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

[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 ?

 

[marcus]

c=1
light travels 1 lightzeit per zeit
that is, by itself without counting the help of expansion.

 

[marcus]

Isn't it true that sinh^2/3(1.5*0.8) is about equal to 1.3?
So we could make it easy for ourselves and just take a(t) to be sinh^2/3(1.5t)/1.3

That is approximately right. Is that OK?

So the distance the light has come, how far it is now from home, measured in lightzeits, is

$$D(0.54) = 1.3\int_.54^.8 \frac{dt}{\sinh^{2/3}(1.5t)}$$

we need to see how to put that into numberempire (I have dropped the c, it is just 1)

 

[RyanH42]

I didnt see your edit.So I made wrong.Now I can see

 

[RyanH42]

0.3042 lightzeit ?

 

[marcus]

actually I never did that integral at numberempire. I don't know if that website has the sinh function. : ^)
We will see. Before now I always used a different version of the integral that involved a calculus trick called "change of variable".
But we should try this version.

What should go in the box? You see the box I mean? It is where you type the function to be integrated. Like in Example 1 the function was x*x

1.3*sinh(1.5*t)^(-2/3)

Let's try that, and make the limits .54 and .8, and in the variable box put t. The default is x but we are using t, either is OK,it does not matter.

I am slow. you already went ahead. Now I want to try

I get
==quote==
Integral of 1.3*sinh(1.5*t)^(-2/3) by t on the interval from .54 to .8:
.3046035045325962
==endquote==

YES! I see that is the same as what you got.

 

[RyanH42]

Why sinh^(-2/3)(1.5t) why there's minus sign front of 2/3 an there will be 1.3/sinh^(2/3)(1.5t) isn't it

 

[RyanH42]

We are the doind same thing so so sorry.

 

[marcus]

What you say works. X^-1 is the same as 1/X
sorry?
Nobody should be sorry! This is very good progress. I am very happy about this.

There is still that "change of variable" integral trick. Maybe we can discuss that later. You may have other things you want to do now. I will get some more coffee.

the point of the "change of variable" is when the light comes in it does not tell us what time (like 0.54) it was emitted and started on its way to us

instead the light tells us HOW MUCH IT HAS BEEN STRETCHED

I think you know how to get the time from the stretch factor, using the natural logarithm, ln.
this is very good. maybe that is the best way to go.

but one can also introduce a variable S = 1.3/a(t) and integrate dS instead of dt.

that seems to me overly complicated right now. But that is how I have been doing it. I'll think about this a little.

 

[RyanH42]

I learned lots of thing today.I guess I finished Part 2 of article

 

[RyanH42]

Thank you I will be online in PF next 3 hours.Whenever you ready or we can do it tomorrow.

I tried to do it myself.I don't know its right but here it is.

$D(t)=1.3ln(a(t))$, $(c=1)$
e^D(t)/1.3=a(t)
So we can put $S=1.3/a(t)$
e^D(t)/1.3=1.3/S

I tried to use another method but it seems its not true.

dS/dt=dS/d(a(t))*d(a(t))/dt=-1.3a(t)^-2a'(t)

$D(t)=1.3∫dt/a(t)$
D(t)=-1.3∫dS/a(t)1.3a(t)^-2a'(t)
D(t)=-∫dS/a(t)^-1a'(t)
$D(t)=-∫dS/H$

I made a mistake somewhere.You can check any time you want
Again thanks

 

[marcus]

I tried to use another method but it seems its not true.

dS/dt=dS/d(a(t))*d(a(t))/dt=-1.3a(t)^-2a'(t)

$D(t)=1.3∫dt/a(t)$
D(t)=-1.3∫dS/a(t)1.3a(t)^-2a'(t)
D(t)=-∫dS/a(t)^-1a'(t)
$D(t)=-∫dS/H$

I made a mistake somewhere.You can check any time you want
Again thanks

I think this is true. This looks right:
$D(t)=-∫dS/H$
Do not worry about the minus sign. It will be taken care of by exchanging the upper and lower limits of the integral.

The fact is that S and t increase in opposite senses. as you go back in time (so t gets smaller) the amount of size increase from t to present increases (so S gets larger)

S = 1 corresponds to the present, and say S=1.4 (I forget actually, something like that) corresponds to the earlier time 0.54

So the distance turns out to be something like ∫₁^1.4 dS/H

I need to check that number 1.4. You see why the minus sign goes away? Because now the integral is working backwards in time. the direction has been reversed, upper and lower limits have been exchanged.

 

[RyanH42]

Is H as a function of S ?

If you had another job just tell me I can ask my questions tomorrow.
I will be not sad.

You are teaching me something online that's hard.

 

[RyanH42]

H=[0.44/a^3+1]^(0.5)
Just I need to replace S=1.3/a(t)

 

[RyanH42]

I understand.I guess today is enough for me I ll check again what I learned today.

Edit:I understand why there's minus term

 

[marcus]

Is H as a function of S ?

YES! Exactly! H can be expressed very nicely as a function of S!

It is the square root of (S/1.3)³ +1

this comes from identities relating the functions sinh cosh and tanh

You can see that (S/1.3)³ must be 1/sinh² (it is just a lot of stupid algebra)

And so we have (1 + sinh²)/sinh²

and it is an identity that 1+sinh² = cosh²

I like the French word for stupid, it is bête which means like an animal, beastly. A lot of beastly algebra. but we live by it.

 

[marcus]

Oh, OK. I see you have had enough to think about, for today. That is good. One learns more comfortably and securely if one has time to assimilate.
I too, like to rest.

 

[RyanH42]

I am checking what we did and I have a problem let's suppose we have send a signal 0.5 zeit and we want to calculate how light traveled 0.1 zeit later(0.6 zeit)

So I made D(t)=∫dt/sinh(1.5*t)^(2/3)/sinh(1.5*0.5)^(2/3) integral between 0.5 to 0.6

Is that true

 

[RyanH42]

And for the last example I used

D(t)=∫dS/[(S/1.3)^3+1]^(1/2) time intervals between 1.27751 and 1.4809(I choose time intervals 0.5 and 0.6 zeit) so S=1.3/a(t) so S for 0.5 zeit will be 1.3/sinh(1.5*0.5)^(2/3)=1.4809
for 0.6 zeit 1.3/sinh(1.5*0.6)^(2/3)=1.27751
I found 0.137
My question is did I forget something
In 188 post I wrote something.Did we need 0.44 there ?

And If my calculations are true then I didnt understand what's this number presents ? Whats the meaning of it ?

 

[RyanH42]

a(t)/1.3 means If the distance of two objects is R now then the distance in t will be a(t)/1.3*R.

But S is just opposite of it.I can't visualize it.

 

[marcus]

There is a reason that S is good. Of course admittedly is simply the reciprocal. If a(t) = 1/2 then as you say the distance back then is half the size it is now.
S = 1/a = 2. It just tells us that the distance now is twice the size it was back then.

We can ask "why have S?" It is simply 1/a, so why do we need it?
Also we can say that a(t) is intuitive. We can graph it and picture it---it is the picture of the universe getting bigger!

So why have S? I will explain

For me, at least, S is a good handle on the world because we observe it.
S is what the light tells us when it comes into our telescope.
At the home galaxy of the light, the hydrogen atoms are emitting light of certain wavelengths. they make a pattern. When the light comes in we see that same pattern but each wavelength is twice as big.
Ahah, we say, S for that light is 2.

 

[RyanH42]

Ahhhh.Yes I got it.Thank you

S=2 we can find a(t) and from there when the light emitted.And from there of course the distance the light traveled.

So we are looking wavelenght and we are deciding S.And that tells us how the universe expands until that time from now

 

[marcus]

I am checking what we did and I have a problem let's suppose we have send a signal 0.5 zeit and we want to calculate how light traveled 0.1 zeit later(0.6 zeit)

So I made D(t)=∫dt/sinh(1.5*t)^(2/3)/sinh(1.5*0.5)^(2/3) integral between 0.5 to 0.6

Is that true

No, it should be
D(t)=∫dt/sinh(1.5*t)^(2/3)/sinh(1.5*0.6)^(2/3) integral between 0.5 to 0.6

And probably we should put parenthesis to make clear what we are dividing
D(t)=∫dt/(sinh(1.5*t)^(2/3)/sinh(1.5*0.6)^(2/3)) integral between 0.5 to 0.6

There are some astronomers who live in time t = 0.6
some light emitted at time 0.5 comes to them
how far has this light traveled AS OF THEIR PRESENT day, as they measure distance in t=0.6?

I suppose that is how we should think of it. We should put ourselves in the place of those astronomers back then at time 0.6

 

[marcus]

Ahhhh.Yes I got it.Thank you

S=2 we can find a(t) and from there when the light emitted.And from there of course the distance the light traveled.

So we are looking wavelenght and we are deciding S.And that tells us how the universe expands until that time from now

By tradition, or some historical accident, Astronomers use the number z = S-1 which they call "redshift". So in many of their formulas and equations you will see z+1,...z+1,...z+1,...all over the place. I find it more mathematically convenient to follow Jorrie's example and use S. When he programmed the math for the Lightcone calculator he used S. It is the actual ratio that distances and wavelengths are magnified by.
Just so you are aware of this aspect of technical language. S = z+1

 

[RyanH42]

The other way is impposible to calculate ? We want to calculate the distance traveled by light from the future.I mean we are emitting light right now.So there must be way to calculate it ?

 

[RyanH42]

Yeah I am seeing everywhere z+1=S everywhere .Really everywhere.