# Illustrating the role of the two doubling times (linear d.t. params in cosmic model)

1. Aug 24, 2012

### marcus

I'll have to edit and add to this later today, have to go out in few minutes.
The standard cosmic model depends strongly on two distance growth rates, the present and the longterm future ones. The easiest most intuitive handle on these growth rates are the two Hubble times = the two percentage growth rates flipped over.

The present value of the Hubble time is about 13.9 billion years and this can be thought of as a linear doubling time. I.e. how long it would take any cosmological distance to grow an amount equal to its own length, if it continued at constant speed: its current instantaneous linear rate (without "compounding")

A simpler way to think of it is that 1% of the Hubble time is how long it would take distances to grow by 1% at their present rate.

So what I want to do in this thread is show in a couple of TOY MODEL FORMULAS how the two Hubbletimes play a role in the model and influence the expansion history distance growth curve.

We can and I think will do a lot better than these toy (hand calculator) formulas. They lose accuracy when used back before about redshift 9 or 10 (the earliest galaxies). But nice thing about explicit formulas is that you see the two Hubbletimes, the main parameters, revealed in a transparent way. This can serve as a kind of introduction to the more complete and accurate online cosmology calculators.

A nice feature here is the use of a new online calculator which (unlike the google calculator) automatically displays formulas in neat LaTex-like form. When you type in a formula all on the same line, with fractions, for instance written (1/a + 1)/b, the calculator keeps your one-line version in the window so you can edit it but also shows a more immediately readable version in the space right above the window.
I'll give an example in the next post.

Last edited: Aug 24, 2012
2. Aug 24, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

To see this feature in operation, please go the online calculator (the first hit if you do a search for "online scientific calculator") http://web2.0calc.com/
and try pasting this formula in:

(((16.3/13.9)^2 - 1)/((tanh(1.5*x/16.3))^-2-1))^(1/3)

Typed on a single line, as I've done, it is cluttered with parentheses and hard to read. However you will see that as soon as it is pasted in, the calculator displays it right above the original version in a neater, easier-to-read form.

$$\left(\frac{(\frac{16.3}{13.9})^2-1}{(tanh(\frac{1.5}{16.3}x))^{-2}-1}\right)^{1/3}$$

Notice the role played by the two Hubbletimes: 13.9 billion years and 16.3 billion years. These parameters control the present, and the eventual longterm, distance growth rates.

What the formula computes, given a time x (in billion years) is the scalefactor a(x)---in other words it tells you the expansion history of a generic distance over time.

To use it, just replace the "x" by a time expressed in billions of years (such as 1, or 2, or 13.759 which is the current expansion age) and press the equals sign.

The answer will appear in the window and you can copy it to clipboard if you want. Then to repeat the calculation, click on the LaTex version of the formula as it appears above the window, and you can substitute something else in for the variable.
========================
1--- 0.1471433... (when the universe was 1 billion years old, distances were 14.7% what they are now)
2--- 0.2342347... (at 2 billion years, distances were about 23% what they are now)
...
...
13.759--- 0.9999836... (at the present age of 13.759 billion years they are of course 100% of their present lengths.:-)

And suppose you wanted to know how much more spread out from each other the galaxies will be in the future, say when the age is 20 billion years. Just put 20 where the placeholder "x" was in the formula and press = again.

20--- 1.5235746... (at age 20 billion years, distances will be 52% bigger than they are today.)

Last edited: Aug 24, 2012
3. Aug 24, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

You can vary the model by putting in different values for the two Hubbletime parameters. But what we are using in this sample (13.9 and 16.3 billion years) are quite close to current best estimates based on WMAP and other data.

The parameter 16.3 billion years is a form of the cosmological constant Λ Our model simply assumes the constant as a slight residual curvature inherent in spacetime geometry. We do not yet know the cause, it may show up as the quantum structure of geometry becomes better understood. We do not need to include any "dark energy" in the inventory of matter and radiation to explain this residual negative curvature. Formally it is equal to
-Λ = - 3(1/16.3 billion light years)2
To repeat somewhat, it is this constant which determines to eventual longterm steady expansion rate of the the universe, towards which the present rate is slowly tending.

The reason why our simple model loses accuracy when you go very far back in time is quite interesting. It has to do with the coefficient 1.5 in the formula, which is correct only as long as the great majority of the density in space is due to matter. If the energy density were all due to radiation, and not to matter, the correct coefficient would be 2.0.
But the balance between radiation and matter (in the makeup of the overall density) changes with expansion. Expansion not only reduces the number of photons per unit volume but also extends the wavelength of each individual photon, reducing its energy.

So if you go back to when the scalefactor was 0.001, and distances were a thousand times shorter than today, you find that matter density is 10003 times greater, but radiation density is 10004 times greater! The balance has shifted in favor of radiation. Each photon has a 1000 times shorter wavelength and so is 1000 times more energetic.

Even though radiation is an almost negligible fraction of the density now, and the 1.5 number is correct (so too the cube and cube root exponents you will see occurring in the formulas!) if you go back to scalefactor a = 0.001 this is no longer the case. Indeed if you go back even farther to a=0.0003 (distances roughly 3300 times shorter than today) the two components of the density, matter and radiation, are roughly equal.

So that is when our formulas, in this simple toy model, go wrong. They don't have a means for changing the coefficient 1.5 and moving towards 2.0 as the balance between matter and radiation shifts in the early universe (or adjusting the relevant exponents either.) Still, they serve rather nicely as long as we don't push back in time too close to the beginning of expansion.

I think at this point we might look at the inverse of the formula that I hope some folks have already tried out. What it did, given a time, like 1, 3, 5 billion years, was output a scalefactor showing the growth of a generic distance over time.
Now suppose we specify a scalefactor a, like a=0.1 when distances were 1/10 what they are at present, and we ask what time did that happen? What was the expansion age when a=0.1?

Last edited: Aug 24, 2012
4. Aug 24, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

To see how to work back from a scalefactor, like 0.1,to the time when distances were that size, please go the online calculator http://web2.0calc.com/
and try pasting this formula in:

(16.3/1.5)atanh((((16.3/13.9)^2 -1)/x^3+1)^-.5)

You will see that as soon as it is pasted in, the calculator displays it in a neater, easier-to-read format.

$$\frac{16.3}{1.5}atanh\left(\left(\frac{(\frac{16.3}{13.9})^2-1}{x^3}+1\right)^{-.5}\right)$$

Notice the role played by the two Hubbletimes: 13.9 billion years and 16.3 billion years. These parameters control the present, and the eventual longterm, distance growth rates. Also there's the coefficient 1.5 appropriate as long as the contribution by radiation is small compared wth that of (dark and ordinary) matter.

What the formula computes, given a scalefactor x (in billion years) is the time t(x) that it occurred.

To use it, just replace the "x" by a a scalefactor, some number between 0 and 1 but I normally choose ones between 0.1 and 1 , the answer will be a time expressed in billions of years

===================
.1--- 0.5608... (when distances were a tenth their today size, expansion was 0.56 billion (= 560 million) years old.)
.2--- 1.5813... (when distances were a fifth of their present size, expansion age was 1.58 billion years. )
...
...
1--- 13.75922... (scalefactor 1 corresponds to the present: expansion age= 13.759 billion years.)

And suppose you wanted to know when distances will be twice what they are now. Just put 2 in for x and press equals.

2--- 24.2829...(scalefactor 2 corresponds to an expansion age of 24.3 billion years, or betweeen 10 billion and 11 billion years from now, in the future. )

Last edited: Aug 24, 2012
5. Aug 24, 2012

### Drakkith

Staff Emeritus
Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

Huh, pretty neat. Thanks Marcus!

6. Aug 24, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

Thanks Drakkith! I'm so glad you like it!

7. Aug 25, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

I was pleasantly surprised just now when I used numerical integration in this model to find the comoving radial distance to a source at redshift z=0.298 (as a check)
and found that my answer, 3.875 billion lightyears,
agreed with Ned Wright's calculator to 4 decimal places.

If you want to compare this 13.9/16.3 model with Wright, the parameters to enter in the latter are (70.3463, 0.7272, 0.2728). Those are the equivalents of the two Hubbletimes, 13.9 and 16.3 billion years, flat case. If you put z=0.298 in Wright's, you get Dnow=3.875 Gly.

Using this simple model, which has a present age of 13.759 Gy, I went back to 10.359 Gy in steps of 0.1 Gy and summed the reciprocal of the scale factor.

Dnow = c∫ 1/a dt.

The idea is that during each small time segment dt, the light travels cdt,
and subsequently that bit of distance is expanded by the reciprocal of a(t) the scalefactor at that time. For example if the scalefactor at that moment were 0.1, then the distance the light traveled would, between that time and now, be enlarged by a factor of 10.

Pretty clearly that has to be the integral that gives the present distance to the source!
The model's formula for 1/a(t) is
$$\left(\frac{(\frac{16.3}{13.9})^2-1}{(tanh(\frac{1.5}{16.3}x))^{-2}-1}\right)^{-1/3}$$
Which in single-file version, for online calculators, reads:
(((16.3/13.9)^2 - 1)/((tanh(1.5*x/16.3))^-2-1))^(-1/3)
So I went along in steps of 0.1 Gy, adding up the 1/a value that was more or less in the middle of the interval. As for example
13.759
13.7 1.00427381939236
13.659

The cumulative total, 38.75...Gly needs to be multiplied by 0.1, the length of the interval, whereupon it reproduces the Wright calculator result to four significant figures.

13.759
13.7 1.00427381939236
13.6 2.0158223801626576838
13.5 3.0347207152108601105
13.4 4.0610452200472518050
13.3 5.0948736910868903526
13.2 6.1362853653242438101
13.1 7.1853609614480919241
13.0 8.2421827224594632509
12.9 9.3068344598587501156
12.8 10.3794015994718037020
12.7 11.4599712289879818610
12.6 12.5486321472875230112
12.5 13.6454749156392562779
12.4 14.7505919108543849240
12.3 15.8640773804864639493
12.2 16.9860275001728142270
12.1 18.1165404332174947785
12.0 19.2557163925220629092
11.9 20.4036577049756485777
11.8 21.5604688784229190185
11.7 22.7262566713342503271
11.6 23.9011301653106810452
11.5 25.0852008405629742440
11.4 26.2785826545127844818
11.3 27.4813921236723567751
11.2 28.6937484089686575906
11.1 29.9157734046877656446
11.0 31.1475918312258113562
10.9 32.3893313318443482323
10.8 33.6411225736401211799
10.7 34.9030993529521699267
10.6 36.1753987054436315944
10.5 37.4581610211097981673
10.4 38.7515301644811633129
10.359 (corresponding to scalefactor 0.77056 or redshift 0.298 )

Last edited: Aug 25, 2012
8. Aug 25, 2012

### Jorrie

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

Cool overall description, Marcus. Just note the wording "... given a scalefactor x (in billion years)", which seems wrong.

9. Aug 25, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

Thanks! It certainly is wrong. Too much in a hurry, I copied and pasted sentences from the previous post #2, thinking I'd just make the necessary changes, and forgot to erase that parenthetical phrase.
"What the formula computes, given a scalefactor x, is the time t(x) that it occurred."

The sentence I copy-pasted from post#2 was:
"What the formula computes, given a time x (in billion years) is the scalefactor a(x)---in other words it tells you the expansion history of a generic distance over time."

Dumb

But anyway, I like the 13.9/16.3 toy model formulas a lot! It will be great when there is an implementation of professional-grade cosmos calculator using the two Hubbletimes as parameters!

Last edited: Aug 25, 2012
10. Aug 28, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

I'm going to try a coarse numerical integration to see how close it gets to the correct presentday distance to a galaxy. Suppose the light from a galaxy tells us that it was emitted when the scalefactor was 0.5. Distances and wavelengths have doubled while the light was in transit. So what it the distance to it NOW? (assuming we could freeze the expansion process to give us time to measure by direct means.)

One idea is to add up light traveltime intervals (multiplied by c to give the distance the light would have traveled on its own in a nonexpanding universe) SCALED UP by the reciprocal scalefactor 1/a that tells how much distances have been expanded since that traveltime interval took place. It is not as bad as it sounds. If you go to the online calculator http://web2.0calc.com/
and try pasting this formula in:

(16.3/1.5)(atanh((((16.3/13.9)^2 -1)/(a+.05)^3+1)^-.5) - atanh((((16.3/13.9)^2 -1)/(a-.05)^3+1)^-.5))/a

here is how the calculator should display it:

$$\frac{16.3}{1.5}\left( atanh\left(\left(\frac{(\frac{16.3}{13.9})^2-1}{(a+.05)^3}+1\right)^{-.5}\right) - atanh\left(\left(\frac{(\frac{16.3}{13.9})^2-1}{(a-.05)^3}+1\right)^{-.5}\right)\right)/a$$

I shall add these up for values of a = .95, .85, .75, .65, .55. The last one, for instance, will give the traveltime interval between a=0.5 and a=0.6, scaled up by a midrange value 1/a = 1/0.55, namely: ( t0.6 - t0.5)/a

Well, it turned out to work pretty well. By this rather crude approximation I got that the presentday distance to the galaxy is 10.879 Gly, and using the equivalent model parameters in Wright calculator one gets 10.901 Gly. That is three-place accuracy and I expect one could easily get much better by making a finer division of the interval.

Last edited: Aug 28, 2012
11. Aug 28, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

In the preceding post the distance calculation just involved adding up a sequence of distances, each expanded by an appropriate factor. For the intervals around .95, .85,..., .55, the expanded presentday distances (given by the calculator in more precision but rounded off) were:
1.506+1.777+2.107+2.506+2.983

One can think of it as integrating c∫(1/a)dt between the limits defined by a=0.5 and a=1.0.
Taking smaller divisions of the overall interval would lead to a more accurate result.

To repeat somewhat, the interval cdt, a time interval multiplied by c, gives a distance traveled by the light on its own (during an interval of time in the past) and that interval must be scaled up by the expansion factor 1/a, to give the corresponding presentday distance.

then the sum of all those segments is the total presentday distance.

1.506+1.777+2.107+2.506+2.983= 10.879 ≈ 10.9 Gly, the desired result.

Last edited: Aug 28, 2012
12. Aug 28, 2012

### marcus

Re: Illustrating the role of the two doubling times (linear d.t. params in cosmic mod

I'll try another coarse numerical integration, this time letting the RECIPROCAL of the scalefactor serve as the integration variable. I still like making the scalefactor a the main variable our simple cosmic model runs on. But just for this distance calculation I want to try summing in steps of s = 1/a, instead of a.

Let's see how close it gets to the correct presentday distance to a galaxy. Suppose the light from a galaxy tells us that it was emitted when the scalefactor was 0.5. Distances and wavelengths have doubled while the light was in transit. So what is the distance to it now? (assuming we could freeze the expansion process giving time to measure by direct means.)

I'll do the same thing as before, which was to add up light traveltime intervals (multiplied by c to give the distance the light would have traveled on its own in a nonexpanding universe) SCALED UP by the reciprocal scalefactor s = 1/a that tells how much distances have expanded since that traveltime interval took place. It's just an algebraic rearrangement of the previous formula. The online calculator is http://web2.0calc.com/
and we paste this formula in:

(16.3*s/1.5)(atanh((((16.3/13.9)^2 -1)(s -.1)^3+1)^-.5) - atanh((((16.3/13.9)^2 -1)(s+.1)^3+1)^-.5))

Here is how the calculator should display it:

$$\frac{16.3s}{1.5}\left( atanh\left(\left( ((\frac{16.3}{13.9})^2-1)(s-.1)^3+1\right)^{-.5}\right) - atanh\left(\left( ((\frac{16.3}{13.9})^2-1)(s+.1)^3+1\right)^{-.5}\right)\right)$$

I shall add these up for values of s = 1.1, 1.3, 1.5, 1.7, 1.9. The last one, for instance, will give the traveltime interval between ts for s=1.8 and s=2.0, scaled up by a midrange value s=1.9 namely: s( t1.8 - t2.0)

Earlier this turned out pretty well. By this rather crude approximation I got that the presentday distance to the galaxy is 10.879 Gly, and using the equivalent model parameters in Wright calculator one gets 10.901 Gly. Let's see how it goes this time.

It worked OK! For the stated s values I got:
2.6725+2.4218+2.1722+1.9388+1.7290 adding up to 10.9343, which is again good to 3 significant figures.