From Aeon to Zeon to Zeit, simplifying the standard cosmic model

[marcus]

Jorrie, Wabbit (Wabbit glad you like the looks of the zeit version!) what we seem to be moving towards is a presentation that might at least put some ideas in the heads of Jorrie's constituency even if they don't take hold and work with it right away, and could be used in other contexts as well, that might go like this:
1. an introductory essay along the lines of what Jorrie already wrote. (but possibly even more explicit about "e-fold", for instance "e-fold is like three-fold except with e=2.718 instead of 3, the e-fold time is the time it takes something to expand by a factor of e. If f(t) is any positive increasing function with instantaneous fractional growth rate f'(t)/f(t) constant then it's a fact of calculus that the reciprocal of that rate, f(t)/f'(t), is the function's e-fold time. And the function must be of the form f(t) = e^rtwhere r equals the constant growth rate f'(t)/f(t).")

2. graphic introduction to the zeit version of Lightcone, particularly with a bunch of curves that illustrate the model.

3. the equations in part 1 have hand-calculable solutions that closely approximate the Lightcone tables and curves. In particular the expansion of the universe follows a hyperbolic sine curve: $$a = \frac{\sinh^{2/3}(\frac{3}{2}t)}{1.3}$$ and that relation of distance size to time can be inverted to tell us time as a function of the scale factor a. $$t = \frac{2}{3} \ln \big((1.3a)^{3/2} + \sqrt{(1.3a)^3 + 1}\big)$$

So here's a sample exercise. The present age (as it says in Jorrie's part 1.) is 0.8 zeit. You fall asleep and when you wake up you discover the cosmic microwave background is much colder. It must be some time in the distant future! The CMB is only one tenth its present temperature. what time is it?

Solution. The scale factor a has increased from 1 (present) to 10. So just put that 10 into the previous equation and calculate: 1.3a = 13
$$t = \frac{2}{3} \ln \big((13)^{3/2} + \sqrt{(13)^3 + 1}\big)$$
t will be the age, in zeits, at that future time.

 

[marcus]

As a check, I put in 2/3*ln(13^(3/2)+(13^3 + 1)^(1/2))
and got an age 3.027123...zeits

Again, I put in (sinh(3/2*3.027123)^(2/3)/1.3 and got 10 (correct to six significant figures : ^) so it seems ok.
As a further check, I went to Lightcone Zeit
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7zeit/LightConeZeit1.html
and put in S_lower = 0.1 and got that the corresponding age was 3.0356...zeit.
Close enough! The calculator uses numerical integration and implements the full LambdaCDM standard cosmic model. With simple formulas we should be happy to get within 1 percent and the agreement here is better than 1%.

then main thing is not to do these exercises (except as a check) but to invent some new exercises to challenge readers with. Can you do this? Or do you have to fall back on Ned Wright cosmocalculator, or Jorries Lightcone? Can you hand-calculate the U expansion history?

the Earth formed 0.26 zeit ago, i.e at expansion age 0.54 zeit. Suppose right now somebody with a telescope is studying our galaxy as it was when the Earth formed. What redshift do they measure?

Solution. $$a = \frac{\sinh^{2/3}(\frac{3}{2}0.54)}{1.3}$$

and z = s-1 = 1/a - 1

I put in (sinh(3/2*0.54)^(2/3)/1.3 and got a = 0.71786...
the stretch s = 1/a = 1.39302.. so redshift .39
I'm experimenting to see how it might be to just give novices one formula, initially, and make the first batch of exercise challenges all be be applications of that one formula (and its inverse).

 

[marcus]

Another sample exercise.
the Earth formed 0.26 zeit ago, i.e at expansion age 0.54 zeit. Suppose right now somebody with a telescope is studying our galaxy as it was when the Earth formed. How far away are they?Solution. use the same hyperbolic sinh formula for the scale factor. c=1 so distance comes out in lightzeits.
$$D_{now} = \int_{.54}^{.8} \frac{1.3dt}{\sinh^{2/3}(\frac{3}{2}t)}$$
Each tiny step cdt along there way gets expanded (between then and now) by a factor of 1/a. And they all get added up.
I googled "definite integral calculator" which got me to numberempire.com and put in
1.3*(sinh(1.5*t))^(-2/3)
for the function to integrate, with t running from 0.54 to 0.8
and it gave 0.30460...
That basically means the astronomer is currently 0.3 lightzeit from us.

 

[marcus]

Another sample problem. You are studying a galaxy at redshift z = s-1 = 2. that is to say wavestretch factor s = 3. Suddenly you notice an exploding supernova! What time in history did it happen? What was the expansion age when that star exploded?

Solution. a = 1/3. Use the inverse of our one-formula-with-many-uses.
$$t = \frac{2}{3} \ln \big((1.3a)^{3/2} + \sqrt{(1.3a)^3 + 1}\big)$$
1.3a = 1.3/3 = 0.4333
putting in 2/3*ln(.4333^(3/2)+(.4333^3 + 1)^(1/2)) gives 0.18765... zeit
Better check that, seems like a long time ago. 0.188 zeit. Over 0.61 zeit ago!
I put S_lower = 3 in LightconeZeit and it said 0.18989...
Not too good but we have to make allowances for the fact that we're using a rough 1.3 instead of a more accurate larger 1.31146... That would make the 0.188 larger and improve the approximation.
Still pretty good considering.

( (17.3/14.4)² - 1)^-1/3 = 1.31146...

 

[marcus]

I'd welcome any comments including critical, at this point. What I see coming out of this discussion is a rough 3-part outline of a "simple introduction to (H_∞) cosmology" for a fairly wide audience, anybody who's interested basically. Part A is something like Jorrie's brief account. Part B is graphic. You show the curves that come out of equations like those in Part A, and give the reader a chance to get hands-on experience with the model via Lightcone.
and part C might involve hand-calculation exercise to convey a sense of empowerment---"look at what you can do on your own, you can hand-calculate facts of expansion history..."

Part C (some calculator challenges) might or might not appeal. I'm not sure at this point.

I'm thinking now that Part A needs one more thing tacked on at the end.
1. We already get that H(t) ≈ coth(1.5 t) and we get Jorrie's equation #4, I think it is: H² - 1 ~ s³

2. From this we could go a step further and prove that a(t) ~ sinh^2/3(1.5 t)

That is because of the identity coth² - 1 = 1/sinh²
which gives you 1/sinh²(1.5 t) ~ s(t)³
You just flip this over and you have the result. The constants in these proportionalities are things like .4433 and 1.3115 which are simply related to each other.
The identity comes from coth² - 1 = (cosh² - sinh²)/sinh² = 1/sinh²
because of that thing about cosh² - sinh² ≡ 1

So from Part A we can get the multipurpose formula a(t) = sinh^2/3(1.5 t)/1.311..., or for rough work simply use 1.3
that is the thing that graphs like a pair of antelope horns and shows a slight inflection between deceleration and acceleration. Nice curve.
So I think it might be nice to base the calculator challenges mostly on that formula. (but that might be wrong, there are simpler formulas one might use that don't involve a 2/3 power.)

 

[Jorrie]

So I think it might be nice to base the calculator challenges mostly on that formula. (but that might be wrong, there are simpler formulas one might use that don't involve a 2/3 power.)

Yea, for simplicity and given that S = z+1 = 1/a, I'm a little biased towards basing everything on

H = 1 + 0.443 S^3 and t = \ln(\frac{H+1}{H-1})/3

probably because that's the way I started my Blog entry. I actually shied away from the hyperbolic functions as far as possible because they do not come naturally to my niche readership. We discussed this briefly in https://www.physicsforums.com/threads/from-aeon-to-zeon-simplifying-the-standard-cosmic-model.811718/page-3#post-5121465.

Values like proper distances might require integration, but that (I think) is unavoidable. Quite a few of the exercises that you mentioned requires only the two equations above.

 

[marcus]

Yea, for simplicity and given that S = z+1 = 1/a, I'm a little biased towards basing everything on

H^2 = 1 + 0.443 S^3 and t = \ln(\frac{H+1}{H-1})/3

probably because that's the way I started my Blog entry. I actually shied away from the hyperbolic functions as far as possible because they do not come naturally to my niche readership. We discussed this briefly in https://www.physicsforums.com/threads/from-aeon-to-zeon-simplifying-the-standard-cosmic-model.811718/page-3#post-5121465.

Values like proper distances might require integration, but that (I think) is unavoidable. Quite a few of the exercises that you mentioned requires only the two equations above.

Hi Jorrie, I put the exponent on the H² in the quote. I'm often forgetting to square the H and things like that but fortunately we can go back and edit. Your judgement about style and notation has been a guide to me recently, about using zeit and e-fold and keeping terminology simple.
Also your understanding of what communicates to that particular audience.
Maybe we should develop two versions, one avoiding the hyperbolic sine (and the 2/3 power) and one employing it.

The two equations you like are remarkably versatile, and the one expressing H as a function of S² is just what we need to do the integral for distance D_now based on an observable quantity S. That simplifies the exposition. We don't observe time, we observe the stretch S.

On the other hand the curve of expansion history a(t) is intuitively appealing to some newcomers (not the engineers necessarily, maybe another niche). Because they have been wondering about it. What is this expansion I've been hearing about? What is this "acceleration" they talk about?
and the a(t) curve shows it. So it has that plus, to balance the off-putting strangeness of the hyperbolic function. Maybe there should be both.

How if you push ahead with a "4engineers" version? Maybe replace "doubling" by "e-fold time" in part A, and show them some graphs with LightconeZeit, as per part B? Add one or two sample problems if you feel like it? You could make a development thread here at PF which would allow Wabbit and me to comment and help, if you want. And transfer to your blog later. Or build it there, in situ, if you prefer.

 

[Jorrie]

On the other hand the curve of expansion history a(t) is intuitively appealing to some newcomers (not the engineers necessarily, maybe another niche). Because they have been wondering about it. What is this expansion I've been hearing about? What is this "acceleration" they talk about?
and the a(t) curve shows it. So it has that plus, to balance the off-putting strangeness of the hyperbolic function. Maybe there should be both.

Hi Marcus, thanks for correcting the H-typo.
One can obviously get a(t) without reverting to the hyperbolic functions, but I agree that it becomes a more intimidating equation, as per this attachment that you have posted before.

Mostly, I will just use the two equations H^2 = 1 + 0.443 S^3 and t= \ln(\frac{H+1}{H-1})/3
without solving them together. I would rather explain how they can be used in an algorithm to calculate specific values of a(t) =1/S and plot expansion curves. My feeling is that for beginners, it takes too much explaining on how the hyperbolic solution is arrived at, but it's just a personal preference.

PS (edit): I changed the Intro text box and the Time tooltip somewhat and then uploaded LightCone7zeit into the place of the original LightCone7z, so you do not have to change your signature link. The 7zeit link will still work for compatibility with older posts.

 

[marcus]

...PS (edit): I changed the Intro text box and the Time tooltip somewhat and then uploaded LightCone7zeit into the place of the original LightCone7z, so you do not have to change your signature link. The 7zeit link will still work for compatibility with older posts.

Jorrie thanks! Lightcone is at the core of this modest little project. I think you were right about a one-syllable name for the unit and am glad you modified Lightcone7z accordingly. In your blog, is it possible that the niche readership could be introduced to the term "e-fold time" to replace doubling (i.e. two-fold) time?

Mostly, I will just use the two equations H^2 = 1 + 0.443 S^3 and t= \ln(\frac{H+1}{H-1})/3
without solving them together. I would rather explain how they can be used in an algorithm to calculate specific values of a(t) =1/S and plot expansion curves. My feeling is that for beginners, it takes too much explaining on how the hyperbolic solution is arrived at, but it's just a personal preference..

In fact I'm eager to see the next chapter. I think your judgment (which you say is just a personal preference : ^) may well be right. Those two equations are simple and not alienating, I think, and they suffice for natural sorts of exercise problems. I hope you pursue development of that approach and I'll try to help with suggestions if I have any, or comment.
It looks to me as if we should have a parallel development. I'm trying to think of ways to make sinh^2/3(1.5t) palatable and not scary.

 

[RyanH42]

You are preparing a PF insight post I guess.I am in high school which it makes hard to understand the things.I see great effort here Marcus and of course Jorrie and the others.

I saw your part 1 LCMD basic math calculations.And it has been deleted.(I don't know why).Or you are working on it.

The problem is the who you are writing to it.You are making just the theory part I guess.

My suggestion is If you get a number or equation please explain how it evolve(If you are doing that then there's no problem)

I just want to say for an amateur I can't understand the main idea.

 

[RyanH42]

Your math is understandble but teaching part is not enough(For insight post).My first language is not english(Maybe that's the reason).
I asked some questions which shows I didnt understand maybe I should wait part 2 and part 3.

 

[Jorrie]

I saw your part 1 LCMD basic math calculations. And it has been deleted.(I don't know why). Or you are working on it.

The Insights post has not been deleted, AFAIK. https://www.physicsforums.com/insights/approximate-lcdm-expansion-simplified-math/

My suggestion is If you get a number or equation please explain how it evolve(If you are doing that then there's no problem)

Ryan, my Insights post is an article, not a tutorial, so it is does not include mathematical derivations. You are welcome to ask questions as you have done in the comments section: https://www.physicsforums.com/threads/approximate-lcdm-expansion-in-simplified-math-comments.823929/

I am contemplating writing an appendix with the reasoning behind the specific equations and the derivations, but that does not belong in an article and will come later. Marcus and Wabbit have provided many of the reasoning in the various threads of PF if you want to search for them, but they are quite scattered. Part of the motive for my Insights series is to gather things together for easier reading and reference.

 

[RyanH42]

I am so so sorry.I will not ask any question anymore.

 

[Jorrie]

I am so so sorry. I will not ask any question anymore.

No, No! You are entitled to ask questions - this is the main purpose of forums like this one. For the Insights posts, stick to questions about the specifics of the thread, like you have done up to now. That's a good way to learn.

You may sometimes be directed to go and read certain answers that have already been covered somewhere else, but if physics questions are to the point and not obvious, they are mostly answered here.

 

[marcus]

This thread might be a good place to ask basic questions. I can try to answer.

If you don't dislike my asking, what is your first language? You said it is not English. Sometimes it helps me to know what the other person's main language is.

I noticed in another thread (where I did not want to jump in, because it was not my thread) that you did not seem to understand the exponential function e^x.

that is how you calculate with an instantaneous growth rate like H.

If you have a distance like 1 km. And H is the growth rate of 10% per year. Then it is not true that after one year the distance is 1.1 km.

The way you calculate the distance after one year is you paste this into google:
e^(.1*1)
that is e^(rate*time) = e^(.1 per year x 1 year)

Do you understand what I am saying?

 

[marcus]

If you use a different calculator we might experience confusion. If you use the google calculator then we are both using the same one and it is easier to talk. But with something simple like e^(.1*1) we should both get 1.10517...

Try it and see.

And try it for longer periods of time, like 5 years and 10 years. What do you get?

If you have a distance of 1 km, and it is growing 0.1 per year (instantaneous growth rate) then how big is it after 5 years?

You should get 1.6487...

If you don't get 1.6487... then we are in big trouble! Big confusion!

 

[RyanH42]

I get that number Marcus.My english level is good actually.I don't know.The problem is I am just cannot understand the idea.Which it seems it my problem.I will wait next articles to understand the iasue better.Today maybe 1 hour later I will going to come and I will going to ask my questions.
Thank ypu

 

[RyanH42]

H is know 1/144% per million year so it mean now distance R grows e^(1/144.t) t in million years.So 100,000 thousand years R will be increase e^(1/144.0.1)=1,00069.So I will multiply R*1,00069=A distance in 100.000 thousand years later.

Time in zeit means t/17.3(unit billion years)
Normalized H means H/17.3(I guess).Which current value is 1.2 zeit^-1.


I understand the other parts.

H=1.2 zeit^-1=1/144% per million year or 1/14.4 billion year.
Is that mean H=1/14.4 billion years ?
Can you check my idea

Thank you.

 

[Jorrie]

If you have a distance like 1 km. And H is the growth rate of 10% per year. Then it is not true that after one year the distance is 1.1 km.

The way you calculate the distance after one year is you paste this into google:
e^(.1*1)
that is e^(rate*time) = e^(.1 per year x 1 year)

Marcus, my feeling is that the e-folding is complicating the understanding part a bit, especially for beginners. I would rather relate it to things that high school students know about, e.g. interest calculation and monetary growth. Granted, it uses an approximation, with discrete periods like months or years, which is not quite accurate for an instantaneous rate. But then, this whole approach is an approximation of the LCDM model, which is again an approximation of reality.

Because a million years is such a small time interval in the cosmological times we are working with here, I see no harm in using the calculation I showed to Ryan in comment: https://www.physicsforums.com/threa...simplified-math-comments.823929/#post-5173700. I agree that we should gradually introduce the e-fold idea of expansion.

 

[marcus]

...
Because a million years is such a small time interval in the cosmological times we are working with here, I see no harm in using the calculation I showed to Ryan in comment: https://www.physicsforums.com/threa...simplified-math-comments.823929/#post-5173700. I agree that we should gradually introduce the e-fold idea of expansion.

OK you are the best judge of the timing, I think. Eventually we have to get to the understanding that H is not tied to a particular interval of time like a year, or a million years, or something else, because it is a continuous growth rate. So the only way to compute it really precisely is e^Ht.
But the key word is "gradually introduce". So I will keep quiet and not mess things up by intruding.

One thing I'm curious about though. Who is the painter of the lovely avatar picture? I have the feeling that it is Spanish of maybe 18th or 19th century. But I don't really know, it could be painted anywhere in the world, in any century---I just have a feeling that it is a portrait of a Spanish lady by a contemporary of Goya or Ingres. Would Ryan be willing to tell us who the painter is, and where he found the image? Or even just the painter's name and I could look it up.

 

[RyanH42]

You did not the answer to my question.

I think my picture is so special isn't it.I found it google plus someone share the picture.And I like it.
I don't know where it came from

 

[RyanH42]

Here the full picture

 

[marcus]

Thank you for the Askar painting! I will look for more Askar paintings.

You did not the answer to my question.

I think my picture is so special isn't it.I found it google plus someone share the picture.And I like it.
I don't know where it came from

What question exactly?

 

[RyanH42]

I shared something.And I want to know its true or not true

Your are welcome

 

[marcus]

Ryan the problem is with timing. eventually you have to learn that with a continuous growth rate, say for simplicity it is constant, the thing grows as
e^Ht

So you have to know how to multiply H and t

If H is 0.07 per billion years
and t is 3 billion years
then if you multiply together, H x t, the "per billion years" and the "billion years" CANCEL
and you just get 0.21

and then you calculate e^0.21 and that is the answer.

The important thing is to keep account of the units, make sure they cancel. don't confuse million and billion.
=======================
However since you are in High School, you say, you may not be used to the function e^x and the idea of a continuous growth rate.
Many people are confused by that kind of growth rate and they want to work with some small fixed unit of time.
If I jump into this conversation and say "e^x" to you then this causes a DANGER OF CONFUSION because Jorrie and I would be saying different things and you might get hopelessly mixed up.

Let's do the same thing with a small fixed unit of time. million years.
Then H is 0.00007 per million years
and t is 3000 million years.

So every million years the thing grows by a fraction 0.00007 and becomes 1.00007 of its previous size. Do you understand?
This happens 3000 times. So at the end, the things size is
(1.00007)³⁰⁰⁰ multiplied by its size at the very beginning. Do you understand?

So you must calculate that. Using google calculator it means putting this into the window:
1.00007^3000
When I put that in the window and press return, I get 1.234. What do you get?

Now the surprise is that this is the same answer you get with e^.21.
If you put e^.21 into google and press return, you also get 1.234

A continuous growth rate of 0.07 per billion years works the same as a discrete step-by-step growth rate of 0.00007 per million years, using a small fixed time interval (instead of e^x) as long as the fixed time interval is small enough.
======================
I think your question was "Is what I said right?" You said some wrong calculation and then said "is that right?" the answer I guess is "no". But the real answer is to look at what I just wrote. It is the right way to calculate the growth at a constant rate of 0.07 per billion years. I don't remember what you said but it was not like this.

I must go look for paintings by the painter Askar. That venus picture is great!

 

[RyanH42]

I get 1.23

 

[RyanH42]

I am confused in the H part.Whats the exactly H, 1/144 per million years 1/14,4 billion years 0.07 ,0.00007 I am confused in H so i can't move forward

 

[RyanH42]

And 1.2 zeit ^-1.Theres so many terms

 

[RyanH42]

you can tell me one of them and I will use it always

 

[RyanH42]

H is always changing.I used in my example 100.000 year which 0.1 million year and H=1/144 % per million year.And then I multiply them.I can't see any wrong in my equation (Except H I guess)

 

[RyanH42]

Or I should write 1/14400*0.1 ?

 

[marcus]

I get 1.23

Did you get that from 1.00007^3000

or did you get it from e^.21 ?

Try the one you didn't try before, to make sure that both ways give 1.234

 

[RyanH42]

Did you get that from 1.00007^3000

or did you get it from e^.21 ?

Try the one you didn't try before, to make sure that both ways give 1.234

Same answer

 

[marcus]

Or I should write 1/14400*0.1 ?

You may be getting the idea! 1/14400 is the same as 1/144 percent! So it looks like you are using the unit of a million years.

so that means we must measure time in units of million years (so the units cancel). Must be very careful about that.

So you are answering the question about growth in 0.1 million years. That is 100,000 years. But that is such a small interval of time!

Why don't you use billion years consistently? Then you should write 1/14.4 * 0.1 and we are talking about 0.1 billion years.

Now you have to get google to calculate e^{1/14.4 * 0.1} but look, 1/14.4 * 0.1 is the same as 1/144
so put this into the window:
e^(1/144)

 

[RyanH42]

1,0069

 

[RyanH42]

I understand H=1/144% per million year=1/14400 per million year=0.007% per million year=0.00007 per million year=1/14.4 per billion year=0.07 per billion year

 

[marcus]

you can tell me one of them and I will use it always

Same answer

Well I will tell you my preference, but you may have to use different language talking with different people. For example Jorrie may want to talk in different terms, not using e^x.

My preference is always use zeit, and never say "percent". I used to say stuff like "1/144 percent per million years" but now I think percent is very confusing to people.

I want always to give the presentday H as 1.2 per zeit
and suppose I want to know the growth in 0.1 zeit at that constant rate. Then I simply say e^1.2*0.1 = e^0.12
So I put this into google:
e^.12

If I want to know what the growth would be in 0.2 zeit at that constant rate then I put this into google:
e^.24

That would be my preference.

 

[marcus]

I understand H=1/144% per million year=1/14400 per million year=0.007% per million year=0.00007 per million year=1/14.4 per billion year=0.07 per billion year

Very good. Now that you understand that clearly you can decide which notation you prefer.
Do you like
1.00007^3000
or do you like e^.21

These are both adequate ways to say what the growth would be at today's rate (if it stayed constant) for 3 billion years.
Or alternatively you could move over to zeit and then 3 billion years would be 3/17.3 zeit
But 1.2*3/17.3 is again 0.21.
So you would again calculate e^.21 to get the answer (this time using zeits)
e^Ht = e^{1.2 per zeit x (3/17.3) zeit} = e^1.2*3/17.3 = e^0.21

 

[RyanH42]

e^.21

 

[RyanH42]

I guess I am the only one who reads the article and makes comment

 

[marcus]

The original painter was William-Adolphe Bouguereau, who was born in 1825. When I guessed a contemporary of Ingres and Goya I got the general period right.

the person called "Askar" (Alexander T. Scaramanga) is someone who has digitized the original painting by Bouguereau and (perhaps as a joke) he claims the digital version as his own work and says that the Aphrodite is "often misattributed to Bouguereau". It is really really beautiful

I'm curious about what your first language is. (I asked earlier.) Would it be all right to say? Or do you wish to keep it secret for some reason?

 

[RyanH42]

Thanks for information.She is really really beautiful as you said.

 

[marcus]

...Now that you understand that clearly you can decide which notation you prefer.
Do you like
1.00007^3000
or do you like e^.21
?

e^.21

 

[RyanH42]

I wanted to keep in secret my first language

 

[marcus]

That's all right, it can be a secret.
Since you are getting along with e^x , I want to introduce the "e^x" analogs of the trigonometric functions sine and cosine.

These are simple functions made of e^x which have the same SYMMETRY as the trig functions. We put a letter h
after the name sin(x) and cos(x) because this is the traditional way to distinguish them from the ordinary trig functions$$\sinh(x) = \frac{e^x - e^{-x}}{2}$$ $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$$$\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$
Did you already meet these functions in high school?
If you did not meet sinh, cosh, and tanh, did you already work with the ordinary trig functions sin(x) etc.?
I assume you did but I ask just to be sure.

 

[RyanH42]

I know them

 

[RyanH42]

a=sinh^2/3(3/2x) then a'=cosh^-1/3(3/2x)
If we divide them a'/a=coth(3/2x)=H

Here x unit is zeit I guess.

D=ra(t).So every distance will grow according to a=sinh^2/3(3/2x) this equation

 

[RyanH42]

That's all right, it can be a secret.
Since you are getting along with e^x , I want to introduce the "e^x" analogs of the trigonometric functions sine and cosine.

These are simple functions made of e^x which have the same SYMMETRY as the trig functions. We put a letter h
after the name sin(x) and cos(x) because this is the traditional way to distinguish them from the ordinary trig functions$$\sinh(x) = \frac{e^x - e^{-x}}{2}$$ $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$$$\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$
Did you already meet these functions in high school?
If you did not meet sinh, cosh, and tanh, did you already work with the ordinary trig functions sin(x) etc.?
I assume you did but I ask just to be sure.

Not too much but I can handle them.Its seems not too hard.

 

[marcus]

a=sinh^2/3(3/2x) then a'=cosh^-1/3(3/2x)
If we divide them a'/a=coth^-1(2/3x)=H

perfect! except
a'=sinh^-1/3(3/2x)cosh(3/2x)
and except for the coth at the very end
a'/a=coth(3/2x)=H

There is something you learn in differential calculus called (in English) "the chain rule" that enables you to take the derivative of NESTED functions like f(g(x)) where you first do g(x) and then put the result of that into f( . )

a=sinh^2/3(3/2x) involves doing sinh and then doing X --> X^2/3
so the functions are nested, one inside the other
taking the derivative involves the chain rule
the derivative of f(g(x)) is f'(g(x)) g'(x)
the derivative of the first multiplied by the derivative of the second.

 

[RyanH42]

If an object 5 billion years from at time 3 billion years ago for us then the equation becames D=ra(t) we get 5.sinh^2/3(3/2.(10.8/17.3))
Wich tells us the galaxy position three billions years ago.Which it now 5 billion year away from us ?