# Actual size of all space

1. Aug 15, 2009

### rasp

This is my 1st post ever, apologies and thanks in advance to all who read or respond. How does the actual size of all space (in linear distance dimensions) relate to the observable universe's event horizon? For example, imagine measuring out from the earth a sphere whose radius is the event horizon (equivalent to the distance light has traveled since the Big Bang?). I imagine we would reach a particle (or observor) on the surface of that sphere, which would itself be able draw a vector length in all directions that light has traveled from it since the big bang. In one of these directions their event horizon will be a linear addition to a radius of our event horizon and will end at a third particle, which can continue the logic of viewing an event horizon in all directions. This process continued would argue for an infinite set of spheres and spatial lengths. As infinity is usually a sure sign of error, can someone explain the correct reasoning to me without using "advanced" (post college alegbra) math? Thanks kindly.

2. Aug 15, 2009

### marcus

I take by actual size in linear distance dimensions you mean in terms of distance now at this moment----if you could temporarily freeze expansion so as to provide an opportunity to carefully measure distances by radar or some other conventional yardstick.

The observable part of the universe is normally considered to be a ball with us as center and radius about 45 billion lightyears. That is, the most distant matter from which we are now receiving light is today (partly because of expansion) slightly over 45 billion lightyears from us in actual distance.

That means if you could freeze expansion it would, starting today, take 45 billion years for a signal from us to reach that matter. We are seeing that matter (as it was over 13 billion years ago when it emitted the light) whenever we map the cosmic microwave background.

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Since you are new to the board (*welcome* by the way!) maybe I should say that astronomers call the actual distance to an object at some given time (freezing expansion at that moment) its proper distance.

There is no simple relation between proper distance and light "travel time", because the rate of expansion has changed a lot in the past.

The jargon term astronomers use for this 45 billion lightyear distance is the "particle horizon". If shortly after the start of expansion our matter had sent out a particle at the speed of light then that particle would now be 45 billion lightyears from us (because of the combined effects of both expansion and the particle's own speed.)

The particle horizon is normally what is called the radius of the observable universe.
That is probably more than enough jargon
So let's get to your main question.
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The standard model cosmology that astronomers use comes in two flavors---infinite space and finite space.

The simplest finite space case is where there is a very slight positive curvature so that space closes on itself like the 3D analog of the 2D surface of a balloon.
The most recent authoritative complete set of cosmo data I know was published January 2009 and is called the WMAP5 data (the 5th year report from a certain NASA mission that measured cosmo parameters the most accurately so far.)
Essentially what they said is that (of course it could be the infinite space case but) in the simple finite case the radius of curvature would be at least 100 billion lightyears, so that the circumference would be at least 2 pi times that---in other words in round numbers at least 600.

That means if you could freeze expansion at this moment, and would set out at the speed of light, like a lightbeam pointed in some direction, it would take you at least 600 billion years to make the full circuit and get back home. But probably more.
We don't have a better estimate, but that gives something to compare with the figure of 45 billion lightyears for the particle horizon.

Last edited: Aug 15, 2009
3. Aug 15, 2009

### sylas

Basically, because the speed of light is finite, you cannot separate the idea of looking a long way without also looking back in time; and because the universe as we know it has a finite age, there's a hard limit to how far you can see.

(As a minor terminology quibble, the term "event horizon" usually means something rather different. It's usually the boundary between us and particles we can never see no matter how long we wait into the future; rather than how far we can see right now. Only certain kinds of expansion, involving "acceleration", actually have an event horizon in this sense.)

Cheers -- sylas

4. Aug 15, 2009

### marcus

Rasp,

If you want some hands-on experience with the standard cosmo model, which doesn't require more than highschool algebra (not even that really), try playing around with Ned Wright's cosmo calculator.

He teaches the cosmology course at UCLA and is one of the leaders in the field, plus he has a great introductory website for the general public!

To get the calculator just google "wright calculator".

Try plugging in some large redshifts (the symbol for redshift is "z")
and see what distances you get.
If you type in some large redshift like 1000 or 10000 into the z box and press the button called "general" it should give you the actual distance to matter which we are seeing now with that redshift (as it was a long time ago, of course).
You will get outputs like around 45 billion lightyears, as you might expect. Good thing to check that kind of thing hands on.

5. Aug 15, 2009

### rasp

Thank you both for your input. Sylas, let me restate what i believe I learned, and please correct me if I'm wrong. 1) The observable universe can be considered as a sphere with earth as its center having a radius of 13 billion light years (p.s. I thought it was closer to 14.5 billion) and that due to inflation there are some objects that are further away such that their light emitted 13 billion years ago has not yet reached us. 2) In a positively curved, closed finite model, the entire universe curves back upon itself such that it has a circumference of 600 billion light years. 3) The particle horizon is the proper distance that would be found if we could measure the distance "now" to those objects which we see emitting light 13 billion years ago and which are at the edge of our observable universe.
If that is the case, there are several questions I might ask, but the first that comes to mind is, "What is the the limit to our observable universe?" Is it the limits of our technology or something more fundamental like the current speed of light? Can I assume the sky goes completely dark in all directions exactly at the 13 billion light year range?

6. Aug 15, 2009

### marcus

The age of the universe is estimated at 13-some billion years. For some years the prevailing estimate was 13.7, now some recent work has something like 13.4.

The presentday radius of the observable is estimated at 45 billion light years.
It has to be a lot more than 13.7 billion light years because the light could have traveled 13.7 billion light years on its own, without the help of expansion.
Expansion has an effect like savings-account interest. Every mile you put behind you gets increased by a little percentage bit each year, so you end up having a lot more miles behind you than expected. It works out in this case to a bit over 45 billion lightyears.

That is how far away, today, the most distant stuff we are seeing is.

The limit of the observable is a mathematical limit. We are very close to it with our presentday instruments. Within a few hundred thousand years of the theoretical limit to how far back in time. Future instruments will be able to probe deeper. The technical problem now is that back at the redshift of the microwave background the universe is a glowing hot fog that light can't get thru. We need instruments that see low energy neutrinos that were able to pass thru the fog and extend our vision that last little way. Mostly the limit is fundamental, but a small remaining part is technical.

No. We can see way past the 13 billion lightyear range and the universe looks pretty much the same out there as here. Same kind of galaxies. Same kind of supernovas. Same kind of quasars, clusters, same average density if you allow for the effect of expansion with the passage of time etc etc.

And although we can't see past 45 billion lightyears, the assumption is that if you could go out there and look, that it would look very much the same now as it does here. Same kinds of stars, galaxies, clusters of galaxies etc etc. We have no reason to suppose it is different from here.

Astronomers have never found any evidence that an "edge" exists, so they normally don't assume that such a thing exists. It would be an unnecessary complication without there being any logical reason for it.

Last edited: Aug 15, 2009
7. Aug 16, 2009

### Chronos

If we had optical instruments of sufficient power, the EM universe would have an 'edge' somewhere near 13.3 billion lights years distant [z~1100].

8. Aug 16, 2009

### Rymer

The surface of last scattering (or CMB) redshift is not really related to distance in the same way as a normal cosmological redshift. In order to talking in terms of distance, there needs to be a specific 'location' separation. For CMB the redshift (as a value of 1100) is time related. (with several assumptions -- ideal reversible gas model -- the distance related redshift would be about the square root of this value -- as a sort of 'average value'. This isn't very good as a number either in some ways -- more of a 'size' than a 'distance')

9. Aug 16, 2009

### Chalnoth

Huh? It most certainly is. It's exactly the same. Just further away than galaxies and whatnot.

10. Aug 16, 2009

### Rymer

Actually, the confusion here seems to stem from assuming 'distance' and 'size' are the same. In general, they are not. One example using the initial question is that its only an assumption that you can infinitely string together these radial separations. What you are calculating is an infinite 'distance' -- which could exist in a finite sized space.

11. Aug 16, 2009

### Rymer

Definitely NOT. One place the surface of last scattering occurred is RIGHT HERE. It includes ALL the universe. Its displacement is in time.

12. Aug 16, 2009

### Chalnoth

That doesn't make the redshift interpretation any different. The surface of last scattering was right here about 13.7 billion years ago, this is true. But the part of the surface which we see today in CMB experiments was some 45 million light years away or so from this location when it was emitted.

13. Aug 16, 2009

### Rymer

Yes, and so were many other places. The location must be specific to 'measure' a distance.
The 1100 number is a Wien Displacement of a blackbody 'peak'. That is not distance related. It is also not a recession velocity effect. It is related to energy (density) -- an adiabatic expansion. The 'size' of the universe would increase with about the square root of this value (depending on a +- 1 in determining z).

Last edited: Aug 16, 2009
14. Aug 16, 2009

### Chalnoth

Every place that was that far away is found within what we see as the CMB. So, pick a spot. Anywhere is fine, it's all the same distance in every direction.

True. This is a difference in how the redshift is measured compared with galaxies, because it is much more accurate to measure the redshifts of galaxies via their spectral lines (galaxies aren't very good black bodies). But it isn't a difference in the physics of how the redshift is produced.

15. Aug 16, 2009

### Rymer

Pick a spot isn't allowed. Need to be much more specific than that.

The difference (if any) isn't in how the radiation is produced -- the issue is WHERE. For CMB it was 'produced' everywhere at that 'time'. This is NO distance reference. All it reflects is the 'cooling' since that 'time'. The 'cooling' is related to a size increase -- which is related the square root of the temperature -- if assumed to be an reversible ideal gas -- or that there is no energy loss in the universe (adiabatic). Specifically this redshift is NOT related directly to a recession velocity -- others are.

16. Aug 16, 2009

### Chalnoth

Why? It doesn't make a difference. Every single place that emitted the light we now see as the CMB came from a shell of matter that was, at that time, about 45 million light years away.

Yes, but we don't see most of that. Most of that light is, today, very far away from us now, either because it was traveling in some other direction, or because it already passed us, or because it hasn't reached us yet. It is only that little bit of light that was emitted 45 million light years away and happened to be traveling in this direction that we see today.

Well, it reflects that too. But we're also seeing the glow of actual matter that has been redshifted, in precisely the way that light from galaxies is redshifted.

Minor nitpick, but the temperature of photons scales linearly with the expansion. There is no square root involved. That is:

$$T \propto \frac{1}{a}$$

Actually, the redshift of other matter isn't directly related to recession velocity. What's important is the history of the expansion from the time the photons were emitted to today, not how fast the object was moving either now or when the light was emitted.

17. Aug 16, 2009

### Rymer

Spot matters greatly. Must be related to a specific spot in co-moving space. CMB is ALL of co-moving space.

All we ever see is the 'little bit of light' that is headed our way -- not specific enough.
(That is all we ever see.)

Not exactly the same way -- normally we do NOT measure redshift from the blackbody radiation peak -- but from emission lines from excited atoms. Much different in source and in precision.

Yes, temperature is linear with Wien Displacement -- but an adiabatic process has constant total energy -- its temperature cools as the square of its increased size -- same as a radiation dropoff -- and as a reversible ideal gas. But size is not necessarily distance.
In this case it definitely isn't related to the apparent recession velocity of co-moving space.

That means the Hubble Law relation DOES NOT apply to CMB redshift since Hubble Law is only really true in co-moving space.

18. Aug 16, 2009

### Rymer

Note: while redshift can be caused by several processes, the one that we call 'cosmological' is the one that obeys Hubble's Law. Hubble's Law is only truly valid in co-moving space -- in fact that could even in many cases be used as a definition of co-moving space.

Hubble's Law relates a recession velocity to distance.

19. Aug 16, 2009

### Chalnoth

Not the CMB that we observe. The CMB that we observe is only a small portion of the background radiation: that portion emitted by matter that was, when it was emitted, around 45 million light years away.

That's not what I was saying, though. What I said was that the physics that generates the redshift is the same. Yes, it's measured differently, because we're talking about different physical systems. The CMB, for instance, comes from a physical system that produced a near perfect black-body spectrum at a very specific temperature. This makes measuring its redshift from its temperature an excellent way to measure it. Most other objects in the universe are very poor black bodies, and even if they are reasonably good ones, we don't necessarily know their intrinsic temperature. So looking for absorption/emission lines in their spectra is a much better way to go.

But this should be contrasted with the physics which produces the redshift in the CMB versus galaxies: it's exactly the same. Exactly. There is no difference in what causes the light of a galaxy to be redshifted and what causes the light of the CMB to be redshifted. And when we look at the CMB, we are actually looking at a specific shell of matter that emitted that light (what we are seeing is what made up the universe at that time: a nearly-uniform bath of cooling plasma, and we see a particular shell of that plasma determined by how long ago it cooled to the point that the photons could flow freely).

I believe you're talking about a non-relativistic gas here. Photons are quite relativistic.

In no case is the cosmological redshift simply related to recession velocity. Sure, for very short distances, before the curvature of space-time becomes terribly important, the interpretation of redshift as recession velocity is reasonably good. But at larger distances it starts to make less and less sense. Most of the galaxies we observe are far beyond the point where approximating redshift as being due to recession velocity is even close to reasonable.

Except it does.

20. Aug 16, 2009

### Rymer

Starting at the end: Except it does NOT. Hubble's Law is ONLY about recession velocity - distance relation -- and is only true exactly in co-moving space. CMB is not identifiable at a particularly place in co-moving space and therefore does not have a specific recession velocity. (Everything over some value isn't good enough.)

The simple inverse square relation can be shown as a basic relation for an adiabatic process -- one that has no energy loss, I don't see how 'relavistic' comes into it for this purpose.

Its definitely NOT just a simple temperature redshift one to relate to distance in the same manner as a recession velocity. The CMB redshift cannot be used in the same way as any other cosmological redshift. It may (or may not -- model dependent) have a similar resulting value -- but it is NOT a Hubble relation -- and therefore NOT a cosmological redshift.