder that density (dark/ordinary matter and radiation) goes as inverse volume, i.e. as inverse cube of scale. When the universe is mainly full of radiation there would be a fourth power on the left. It would say a⁴ instead of a³. But that radiation era is very brief and hardly shows up in the broad outline picture.

And we can let $x = H_\infty T$ and adopt the universe's time scale instead of using our billions of years, so that the present is x = 0.8.

$$a(x)^3 = \sinh^2(\frac{3}{2}x)$$

If you differentiate this equation d/dx, and then divide by the same equation, you get one for a'/a = H and find that
$$ H(x) = \coth(\frac{3}{2}x)$$

Another thing that would be nice to derive would be an equation for D_now(x_em)
that is the proper distance now (i.e. the comoving distance) of a galaxy whose light was emitted at time x_em and received today x_now=0.8.

Essentially it is the integral of cdx over a(x) between x_em and x_now
because cdx is the little interval that the light actually travels and dividing by a(x) = sinh^2/3(1.5x) shows how much it has expanded. the integral should be multiplied by a(x_now) = 0.8 and it should be between x_em and x_now

 

[wabbit]

Another thing is, after playing with the FRLW diff equation, I am now a convert to the use f a/S/z as a parameter, rather than T... though here both are equally simple, in general T(a) is much easier than a(T). No I'm not suggesting to redo the whole thread that way

 

[wabbit]

About that time scale...there are others, all related to the ratios of matter, energy, and lambda densities. Without the formulas here, a0 being the end of inflation, they define the successive eras.
Below, which components are sufficient in the FLRW model at each period (when only one is listed you can also of course add any second one if you like, these are the transition eras where two 2-component models overlap):

With 10 % precision (1 % would work too, too many choices...)

a0 - 0.000 03 radiation (30-60 e-folds)
0.000 03 - 0.003 radiation + matter (5 e-folds)
0.003 - 0.3 matter (7 e-folds)
0.3 - 3 matter + lambda (2 e-folds) <------- this is us !
>3 lambda

Maybe I'll make a new thread about these, several interesting things happening there.

 

[marcus]

Another thing is, after playing with the FRLW diff equation, I am now a convert to the use f a/S/z as a parameter, rather than T... though here both are equally simple, in general T(a) is much easier than a(T). No I'm not suggesting to redo the whole thread that way

I had a similar perception. I like the way it works out using S.
Newcomers normally think in terms of time and may find time-evolution formulas more cogent---immediately understandable, convincing.
I am going to continue for a while trying out an expository approach that uses a simplified time x = H_∞T

$$D_{now}(x_{em}) = a(x_{now})\int_{x_{em}}^{x_{now}}\frac{cdx}{sinh^{2/3}(1.5x)}$$

x_now = H_∞T = 1.832 attohertz*13.787 billion years =0.797...≈ 0.8
a(x_now) = (sinh(1.5*0.797)^(2/3) = 1.311 ≈ 1.3

 

[wabbit]

Yes I agree with that, when you start reading about these, expressions in terms of z seem bizarre - and even in term of a, unnatural. It's only with familiarity with the equations that these become natural. For me the switch is very recent (like, two posts ago )

 

[wabbit]

Now I believe that time scale we were wondering about is closely related to this matter-lambda balance time

Bit late for a correction but I was talking rubbish. $1/H\infty$ is of course only an expression of the CC and has nothing to do with matter-lambda balance, sorry about that.

 

[wabbit]

$$D_{now}(x_{em}) = a(x_{now})\int_{x_{em}}^{x_{now}}\frac{cdx}{sinh^{2/3}(1.5x)}$$

OK let's see if I can interpret this distance correctly. Locally it says how much light has traveled in the frame of a comoving observer - say, our frame. So I can say it is the total distance light has traveled from emission. I would conclude that it is the distance in our (comoving, intertial) frame to the source of the light (or to where it was then, but let's just say the source is also comoving).

The equivalent expression in terms of a is $$ D_{now}(a_{em})=c\int_{a_{em}}^{a_{now}}H(a)da $$
Either way I don't think we can escape numerical integration for that one.

 

[marcus]

...
Either way I don't think we can escape numerical integration for that one.

I think you are right. BTW Lineweaver's 2003 tutorial "Inflation and the CMB" was an important article for me and it has a kind of iconic plot of a(T), I think it might be figure 14
http://ned.ipac.caltech.edu/level5/March03/Lineweaver/Figures/figure14.jpg

this is the Desmos plot of |sinh(1.5x)|^2/3

https://www.desmos.com/calculator

 

[wabbit]

I don't recall reading it, is it this one ?
http://arxiv.org/abs/astro-ph/0305179
Inflation and the Cosmic Microwave Background
Charles H. Lineweaver (School of Physics, University of New South Wales, Sydney, Australia)
(Submitted on 12 May 2003)

Edit : had a quick look, seems pretty cool, I ll set it aside for a good read.

 

[marcus]

Yeah. http://arxiv.org/abs/astro-ph/0305179 Perhaps in certain ways it is dated. Ashtetkar's suggestion of H² = const. ρ(1-ρ/ρ_crit) which changes the
H(x) = coth(x) into a continuous bounded function (resolves singularity), was later, 2007, 2008 I think.

Cai&W-E was later. maybe inflation is not needed and so on. But Lineweaver tutorial is still good in a lot of ways I think.
In case others are reading thread, the plot of the "coth-like function" is from Cai&Wilson-Ewing Dec 2014 "LambdaCDM bounce" paper.