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

[RyanH42]

Wait I will going to fimd it myself then I will take an integral so D(t)=1.3∫dt/sinh(1.5t)^(2/3) and 0.338 to 0.8

 

[marcus]

I have an errand, I will be back in an hour or slightly more. Have to run.

 

[RyanH42]

Integral answer is 0.6389

 

[RyanH42]

Ok,No problem

I find the solution in 20min that's bad.But I wasnt sure enough that I am doing right or wrong.My mistake

 

[RyanH42]

I am trying to find more easy way
D(t)=-∫dS/H the integral between S=1(Now) and S=2 H will be as you said H=[(S/1.3)^3+1]^(1/2)

I made and I found 0.6347

It took time to realize the question and prepare an answer.

Did I pass ?

And also I checked my answer to use lightcone(Which I learned I guess) and it seems my answer is true.

 

[marcus]

Hi, I just got back. The errand was more work than I thought.
Before I forget it, I will say a problem that came to mind while I was out. I am still learning how to think up cosmology quiz questions.

In ancient times there was a race of giant stars called "PopIII" stars. The name is an historical accident and meaningless. They lived when distances were about a TENTH the present size. These stars are very interesting because they formed before there were heavier elements---they were made of H and He (with traces of Li) and their light had the wavelengths of hot H and He but not other gases.

It is very hard to find the small protogalaxies or regions with these PopIII stars. They were 100 to 1000 more massive than Sun and so burned very hot and had short lives. Sometimes astronomers have detected these ancient stars. How far away is the matter which once formed them?
Of course they are no longer. The actual stars exploded in SNe long ago, but the matter is still there, that made the light.

Suppose you are an astronomer who is just now getting some S=10 light from a region with these giant stars. How far away is that?

 

[marcus]

...

And also I checked my answer to use lightcone(Which I learned I guess) and it seems my answer is true.

: ^) Yes it seems it is.

 

[RyanH42]

D(t)=-∫dS/H integral between 1 and 10
H=[(S/1.3)^3+1]^(1/2) so the answer is 1.7564

 

[RyanH42]

Great story

 

[marcus]

I just had lunch. It is getting near 3pm pacific when you end the day, I think. There may be no more time. until tomorrow.
Here is a quick variation on an earlier problem.

You are transported to some unknown place and time in the future. You measure and find that the CMB is HALF the temperature it is today.
What time is it?

By an amazing coincidence, while you are listening to the radio you receive a message that was sent by us at PF today. What distance are you from Earth?

 

[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 wavelength .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 wavelength .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 calculator. 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