What can you learn about cosmology from Google calculator?

[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.

 

[wabbit]

Yes I'm reading another one now on inflation which was suggested by Chalnoth - Lesgourgues' lecture notes. Quite good, but the latter part on high energy physics is beyond me. Might switch to Lineweaver, he seems more gentle.

I saw in Wilson-Ewing that he refers to an earlier paper for a fuller LQC treatment of the bounce (just bounce plus radiation era)
http://arxiv.org/abs/1404.4036
T. Pawlowski, R. Pierini, and E. Wilson-Ewing, “Loop quantum cosmology of a radiation-dominated flat FLRW universe"

 

[marcus]

If you look at x = -0.45 on the Desmos plot in post #88 you see an inflection where deceleration changes to acceleration.
Up to that point the contraction is decelerating (as if under the influence of Lambda ) and after that point, contraction is accelerating (as if under the influence of matter.)

But that is just if you want to view the whole curve as being of cosmological interest (it shows the Cai&Wilson_Ewing picture, in broad outline.) Otherwise we just refer to the right half (the positive x side).

On the right, keep in mind that the present day corresponds to x_now = 0.8.
For simplicity we are using the H_∞ time scale
H_∞T_now = 1.832 attohertz*13.787 billion years = 0.797 ≈ 0.8
We know that on the conventional timescale the inflection (in distance growth) where acceleration kicks in, happens around year 8 billion, perhaps slightly before.
x_infl = H_∞T_infl = 1.832attohertz*8 billion years ≈ 0.45.

If you look on the expansion side of the picture, at x = 0.45, it is the same story. There is an inflection. As long as matter dominates, expansion decelerates, and then as soon as Lambda dominates it begins to accelerate.

The natural sinh^2/3(1.5x) time x = 0.45 corresponds to year 8 billion, just as x = 0.8 corresponds to the present year 13.787 billion.
==================
You get the same story in Lineweaver's Figure 14, but the point on the x-axis to look for is "-6 billion" that is 6 billion years before present. IOW around 8 billion.
==================
I'm thinking of how a newcomer might best be introduced to cosmology. The first question is the relation of time to distance: the expansion history. As so many very naturally ask, what about this ACCELERATION I've heard about?

That Figure 14 shape IS the most characteristic thing about our universe expansion history. It shows growth of scale factor over time. And the
$$a(x) = \sinh^{2/3}(1.5x)$$
shape is the most characteristic thing about it.
The fact that the distance growth curve has an inflection point is what the cosmology folks mean when they talk about "acceleration".

that curve could be step one of an exposition. And the timescale H_∞ = 1.832 attohertz could be step two, because that is how you translate conventional year numbers (as we humans measure time) into the x time that the sinh^2/3 function likes. Google calculator helps here:
1.832 attohertz*13.787 billion years ≈ 0.8
1.832 attohertz*8 billion years ≈ 0.45
Step three might be the idea of NORMALIZING the a(...) function so that it equals one at present. E.g. by deciding to always divide a(x) by a(.8)

Or maybe one just leaves it as it is, un-normalized, since with the scale factor all we really care about is RATIOS, of scales at various times. It could be argued that normalizing it to unity at the present is somewhat "presento-centric". But it's both convenient and conventional so maybe we should normalize.

 

[wabbit]

About that distance to a past light emission event - while we can't solve it explicitly for a matter-lambda mix, there are formulas for lambda-only / de Sitter space and for matter-only expansion that might perhaps be useful for exposition, the actual behavior falling somewhere in between.

Let's call $D(t,t_0)$ the distance from the light emission at t, as measured by the observer at $t_0$ .

No expansion :$$ D(t,t_0) =c(t_0-t)=_{def}d $$
De Sitter expansion :$$ D(t,t_0) =\frac{c}{H}\left(e^{H(t_0-t)}-1\right)$$
Matter expansion : $$ D(t,t_0) = c t_0\cdot 3\left(1-\left(\frac{t}{t_0}\right)^{1/3}\right)
=\frac{3c(t_0-t)}{1+\left(\frac{t}{t_0}\right)^{1/3}+\left(\frac{t}{t_0}\right)^{2/3}}$$

 

[wabbit]

A couple more - let's compute the ratio $\xi=D(t,t_0)/d$

No expansion :$$ \xi=1 $$
De Sitter :$$ \xi=\frac{e^{Hd/c}-1}{Hd/c}$$
Matter :$$ \xi=\frac{3}{1+\left(\frac{t}{t_0}\right)^{1/3}+\left(\frac{t}{t_0}\right)^{2/3}}$$

 

[marcus]

"Nothing exists except fields and geometry; everything else is opinion."

 

[wabbit]

I still find it difficult to figure what $D(t_0,t)$ really is - I said before we're measuring a spatial distance between two events from the point of view of the comoving observer at t0, but what is his point of view ? i.e. which coordinates is he using and in what sense is this a distance ? It is not the distance to the current position of the comoving light source, and it is only defined as an integral...

Well it is the comoving distance at the time of emission between the source and us. Whatever that means. Or the comoving distance to the emission event. Still doesn't register.

 

[wabbit]

"Nothing exists except fields and geometry; everything else is opinion."

But geometry is nothing but fields, so nothing exists except fields.

... actually, nothing exists except information, everything else is opinion - and that is information too, so :

Nothing exists except information.

So we did make some progress over 2500 years

 

[marcus]

We seem to be getting the gist of Jorrie's Lightcone calculator just using Google calculator to evaluate some fairly simple formulas. There is some satisfaction at being able to "do it yourself" or "cook from scratch" and maybe I understand the standard cosmic model a little better because of it.
I still think Lightcone tables are a great hands-on way of getting acquainted with LambdaCDM model and past+future cosmic history.

In this approach, we are using the eventual Hubble radius (a version of cosmological constant) as a time and distance scale.
It makes the formulas much neater and easier to type to measure time in multiples of 17.3 billion years and distance in multiples of 17.3 billion light years. So our time variable, x = T/17.3 billion years, is just our conventional time divided by R_∞ = 1/H_∞

I want some sample times to use in trying out our formulas so I'll list some, along with normalized scale factors S = 1.311 sinh(1.5x)^(2/3)

x-time  billion years   S
.1         1.73               4.632
.2         3.46               2.896
.3         5.19               2.183
.4         6.92               1.771
.5         8.65               1.494
.6        10.38               1.288
.7        12.11               1.127
.797      13.787              1.000

These should check with Lightcone (to within a percent or so) in the sense that if you type an upper S value into Lightcone, like S = 1.494 you should get the time T = 8.65 billion years. Or close to it. These S values are google-calculator ones.
Now if you put S=1.494 into Lightcone, for example, you get a bunch of other information. Like D_now(S). This is the proper distance NOW of a galaxy whose light was emitted back in year 8.65 billion and comes to us today stretched out by a factor of 1.494. wavelengths are one and a half times longer than they were when the light was emitted.
Our formula for the distance is
$$D_{now} = 1.311 \int_{x_{em}}^.797 |\sinh(\frac{3}{2}x)|^{-2/3} dx$$
That gives it to us in our terms and to convert to years we just multiply by 17.3 billion light years.
The result should agree with the D_now from Lightcone calculator. I'll use my laptop's integration utility. I put a couple of Lightcone values in for comparison.

x-time  billion years       S            Dnow billion light years
.1         1.73               4.632      23.03 (23.014)
.2         3.46               2.896      16.80
.3         5.19               2.183      12.47
.4         6.92               1.771       9.08
.5         8.65               1.494       6.26
.6        10.38               1.288       3.87
.7        12.11               1.127       1.78  (1.759)
.797      13.787              1.000

Hey the Mac laptop integration utility is really easy to use! It's hardly any more trouble than just typing in the formula and the upper and lower limits of integration. Scarcely any more bother than using the google calculator. You just have to go to it (applications --> utilities --> grapher). It makes a graph of what it's integrating.

 

[wabbit]

Dnow(S). This is the proper distance NOW of a galaxy whose light was emitted back in year 8.65 billion and comes to us today stretched out by a factor of 1.494.

But is it ? I've been struggling with this. If $d_{em}$ was our distance to that galaxy at the time of emission, its distance now is just $d_{em}\frac{a}{a_{em}}$ but the light traveled $D=\int \frac{cdt}{a(t)}$ and I don't see why the two should be equal - I read D as the spatial distance to the emission event but this doesn't make much concrete sense to me. So far I can only understand it as "how much the light travelled" but even that is a bit hairy...

Edit : no I see it now you're right sorry, $D=d_{em}$ since $D=\int_0^d dx$. And you must have included the factor $\frac{a}{a_{em}}$ in your definition of D. Is that right, your D is actually $\frac{a}{a_{em}}\int \frac{cdt}{a(t)}$?

 

[marcus]

Hi Wabbit,
You probably recognize the 1.311 factor as sinh^2/3(1.5* 0.797). When we drew the curve for how a(x) increases over time, the present was around .8 and its a(x) = a(.8)was around 1.3. that lovely antelope-horns graph.

So S(x) the factor the wavelength of bit of light will get stretched if it is emitted at time x and received today, is the ratio
1.311/sinh^2/3(1.5* x)

and if a bit of light is traveling already and between time x and x+dx it travels a certain amount cdx (we can ignore c = 1) then that little interval will ALSO get expanded along with everything else by the ratio S(x) by the time it arrives.

So what the integral formula does is simple. It just adds up all the little bits of distance the light traveled and magnifies each one by the appropriate stretch factor.

We are using the same definition and concept of Dnow as Jorrie's calculator. We just have a kind of naive basic formula for calculating a (fairly good) approximation to it. I took the numbers in parens from the Dnow column of the Lightcone calculator.

I picture Dnow as the distance light travels between time x_em and time x_now=.797.

the formulas can be fairly light, easy to type because scaled for simplicity. If you come in with a conventional time T in years,
you have to divide by 17.3 billion years to get x.
If your formula gives you a distance you then have to multiply by 17.3 billion light years to get the conventional distance in light years.

 

[marcus]

I forget if you have a Mac. You may have said. The numerical integration utility is so easy to use! It's refreshing.

You don't even have to go through the formality of typing an integral sign. You just type in the function you want integrated and type in the limits of integration. It draws a picture of the function with a shaded area between the upper and lower limits, and tells you the answer.

 

[wabbit]

To be honest I'm having a hard time following now, with the change of units and coefficients. Each time I see x I need to remind myself of what it is, and same for 1.31 or so.

 

[wabbit]

I forget if you have a Mac. You may have said. The numerical integration utility is so easy to use! It's refreshing

Nope, no Mac. I integrate with pen and paper Sir, like our forebears : )(Oh if I really need to, I use a lowly PC )

 

[marcus]

To be honest I'm having a hard time following now, with the change of units and coefficients. Each time I see x I need to remind myself of what it is, and same for 1.31 or so.

Thanks for the heads-up! It's time to pause, straighten out the notation, summarize. We've been discovering relations and possibilities as we went along. Tine to review, simplify and make consistent. I think presented in orderly fashion none of it will be very hard, and it will match the standard model (e.g. Jorrie's calculator) pretty well.

 

[wabbit]

About that, to me at least, it's much easier when the units are explicit (i.e. everything is expressed as a ratio or similar, like $H_{\infty}T$ or (easiest) $\frac{T}{T_\infty}$) rather than implicit like $x$. Of course the formulas becomes lighter with just x but in any case to plug them into a calculator you need to substitute with a ratio. Another possibility is also to formulate with x but frequently add a reminder "(where $x=\frac{T}{T_\infty}$)".