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

[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 ?

 

[RyanH42]

So If I try to calculate that objects position 0.1 billion years later position.Then the same thing 5sinh^2/3(3/2(13.9/17.3)) and that's equal(I guess) e^Ht=e^(0.07.0.1)

 

[RyanH42]

I go too fast I guess.

 

[marcus]

Let's try to talk only in zeits (for time) and lightzeits( for distance). Talking the time time about "billions of lightyears" is a bother, I think.
Let's use easy numbers.

The present is 0.8 zeit.
How much did distances expand between the time the Earth was forming (around 0.54 zeit) and now? By what factor did they expand?

That's easy. You just have to calculate the ratio a(.8)/a(.54)

Here is something you can paste into google:

sinh(1.5*0.8)^(2/3)/sinh(1.5*0.54)^(2/3)

Can you think of an even simpler example to work? Simple examples are good.

 

[marcus]

By what factor will distances expand between NOW and ONE ZEIT FROM NOW?
That is another very easy exercise. It is good to do several for practice. now = 0.8 and one zeit from now in the future is 1.8
sinh(1.5*1.8)^(2/3)/sinh(1.5*0.8)^(2/3)

Distances will be almost 3 times what they are at present. You can find the more precise figure.

 

[marcus]

Our galaxy disk formed around time t = 0.29 zeit and the Earth formed much later at t = 0.54 zeit.

By what factor did distances expand in the time between those two events?

 

[RyanH42]

sinh(1.5*0.54)^(2/3)/sinh(1.5*0.29)^(2/3)

I found 1.591.This number means If we call scale factor 1 at 0.29 zeit in 0.54 zeit scale factor will be 1.591.So distance R in 0.29 zeit will be R*1.591in 0.54 zeit.

Discovering universe is the greatest thing.
(I hope my idea is true )

 

[RyanH42]

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.

I know derivatives,simple partial derivatives(very simple ones) and integral(Not too much but enough to understand many applications)
If I made a mistake here probably that's reason is I calculate something wrong,The reason cannot be knowladge. I am curious person and I want to everything about cosmology.Problem I am learning too fast and that causes sometimes wrong results.

 

[marcus]

sinh(1.5*0.54)^(2/3)/sinh(1.5*0.29)^(2/3)

I found 1.591.This number means If we call scale factor 1 at 0.29 zeit in 0.54 zeit scale factor will be 1.591.So distance R in 0.29 zeit will be R*1.591in 0.54 zeit.

Discovering universe is the greatest thing.
(I hope my idea is true )

YES!
I fell asleep early last night around 10 pm pacific time and did not see your posts. I just woke up and came downstairs, it is around 6 am pacific. I am very happy to see several Ryan posts! 1.59 is exactly right. The galaxy disk formed at around 0.29 zeit (we think) and then later when the Earth formed, large-scale distances (not in solar system or within our local group of galaxies which is held together by gravity but REALLY large-scale distances) had grown to about 1.6 times their earlier size.

 

[marcus]

...

I am curious person and I want to everything about cosmology.Problem I am learning too fast and that causes sometimes wrong results.

Yes, I understand that. It is good to be curious and learn fast.

Let f(x) = x^2/3
then f'(x) = (2/3) x^-1/3

this is just an application of the general rule that the derivative of xⁿ is nx^n-1

now the chain rule says f(g(x)) derivative is f'(g(x))g'(x)

so the derivative of (g(x))^2/3 is (2/3)g(x) g'(x)

there is one other detail. In this case (the distance growth in universe) the function g(t) is sinh(1.5t) so again by chain rule we have
g'(t) = 1.5 cosh(1.5t)
a factor of 1.5 comes out when we take the derivative
so that 1.5 cancels the 2/3 that appeared earlier.
there are really two applications of the chain rule here

 

[RyanH42]

I am very happy know.

 

[marcus]

I also am happy. It was nice to find these posts when I woke up this morning. Now I will go get some coffee.

 

[RyanH42]

Thats great.Have a nice day

Finally I learned and understand the idea.

Thank you.

 

[marcus]

I got some coffee and am back. I would like to discuss something else, and proceed slowly. You said you had some integral calculus.
Please take a look at the numberempire.com web page where they have a "definite integral calculator" and see if you understand how to use it
http://www.numberempire.com/definiteintegralcalculator.php

 

[marcus]

It's good to start with very simple examples
http://www.numberempire.com/definiteintegralcalculator.php

When you go there, if you scroll down the page to where it says EXAMPLES there is a box you can click on that says "Example 1"
If you click on this it will show the first simple example, how to calculate the definite integral from 0 to 4 of the function x² using the computer. Actually that is such a simple problem that you don't need to use the computer! you would simply evaluate (1/3)x³ at x=4
but integrating more complicated functions sometimes requires using computer, so it is good to know how to do this

 

[RyanH42]

Yeah,I lookd

 

[RyanH42]

It's good to start with very simple examples
http://www.numberempire.com/definiteintegralcalculator.php

When you go there, if you scroll down the page to where it says EXAMPLES there is a box you can click on that says "Example 1"
If you click on this it will show the first simple example, how to calculate the definite integral from 0 to 4 of the function x²

I know that x^3/3 then 4^3/3-0

 

[RyanH42]

It's good to start with very simple examples
http://www.numberempire.com/definiteintegralcalculator.php

When you go there, if you scroll down the page to where it says EXAMPLES there is a box you can click on that says "Example 1"
If you click on this it will show the first simple example, how to calculate the definite integral from 0 to 4 of the function x²

I know that x^3/3 then 4^3/3-0

 

[RyanH42]

I am busy right know can we start 1 hour later.

 

[marcus]

I know that x^3/3 then 4^3/3-0

Good! It is a case where the function is so simple we do not need "numerical integration"---that is we do not need the computer.

but when you do need the computer in more complicated cases it is good to know how.

You type the function to be integrated into the box, at the top of the page. And you type the limits (like 0 and 4). And you press "calculate".

 

[marcus]

I am busy right know can we start 1 hour later.

Sure.

 

[RyanH42]

I am here now

 

[marcus]

Oh, there you are! BTW there is no rush. No need to hurry on to a new topic.
I only wanted to LAY OUT the next topic for when you might want to proceed. It is distance. Calculating the distance that some light is now, from its source galaxy, (how far it has come,with the help of expansion and its own speed)---this requires getting numberempire or some other tool to calculate a definite integral.

For example, you remember the Earth formed about .54 zeit and the present day is .8 zeit. How far would some light travel in that time?

By itself, without help by expansion it could only go .26 lightzeit.

But now that we have the sinh^2/3 function we can also factor in how much each little step the light takes will be enlarged by expansion.

You should not feel any pressure to proceed. Only when you feel curious about this and are ready, and are not too busy with other work. But I only want to set this topic out, so we know what the next thing is.

 

[marcus]

$$D(0.54) = \int_.54^.8 \frac{cdt}{a(t)}$$

here t is some moment in time in the interval between 0.54 and 0.8
and cdt is a little step that the light takes at time t, a little distance.
and a(t) is how much smaller the distances were then, at time t, than they are now.

So that dividing by a(t) scales the little step up to its size now. IOW 1/a is the factor by which the little step at time t is magnified, between time t and the present.

a(t) = sinh^2/3(1.5t)/sinh^2/3(1.5*0.8)

And D(0.54) is how far the light (that was emitted by its source galaxy at time .54) has traveled. How far away it is from home now.

 

[RyanH42]

0.26 lightzeit come from (0.8zeit-0.54zeit)*17.3=0.26 lightzeit.

I do the integral and I found ln(a(0.8))-ln(a(0.54)) will be the answer.I can do that just a second

 

[RyanH42]

0.234 I forget c.c*0.234 lightzeit.But c unit must be in zeit isn't it ?