# Size of the CMB sphere

1. Jan 26, 2013

### jety89

Hi

What was the distance between our point in space and the points from which the Cosmic Microwave Background radiation emanated that we detect today at the moment that that radiation was created.
Or similarly, what was the diameter of the sphere, the surface of which we detect today as CMB.

2. Jan 26, 2013

### phinds

As I recall, it was about 47 million light years and has since expanded by a factor of about 1100.

Really, this is the kind of thing that you should be able to find with Google quite easily.

3. Jan 26, 2013

### Chronos

Using Jorrie's cosmo-calculator, http://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm, The CMB was at a distance of 42.17 million light years when the photons we now observe were emitted. The diameter of this 'sphere' is unknown. It could even be infinite. The age of the universe is estimated to be about 380,000 years at that time, so you could attempt to calculate it on that basis, but, the results are misleading. It requires unsupported assumptions to attempt such a calculation.

4. Jan 26, 2013

### jety89

Thanks for the quick responses.

As I found out, this post actually answers my question,
https://www.physicsforums.com/showpost.php?p=1899571&postcount=11

but I still feel that it might be valuable to put it out as a separate thread.

As for the diameter of the sphere of CMB, it must have been 82.34 million light years back then, and 90 billion light years today.

5. Jan 26, 2013

### Chronos

The distance to the CMB is not the radius of the CMB 'sphere'.

6. Jan 26, 2013

### jety89

Then what is it?

Chronos, could you please elaborate on the Why?
You statement seems wildly counter-intuitive to me.
Since the distances to the CMB, are the same in every direction, they add up to a sphere, at least that is my understanding.

Last edited: Jan 26, 2013
7. Jan 26, 2013

### cepheid

Staff Emeritus
I am also puzzled. If not the surface of last scattering, then what "sphere" are you referring to?

8. Jan 26, 2013

### jety89

That is exactly what I am referring to.
So what Is the source of my misunderstanding?
Nonlinear geometry of spacetime, perhaps?
You seem to know something that I don't.

9. Jan 26, 2013

### cepheid

Staff Emeritus
Nope. I was asking *Chronos* what he meant by "cmb sphere" if not the last scattering surface. Edit: the comoving distance to the last scattering surface is definitely around 46 Billion ly in the concordance model of cosmology.

10. Jan 26, 2013

### marcus

I also was puzzled by Chronos' answer, but what you have said here makes complete sense to me. So I would say just stick with that and not to worry: the proper distance (at emission) is estimated at around 42 million ly so the diameter was around 84 million ly. Depending on whose calculator/which precise parameters you use. Your 82 is just as good an estimate.

The distance to that same sphere of material NOW is put at around 46 billion ly, making the diameter NOW 92 billion ly. The distance now can be called either the proper distance at the present moment or the comoving distance. In either case it is what you would measure by any ordinary means if you could stop the expansion process today to give time to measure.

46 billion was what Cepheid said so let's take that as benchmark---then just as a check, the expansion ratio z+1 that pretty much everybody agrees on for the CMB is z+1 = 1090.
So if you divide 46 billion by 1090, you get 42.2 million. Round off to 42 so as not to seem too precise.

Last edited: Jan 26, 2013
11. Jan 26, 2013

### jety89

I've looked at the cosmological calculator in some greater detail, and played with it a little bit. I took the omega values and hubble constant it gave for "the moment the universe was lit up" and fed it back into the calculator, using the same redshift value, z=1088. Apparently, the proper distance to a hypothetical last scattering surface back "then" then would have been a mere 800 light years. This is fun!
I've tried to do it twice in a row, but the result was just 0.

12. Jan 26, 2013

### marcus

The calculator most people have heard of is Ned Wright's

google "wright calculator" and put in 1088, see what you get.

I get 0.0419 Gly (look down to "angular size distance")

That is the same as 41.9 million ly.

Ned is one of the principal people in studying CMB, teaches cosmology at UCLA.

Or take what he says for "comoving distance" and divide it by 1089...same answer.

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

The link to Jorrie's latest calculator (version A27) is in my signature
http://www.einsteins-theory-of-relativity-4engineers.com/CosmoLean_A27.html
The quick way to use it: go there, in "upper limit" put 1089
in "step size" put -1, press calculate.

putting step = -1 gets you a table with 2 rows. Tabular output can be really interesting but it's not what you are looking for.
You will see the same two distance numbers: essentially 46 Gly for Dnow (distance now) and 42 Mly for Dthen (distance then).
====================

What calculator were you using? You seem to have gotten some strange results twice in a row.

Last edited: Jan 26, 2013
13. Jan 26, 2013

### Tanelorn

Incredible to think that the radius of the entire observable unverse, in other words the radius of the CMBR sphere, and what was to become 100s of billions of galaxies, was originally just 400 times the diameter of the milky way.

Last edited: Jan 26, 2013
14. Jan 26, 2013

### jety89

The answer to my original question is 42 MILLION light years, NOT 800 light years. I got the second figure by fooling around with this calculator:
http://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm

... sorry for the confusing statement.

15. Jan 26, 2013

### Tanelorn

jety89, 400 times 100,000 light years is 40Million light years

Last edited: Jan 26, 2013
16. Jan 26, 2013

### Chronos

I used the word 'sphere' instead of sphere because the topology of the surface of last scattering is unknown. Using Jorrie's calculator with z=1088, tells you the distance to the surface of last scattering at the time the photons we now observe was 42.17 Mly. It tells you nothing about the radius of the surface of last scattering. Deducing the radius of the surface of last scattering to then be 42.17 Mly makes as much sense as deducing the radius of the sun is 93 million miles.

17. Jan 26, 2013

### marcus

Have to think about this, Chronos. I have a hard time picturing how you are imagining this.
Maybe someone else will clarify (or we can get back to it in the morning.)

18. Jan 26, 2013

### marcus

In the meantime, for Jety89 the OP who asked the question, and others thinking about the surface of last scattering in the usual way, it is a hollow sphere with us as center. The radius now is estimated to be 46 Gly. We've been thru this several times already in this thread (and in other threads.)

That is the socalled "comoving distance" to the matter which gave us the CMB we are now receiving---emitted the CMB photons back in year 380,000. It is also the proper distance at the present time.

Back in year 380,000 the proper distance to that same matter was about 46 Gly/1090 = 42 Mly. In other words the distance has expanded by a factor of 1090 and the wavelengths of the ancient light have been stretched out by a factor of 1090. Or say 1089 if you prefer. So the radius of the sphere (with us, or the matter that became us, as center) would have been 42 Mly.

Last edited: Jan 26, 2013
19. Jan 27, 2013

### jety89

Looks like an open question in science -- the topology of the last scattering surface. My understanding is, if spacetime is flat, then it makes sense to talk about a sphere intead of a 'sphere', but that is an unconfirmed assumption -- or is it?

20. Jan 27, 2013

### jety89

21. Jan 27, 2013

### Chronos

Apologies. I agree all observers share the illusion of being uniformly surrounded by a cmb that was 42 mly distant when the universe was 380,000 years old. Claiming this tells us nothing about the 'true' size or topology of ther cmb was confusing.

22. Jan 27, 2013

### jety89

That about concludes this topic, i guess.