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

[RyanH42]

First qustion answer
We are in the future that's certain.So Let's call the time T.In this time we measure the CMB and we saw that it was half of its tempature today(Here we can use Wien Law Half of tempature means 2 times wavelenght .Then the equation becomes $2=a(T)/a(t_0)$
So $2=a(T)/1.3$
$a(T)=2.6$ then what will be T sinh(1.5T)^(2/3)=2.6
sinh(1.5T)=2.6^(3/2)
$sinh(1.5T)=4.19237$
$T=1.427$ zeit

If I find a better solution I will going to write it

 

[RyanH42]

D(1.427)=∫dtsinh(1.5*1.427)^(2/3)/sinh(1.5*t)^(2/3) integral 0.8 to 1.427
1.301 lightzeit

 

[RyanH42]

I didnt underatand first and second question is related first.And I can assure you I have never checked what we did before (The equations).If my answers are true then I am certain that you teach me.

 

[marcus]

First qustion answer
We are in the future that's certain.So Let's call the time T.In this time we measure the CMB and we saw that it was half of its tempature today(Here we can use Wien Law Half of tempature means 2 times wavelenght .Then the equation becomes $2=a(T)/a(t_0)$
So $2=a(T)/1.3$
$a(T)=2.6$ then what will be T sinh(1.5T)^(2/3)=2.6
sinh(1.5T)=2.6^(3/2)
$sinh(1.5T)=4.19237$
$T=1.427$ zeit
...

This is correct.
Remember that we are using approximations---for the present I am always saying 0.8 instead of 0.797. And I use the approximation 1.3
instead of something like 1.3115...
So our answers will not agree exactly with Jorrie's calculator
Here I put in S_upper = .5 and number of steps = 0 (just to get one row of the table) and selected D_now
{\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (zeit)&D_{now} (lzeit)&D_{then}(lzeit) \\ \hline 2.000&0.500&1.435182&0.457231&0.914463\\ \hline \end{array}}

So you were transported to a time 1.435 (and we got 1.427 which is close enough since we use approximations like 0.8 for the present.
And you landed on a planet which is NOW 0.457 lightzeit from us.
But it is at a time when distances are TWICE what they are now, so that planet is then 0.914 lightzeit from us.

For the second part you should get around 0.9 lightzeit---because it is an approximation anything near to 0.9 is good. and you showed the correct integral in your answer (but for some reason there was a numerical error.)

If you do the integral again, that you wrote, I think you would get about 0.89 which is close to 0.9.

 

[marcus]

D(1.427)=∫dtsinh(1.5*1.427)^(2/3)/sinh(1.5*t)^(2/3) integral 0.8 to 1.427
...

I tried sinh(1.5*1.427)^(2/3)/sinh(1.5*t)^(2/3) integral 0.8 to 1.427 in numberempire
and got 0.89

 

[RyanH42]

Actually my second part is also true.But I made a little mistake If you put this equation into numberempire you get close to 0.89 (If you put the sinh(1.5*1.427)^(2/3)/sinh(1.5*t)^(2/3))
Its close to 0.9 cause there's some erros as you said.
But I put numberempire
sinh(1.5*1.8)^(2/3)/sinh(1.5*t)^(2/3)
Which the reason why I get 1.3.
I made copy one of the equations which we write here and I am changing the numbers so I forget to change that number.If you put my equation in #252 you will get a close answer(due to some errors)

 

[marcus]

I see! I often do that and make the same typo error. I copy-paste something to save typing and then forget to change one of the numbers in it. : ^)

 

[marcus]

"typo" is short for "typographical error". It's when you know the right thing but type it wrong. Everybody makes typos. I'm impressed by your good answers. Also putting in Wien's Law. I'm happy.

We don't need to go further right now, but I will lay out one or two ideas that we could advance to when you feel like it. there is no hurry. there is the "(cosmic) event horizon" and the "particle horizon"
they are easy to calculate with what you have now, and they are standard cosmology concepts.

 

[RyanH42]

Oh,I know and good morning :)

 

[marcus]

And good afternoon to you :^)

 

[marcus]

Let's try integrating that same thing but changed to give the disgtance now sinh(1.5*0.8)^(2/3)/sinh(1.5*t)^(2/3) integral 0.8 to 1.427 in numberempire

 

[RyanH42]

0.454

 

[marcus]

that gives the distance now, to a galaxy which I could send a signal to today which would get there at time t = 1.427

now let's change that time to infinity
in numberempire you can write "inf" to get the integral out to infinity (the 4th example on their page shows this)

I think this will give me the distance NOW to the farthest galaxy my signal (which I send today) can ever reach. (I make 1.427 very large, in other words)

 

[RyanH42]

I can't write infinity.It makes error in calculator.

I am typing to use my phone.So I am a bit slow

 

[marcus]

I get an error too! I can integrate to 100, and 200 but not to 500!
I see that it is approaching a limit, because it changes very little between 100 and 200

So I was wrong. We cannot simply put "inf" into numberempire and get the integral to infinity.
Numberempire is disappointing me in that respect.

Let's see if we can do it this way

$$D(S) = \int_S^1 ((s/1.3)^3 + 1)^{-1/2}ds$$

Maybe I should get a second cup of coffee first.

 

[RyanH42]

Probably yeah we need another calculater. I will try Symbolab

 

[RyanH42]

No result in Symbolab

 

[marcus]

Wait! it is all right.
I integrate ((s/1.3)^3+1)^(-1/2) from 0.001 to 1 and I get the right thing. D = 0.9508

Let me try from 0 to 1
Yes. It gives 0.952this is the cosmic event horizon. It is the distance now of the farthest galaxy I can reach with a signal I send today
and the distance now of the farthest that can ever have a causal effect on us by what happens there today.
beyond that horizon, no signal they send today can ever reach us

$$ \int_0^1 ((s/1.3)^3 + 1)^{-1/2}ds$$

{\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (zeit)&D_{now} (lzeit)&D_{then}(lzeit)&D_{hor}(lzeit) \\ \hline 1.000&1.000&0.796948&0.000000&0.000000&0.952155\\ \hline \end{array}}

 

[RyanH42]

I have some problems with my mom
She thinks I am spending a lots of time in phone but that's not true.I am talking with you and I am learning something .

Anyway I understand that.If we send a signal now that signal can go further a galaxy which 0.952 lightzeit distance from us.And also that's true for who lives that galaxy.So If I send I am here signal thus signal will reach anywhere but not beyond the 0.952 lightzeit.

 

[marcus]

It seems like you understand the cosmic event horizon idea!
Maybe I will do some other things for an hour or two. It's good to keep the Mom happy and it's good to get exercise and do real world things

 

[RyanH42]

Ok.I am in holiday and here there's really nothing to do.Nothing

 

[marcus]

I am guessing that your time zone is near UTC or european central. It is 8am pacific, so maybe where you live is 16h, or 4pm? or maybe 5 pm?
I don;t have anything I need to do for the next 2 hours.
We could take up another standard concept---the "particle horizon".
but I want your Mom to be happy.

 

[RyanH42]

S=0 means infinty isn't it.I means S=1/a(t) If a(t) goes infinity S goes zero.

 

[marcus]

Yes that's right, it is just another way to do the same integral, but without having to say "sinh"
Real world human relationships are obviously very very important. Please ignore what I say about particle horizon if there is a risk of making your mother unhappy. Any of this "horizon" business can wait till tomorrow or another day.

Let's try
((s/1.3)^3+1)^(-1/2) between 1 and inf
That is going back in time, from the present, back to very early (large S)

Yes! that works. That gives what cosmologists call the present "radius of the observable universe"
You can see it is 2.69.

It is the distance from us NOW of the farthest matter which could have sent us a signal, some light, some neutrinos, some gravity waves, anything, which is arriving today.

this distance is also called "particle horizon" because it is the farthest distance any particle can have traveled (helped by expansion) starting at the earliest time, at the beginning of expansion.

{\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (zeit)&D_{now} (lzeit)&D_{then}(lzeit)&D_{hor}(lzeit)&D_{par}(lzeit) \\ \hline 1.000&1.000&0.796948&0.000000&0.000000&0.952155&2.675083\\ \hline \end{array}}

 

[RyanH42]

Yes We can do

 

[RyanH42]

The problem is am a shy person .I don't have any friends.

 

[RyanH42]

Ok.Lets do this not today.Maybe tomorrow.You can write what you need to write (so you can ready to teach me)

 

[marcus]

Tomorrow is good. I live at the edge of the city near some open country, and some hills with only trees, grass, birds, deer,...
I walk.
It is good exercise. I sometimes get my neighbors to walk with me. It is easy to be with them when we are walking in Nature. looking at the sky and the hills, and the water. The purpose is to get some exercise. The neighbor knows it is for her good health or his good health. So there is a purpose to going on the walk. Either alone or with a companion. I will go walking this morning, with a neighbor who is a retired university professor. She used to teach language, or linguistics. It is an easy way to be with people, when there is a purpose, and a natural environment.