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

[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.

 

[RyanH42]

I thought we live in 0.5zeit and we want to calculate how much distance traveled light when times come 0.6zeit.The opposite idea...Well I understand that.Today will be 0.6zeit and we are calculating past time.We did the same thing before

 

[marcus]

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 ?

Yes we can calculate that the same way. there are two possible answers. One is the distance the light has traveled as seen from THEIR perspective, when they receive it (that would be called the "proper distance" at that future time. And also we can scale that distance down by the S factor to be the distance NOW to that galaxy we are sending the light to---that will receive our message at some time in the future.

I believe it is good to first calculate what the proper distance to the galaxy will be when they receive the light.
Let's say they are to receive the light at time t=1.8.

OOPS sorry I have to leave. We are out of food and I have to go to the store.

 

[RyanH42]

Ok,I will go to sleep too.Maybe tomoorow

 

[RyanH42]

You don't have to teach me.Really.I think you are suffering to write these things.If you really want to teach me something that's ok.But If you don't want to teach me "really".Then don't do that please.You are writing so much and I am feeling sad about it.Maybe I can learn these things later.Cause I am feeling I am keeping you busy and you cannot do your other jobs.And that's makes me sad of course

I'll be waiting an answer in next 8 hours.
Thanks for everything.
Ryan

 

[marcus]

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 ?

Yes, I think I understand and there IS a way to calculate directly---I think it is the way you were thinking. The idea is to find the distance NOW to the galaxy which will receive our message at a certain time in the future.

Let me use the two times 0.8 (for now) and 1.8 in future. We can replace them with whatever you want later. I think maybe I don't even have to say the integral because you have already discovered what it must be.

 

[marcus]

...Cause I am feeling I am keeping you busy and you cannot do your other jobs.And that's makes me sad of course...

No, on the contrary you are helping me by your interest and by doing the calculations along with me. I think this approach to cosmology---with zeit and lightzeit units---is potentially a good one because it makes the formulas simple and transparent
one can actually calculate things quickly and easily. So I am interested in this approach. I think it is worthwhile that is worth the time spent on it

Also for a high school or college student it is a good way to practice calculus. You get to know the chain rule better, and change of variable, and the hyperbolic functions sinh and cosh.

So it seems to me worthwhile to find a good way to introduce and explain this approach to cosmology. Also it's just nice to have a hands-on contact with the universe and its expansion process and the light that travels between the galaxies. It is a more direct contact than one gets when one depends entirely on a calculator like Lightcone. (But Lightcone is good too, perhaps you should get some experience using it, and having it draw curves.)

I am on pacific daylight time which I think is about 8 hours earlier than UTC. You seem to sleep between 3pm and 11pm pacific time. I will check again around 7h UTC.

 

[RyanH42]

I am here again.Its really good.If you are ready to teach me then I am ready to learn.Whenever you want to start.Cause it seems times is there 10 pm.

 

[RyanH42]

I don't know lightcone.Maybe I can do that

 

[RyanH42]

Yes, I think I understand and there IS a way to calculate directly---I think it is the way you were thinking. The idea is to find the distance NOW to the galaxy which will receive our message at a certain time in the future.

Let me use the two times 0.8 (for now) and 1.8 in future. We can replace them with whatever you want later. I think maybe I don't even have to say the integral because you have already discovered what it must be.

I will use
1///D(t)=∫sinh(1.5*0.8)^(2/3)dt/sinh(1.5*t)^(2/3) by x on the interval from 0.8 to1.8
And
2///D(t)=sinh(1.5*1.8)^(2/3)/sinh(1.5*x)^(2/3) by x on the interval from 0.8 to 1.8
First equation gave me 0.607
Second equation gave me 1.755
unit is lightzeit and time is zeit.
Light cannot travel 1.7lightzeit in 1 zeit so my first equation becames true.(And also my example which I wrote before and later you also claimed that)
So if we take time for now (t=0.8) then the integral will be always

D(t)=∫1.3dt/sinh(1.5*t)^(2/3)by t on the interval from when t>0.8 → 0.8 to t
when t<0.8 →t to 0.8

When time is not now then we need to pick a time to normalize the scale .The only change , there will be no 1.3 so we need to use

D(t)=∫sinh(1.5*T)^(2/3)dt/sinh(1.5*t)^(2/3)
Again T is here the time which we call now.And the t is just normal time
Again
when t>T→ T to t
when t<T→t to T

This is the way I guess.

 

[RyanH42]

No, on the contrary you are helping me by your interest and by doing the calculations along with me. I think this approach to cosmology---with zeit and lightzeit units---is potentially a good one because it makes the formulas simple and transparent
one can actually calculate things quickly and easily. So I am interested in this approach. I think it is worthwhile that is worth the time spent on it

Also for a high school or college student it is a good way to practice calculus. You get to know the chain rule better, and change of variable, and the hyperbolic functions sinh and cosh.

So it seems to me worthwhile to find a good way to introduce and explain this approach to cosmology. Also it's just nice to have a hands-on contact with the universe and its expansion process and the light that travels between the galaxies. It is a more direct contact than one gets when one depends entirely on a calculator like Lightcone. (But Lightcone is good too, perhaps you should get some experience using it, and having it draw curves.)

I am on pacific daylight time which I think is about 8 hours earlier than UTC. You seem to sleep between 3pm and 11pm pacific time. I will check again around 7h UTC.

Sorry I sleep 9 hours.I woke up 12 pm pacific time.I don't know sometimes I sleep 7 hours sometimes 8 sometimes 9.Its good to learning cosmology.