What can you learn about cosmology from Google calculator?

[marcus]

$$S(T) = (1.5/sinh(\frac{3}{2}H_\infty T))^{2/3}$$
So let's do a simple example. Some light is emitted in year 2.15 billion, in our direction, and arrives today. What is the stretch factor S?
Standard model parameters so H_∞ = 1.83 attohertz. What do we paste into google?

( 1.5/sinh(3/2*1.83attohertz*2.15 billion years) )^(2/3)

Google calculator gives back 4.00

So S = 4 and redshift z = S-1 = 3, and the scale factor a = 1/S = 0.25. Back in year 2.15 billion, distances were 25% their present size.
Maybe that is the form of S(T) the calculator likes.

 

[wabbit]

I'm trying to explore how doing some hands-on calculation with google or other calculator could help get comfortable with quantitative cosmology

Watching you do so certainly does, even though I have so far been too lazy to try it myself

 

[marcus]

$$S(T) = (1.5/sinh(\frac{3}{2}H_\infty T))^{2/3}$$

For aesthetic reasons let's convert this to a formula for a(T), the scale factor at some given cosmic time T.

$$a(T) = (sinh(\frac{3}{2}H_\infty T)/1.5)^{2/3}$$Since that number 1.5 is actually $(H_0^2/H_\infty^2 - 1)^{-1/2}$ = (2.20²/1.83² - 1)^-1/2, dividing by 1.5^2/3 is really the same as multiplying by $(H_0^2/H_\infty^2 - 1)^{1/3}$ = (2.20²/1.83² - 1)^1/3 = ((2.20/1.83)^2 - 1)^(1/3)
which when I paste it in, google says is 0.7636

$$a(T) = (H_0^2/H_\infty^2 - 1)^{1/3}(sinh(\frac{3}{2}H_\infty T))^{2/3}$$

Once we have fixed the two main model parameters, this coefficient out front is just 0.76 so we can try out a practical formula for a(T)
a(T) = 0.76(sinh(3/2*1.83 attohertz*T)^(2/3)
Let's try that out for T = 2.15 billion years.
a(T) = 0.76(sinh(3/2*1.83 attohertz*2.15 billion years)^(2/3)
Good.
0.76(sinh(3/2*1.83 attohertz*2.15 billion years)^(2/3) gives back 0.2488..., close enough to a=0.25

If I put in 0.7636 it comes out 0.249989

 

[marcus]

What I'm finding is that this quantity H_∞ which is a version of the cosmological constant is entering in everywhere as a distance scale or time scale, or distance growth rate scale, in other words a spacetime curvature scale rather than, say, an energy scale.
This H_∞ the eventual Hubble rate, or aliases like 17.3 billion years the eventual Hubble time.
Or the square of H_∞ which is actually equal to the cosmological constant itself, divided by 3.
And is a curvature.
All these various forms of Lambda keep coming in as basic time distance and geometric scales.

And google seems to like to call Lambda "10 square attohertz"
or if you want a little more precision "10.07 square attohertz"

(Λ/3)^1/2 = (10.07/3)^1/2 attohz ≈ 1.83 attohz, the eventual growth rate

Actually 1.832...
I should probably be using 1.832 attohz and then I would get more exact agreement with what Jorrie's calculator says using its basic default parameters 14.4 and 17.3 billion lightyears. But the agreement is already pretty good.

 

[marcus]

A better value of the coefficient 0.76 is (from a couple of posts back)
$(H_0^2/H_\infty^2 - 1)^{1/3}$ = (2.20²/1.83² - 1)^1/3 = ((2.20/1.83)^2 - 1)^(1/3) = 0.7636...

Let's use it, for year 13.787 billion:
a(T_now) = a(13.787 billion years) = 0.7636(sinh(3/2*1.83 attohertz*13.787 billion years)^(2/3) = 1.0002...

0.7636 * (sinh((3 / 2) * (1.83 attohertz) * (13.787 billion years))^(2 / 3)) = 1.00021314875

 

[marcus]

I'm trying to explore how doing some hands-on calculation with google or other calculator could help get comfortable with quantitative cosmology ...

Watching you do so certainly does, even though I have so far been too lazy to try it myself

If watching does it, that's efficient. Sometimes what is called laziness can be a sign of creative intelligence.

 

[marcus]

Since we turned a page, maybe I should try to summarize some main points.
Sometimes people come to PF and ask questions like "how do scientists calculate the size of the universe at anyone given time?" Quarlep was basically asking that just recently: basically "how do you plot the growth?"
Quantitative questions: how does the standard cosmic model WORK. Then you can ask how is it justified, how well does it fit the data etc.
I think Jorrie's "Lightcone" calculator is great. And it's basically easy to use once you get acquainted with the two basic parameters R₀ and R_∞ the now and future Hubble radii, and learn how to control the range of stretch S-values you want the table to cover.
The calculator embodies the basic (spatial flat) standard model in a hands-on way and gives you past and future cosmic histories that you can experiment with by varying the two parameters.

So I want to see how an average person could IMITATE that and get standard cosmic model numbers (reasonably close) on their own accord using the google calculator. There are other paths to understanding--this is not the only way to go. It depends on what you find fun and interesting. You could go top down--study GR, see how Friedmann equation is derived from GR, a simplified version of GR, see how Friedmann equation is solved numerically turns out tables like Jorrie's.

ANYWAY for better or worse, here's the path we took and what we did so far in this thread.
We want to imitate Jorrie's numbers so we start with the basic parameters 14.4 and 17.3 billion light years, corresponding to Hubble times 1/H₀ and 1/H_∞ of 14.4 and 17.3 billion years.

And we put 1/14.4 billion years, and 1/17.3 billion years into the google box and out hop two Hubble growth rates:
H₀ = 2.20 attohz
H_∞ = 1.83 attohz
Then we take a look at the Friedmann equation
H(t)² - H_∞² = [Friedmann constant] ρ(t)
where ρ is the combined (mass-equivalent) energy density of dark and ordinary matter plus radiation, which google likes to quantify in pascals and which turns out to be 0.24 nanopascal at present. Thinning out as distances and volumes grow, of course.
The lefthand side is measured in square attohz. It turns out that the cosmological curvature constant Lambda is actually 10.07 square attohz, and by definition Λ = 3H_∞².

The Friedmann constant 8πG/(3c²) = 6.22 square attohz per nanopascal
converts between the energy density on the righthand and the reduced square growth rate on the left.

The equation can be solved using what we know about the behavior of energy density during expansion. (Though the solution is made number-crunchy by the fact that matter and radiation densities attenuate differently with expansion, one as the cube and one as the fourth power of distance. No simple formula if you include a substantial amount of radiation.)

Since after the first million years or so, radiation is a small part of the total, we found simplified closed formulas for how the universe size grows, that work as long as matter >> radiation and approximate essential parts of Jorrie's tables. Here is one:
$$a(T) = (H_0^2/H_\infty^2 - 1)^{1/3}(sinh(\frac{3}{2}H_\infty T))^{2/3}$$

As long as we are using 2.20 and 1.83 attohz for the two main model parameters, we can just calculate a value for the first term once and for all and have a more convenient formula:
$$a(T) = 0.7636 (sinh(\frac{3}{2}H_\infty T))^{2/3}$$

I think 0.76 would be fine for most purposes.
So that shows how the "size of the universe" grows with time. a(T) is the size of a generic distance normalized to equal 1 at present.

 

[wabbit]

If I may, just one question. Your formula above has $a(T)\propto(\sinh(kT))^{2/3}$; I saw elsewhere (http://grwiki.physics.ncsu.edu/wiki/FLRW) for the flat FLRW case $a(T)\propto(\sinh(kT))^{1/2}$ : what is the different exponent due to?

Edit: the relation between LCDM and FLRW isn't that clear to me. I thought LCDM could be read as a special case of FLRW where a(T) is derived from the matter/energy/DE densities.

 

[marcus]

If you may
you may. I'll go look at that.
I think LCDM is indeed a special case of FLRW.

I looked at that GRwiki page and couldn't find an equation like that for a(T)
maybe I'm missing something and you can point me to the right place.
or there is another page.

I could be wrong but I think I'm doing LCDM (with the eponymous Lambda as curvature on the LHS).
Jorrie's calculations are conventional and these preliminary results seem to check with his.

 

[PeterDonis]

I think LCDM is indeed a special case of FLRW.

It is. This can be harder to see if you only look at the very simple FLRW models--the ones that only have one kind of "stuff" (ordinary matter, radiation, or dark energy) in them. In those simple models, the scale factor is a simple closed-form function of time. In the LCDM model, there is no such single function that covers the entire history of the universe, because there isn't just one kind of "stuff" present. But the general FLRW family of models covers that possibility.

 

[wabbit]

I looked at that GRwiki page and couldn't find an equation like that for a(T)
maybe I'm missing something and you can point me to the right place.
or there is another page.

It isn't stated as an equation (almost though, it's a special case of the equation for f(ct), which denotes a²(ct) there; their T is your T₀ and their t is your T) but implied (or so I thought) in the expression for the line element :

Putting the big bang at t=0 yields B=0, and choosing A appropriately results in the flat FLRW line element for a positive cosmological constant as $ds^2 = dct^2 - \left(\frac{sinh\left(2\sqrt{\frac{\lambda}{3}}ct\right)}{sinh\left(2\sqrt{\frac{\lambda}{3}}cT\right)}\right)\left(d\rho ^2 + \rho ^2 d\theta ^2 + \rho ^2 sin^2 \theta d\phi ^2 \right)$

I wonder if the different power might relate to what PeterDonis said, 1/2 corresponding to the simple FLRW case and 2/3 to the LCDM more complex case, possibly as a result of averaging over several successive "simple FLRW" epochs.

 

[marcus]

... Your formula above has $a(T)\propto(\sinh(kT))^{2/3}$; I saw elsewhere (http://grwiki.physics.ncsu.edu/wiki/FLRW) for the flat FLRW case $a(T)\propto(\sinh(kT))^{1/2}$ : what is the different exponent due to?

Edit: the relation between LCDM and FLRW isn't that clear to me. I thought LCDM could be read as a special case of FLRW where a(T) is derived from the matter/energy/DE densities.

It isn't stated as an equation (almost though, it's a special case of the equation for f(ct), which denotes a²(ct) there; their T is your T₀ and their t is your T) but implied (or so I thought) in the expression for the line element :

In the same section, a little past where the line element is found, I see that what they are studying is the radiation-dominated early universe case.
==quote==
The behavior of

was derived for a Ricci-scalar equation above modeling the universe to contain only dark energy and electromagnetic radiation. In the early history of the real universe, much of the electromagnetic radiation made a phase transition to ordinary matter and dark matter. As such the amount of electromagnetic radiation actually observed left over in the cosmic microwave background radiation is only about 10^-5 of the electromagnetic radiation energy density represented in this model.
==endquote==
In the analysis of the radiation dominated case, a 4 replaces the 3, so it is not surprising that 2/3 would change to 2/4. I haven't gone through this but I think it is probably OK.
When you solve the Friedmann equation, the LHS determines the expansion, and the expansion feeds back on the density on the RHS. Radiation goes down as the 4th power of distance and matter density goes down as the 3rd power. So expansion has a more potent effect on radiation energy density.

I think their equation would be best applied to the universe before "recombination" (before year 380,000 if I remember right).
However this is certainly interesting. And I'd be happy if anyone can correct my interpretation.

There is a parameter S_eq which is the S = 1+z factor pointing to the time when radiation = matter energy density. So-called "radiation matter equality". Jorrie makes the default S_eq = 3400. What that means is that at present the radiation energy density is only about 1/3400 of the matter. As you go back in time the former increases as the 4th power of S and the latter as the 3rd. So by S=3400 they are equal. From that point on, as redshift increases, you are getting in the radiation-dominate era, where matter can even be neglected.

Like these people do, in the GRwiki. They do not include any matter because it is insignificant in the era that they are analyzing. Or so I think, anyway.

The formula for a(T) that I gave should only be used matter-dominated era, say after year 1 million, very roughly. But still it covers most of the history.

 

[wabbit]

Ah thanks for the clarification, yes this makes sense. I hadn't paid due attention to that radiation mention, nor to the dependence of the power to the energy/matter balance. As you can see from my last edit above I was laboriously starting to approach a guess along those lines, but that was a slow process...

And your (or Jorrie's) formula is an approximation because it ignores 0.01% of history, I suppose that's forgivable - at least more so than my using the 1/2, which ignores 99.99% of history :-)

I now think the exact value of a(T) is an average over different epochs, during each of which the evolution equation has a different parameter, but since each epoch is much shorter than the next one, one can ignore previous epochs and just use for each epoch a single closed form corresponding to the approriate patameter, early 1/2, late 2/3. And very very early something else.

 

[marcus]

Wabbit, I think that's right. And if so, the upshot is we seem to have this formula for how the scale factor increases over time, applicable say after year 1 million which is pretty early. I'll quote part of the earlier post:
==quote==
Since after the first million years or so, radiation is a small part of the total, we found simplified closed formulas for how the universe size grows, that work as long as matter >> radiation and approximate essential parts of Jorrie's tables. Here is one:
$$a(T) = (H_0^2/H_\infty^2 - 1)^{1/3}(sinh(\frac{3}{2}H_\infty T))^{2/3}$$
As long as we are using 2.20 and 1.83 attohz for the two main model parameters, we can just calculate a value for the first term once and for all, and have a more convenient formula:
$$a(T) = 0.7636 (sinh(\frac{3}{2}H_\infty T))^{2/3}$$
I think 0.76 would be fine for most purposes.
So that shows how the "size of the universe" grows with time. a(T) is the size of a generic distance normalized to equal 1 at present.
==endquote==

 

[marcus]

The 0.7636, the first factor in the above equation, is a normalization factor which assures that a(T₀) = 1.

If we set a = 1, we can solve for H₀ as a function of T₀ = 13.787 billion years.

$$H_0^2 = H_\infty^2 (1 + 1/(sinh (\frac{3}{2}H_\infty T_0))^2)$$

Maybe the google-ator will help us calculate the present-day Hubble growth rate, given the present cosmic time.

1.83^2(1+1/(sinh(3/2*1.83 attohertz*13.787 billion years))^2)

it should come out 2.20^2 = 4.84
WOW! It comes out right! 4.839...
It comes out 4.839

So now we have a google formula that depends only on...this is very strange. I may have done something careless. It looks like the formula depends only on the cosmological constant and on how we measure cosmic time, and it gives us the other main parameter of the model. I'd be happy if anyone wants to step in and resolve my confusion about this. Maybe I just didn't get enough sleep last night.

 

[marcus]

Well that seemed a little surprising, but maybe it is all right. We can take, as our two main model parameters not H₀ and H_∞, but T₀ and H_∞---the cosmological constant and the present age of the universe. Of course, it's OK. And it's better to take the two Hubble rates as your basic parameters because they are more directly observable. the age of the universe is a complicated encoding of the same information. Still, it's nice. We have a simple formula now for what the Hubble rate H(T) was in past times.

And taking the reciprocal c/H(T) we will have a formula for what the Hubble radius R(T) was in past times.

I was wanting to get a formula for R(T) because of the kind of question that comes up here at PF from time to time: "how do they calculate this-that?"
And unexpectedly the formula will only depend on the cosmological curvature constant Λ.
$$H_0^2 = H_\infty^2 (1 + 1/(sinh (\frac{3}{2}H_\infty T_0))^2)$$
$$H_0 = H_\infty (1 + 1/(sinh (\frac{3}{2}H_\infty T_0))^2)^{1/2}$$
1.83 attohertz*(1+1/(sinh(3/2*1.83 attohertz*13.787 billion years))^2)^(1/2)
Given this, google gives back today's Hubble rate of 2.20 attohertz.

So now we can get H(T) the Hubble growth rate for any year (say after year 1 million, before that radiation plays a major role in the energy density and the approximation is not so good).

Let's try it for year 5.854 billion (the S=2 year when distances were half todays')
1.83 attohertz*(1+1/(sinh(3/2*1.83 attohertz*5.864 billion years))^2)^(1/2)
Google says H(then) = 3.907 attohertz.
Let's check with Jorrie's calculator.

 

[wabbit]

Neat. We can even simplify it a bit more by using $1+1/\sinh^2=\coth^2$, so
$$H_0=H_\infty \coth(\frac{3}{2} H_\infty T_0)$$
$$R_0=R_\infty \tanh(\frac{3}{2}\frac{ c T_0}{R_\infty})$$
$$a(T)=\left(\frac{\sinh(H_\infty T)}{\sinh(H_\infty T_0)}\right)^{2/3}\simeq 0.7636(\sinh(H_\infty T))^{2/3}$$

 

[marcus]

Jorrie's calculator says that back in S=2, R was 8.1124 billion light years. So Hubble time was 8.1124 billion years.
and when we type 1/(8.1124 billion years) to google to get H(then)
Google says 3.906 attohertz.
That is pretty good agreement with 3.907 attohertz.
So we are in business. We can calculate the scale factor a(T) and the Hubble rate H(T) for past years. The latter formula being
$$H(T) = H_\infty (1 + 1/(sinh (\frac{3}{2}H_\infty T))^2)^{1/2}$$
1.83 attohertz*(1+1/(sinh(3/2*1.83 attohertz*T))^2)^(1/2)

 

[wabbit]

One thing I notice from this (should have been obvious I suppose from R=c/H, but it didnt strike me somehow) is that $\lim_{T\rightarrow 0}H_T=+\infty$

 

[marcus]

Neat. We can even simplify it a bit more by using $1+1/\sinh^2=\coth^2$, so...

Thanks Wabbit. It gets even nicer:
$$H(T)=H_\infty \coth(\frac{3}{2} H_\infty T)$$
$$R(T)=R_\infty \tanh(\frac{3}{2}\frac{ c T}{R_\infty})$$
$$a(T)=\left(\frac{\sinh(\frac{3}{2}H_\infty T)}{\sinh(\frac{3}{2}H_\infty T_0)}\right)^{2/3}\simeq0.7636\cdot\sinh(\frac{3}{2}H_\infty T)^{2/3}$$
So we have several things we want expressed simply in terms of the cosmological constant and the cosmic time T. In the formula for a(T) the term involving T₀ is now very clearly there as a normalization, to make sure that a(now) = 1. Otherwise everything comes simply from knowing the year, and the cosmological curvature constant.
EDIT: I fixed an error in the formula for a(T) by simply copying a line from Wabbit's post #63 where the mistake was corrected. I had carelessly omitted a factor of 3/2.

 

[wabbit]

The more you look at FLRW, the simpler it gets:)

 

[marcus]

One thing I notice from this (should have been obvious I suppose from R=c/H, but it didnt strike me somehow) is that $\lim_{T\rightarrow 0}H_T=+\infty$

What would really be nice would be google-calculable formulas that go back to the radiation-era and the start of expansion. No hope, just dream.

Wait. Ashtekar has a simple modification of the Friedmann equation with a density dependent term (suppressed at Planck scale) that only comes into play at extreme energy density. We should look at it.

BTW Cai and Wilson-Ewing put a slightly longer version 2 of their "LambdaCDM bounce" paper on arXiv around beginning of February. The paper has been published in JCAP. As I see it, that is the leading candidate for a simple cosmic model. no inflaton needed, no multiverse implied, no fluctuation ex nihilo. Instead of mythical add-ons, a bounce of something already familiar to us (that we can see and for which we already have decent equations) the ΛCDM

http://arxiv.org/abs/1412.2914
A ΛCDM bounce scenario
Yi-Fu Cai, Edward Wilson-Ewing
(Submitted on 9 Dec 2014, revised 28 Jan 2015)
We study a contracting universe composed of cold dark matter and radiation, and with a positive cosmological constant. As is well known from standard cosmological perturbation theory, under the assumption of initial quantum vacuum fluctuations the Fourier modes of the comoving curvature perturbation that exit the (sound) Hubble radius in such a contracting universe at a time of matter-domination will be nearly scale-invariant. Furthermore, the modes that exit the (sound) Hubble radius when the effective equation of state is slightly negative due to the cosmological constant will have a slight red tilt, in agreement with observations. We assume that loop quantum cosmology captures the correct high-curvature dynamics of the space-time, and this ensures that the big-bang singularity is resolved and is replaced by a bounce. We calculate the evolution of the perturbations through the bounce and find that they remain nearly scale-invariant. We also show that the amplitude of the scalar perturbations in this cosmology depends on a combination of the sound speed of cold dark matter, the Hubble rate in the contracting branch at the time of equality of the energy densities of cold dark matter and radiation, and the curvature scale that the loop quantum cosmology bounce occurs at. Importantly, as this scenario predicts a positive running of the scalar index, observations can potentially differentiate between it and inflationary models. Finally, for a small sound speed of cold dark matter, this scenario predicts a small tensor-to-scalar ratio.
14 pages, 8 figures, published in JCAP March 2015

 

[wabbit]

Noted, I might've had an older one, will check this. Yes I must say if this really gets rid of inflation and replaces it with a natural, necessary mechanism, plus a "quantum washing machine" at the bounce to smooth out all the wrinkles in a prior contracting universe, what more could one ask for? (Well, some ordinary matter would be nice, so far they have only CDM, right?) . What is the opinion of the cosmology community on this?

LQC still seems speculative though (I found Bojowald's review paper http://arxiv.org/abs/1501.04899 really interesting in this respect though I don't know whether it reflects a consensus or not) - so maybe still early to declare a winner:)

 

[wabbit]

Oh and I finally used google calculator, to compute the current radius of curvature of the universe - I got 1bn ly :-)) - no fault of google though, my formula is obviously wrong here.

 

[marcus]

WMAP used to report a confidence interval for Ω_k which by an odd convention was minus what you expected.
If Ω was 1.01 then it was -0.01

And in the case of positive spatial curvature, the radius of curvature was considered to be the Hubble radius
divided by the square root of |Ω_k|

As I recall that was explained in the 5th year WMAP report dealing with cosmology, giving the results for various parameters etc.

So in that case the square root would be 0.1
and dividing by that would be like multiplying by 10
and if R = 14.4 billion light years then the radius of curvature would be 144 billion light years.

In practice they would give a confidence interval for Ω like [0.99, 1.01] and you would say what is the SMALLEST the universe could be, that would be if Ω were at the upper limit 1.01 and then the square root would be 0.1 and the radius curvature would be 144 Gly and you could calculate the CIRCUMFERENCE, a real distance. The radius of curvature has just a formal mathematical existence but the circumference is somehow real. If the universe would somehow stop expanding then a very long-lived being could in principle travel it and get back to point of departure.

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.

 

[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.
$${\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}$$ $${\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (Gy)&R (Gly) \\ \hline 0.167&6.000&1.1738&1.76\\ \hline 0.200&5.000&1.5417&2.30\\ \hline 0.250&4.000&2.1494&3.19\\ \hline 0.333&3.000&3.2851&4.80\\ \hline 0.500&2.000&5.8636&8.11\\ \hline 1.000&1.000&13.7872&14.40\\ \hline \end{array}}$$
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) ⇒