Effort to get us all on the same page (balloon analogy)

[Jorrie]

In other words, what 0.728 actually tells you is how big _now is compared with H_now.

0.728 tells you what the long-term growth rate is, compared to the present growth rate.

I agree with what you said, but sticking in a numerical value like 0.728 is not so good; I guess you should then give it another symbol and say that it is determined experimentally and is bound to change. It is essentially \Omega_{\Lambda(now)} and also changes over cosmic time, like H.

I understand your drive based on a fundamental constant like \lambda and get away from 'energy density of the vacuum', which is a slippery concept to explain to grandmother.

Maybe give the ratio (H_∞/H_{now})^2 a new symbol?

 

[marcus]

I agree with what you said, but sticking in a numerical value like 0.728 is not so good; I guess you should then give it another symbol and say that it is determined experimentally and is bound to change. It is essentially \Omega_{\Lambda(now)} and also changes over cosmic time, like H.

I understand your drive based on a fundamental constant like \lambda and get away from 'energy density of the vacuum', which is a slippery concept to explain to grandmother.

Maybe give the ratio (H_∞/H_{now})^2 a new symbol?

I understand your point now! Duh! Of course. We can't use the raw number we need a symbolic expression for it. Exactly for the reason you said. It might change.
Why didn't I see what you were driving at immediately? OK I'll think about that and try to come up with something.

The normal thing to do is grab a Greek letter, but let's see how this looks

Y_now:∞= Y_now/Y_∞
Y_now/∞= Y_now/Y_∞
and in cosmology the present value is tagged with a subscript zero, like H₀ is standard notation for H_now
So how about
Y_o∞= Y_o/Y_∞
Y_o:∞= Y_o/Y_∞
Y_o/∞= Y_o/Y_∞
Or how about the letter ξ which I don't recall seeing much of lately?
ξ = Y_o/Y_∞

At the moment I'm leaning towards this way of handling it:
$$(1+z)^3 = {\left({\left(\frac {Y_∞}{Y} \right)}^2 -1\right)}/ {\left({\left(\frac {Y_∞}{Y_{now}} \right)}^2 -1\right)}$$
The current estimated value of the denominator in this expression is 0.3736 = 1/0.728 - 1.
So for practical purposes, until the estimate 0.728 changes, all we do is divide by 0.3736.
But here's another idea:
$$(1+z(t))^3 = (Y_{∞/t}^2 -1)/(Y_{∞/now}^2 -1)$$

That is, create symbols for the ratios of Hubble times. Several equations only depend on the times through their ratios, so having symbols for the ratios of the times would make the equations more compact. How would it go down with second year undergraduates, I wonder?

 

[Jorrie]

At the moment I'm leaning towards this way of handling it:
$$(1+z)^3 = {\left({\left(\frac {Y_∞}{Y} \right)}^2 -1\right)}/ {\left({\left(\frac {Y_∞}{Y_{now}} \right)}^2 -1\right)}$$
The current estimated value of the denominator in this expression is 0.3736 = 1/0.728 - 1.
So for practical purposes, until the estimate 0.728 changes, all we do is divide by 0.3736.
But here's another idea:
$$(1+z(t))^3 = (Y_{∞/t}^2 -1)/(Y_{∞/now}^2 -1)$$

That is, create symbols for the ratios of Hubble times. Several equations only depend on the times through their ratios, so having symbols for the ratios of the times would make the equations more compact. How would it go down with second year undergraduates, I wonder?

It starts to look all the more promising to me. I would prefer the latter of the two, because when you define Y, you can just as well define all the different subscripts and combinations. The fact that it is the ratio of two Hubble constants for different times, makes it appealing to me.

I also think 2nd-years will appreciate the more compact and easier to remember/recognize (and write) equation. It may be more of a problem for established cosmologists, where too much simplification/normalization becomes a hindrance. But since that's not your usage, it should be fine.

 

[marcus]

It's encouraging that you see the notation as moving in the right direction. I want to get back to something you said earlier (my original response was out of order) about the Y'-prime equation.

...Maybe <br /> Y&#039;= \frac{3}{2} {\left(1- {\left(\frac {Y}{Y_∞} \right)}^2 \right)}
would have looked more awkward in 1/H form. (?)

Right. I think you spotted the key factor. Try writing a differential equation for H. It gets messy. Constants like G and 8π and c come in, units come in. H' is dimensionful. The units are 1 over time squared. The density may get into the picture. So you have more things to think about.

By contrast, the Y' equation needs only the one constant, and the equation is dimensionless. Y' is simply a number. the units cancel out. change in time, per time.
In the fairly early universe Y' hovers near 1.5 for the greater part of a billion years and then drifts down to present value around 0.4.
The units play no role. Saying 0.4 years per year is the same as saying 0.4 century per century.

That equation does the work of the conventional Friedmann and conventional Raychaudhuri equations, in generating a curve from which you can calculate the other parts of the expansion history. and it is simpler than either conventional equation.

That is why I'm thinking that it might be a good basis for a beginner's tutorial.
For ADVANCED students it's a different matter. they can handle messier equations, spatial curvature term, radiation era etc etc. So of course let them wrestle with the complete unsmplified animal.

Since we're experimenting with the Y_now/∞ notation for ratios of Hubble times, let's try using that ratio symbol in the basic equation.

<br /> Y&#039;(t) = \frac{3}{2} {\left(1- {\left(\frac {Y(t)}{Y_∞} \right)}^2 \right)}

<br /> Y&#039;(t) = \frac{3}{2} {\left(1- Y_{t/∞}^2 \right)}

If that seems too compressed, one can always temporarily expand it back to the version on previous line, to explain something. Your point about compactness (at end of post #393) is well taken.

 

[George Jones]

<br /> Y&#039;(t) = \frac{3}{2} {\left(1- {\left(\frac {Y(t)}{Y_∞} \right)}^2 \right)}

<br /> Y&#039;(t) = \frac{3}{2} {\left(1- Y_{t/∞}^2 \right)}

I haven't been following this thread, so I might have missed some things. Did anyone write down the easy analytical solution to this separable differential equation?

<br /> \int^{Y\left(t\right)}_{Y\left(0\right)} \frac{dY}{1 - \left( \frac{Y}{Y_\infty} \right)^2} = \frac{3}{2} \int^t_0 dt&#039;<br />
gives

<br /> Y\left(t\right) = Y_\infty \tanh\left( \frac{3t}{2Y_\infty} \right)<br />
for Y\left(0\right) = 0.

 

[marcus]

Beautiful! Thanks, George! Now we are cooking

Following the discussion we just had, I'll emend the skeleton outline for a cosmo tutorial I sketched out in post#385.

We want to understand the expansion history and if possible reproduce it with some simple equations. The Hubble time Y(t) gives a handle on the rate distances are expanding at present or at some moment in the past.
One percent of Y(t) is how long it would take for a distance (at its current rate) to grow by one percent.

Two important ones to know (expressed in billions of years, abbr. Gy) are Y_now = 13.9 Gy
and the longterm future value that Y(t) is tending towards
Y_∞ = 16.3 Gy.

Every so often it's convenient to have a slightly smaller unit of time, a tenth of a billion, so we'll occasionally use d=10⁸ years, as an auxilliary unit. In those terms:
Y_now = 139 d
Y_∞ = 163 d.
One percent of Y_now = 139 million years. So, recalling what was just said, that is how long it would take a given cosmological distance to grow by one percent (at its current rate.) Alternatively one can think of distances currently growing 1/139 of a percent per million years.

A couple of equations we use depend on RATIOS of Hubble times, so as a compact notation we will write:
$$Y_{now/∞} = \frac {Y_{now}}{Y_∞} $$ The square of this ratio is currently estimated by cosmologists to be about 0.728. It's a key number and may change slightly in future, with increasingly accurate observations. For the time being we write:
$$Y_{now/∞}^2 = 0.728$$ Given these data we can plot the curve Y(t) of Hubble time all the way back to a lookback time of 13.7 Gy. The standard model age is 13.757 Gy, so we come within 57 million years of start of expansion, close enough for an introductory treatment. Lookback time is a negative or "backward" timescale and a bit awkward at first--it takes a little getting used to, but one copes. Often the minus sign is omitted.
Y&#039;(t) = \frac{3}{2} {\left(1- {\left(\frac {Y(t)}{Y_∞} \right)}^2 \right)}
Y&#039;(t) = \frac{3}{2} {\left(1- Y_{t/∞}^2 \right)}
Once we have Y(t) for lookback times from 13.7 Gy to present, we can calculate the redshift z(t) corresponding to any given lookback time t (i.e. how far in the past the light we are now receiving was emitted.)
$$(1+z(t))^3 = {\left({\left(\frac {Y_∞}{Y(t)} \right)}^2 -1\right)}/ {\left({\left(\frac {Y_∞}{Y_{now}} \right)}^2 -1\right)}$$
$$(1+z(t))^3 = {\left(Y_{∞/t}^2 -1\right)}/ {\left(Y_{∞/now}^2 -1\right)}$$
The current estimated value of the denominator in this expression is 0.374 = 1/0.728 - 1.

Various other things can also be calculated, knowing the Hubble time Y(t) in the past, for example the matter density (dark, ordinary, plus a minor contribution from radiation)

$$\rho = \frac {3}{8\pi G}{\left(\frac{1}{Y^2} -\frac {1}{Y_\infty ^2 }\right)}$$

As a check, the calculated matter density at some time in the past should agree with the redshift of light from a given moment in the past.

$$(1+z)^3 = \frac {\rho (then)}{\rho (now)}$$

 

[marcus]

Using George's suggestion of tanh solution for Y(t) involves adapting the timescale. Maybe I should have used forwards time all along, rather than lookback.
To use the tanh solution we need t, measured from the start of expansion (say in convenient d units), and we need the quantity 3t/2Y_∞ = 3t/326
Here 326 is simply twice our familiar 163 number for the longterm Hubble time.

Lookback time [Gy]        13.7     13.6     13.5  ...   2       1        0
Time since start [d]      0.573    1.57     2.57  ... 117.57  127.57   137.57
Y(t)=163tanh(3t/326) [d]  0.859    2.355    3.854 ... 129.41  134.56   138.99 
tanh estimate for z      44.6     22.3     15.8   ...   0.162   0.077    0.001

As a practical matter, to calculate z(t), using current estimates of the parameters, one could use
1 + z = [(coth(3t/326)² - 1)/0.374]^1/3
But the google calculator does not do hyperbolic cotangent, so instead of coth² one can use tanh^-2 and paste this in:
((tanh (.573*3/326)^-2 - 1)/.374)^.333
It gives 45.6 which is roughly right for 1+z, making z about 44.6

 

[Jorrie]

As a practical matter, to calculate z(t), using current estimates of the parameters, one could use
1 + z = [(coth(3t/326)2 - 1)/0.374]1/3
But the google calculator does not do hyperbolic cotangent, so instead of coth² one can use tanh^-2 and paste this in:
((tanh (.573*3/326)^-2 - 1)/.374)^.333
It gives 45.6 which is roughly right for 1+z, making z about 44.6

This is a nice relationship, for although the cosmo-calculator gives time t for a given redshift z, one often needs it in the other direction, which one must find by manual iteration.

BTW, do you know of a way of finding the event horizon for any cosmic time t, other than the integration to t_{infinity} that e.g. Tamara Davis used in eq. A.20 of her "Fundamental Aspects" paper:

\chi_c(t) = c \int_{t}^{t_{end}} \frac{dt&#039;}{R(t&#039;)}

This is comoving distance and her 'R' is the expansion factor 'a'. I suppose one can stop after some time, since the horizon distance will become practically a constant, but how far into the future is required?

 

[marcus]

This is a nice relationship, for although the cosmo-calculator gives time t for a given redshift z, one often needs it in the other direction, which one must find by manual iteration.

BTW, do you know of a way of finding the event horizon for any cosmic time t, other than the integration to t_{infinity} that e.g. Tamara Davis used in eq. A.20 of her "Fundamental Aspects" paper:

\chi_c(t) = c \int_{t}^{t_{end}} \frac{dt&#039;}{R(t&#039;)}

This is comoving distance and her 'R' is the expansion factor 'a'. I suppose one can stop after some time, since the horizon distance will become practically a constant, but how far into the future is required?

Tamara's thesis! I remember delving into that (as curious amateur) when it came out! What a kick! She helps one concretely visualize stuff far in future, like galaxes on their way out thru the CEH. I think she must be Lineweavers pride and joy. No! I don't know of a better way to find the CEH. I always just use the eventual limiting value of 16 billion ly. Proper.
George Jones might be willing to go into this with you. Or heck! how about writing Tamara Davis? Give her a link to your calculator (which she will recognize as a community service) and ask her a concise question. She has an automatic interest in helping you because it's good for the community to have accurate cosmo calculators online.
I think she is at Copenhagen now. Don't know her email but if you want to write and have any trouble finding it let me know.

I used your calculator to estimate the expansion age if one uses the two numbers 139 and 163 as basic input, as if they were exact. And your figure of 0.0000812 for omega_rad.
It came to 13.7574 billion
So then I calculated the redshift for the PRESENT, which should come out exactly zero
((tanh (137.574*3/326)^-2 - 1)/(1/.7272-1))^(1/3)

The reason for the .7272 is that the ratio 139/163 is almost precisely sqrt.7272.

I like the idea of a cosmic calculator where your inputs are the two relevant Hubble times,
Y_now and Y_∞
In a clumsy way, I can implement that with your calculator.
The condition of flatness is:
.7272
.2727188
.0000812
and 139 corresponds to a conventional Hubble rate number of 70.3463274

that is what google gives if you say "1/(13.9 billion years) in km/s per Mpc"
It will convert 1/(13.9 billion years) into conventional expansion rate units.

So for fun I put those three omega numbers, and that conventional Hubble rate number into your calculator and it said the expansion age is 137.574

So I calculated the redshift for that expansion age:
((tanh (137.574*3/326)^-2 - 1)/(1/.7272-1))^(1/3)
and it came out pretty close to zero, which is nice.
1+z came out 1.00013003035, according to google.

You could say it did credit to your calculator---that there is, somehow, not a lot of unnecessary roundoff error, and that the number 0.0000812 for radiation is OK. Neat.

I mean neat because, if you give your calculator inputs corresponding "exactly" to those two Hubble times (139 and 163) then it gives an age 137.574 which, if you check, MATCHES very well those two exact inputs. Using George's tanh and the google calculator to check the match. I think you understood what I was saying but I'm repeating in case other readers might not have.

It's really very nice. It means that if I want to use my juniorhigh school Hubble times as cosmic model inputs I simply have to prime your calculator with the numbers
.7272
.2727188
.0000812
70.3463274
and then use an expansion age of 137.574

 

[marcus]

Here's a cosmic history tabulation based on treating now and future Hubble times 13.9 and 16.3 billion years as exact. In the following calculations, 0.7272 is the square of their ratio, and 326 is twice 163.
Google codes used are:
z_t = ((tanh (t*3/326)^-2 - 1)/(1/.7272-1))^(1/3) - 1
Y_t = 163 tanh(t*3/326)
To calculate with a code, paste into google, replace t by an expansion age, and press = or return.

EXPANSION HISTORY, 139/163 MODEL.
Hubble time Y=1/H and age are shown in units of d = 10⁸ years.
Age at present 137.574 d.

   Age          Redshift         Hubble time 
   t (d)           z[SUB]t[/SUB]             Y[SUB]t[/SUB] (d)    
   0.0049         1093           0.00735
   1            30.574            1.5000 
   2            18.890            2.9997
   3            14.178            4.4989
   4            11.528            5.9973      
   5             9.796            7.4947
   6             8.559            8.9909
   7             7.625           10.4855
   8             6.889           11.9784
   9             6.292           13.4692
  10             5.796           14.9578
  20             3.269           29.6658
  30             2.243           43.8906
  40             1.659           57.4293
  50             1.273           70.1200
  60             0.992           81.8468
  70             0.776           92.5405
  80             0.603          102.1752
  90             0.459          110.7617
 100             0.337          118.3402
 110             0.232          124.9720
 120             0.139          130.7319
 130             0.057          135.7021
 131             0.049          136.1590
 132             0.041          136.6088
 133             0.034          137.0518
 134             0.026          137.4880
 135             0.019          137.9174
 136             0.012          138.3402
 137             0.004          138.7565
 137.574         0.000          138.9925

z : fractional amount distances and wavelengths have increased while light was in transit. Arriving wavelength is 1+z times original.
H=1/Y: Hubble expansion rate. Distances between stationary observers grow at this fractional rate--a certain fraction or percentage of their length per unit time.
H(per d) : fractional increase per convenient unit of time d = 10⁸ years.
Y=1/H: Hubble time. 1% of the current Hubble time is how long it would take for distances to increase by 1%, growing at current rate. At present, Y is 139 d = 13.9 billion years.
Hubble time is proportional to the Hubble radius = c/H: distances smaller than this grow slower than the speed of light. At present, the Hubble radius is 13.9 billion ly (proper distance)

The Hubble law describes the expansion of distances between observers at rest with respect to the background of ancient light and the expansion process itself: Observers who see the ancient light and the expansion process approximately the same in all directions, e.g. no Doppler hotspots.
The field of an observer's view can be thought of as pear-shape because distances were shorter back then. Here is a picture of an Anjou pear.
http://carrotsareorange.com/wp-content/uploads/2010/05/pear-anjou.jpg
Here is Lineweaver's spacetime diagram:
http://ned.ipac.caltech.edu/level5/March03/Lineweaver/Figures/figure1.jpg
The upperstory figure, with horizontal scale in proper distance, shows the lightpear outline.
Here is Lineweavers plot of the growth of the scalefactor R(t), which models the growth of all distances between observers at universe-rest (at rest with respect to background.)
http://ned.ipac.caltech.edu/level5/March03/Lineweaver/Figures/figure14.jpg
The dark solid line is according to standard model parameters. Various other cases are shown as well.

 

[Jorrie]

Here's a cosmic history tabulation based on treating the now and future Hubble times 13.9 and 16.3 billion years as exact. In the following calculations, 0.7272 is the square of their ratio, and 326 is twice 163.

It is interesting to note that for time t > 1d, the errors for the assumption is less than 2%, while at t = 0.0049d, it grows to 10%. The 'real' redshift for t = .0049d is z = 950. I think the error is due to the assumption to ignore radiation density in the 139/163 model. It plays a significant role for z > 1000. At z ~3300, past radiation and matter energy densities were the same.

As long as we keep the redshift low enough, the 139/163 model is an excellent approximation and maybe worth a little calculator on its own...

 

[marcus]

YIPPEE! I would like to see such a thing!

I think you are right about the cause of the trouble at very early times and high redshifts.
The Y' prime equation (for which the tanh formula is a solution) was derived assuming matter-dominated era. The different behavior of radiation, during expansion, was not taken into account.

 

[marcus]

I guess what I would find most interesting would be a calculator where you do not put in a value for H_now or Omega_Lambda
but rather you give it two Hubble times to work with:
Y_now and Y_∞

But to improve performance at early times, what would be great is if you could put
in a third parameter as well, something like the present ratio of radiation to (dark and ordinary) matter.
And have the calculator take account of the balance shifting towards radiation in early times.

Basically I don't care so much about accuracy for z > 1000, but it would be nice to stay reasonably accurate back as far as z = 1000.
I noticed that for the time the CMB was emitted ( last scattering time), it was giving t = 490,000 years instead of t = 390,000 years.

I don't know how easy or hard it would be to implement.

I see your present default ratio of radiation to matter is 0.0000812/0.272 = 0.0002985

What I imagine being able to do is to decide on three parameters:
say 13.9 billion years
16.3 billion years
and present rad/mat ratio 0.0002985

So I enter those three parameters. And then being able to calculate a time from a redshift,
or (if it was the other way around) a redshift from a time.
Or maybe two outputs: an age and a Y, from a redshift.
or a redshift and a Y, from an age.

I wouldn't care about having the calculator output values of the Hubble rate, because I can always flip the Hubble time over and interpret 13.9 billion years as "1/139 percent per million years."
That's about as far as I've gone, thinking about it.

Anyway, it's an exciting idea. Any cosmo calculator that you prime basically by just putting in two Hubble times.

 

[marcus]

My viewpoint is more or less as follows: All distances between pairs of stationary observers have an intrinsic tendency to grow at a certain percentage rate, which we think is about 1/163 % per million years. Why this should be is an interesting question probably having to do with the microscopic structure of geometry, geometry at quantum level. This is simply observed, and speculated about, but not yet understood.

Geometry has its own dynamic rules and, once some event starts it expanding, tends to continue, subject to gradual slowing by matter. The present rate is larger than the intrinsic "vacuum" rate of expansion. We believe it is currently 1/139 % per million years. The excess is gradually being reduced, and to our best knowledge in the distant future, say 20 billion years hence, expansion will nearly be stabilized at 1/163 %.

The question on my mind is how to describe this to general audience, how to introduce expansion cosmology in a way that does not give people misleading pictures.

Like galaxies hurtling thru space at fantastic speeds, or like a mysterious "energy" pushing on them so that they accelerate. Such pictures, once gotten into the head, interfere with understanding and are difficult to root out.

I am exploring ways to introduce cosmology without ever showing the reader the Hubble constant in conventional km/s per Mpc terms. Because, when someone tries to get an intuitive grasp of H in those terms they are apt to fall into the trap of imagining it as the speed (km/s) of something moving thru space. It's the original pedagogical blunder.

People could learn to think about expansion of various distances as similar to the percentage growth of various savings accounts at bank. Different size accounts grow by different annual amounts--all growing at a certain interest rate.

Once you learn the Hubble growth rate H as 1/139 % per million years then LATER you might work with it as the speed (e.g. 70.35 km/s) with which a Megaparsec distance is growing. Nothing is moving thru space, but the distance between two stationary observers who are one Mpc apart is increasing at that speed. (One of the things dynamic geometry can do, just as it can make corners of a triangle add up to more or less than 180 degrees, depending.)

So suppose we want to put the cosmo-intro focus on the dynamics of a distance growth rate. How do we do this?

Well one nice way is to focus on the evolution of Y=1/H the Hubble time. I'm using uppercase Y rather than a T with subscript to keep notation simple. Uppercase Y looks like T with arms bent up.

Y grows according to a simple differential equation Y' = (3/2) (1- Y²/Y_∞²) involving dimensionless quantities, which applies back nearly to the start, assuming spatial flatness. It isn't valid back in the initial radiation dominated era but that is comparatively brief, so it works over much of expansion history. And the equation has an explicit solution Y(t) = 163 tanh(3t/326).

The Hubble time constant 16.3 billion years carries the same information as the usual cosmological constant, and in the above formula the 163 appears twice. Once as coefficient out in front of the tanh and again, doubled, as 326 in the argument. The pedagogical idea here is that beginners get hands-on experience with this cosmo constant information as a feature of geometry that they can calculate with---a definite and inherent geometric feature of spacetime that they can use with a calculator, and not as some mysterious "dark energy" constituting a possibly fictitious fraction of the "critical density".

Thanks to G. Jones, Jorrie, and Mark M. for help and encouragement with this.

 

[Jorrie]

So suppose we want to put the cosmo-intro focus on the dynamics of a distance growth rate. How do we do this?

Well one nice way is to focus on the evolution of Y=1/H the Hubble time. I'm using uppercase Y rather than a T with subscript to keep notation simple. Uppercase Y looks like T with arms bent up.

Y grows according to a simple differential equation Y' = (3/2) (1- Y²/Y_∞²) involving dimensionless quantities, which applies back nearly to the start, assuming spatial flatness. It isn't valid back in the initial radiation dominated era but that is comparatively brief, so it works over much of expansion history. And the equation has an explicit solution Y(t) = 163 tanh(3t/326).

Well motivated, Marcus.

One thing that still bothers me a little is the units d=10^8 years that you are using. It's still somewhat confusing and I cannot quite remember your motivation for not using standard 10^6 or 10^9 years. Things should work equally well with 13.9 and 16.3 Gy, or perhaps 13900 and 16900 My.

I think it should be simple to include radiation density, but I will experiment a bit. I suspect that the growth rate law during the radiation dominated phase changes to something like
Y&#039; = 2 (1- Y_{rad}^2/Y^2), but I need to verify this.

 

[marcus]

Matter era: Y&#039; = \frac{3}{2}(1- Y^2/Y_\infty^2)
Radiation era: Y&#039; = 2 (1- Y^2/Y_\infty^2)
I will double check that in the next couple of posts.
Jorrie you are right about the odd time unit. It is an ignominious practice to introduce a working time unit. I'm thinking "d" stands for "deci" as in deciBell, deciLiter, deciMeter. I just find it very comfortable to work in that size (a tenth of a billion years) time unit while we are developing this approach but in the end it will probably be changed to giga year (Gy) or mega year.

Now we've reached the point where we need to consider radiation density. So I will quote the overly verbose post #360 which has some stuff about deriving the Y' equation and we will see how it is different in the radiation-era, compared with the matter-era. I think your coefficient of 2 in the preceding post, instead of 3/2 is right. In fact you must be somewhat of an expert in this kind of thing, I expect, having built your calculator. But I want to go thru it myself. First I will quote this incredibly long post
Scroll down to the RED highlight.
===quote post #360===
For people who would like to see the (elementary calculus) way the equation for Y' is derived:

Y' = (1/H)' = - H'/H² = 4πGρ/H² = 4πGρY²

All this uses is H' = - 4πGρ, which we know from a previous post. And then we use the Friedmann equation to get an expression for ρ, and substitute it into the above.

ρ = (3/8πG)[1/Y² - 1/Y_∞²]

Y' = (3/2)[1/Y² - 1/Y_∞²]Y²
= (3/2)[1- Y²/Y_∞²]

The square of the ratio 139/163 is a familiar model parameter that is often quoted, namely 0.728.
Here we give it a new significance as determining the current rate of increase of the Hubble time.
One minus 0.728, namely 0.272, multiplied by 3/2, is this number 0.41... we're talking about.

The current value of the Hubble time is increasing 0.4 year per year. Or 0.4 century per century. Or 0.4 Gy per Gy. That is (if the rate were steady it would result in) an increase from 13.9 billion years to 14.3 billion years in a billion year interval...

Since the equations here are based on introductory work in post#313, which is several pages back, I will bring forward part of that post:

=====quote post#313======
By definition H = a'/a, the fractional rate of increase of the scalefactor.

We'll use ρ to stand for the combined mass density of dark matter, ordinary matter and radiation. In the early universe radiation played a dominant role but for most of expansion history the density has been matter-dominated with radiation making only a very small contribution to the total. Because of this, ρ goes as the reciprocal of volume. It's equal to some constant M divided by the cube of the scalefactor: M/a³.
Differentiating, we get an important formula for the change in density, namely ρ'.
ρ' = (M/a³)' = -3(M/a⁴)a' = -3ρ(a'/a) = -3ρH
The last step is by definition of H, which equals a'/a

Next comes the Friedmann equation conditioned on spatial flatness.
H² - H_∞² = (8πG/3)ρ
Differentiating, the constant term drops out.
2HH' = (8πG/3)ρ'
Then we use our formula for the density change:
2HH' = (8πG/3)(-3ρH) = - 8πGρH, and we can cancel 2H to get the change in H, namely H':

H' = - 4πGρ
====endquote====
...
==endquote==

 

[marcus]

So we have to change this:

Differentiating, we get an important formula for the change in density, namely ρ'.
ρ' = (M/a³)' = -3(M/a⁴)a' = -3ρ(a'/a) = -3ρH
​

into this:

Differentiating, we get an important formula for the change in density, namely ρ'.
ρ' = (M/a⁴)' = -4(M/a⁵)a' = -4ρ(a'/a) = -4ρH
​

And we have to change this:

Next comes the Friedmann equation conditioned on spatial flatness.
H² - H_∞² = (8πG/3)ρ
Differentiating, the constant term drops out.
2HH' = (8πG/3)ρ'
Then we use our formula for the density change:
2HH' = (8πG/3)(-3ρH) = - 8πGρH, and we can cancel 2H to get the change in H, namely H':

H' = - 4πGρ
​

into this:

Next comes the Friedmann equation conditioned on spatial flatness.
H² - H_∞² = (8πG/3)ρ
Differentiating, the constant term drops out.
2HH' = (8πG/3)ρ'
Then we use our formula for the density change:
2HH' = (8πG/3)(-4ρH) = - (32/3)πGρH, and we can cancel 2H to get the change in H, namely H':

H' = - (16/3)πGρ
​

And finally, we have to change this:

Y' = (1/H)' = - H'/H² = 4πGρ/H² = 4πGρY²

All this uses is H' = - 4πGρ, which we know from a previous post. And then we use the Friedmann equation to get an expression for ρ, and substitute it into the above.

ρ = (3/8πG)[1/Y² - 1/Y_∞²]

Y' = (3/2)[1/Y² - 1/Y_∞²]Y²
= (3/2)[1- Y²/Y_∞²]
​

into this:

Y' = (1/H)' = - H'/H² = (16/3)πGρ/H² = (16/3)πGρY²

All this uses is H' = - (16/3)πGρ, which we know from a previous post. And then we use the Friedmann equation to get an expression for ρ, and substitute it into the above.

ρ = (3/8πG)[1/Y² - 1/Y_∞²]

Y' = (16/3)πG (3/8πG)[1/Y² - 1/Y_∞²]Y²


Y' = 2[1/Y² - 1/Y_∞²]Y²
= 2[1- Y²/Y_∞²]​

=======================

Yes, you were right about the coefficient. The arithmetic is simply that 16/3 x 3/8 = 2.

Hot dog! (sorry about long-windedness, old guys have to proceed deliberately, so easy to make mistakes )

The real hard thing, I think, is when the radiation and matter densities are around the same order of magnitude, neither one dominant. Then it seems like a BLEND of the two different Y' equations.

 

[marcus]

I guess one way to do it would be simply to DECLARE that the radiation era lasted until, say, 300,000 years, and up to that point go with Y(t )= 163 tanh(t*2/163)

and then at that point in time, which I'm still calling 0.003 d, you switch over to using
Y(t )= 163 tanh((t+.001)*1.5/163)
where the time advance of 0.001 d is for continuity. So that one function takes over where the other left off. They match, at transition time, with that adjustment.
Y(t )= 163 tanh((.003+.001)*1.5/163) = 163 tanh((.003)*2/163)
So then let's see how things look at the last scattering time of 390,000 years (0.0039 d):

Y(.0039 )= 163 tanh((.0039+.001)*1.5/163) = 0.0735

Not bad! It's sort of an OK Hubbletime for last scatter.
===================
And here's the corresponding redshift:
((tanh ((.0039+.001)*1.5/163)^-2 - 1)/(1/.7272-1))^(1/3)
gives 1094
===================
It's not really satisfactory. The equation is really too simple to deal properly with early universe. Conventionally I think one says radiation era lasts briefer, e.g. to 54,000 years. Presumably there is a gradual transition with the large amount of energy in the form of light playing a significant role.
So I see no clearcut place where you change over from coefficient 2 to 1.5.
Just declaring a transition at time 300,000 is a kludge. But with such simple tools, and limited possibilities, it might be the best way out.

 

[Jorrie]

I guess one way to do it would be simply to DECLARE that the radiation era lasted until, say, 300,000 years, and up to that point go with Y(t )= 163 tanh(t*2/163)

and then at that point in time, which I'm still calling 0.003 d, you switch over to using
Y(t )= 163 tanh((t+.001)*1.5/163)
...
It's not really satisfactory. The equation is really too simple to deal properly with early universe. Conventionally I think one says radiation era lasts briefer, e.g. to 54,000 years. Presumably there is a gradual transition with the large amount of energy in the form of light playing a significant role.
So I see no clearcut place where you change over from coefficient 2 to 1.5.
Just declaring a transition at time 300,000 is a kludge. But with such simple tools, and limited possibilities, it might be the best way out.

I also experimented a bit on existing spreadsheets as a reference, but with similar mixed success. Could get it close to right for the CMB-era and for Now, but with uncomfortable deviations en-route.

I woke up (yes, it's rise and shine time here already), with the subconscious telling me the following: convert the new inputs to the old inputs behind the scenes and perform the proper Friedman calculation à la the old calculator. Then pump out a simplified set of results and give your approximation equations in info popups, with some caveats.

How would this sit with you?

PS: It's also a lot less work... ;-)

 

[marcus]

I imagine I would be delighted.
The main requirement is that the project make sense to you (the calculator builder) and that you be satisfied with the results.

I was intrigued by the idea of a calculator (possibly quite simple) that would not outwardly involve H and Omega_Lambda. It would let you control the current growth rate and the future growth rate by entering two Hubbletimes, instead.

Anything that does this, however it does it, seems like an interesting pedagogical tool. No outward reference to "km/s" and "dark energy". Instead: a current percentage growth rate and a future asymptotic one.
I think Hubbletimes are probably the easiest handles to use, to specify current and future growth rates.
So I immediately think of being able to input, say, 13.9 Gy and 16.3 Gy. and maybe the presentday ratio of matter to radiation, and that's it. After that I can convert any redshift to several outputs, or an expansion age.

But you are the one who has worked on cosmology calculators so you will have your own criteria and ways to reckon how well things will communicate to the user. You're the one with experience, so you be the judge.

As I recall I found that 13.9 and 16.3 corresponded to something like 70.35 km/s per Mpc and
.7272 for Omega_Lambda.
Then Omega matter was 0.2727188
And Omega radiation was 0.0000812
So they added up to 0.2728, giving flatness.

That means the matter/radiation ratio was 3359*. (Which is why you need to go back to a redshift of around 3350 or 3360 in order for them to be on par with each other.)

So if I was using your calculator I would like to be able to input 13.9 billion and 16.3 billion and a number like 3360.
Then the calculator would secretly change (13.9, 16.3, 3360) inputs for the existing program and proceed from there.
I'm getting curious to see how this takes shape! It is like constructing a new "front end" for something you already have that runs well.

Let me check that my memory was accurate about that 70.35...
I paste this into google:
1/(13.9 billion years) in km/s per Mpc
Yes! it immediately comes back with "70.3463274 (km/s) per Mpc"
Since it is internal, perhaps better to use 70.34633 or something like that.
And google also tells me that 13.9^2/16.3^2 = 0.727200873

*calculated from 2727188/812=3358.6...

 

[marcus]

Updated expansion table based on treating now and future Hubbletimes 13.9 and 16.3 billion years as exact. In the following calculations, 0.7272 is the square of their ratio, and a time unit (d) is used which is a tenth of a billion years. Age at present 137.574 d. The model has to be considered a "toy" because the simplified equations give only rough approximation at times before 1 d. I'm trying out an additional column for the scalefactor a_t = 1/(1+z_t) which shows the growth of a generic distance.
Google codes used are:
a_t = (1/.7272-1))^(1/3)/((tanh((t+.001)*1.5/163))^-2 - 1)^(1/3)
z_t = ((tanh((t+.001)*1.5/163))^-2 - 1)/(1/.7272-1))^(1/3) - 1
Y_t = 163 tanh((t+.001)*1.5/163)
To calculate with a code, paste blue expression into google, replace t by an expansion age, and press =.

EXPANSION HISTORY, 139/163 MODEL.

   Age          Redshift        Hubble time        Scale factor
   t (d)           z[SUB]t[/SUB]             Y[SUB]t[/SUB] (d)               a[SUB]t[/SUB] 
   0.0030         1252           0.00600             0.0008  
   0.0039         1093           0.00735             0.0009
   1            30.553            1.501              0.032
   2            18.883            3.001              0.050
   3            14.175            4.500              0.066
   4            11.526            5.999              0.080
   5             9.794            7.496              0.093         
   6             8.558            8.992              0.105
   7             7.624           10.487              0.116
   8             6.888           11.980              0.127
   9             6.291           13.471              0.137
  10             5.796           14.959              0.147
  20             3.269           29.667              0.234
  30             2.243           43.892              0.308
  40             1.659           57.431              0.376
  50             1.273           70.121              0.440
  60             0.992           81.848              0.502
  70             0.776           92.542              0.563
  80             0.603          102.176              0.624
  90             0.459          110.762              0.685
 100             0.337          118.341              0.748
 110             0.232          124.973              0.812
 120             0.139          130.732              0.878
 130             0.057          135.703              0.946
 131             0.049          136.159              0.953
 132             0.041          136.609              0.960
 133             0.034          137.052              0.967
 134             0.026          137.488              0.974
 135             0.019          137.918              0.981
 136             0.012          138.341              0.989
 137             0.004          138.757              0.996
 137.574         0.000          138.993              1.000

For times earlier than 0.0030 d (before year 300,000) these "radiation era" google codes are preferable:
z_t = ((tanh(t*2/163)^-2 - 1)/(1/.7272-1))^(1/3) - 1
Y_t = 163 tanh(t*2/163)
The same caveat applies. Only a rough approximation to early universe behavior.
Some notes on the table:
z : fractional amount distances and wavelengths have increased while light was in transit. Arriving wavelength is 1+z times original.
H=1/Y: Hubble expansion rate. Distances between stationary observers grow at this fractional rate--a certain fraction or percentage of their length per unit time.
H(per d) : fractional increase per convenient unit of time d = 10⁸ years.
Y=1/H: Hubble time. 1% of the current Hubble time is how long it would take for distances to increase by 1%, growing at current rate. At present, Y is 139 d = 13.9 billion years.
Hubble time is proportional to the Hubble radius = c/H: distances smaller than this grow slower than the speed of light. At present, the Hubble radius is 13.9 billion ly (proper distance)

The Hubble law describes the expansion of distances between observers at rest with respect to the background of ancient light and the expansion process itself: Observers who see the ancient light and the expansion process approximately the same in all directions, e.g. no Doppler hotspots.
The field of an observer's view can be thought of as pear-shape because distances were shorter back then. Here is a picture of an Anjou pear.
http://carrotsareorange.com/wp-content/uploads/2010/05/pear-anjou.jpg
Here is Lineweaver's spacetime diagram:
http://ned.ipac.caltech.edu/level5/March03/Lineweaver/Figures/figure1.jpg
The upperstory figure, with horizontal scale in proper distance, shows the lightpear outline.
Here is Lineweavers plot of the growth of the scalefactor R(t), which models the growth of all distances between observers at universe-rest (at rest with respect to background.)
http://ned.ipac.caltech.edu/level5/March03/Lineweaver/Figures/figure14.jpg
The dark solid line is according to standard model parameters. Various other cases are shown as well.

 

[Jorrie]

So if I was using your calculator I would like to be able to input 13.9 billion and 16.3 billion and a number like 3360.
Then the calculator would secretly change (13.9, 16.3, 3360) inputs for the existing program and proceed from there.
I'm getting curious to see how this takes shape! It is like constructing a new "front end" for something you already have that runs well.

Yes, I have similar ideas, but I now tend towards using three Hubble times: start of the matter era (Y_m=0.1 My)*, the present day (Y_0=13900 My) and the maximal Hubble time (Y_inf=16300 My) as default constants (changeable by user). Then let the user specify a range of either z or t, with a required incremental step. The calculator then to produce one of your tables automatically. The project requires redesigning and programming new input and output ends for the existing 'LCDM engine'. Will see how far this takes us...

*Start of the matter era is at radiation/matter equalization around z=3300, with t=0.057 My and Hubble time 0.1 My. This is enough info to bring in the radiation energy effect accurately.

 

[marcus]

Yes, I have similar ideas, but I now tend towards using three Hubble times: start of the matter era (Y_m=0.1 My)*, the present day (Y_0=13900 My) and the maximal Hubble time (Y_inf=16300 My) as default constants (changeable by user). Then let the user specify a range of either z or t, with a required incremental step. The calculator then to produce one of your tables automatically. The project requires redesigning and programming new input and output ends for the existing 'LCDM engine'. Will see how far this takes us...

*Start of the matter era is at radiation/matter equalization around z=3300, with t=0.057 My and Hubble time 0.1 My. This is enough info to bring in the radiation energy effect accurately.

An extremely interesting idea. IMHO you could design a whole semester course around that kind of teaching/learning resource.

It is intriguing to think of a calculator that you put 3 model parameters into and it then generates a table, going along the t-scale step by step.

I would experiment with using the scalefactor as an alternative lefthandcolumn variable, instead of z.

The thing is, when someone says we see this galaxy with redshift 4, if you want to look it up you could just think: scalefactor a = 1/(1+z) = 1/5 = 0.2
I am looking at this galaxy as it was when distances were 20% of what they are today.
that galaxy I'm looking at is back in the days of scalefactor 0.2

I would want undergrad students to be familiar with converting z that they read into scalefactor, and then putting scalefactor into calculator.

So I would put the scalefactor along the lefthand column of the second option table. And next to it the time (derived from that scalefactor).

IMO we observe the scalefactor just as directly (from the spectrum of incoming light) as we observe the z. they are just different algebraic versions of the same basic datum.

And a is increasing, it is a lot more like t. You've got to follow your own craft-sense. But I think I'll try making a table with increments of scalefactor a and see what it looks like.
Maybe it's a bad idea for some reason I don't see yet.

Your idea of making something that will accept 3 inputs like (13900, 16300, 0.1 My) and from those 3 inputs crank out a table (even a small table, with specified range and stepsize) is terrific.
==================
EDIT: have to go to the trainstation but just want to write down this google code (no time to check it)
t+0.001 = (163/1.5)arctanh sqrt(a^3/(a^3 -1 + 1/.7272))

EDIT: google calculator does not have arctanh, or artanh, the inverse of tanh. I will try to implement using the analytical expression for arctanh, which employs the natural logarithm ln(x)
t+0.001 = (163/3)ln(1+(1+(1/.7272-1)/a^3)^-.5) - (163/3)ln(1-(1+(1/.7272-1)/a^3)^-.5)

When I try this with a = .5 I get that t+.001 = about 59.7. Seems right, so I'll make a table based on the scalefactor.

Scalefactor   Age Gy
.1                0.56
.2                1.58
.3                2.88
.4                4.37 
.5                5.97
.6                7.61
.7                9.24 
.8               10.82  
.9               12.33
1.0              13.759             
1.1              15.11
1.2              16.38
1.3              17.57
1.4              18.70

Scalefactor 1.4 refers to a time in the future when they will observe OUR light with wavelengths 140% of what they were when our stars emitted the light, today. Somewhere in some galaxy they will point a telescope at the Milkyway and see light emitted by the sun and other stars today. And the wavelength will be extended by a factor of 1.4.
The table shows that that will happen about 5 billion years from now.

 

[Jorrie]

I would experiment with using the scalefactor as an alternative lefthandcolumn variable, instead of z.
...
EDIT: google calculator does not have arctanh, or artanh, the inverse of tanh. I will try to implement using the analytical expression for arctanh, which employs the natural logarithm ln(x)
t+0.001 = (163/3)ln(1+(1+(1/.7272-1)/a^3)^-.5) - (163/3)ln(1-(1+(1/.7272-1)/a^3)^-.5)

When I try this with a = .5 I get that t+.001 = about 59.7. Seems right, so I'll make a table based on the scalefactor.

Scalefactor   Age Gy
.1                0.56
.2                1.58
.3                2.88
.4                4.37 
.5                5.97
.6                7.61
.7                9.24 
.8               10.82  
.9               12.33
1.0              13.759             
1.1              15.11
1.2              16.38
1.3              17.57
1.4              18.70

As a matter of fact, most cosmo-calculators use a as the core independent variable that they ramp up or down, normally for a from ~0 to 1, i.e. the best part of post-inflation expansion history. My calculators do the same. It is simply easier to set it up so that a=1 (identically) comes out of the numerical integration.

The equation that you use will again be very close for a > .01, but will start to deviate for smaller a, due to the hotter radiation at early times. Our proposed simplified calculator should be fairly accurate for a down to around one millionth or so, provided that we get the radiation component in correctly.

 

[marcus]

As a matter of fact, most cosmo-calculators use a as the core independent variable that they ramp up or down, normally for a from ~0 to 1, i.e. the best part of post-inflation expansion history. My calculators do the same...

...down to around one millionth or so, provided that we get the radiation component in correctly.

That's good news. As I see it a is a directly observed quantity. z is just an algebraic variant of a. When you look at the hydrogen line in a spectrograph and see by what ratio the wavelength is enlarged you could just as well consider that you are reading a off the instrument as think of it as reading z, which is just z = 1/a - 1.

So a is a directly observed (not model dependent) quantity, and it is also a key variable in the calculation. The fact that it's this way gives IMO a solid empirical feel to the situation.

I think in my dream calculator you would have a box for 1+z, and a box for a. They are reciprocals of each other, and putting a number in either would work. It wouldn't have a box for z. If a student reads somewhere that a galaxy was observed with redshift 3, then he or she should know to put in 4, or mentally convert that to 1/(3+1) = 0.25 and put 0.25 into the a box.

It's getting late here. Maybe some fresh ideas in the morning. I should try to make this more compact:

t+0.001 = (163/3)ln(1+(1+(1/.7272-1)/a^3)^-.5) - (163/3)ln(1-(1+(1/.7272-1)/a^3)^-.5)

t_a = (16.3/3)ln((1+(1+(1/.7272-1)/a^3)^-.5)/(1-(1+(1/.7272-1)/a^3)^-.5))

I've dropped the little time adjustment of .1 million year, and put a decimal point into the 16.3 so it gives answers in billions of years as in that brief table.
So now, associated with every directly measurable scalefactor a we have the estimated expansion age TIME when distances were that size, or when the light was emitted.
And our handle on how fast the world was expanding at that epoch is the Hubbletime. Basically a sort of linear "doubling time" for distance growth. To every scalefactor a in the past there should be an associated growthrate.
Let's add Hubbletime Y_a to that brief table:
Y_a= 16.3(1+(1/.7272-1)/a^3)^-.5

Scalefactor     Age (Gy)          Hubbletime (Gy)        ?
 a                 t[SUB]a[/SUB]                  Y[SUB]a[/SUB]                Δ[SUB]a[/SUB]
.1                0.56                 0.84              5.38  
.2                1.58                 2.36              3.46
.3                2.88                 4.22              2.61         
.4                4.37                 6.22              2.11
.5                5.97                 8.15              1.77
.6                7.61                 9.85              1.53
.7                9.24                11.26              1.35
.8               10.82                12.38              1.21
.9               12.33                13.24              1.09
1.0              13.759               13.900             1.00        
1.1              15.11                14.40  
1.2              16.38                14.77
1.3              17.57                15.06
1.4              18.70                15.29

So, to read something off the table, it says that a little over 2 billion years from now there will be people in another galaxy looking at our Milkyway galaxy with their telescope and they will observe that the hydrogen wavelengths are 30% longer (than hot hydrogen rainbow wavelengths in their lab) and they will say "Hmmm, distances back then when the light was emitted were 1/1.3 what they are today..." And they will be wondering how long ago that was rapidly distances were expanding back then so they will look at their table and say "Hmmm, that was 2.2 billion years ago, and in those days it took only 139 million years for a distance to grow 1%, whereas now it takes 150.06 million years, so expansion was more rapid back then."

While I can still edit I'll try adding an interesting incremental distance number that can be calculated at each scalefactor a:

Δ_a = 1/sqrt(.2728*a + .7272*a^2)
It may turn out to have no use, but the table has room for another column

 

[Jorrie]

*Start of the matter era is at radiation/matter equalization around z=3300, with t=0.057 My and Hubble time 0.1 My. This is enough info to bring in the radiation energy effect accurately.

I had a look at this from the Friedman POV. All we need are the three Hubble times: rad/matter equality, Y_eq = 0.1 My, Y_now = 13900 My and Y_inf = 16300 My, plus the redshift for equality, z_eq = 3350. Assuming flatness, all three present energy densities are then calculable. The rest is just a matter of standard calculation and presentation.

The most troublesome one to find from the inputs is the present radiation energy density, but because it is very small (relatively), the following seems to work well:
\Omega_r = \left(\frac{Y_{now}}{Y_{eq}}\right)^2 a^4 - \Omega_m a - \Omega_\Lambda a^4
where: a=(1+z_{eq})^{-1}, \Omega_\Lambda = (Y_{now}/Y_{inf})^2, \Omega_m\approx 1-\Omega_\Lambda (provided \Omega_r \ll 1).

I have checked this by means of a spreadsheet and it looks promising, with errors far below the input accuracies throughout the redshift range of interest, zero to 3350.

Edit: Surprisingly, the rather complex equation may be unnecessary, because a simple \Omega_r \approx \Omega_m/z_{eq} seems to be just as accurate.
Marcus mentioned this relationship in a prior reply.

 

[marcus]

Jorrie, I didn't see your last post (#416) until just now when I posted mine. I'm glad to hear this, it's looking good!

I had a look at this from the Friedman POV. All we need are the three Hubble times: rad/matter equality, Y_eq = 0.1 My, Y_now = 13900 My and Y_inf = 16300 My, plus the redshift for equality, z_eq = 3350. Assuming flatness, all three present energy densities are then calculable. The rest is just a matter of standard calculation and presentation.

The most troublesome one to find from the inputs is the present radiation energy density, but because it is very small (relatively), the following seems to work well:
\Omega_r = \left(\frac{Y_{now}}{Y_{eq}}\right)^2 a^4 - \Omega_m a - \Omega_\Lambda a^4
where: a=(1+z)^{-1}, \Omega_\Lambda = (Y_{now}/Y_{inf})^2, \Omega_m\approx 1-\Omega_\Lambda (provided \Omega_r \ll 1).

I have checked this by means of a spreadsheet and it looks promising, with errors far below the input accuracies throughout the redshift range of interest, zero to 3350.

I added to that brief table based on the scalefactor.
It's intended to be the 13.9/16.3 model we've been concentrating on and I think the first three columns are right, but am not sure about the accuracy of the last two, the distances to a source at the given scalefactor.
EDIT: For clarity I will write out the google calculator expression for t_a in LaTex:
t_a = \frac{16.3}{3}ln\left((1+(1+(1/.7272-1)/a^3)^{-.5})/(1-(1+(1/.7272-1)/a^3)^{-.5}) \right)
Here's the expression as used in the calculator:
t_a = (16.3/3)ln((1+(1+(1/.7272-1)/a^3)^-.5)/(1-(1+(1/.7272-1)/a^3)^-.5))
Here's the expression for the Hubble time:
Y_a= 16.3(1+(1/.7272-1)/a^3)^-.5

Scalefactor  Age (Gy)    Hubbletime (Gy)   Proper distance to source (Gly)
 a    1/a-1    t[SUB]a[/SUB]            Y[SUB]a[/SUB]               D[SUB]now[/SUB]         D[SUB]then[/SUB]
.1    9.0     0.56          0.84              30.9         3.09
.2    4.0     1.58          2.36              24.0         4.79
.3    2.333   2.88          4.22              18.7         5.62       
.4    1.5     4.37          6.22              14.5         5.79
.5    1.0     5.97          8.15              10.9         5.45
.6    0.666   7.61          9.85               7.9         4.74
.7    0.428*  9.24         11.26               5.4         3.78
.8    0.25   10.82         12.38               3.3         2.63
.9    0.111  12.33         13.24               1.5         1.36
1.0   0.0    13.759        13.900              0.00        0.00  
1.1          15.11         14.40  
1.2          16.38         14.77
1.3          17.57         15.06
1.4          18.70         15.29

*0.428571429
(13.9 Gy, 16.3 Gy, flat) → (70.3463, 0.7272, 0.2728)

 

[Jorrie]

It's intended to be the 13.9/16.3 model we've been concentrating on and I think the first three columns are right, but am not sure about the accuracy of the last two, the distances to a source at the given scalefactor.

The "13.9/16.3 model" is perfectly accurate for the scalefactors that you have shown. It's only from a < 0.01 that accuracy becomes an issue due to radiation density.

I also like the scalefactor "input column", because one can go as far into the future as desired ( a > 1). If we use redshift, it would be negative for the future and that's an awkward concept.

PS: look at the edit I've made to post #416.

 

[marcus]

I found out a minor detail about Wright's calculator. when you tell it .7272 and .2728 it actually uses those values, although it REPORTS that it is using ..727 and .273.

IOW it rounds off what it says the model parameters are that it is using, but you can see the difference in the results. It's just a minor thing, but it's convenient.

You can actually get that calculator to use (70.3463, .7272, .2728) even though it may look as if you can't (because of this rounding off.)

I saw the edit in #416, thanks for the mention :-)
it makes sense. That aspect (getting the right radiation component) looks very hopeful.
What I'm not sure about is how you will be able to build a different "front end"

EDIT: For clarity I will write out the google calculator expression for t_a in LaTex:
t_a = \frac{16.3}{3}ln \frac{1+(1+\frac{1/.7272-1}{a^3})^{-.5}}{1-(1+\frac{1/.7272-1}{a^3})^{-.5}}
t_a = \frac{16.3}{3}ln \frac{1+(1+(1/.7272-1)/a^3)^{-.5}}{1-(1+(1/.7272-1)/a^3)^{-.5}}
t_a = \frac{16.3}{3}ln\left(\frac{(1+(1+(1/.7272-1)/a^3)^{-.5})}{(1-(1+(1/.7272-1)/a^3)^{-.5})} \right)
t_a = \frac{16.3}{3}ln\left((1+(1+(1/.7272-1)/a^3)^{-.5})/(1-(1+(1/.7272-1)/a^3)^{-.5}) \right)

Finally, here's the expression I paste into google calculator for t_a:
(16.3/3)ln((1+(1+(1/.7272-1)/a^3)^-.5)/(1-(1+(1/.7272-1)/a^3)^-.5))

Here's the corresponding expression for the Hubble time Y_a:
16.3(1+(1/.7272-1)/a^3)^-.5

 

[Jorrie]

That aspect (getting the right radiation component) looks very hopeful.
What I'm not sure about is how you will be able to build a different "front end"

If one ignores the small curvature caused by the present radiation density when you determine the radiation density parameter for matter equality, the 'front-end' is actually straightforward. From Y_now, Y_inf and z_eq, the three energy densities are as before:
\Omega_\Lambda = (Y_{now}/Y_{inf})^2; \Omega_m \approx 1-\Omega_\Lambda; \Omega_r\approx \Omega_m /(z_{eq}+1) and H_0 = 1/Y_{now} of course.
This we send to the full version's numerical integration module. Strictly speaking, we should also input the cosmic time (t) for r-m equality, but provided we start the integration early enough (well before r-m equality), we can set the starting time to zero.

EDIT: For clarity I will write out the google calculator expression for t_a in LaTex:
t_a = \frac{16.3}{3}ln\left((1+(1+(1/.7272-1)/a^3)^{-.5})/(1-(1+(1/.7272-1)/a^3)^{-.5}) \right)

Finally, here's the expression I paste into google calculator for t_a:
(16.3/3)ln((1+(1+(1/.7272-1)/a^3)^-.5)/(1-(1+(1/.7272-1)/a^3)^-.5))

Here's the corresponding expression for the Hubble time Y_a:
16.3(1+(1/.7272-1)/a^3)^-.5

I've included this in the draft spreadsheet for the "lean model" that I mailed to you for comment. The spreadsheet shows your time approximations to be within 1.5% for z < 100, 15% for z < 1100 and 40% at r-m equality, good enough for early learning purposes.