Locacal coordinate system in GR observed?

In summary: NOW is about 46 billion light years.I think you should begin with this question.What is the present distance to the last-scattering-surface?In summary, the conversation discusses the concept of a local coordinate system in General Relativity and how it relates to an observer's perspective in the observable universe. The conversation also touches on the concept of a reference system and how it changes in an expanding universe. The main question raised is how to mathematically relate different reference systems. One piece of advice given is to start with simpler questions and use tools like Wright's calculator to better understand distances in the observable universe.
  • #1
hurk4
132
0
Local coordinate system in GR observed.

If one considers coordinate systems in General Relativity.
I think, if I understand it well enough, that in GR space & time and or in space-time a coordinate system depends on locality, e.g. on an average distance between galaxies an the total amount of mass, DM & DE all together concerned in such a ‘local environment or better the local density of that ‘local’ environment
Can or will any of our specialists (e.g. ST, Chronos, Marcus, Garth) give a mathematical relation for an observer who finds himself in another local environment (with another density, at a distance measured in his own coordinate system and with a relative speed als expressed in his own local environment). For myself I see to many pitfalls even to make an attempt to try this.
Many thanks in advance.:confused:
 
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  • #2
local reference system

Apart from the typing fault I made with the word local, I meant local reference system. Excuse me.::redface:
 
  • #3
Local references in our observable universe

Maybe sometimes no reaction can be the best answer one can get. My question is probably not well posed or even it might be a bad question. I am trying to do better now. First I tried to make a drawing for a simple example I made. This example was comparing a 7 GLY observable universe comparing it with a 14GLY observable (nowadays) universe (with the observer in its centres). Then, setting up this post, lucky enough the PF system did not allow for drawings. So then I had to rethink again. I then realized myself that in fact (if I am right?) that in an expanding universe the reference system (if one can really talk about that?) of the observer is continually changing because of the decreasing density in the universe. (The distances of clusters etc are increasing). While cosmic radiation from the background after the decoupling is traveling towards the observer with its constant speed c, the reference system of our observer was changing so that the temperature of the radiation (because of the wavelength elongation) decreased from a few thousand K to 2.73K. So If the observer looks back towards the CMB in his own reference system he concludes that the distance to it is 13.6GLY and that that distance to the 13.6GLY distance CMB always was 13.6 GLY even long time before he could observe the 2.73 K radiation. However the 13.6GLY some 300k years after BB, back extrapolated, was very tiny and that’s, I think, a question of comparing reference systems. (Indeed the, to be considered, radius of the nowadays observable universe (13.6 GLY) was then in reality very small. My initial question in this thread was raised because of how to mathematically relate reference systems. What to me is interesting to note is that how relative this all might be, the temperature of the CMB initially was high while it is low as we receive it now. I think if I had better studied the excellent books of S.Weinberg and of J.Peebles I could have had a better understanding of GR and not have had any reason to start this thread. If I get no further answers I will soon close this thread.
Kind regards.
 
  • #4
hurk4 said:
Maybe sometimes no reaction can be the best answer one can get. My question is probably not well posed or even it might be a bad question. I am trying to do better now.
Hi Hurk4, I'm no specialist, but maybe I can give you some advice:
1) Your question is still rather murky - may be that is why the responses are not quite rushing in.:wink:
2) Rather ask a few simple questions first and take it further in follow-on questions.

I see you are from the Netherlands: "Een of twee eenvoudige vrae op 'n keer!" (jammer, dis Afrikaans, nie Nederlands!:smile: )
 
  • #5
Good advice!

Jorrie said:
Hi Hurk4, I'm no specialist, but maybe I can give you some advice:
1) Your question is still rather murky - may be that is why the responses are not quite rushing in.:wink:
2) Rather ask a few simple questions first and take it further in follow-on questions.

I see you are from the Netherlands: "Een of twee eenvoudige vrae op 'n keer!" (jammer, dis Afrikaans, nie Nederlands!:smile: )

Hi Jorrie.
I see we both are not far away from Greenwich time nor from language.
Thank you for your good advice, I will think it over and see what I can do with it. (Dit is niet jammer, dit is juist leuk!:smile: )
 
  • #6
hurk4 said:
... concludes that the distance to it is 13.6GLY and that that distance to the 13.6GLY distance CMB always was 13.6 GLY even long time before he could observe the 2.73 K radiation. ..

hurk, I agree with Jorrie that it might be helpful if you begin with simpler questions----maybe even with one simple question.

It sounds like you are interested in DISTANCES. And so you could ask, for instance:

What is the present distance to the last-scattering-surface?

the last-scattering-surface, also called the surface-of-last-scattering, is where the MATTER is that emitted the CMB that we are NOW SEEING.

So what you seem to be asking about, namely the "distance to the CMB", is actually the distance to the surface of last scattering.

Actually the "distance to the CMB" is ZERO because those microwaves are right now here with us. So you can't mean that literally. You must mean to ask about the distance to the MATTER that emitted the CMB we are now detecting.

THE PRESENT DISTANCE TO THAT MATTER is not the figure of 13.6 billion LY that you give. So I think the first thing you should do is learn to use Wright's calculator

http://www.astro.ucla.edu/~wright/CosmoCalc.html

Just plug z=1100 in for the redshift.

the redshift of the CMB is 1100

so if you type that in, the calculator will tell you both the TIME since last scatter
and also the present distance (the Hubble law distance) to the matter which we are "seeing" when we look at CMB.

it should say that the light travel time was about 13.6 billion year
(that is the TIME since last scatter)

it should also say that the PRESENT DISTANCE is 45.6 billion LY.
(it is longer than you might expect because distances naturally get stretched out)

I suggest you leave the numbers in the calculator the same except for putting in different z-------and press the "general" button to make it compute.
If you change the other numbers that amounts to putting in different parameters for the model.

there is another online calculator with some different features
http://www.earth.uni.edu/~morgan/ajjar/Cosmology/cosmos.html

but with that one you have to type in the parameters
Omega = 0.27
Lambda = 0.73
Hubble = 71

and then put in your z = 1100 or whatever redshift you want.

=================

I see the Morgan Calculator has a neat feature that let's you compare DISTANCE THEN (when light was emitted) to DISTANCE NOW (when we receive the light)

For example if you put in z= 1100 to find out about the CMB, then you find that the DISTANCE THEN (at the time of last scattering when the light originated) was 0.04 billion LY-------this is to say around 40 million LY.
And the DISTANCE NOW of that matter that scattered the light to us is, as we already know from the other calculator, 45.6 billion LY. Actually Morgan says 45.5 billion LY but approximate agreement.

So when we look at the CMB, the matter that sent us the light USED TO BE 40 million LY away from us, at the moment that it sent us the light.
And it is NOW some 45.6 billion LY away from us, at the moment that we are receiving the light it sent.

I hope this helps with visualizing the picture. Also it will help to become familiar with the calculators. Hellfire, here at PF, has programmed his own, which can always be compared to the other two online ones.
 
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  • #7
Distances, calculators etc

marcus said:
hurk, I agree with Jorrie that it might be helpful if you begin with simpler questions----maybe even with one simple question.

It sounds like you are interested in DISTANCES. And so you could ask, for instance:

What is the present distance to the last-scattering-surface?

the last-scattering-surface, also called the surface-of-last-scattering, is where the MATTER is that emitted the CMB that we are NOW SEEING.

Thank you Marcus for your help.
After a week or so I will have time to exercise the calculators and it will be well possible that I will come then with some more questions or remarks which I hope will be clear enough.
 
  • #8
The most compelling indicator, IMO, is the CMB used to be hotter than it is now:

Molecular Hydrogen in a Damped Lyman-alpha System at z_abs=4.224
http://www.arxiv.org/abs/astro-ph/0602212
. . . The high excitation of neutral carbon in one of the components can be explained if the temperature of the Cosmic Microwave Background Radiation has the value expected at the absorber redshift, T=14.2 K.
 
  • #9
Translations

Thank you Chronos,
You understood that I am trying to imagine myself beeing an observer in situations of a BB environment, that, of course, takes some translations.
After some holydays, probably, I will come back with further questons, remarks, thoughts. :smile:
Kind regards.
 
  • #10
Wright's and Morgan's calculator

marcus said:
So I think the first thing you should do is learn to use Wright's calculator

Indeed I easily learned to use Wrights calculator. I could not work with Morgan's since I have no netscape brouwser installed.
Wright's calculator (as such) works fine and I also took advantage of printing Ned Wright's tutorials. I can see he really is one of the experts in Cobe, WMAP etc and of course in GR.
Now, as a layman, I can't help having some questions.
1) I am interested to see the algorithms used, are they available somewhere?
2) What is the applicability region/area of Wrights calculator?
I went down to z=0.001, below this value z jumped to 0 but I wonder if that makes sense. At a certain point it must become nonsense from a physics point of view, but where is that point?
3) I think it is reasonable to express z+1 as a quotient of two temperatures:
z(n)+1= Te/Tn,(e=emission; n=now), where Te/Tn~a(tn)/a(te). (See: "Principles of Physical Cosmology" by J.Peebles,(5.45),(6.3)).
Taking Te=3000K, I could then relate 'detected' past temperatures or future temperatures with distances then. Using z, in this calculator facility, does that mean that always Te =3000K is understood?
4) But, as time develops the temperature of radiation of emitting stuff (T-last-scatering-surface?) is going down, so it is not reasonable to put Te=3000K for all (time)cases!? Furthermore the emitted radiation being stretched during its travel, through the Hubble flow, also cools down.
So you see I still have difficulties with results I can get from Wright's calculator.
(As) soon (these questions are clarified for me) I will probably come with more questions, but I will then post those in the thread "U could be bigger and older..." My interests are not just only distances or times, but they also have much to do with the physics of the “thick wall” of the Planck radiation black box which serves as a moving horizon of our observable universe. Finally I want/hope to “travel while experiencing” through this wall “escaping” the observable universe if possible.:smile:
 
  • #11
hurk4 said:
1) I am interested to see the algorithms used, are they available somewhere?

2) What is the applicability region/area of Wrights calculator?
I went down to z=0.001, below this value z jumped to 0 but I wonder if that makes sense. At a certain point it must become nonsense from a physics point of view, but where is that point?

3) I think it is reasonable to express z+1 as a quotient of two temperatures:
z(n)+1= Te/Tn,(e=emission; n=now), where Te/Tn~a(tn)/a(te). (See: "Principles of Physical Cosmology" by J.Peebles,(5.45),(6.3)).
Taking Te=3000K, I could then relate 'detected' past temperatures or future temperatures with distances then. Using z, in this calculator facility, does that mean that always Te =3000K is understood?

4) But, as time develops the temperature of radiation of emitting stuff (T-last-scatering-surface?) is going down, so it is not reasonable to put Te=3000K for all (time)cases!? Furthermore the emitted radiation being stretched during its travel, through the Hubble flow, also cools down.

Other PF posters will have to step in. I am not going to be able to reply adequately to all your questions. will however take a stab at some:

1) Hellfire has programmed his own cosmology calculator. You could write him a PM (private message) or post a thread asking for help with the cosmology calculator programme.
I understand that the monthly magazine Sky and Telescope published the numerical steps---maybe they published in the form of a computer programme in the "BASIC" language. I don't know when---maybe in the timeframe 2000-2003 but that is just a guess.

2) I don't know the range. Any program like that will have round-off error where when something gets close enough to zero they will just call it zero---since there aren't enough decimal places---and stuff like that. The worst problem with finite-accuracy would be at high z, I would guess.
I expect that past z=10 it is not quite sober and
past z=1000 it staggers like a drunken sailor

but, you know, that's good enough for me. I only want a rough idea of things at such high z.

3) You are right about z+1 being a ratio of temperatures
whee z=0 is the present and corresponds to 2.728 kelvin or whatever the present CMB temperature is.

to find the CMB temperature at any past time just multiply by z+1
for example you get the temp at z=10 by multiplying 2.728 by 11

you get the temperature at last scattering, or "recombination", which was redshift 1100 (approx) by multiplying 2.728 by 1101.

for large z, people forget to add one because it makes hardly any difference percent-wise to the answer.

the CMB temperature NOW is what we can measure with instruments like WMAP and COBE. so it is better to peg one's calculations to that known thing 2.728 kelvin (or whatever it is, have to look it up)

the CMB temp at last scatter is only roughly estimated to be 3000 kelvin by knowledge of properties of hot, partially ionized, hydrogen gas. they can know it pretty close but it is still more of an estimate.

so make present CMB temperature the keystone of your arch
 
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  • #12
I will ask Hellfire

Hi Marcus

marcus said:
Other PF posters will have to step in. I am not going to be able to reply adequately to all your questions. will however take a stab at some:

1) Hellfire has programmed his own cosmology calculator. You could write him a PM (private message) or post a thread asking for help with the cosmology calculator programme.

2) I don't know the range.
but, you know, that's good enough for me. I only want a rough idea of things at such high z.

3) You are right about z+1 being a ratio of temperatures
whee z=0 is the present and corresponds to 2.728 kelvin or whatever the present CMB temperature is.


the CMB temperature NOW is what we can measure with instruments like WMAP and COBE. so it is better to peg one's calculations to that known thing 2.728 kelvin (or whatever it is, have to look it up)


so make present CMB temperature the keystone of your arch

1) I will send hellfire an e-mail
2) I tried negative z down to 0.999. this gave me negative values, which I don't understand.
3) I understood now bij exercise that T=2.728 is inhearently put in z as a reasonal condition where in z is the result of 2 values. Surely the model wants only freedom for one value.

I proceed now with a new thread on Planck blackbox radiation and the universe. I hope it will be good enough so that experts are not invited to give me a silent treatment sending me back in my own labyrinth.

Thank you again for all your help sofar.
 
  • #13
hurk4 said:
1) I am interested to see the algorithms used, are they available somewhere?
I can provide you the code of the Sky and Telescope and Ned Wrights calculators if you want. These are in BASIC and not very easy to follow. To get my code just go to this url. This is javascript and I have made an effort to keep it as simple and intuitive as possible. You are free to use the code. (By the way, the output for the "size of the observable universe then" is currently incorrect).

hurk4 said:
2) What is the applicability region/area of Wrights calculator?
I went down to z=0.001, below this value z jumped to 0 but I wonder if that makes sense. At a certain point it must become nonsense from a physics point of view, but where is that point?
The finite number of integration steps makes it difficult to deal with such small z numbers. Moreover there is probably a long list of limitations of these calculators. For example, eternal de-Sitter models ([itex]\Omega_{\Lambda} = \Omega = 1[/itex]) are not correctly calculated, outputs for very high redshifts are sometimes incorrect, etc.

hurk4 said:
3) I think it is reasonable to express z+1 as a quotient of two temperatures:
z(n)+1= Te/Tn,(e=emission; n=now), where Te/Tn~a(tn)/a(te). (See: "Principles of Physical Cosmology" by J.Peebles,(5.45),(6.3)).
Taking Te=3000K, I could then relate 'detected' past temperatures or future temperatures with distances then. Using z, in this calculator facility, does that mean that always Te =3000K is understood?
This seams correct. However, Te depends on the photon to baryon ratio [itex]\eta[/itex], i.e. on the [itex]\Omega_{R}[/itex] and [itex]\Omega_{B}[/itex] values, which usually are not an input of these calculators ([itex]\Omega_{R}[/itex] is assumed 0.0005 [itex]\Omega_{M}[/itex] and [itex]\Omega_{B}[/itex] has no special impact on the output and is included within [itex]\Omega_{M}[/itex]). You shoud use the Saha equation to calculate Te, see for example the slides 14 and 15 of www.ita.uni-heidelberg.de/~msb/Lectures/CMB_Lect.pdf[/URL]. As soon as you have Te for a cosmological model, you can calculate then the temperature of the CMB for any later time with the linear formula you wrote.
 
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1. What is the local coordinate system in General Relativity (GR)?

The local coordinate system in General Relativity is a system of coordinates that is used to describe the motion and position of objects in a specific region of spacetime. It is used to describe the curvature of spacetime in a local region, rather than the entire universe.

2. How is the local coordinate system in GR observed?

The local coordinate system in GR is observed through the use of measuring devices, such as rulers and clocks, which are used to determine the positions and motions of objects within the local region of spacetime. These measurements are then used to construct a coordinate system that describes the curvature of spacetime in that region.

3. What is the importance of the local coordinate system in GR?

The local coordinate system in GR is important because it allows us to describe and understand the effects of gravity on objects in a specific region of spacetime. It also allows us to make precise predictions and calculations about the behavior of objects in that region, which is essential for understanding the universe on both small and large scales.

4. How does the local coordinate system in GR relate to Einstein's theory of relativity?

The local coordinate system in GR is a fundamental part of Einstein's theory of relativity. It is based on the principle of general covariance, which states that the laws of physics must be the same for all observers, regardless of their relative motion or position. The local coordinate system allows us to apply this principle to the local region of spacetime, where the effects of gravity are most prominent.

5. Can the local coordinate system in GR be applied to all regions of spacetime?

No, the local coordinate system in GR is only applicable to small regions of spacetime. It is not a global coordinate system that can describe the entire universe. Instead, it is used in conjunction with other coordinate systems, such as the global coordinate system in GR, to provide a complete understanding of the universe.

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