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

[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.there's 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))
which tells us the galaxy position three billions years ago.Which it now 5 billion year away from us ?