Couldn't the universe be finite if Omega =1?

[marcus]

Timdeeg,
I found the helpful page in Lineweaver 2003:

It is page 11 of http://arxiv.org/abs/astro-ph/0305179

0.95 < Ω[SUB]o[/SUB](z = 0) < 1.05                         (33)
0.99995 < Ω(z = 10[SUP]3[/SUP]) < 1.00005                  (31) 
0.9999999999995 < Ω(z = 10[SUP]11[/SUP]) < 1.0000000000005 (32)

In order to have present-day deviation no larger than 0.05
you need, back in the time of redshift z, to have had deviation no larger than 0.05/(1+z)
so at recombination, i.e. 1+z ~ 1000 the devi must have been less than 0.05/1000 = 0.00005
and about 1 second after start, shortly after inflation, say 1+z ~ 10¹¹, the devi must have been less than 0.05/10¹¹, so twelve zeros before the 5.

 

[timmdeeg]

So just to have a pedagoguish example, multiply 14.4 by 1000 and there you are.

From this point of view, oh yes.

BTW did you ever look at Charles Lineweaver's 2003 paper called "Inflation and the CMB"? It has a page or so discussing how the "deviation" (as you and I call it) changes over time. Inflation can pull it down to be very very small, but then it can slowly creep back up again.

Yes, thanks, it belongs to some selected papers, which I have printed out some time ago. The important term is $\rho a^2$. Because according to the Friedmann equation this term increases enormously during inflation and thus drives Ω to 1, but decreases after inflation so that "our deviation" "creeps back again". I guess the latter effect doesn't change too much if one considers the first one. But I am not aware of any figures. Anyhow, this solution of the flatness problem is really very impressive and I do hope that the Planck mission will confirm the expected primordial gravitational waves!

 

[timmdeeg]

In order to have present-day deviation no larger than 0.05
you need, back in the time of redshift z, to have had deviation no larger than 0.05/(1+z)
so at recombination, i.e. 1+z ~ 1000 the devi must have been less than 0.05/1000 = 0.00005
and about 1 second after start, shortly after inflation, say 1+z ~ 10¹¹, the devi must have been less than 0.05/10¹¹, so twelve zeros before the 5.

Yes, that's the 'pencil on its point' problem. I remember my excitement when I read Alan Guth's "The inflationary universe".

 

[bobie]

My favorites are Anaximander (b. circa 600 B)
Great! I hope you do! and that you find the explanation clear enough and somewhat helpful.

Hi marcus, so I was right (apart from poor memory). It struck me, last year when I read it, that a scientist can appreciate and understand philosophy. That's why I like you. A philosopher has an edge on a scientist, as he knows some truths, has some tenets. Even a student philosopher has an edge () on a cosmologist, as he knows for sure there is an edge, that U is finite and spherical, etc...
But, again, you didn't answer my questions!

I found your thread, it's a good idea, but if you want to do a really great job (for students) you should start from scratch and listen to students*, so that you can improve what is obscure (a seminar, work in progress). Then you can re-write it and make a useful 'sticky' (that'll save you hundreds of repetitive posts) or write an excellent article in wikipedia or even publish a successful divulgative booklet. At the end you might even gain a better insight into your own theory!

The main ingredients are simplicity and clarity: always choose one and the simplest option*, therefore U must be observable U ( else nobody will follow you), give first the basic data at BB and now, explain the main ideas of the theory, give a concrete example (how from the redshift of a galaxy you derive all data with your calculator (not everybody is able to deduce that) , and then you can explain the Friedman equation. It's an interesting and ambitious project. Good luck!

If you are intersted in my opinion I'd be glad to help you, send me a PM, since I do not wish to encumber your thread.
* U is infinite/finite/ we don't know , at BB space was finite/ infinite, U is flat/ has a tiny curvature, etc
**experts cannot understand students. Right now you can see there are two threads asking the same question about expansion>C. After 57 posts in my thread, I gave up hope that there is someone who wants or can listen.

 

[bobie]

any deviation, however small, from absolute flat leads to a circle , however huge, I suppose. Or not?

... we talk at least about a big philosophical issue. Any deviation from Ω = 1 related to $k = +1$ supports a finite universe. .

Hi timmdeeg, you are right, and it is not a philosophical or abstract issue , but a concrete geometric issue: a line is straight if and only if it is always/ absolutely straight. It is an undeniable truth. You call it Ω, let's call it angle. The angle between any two adjacent segments must be 0°. Even if it is 57..°/10³¹ the resulting figure is finite and a circle/sphere (with a huge radius, of course).

Please tell me: Friedman equation is founded on the fact that matter/gravity curves spacetime (am I wrong?), doesn't that exclude that U can be flat, a priori?

 

[bobie]

Hi Marcus, I just realized that you replied to me in the other thread:

And Bobie asked how ..So in the other post I took the case that the upper end of the confidence interval was Ω = 1.000001 and the positive curvature number was 0.000001. So then the square root was 0.001 which is a thousandth. And you multiply the Hubble radius 14.4 Gly by a thousand to get 14400 Gly.

I'll continue this discussion here, since I do not wish to spoil your nice thread:
Probably you missed this post

I do not particularly want to know how to find the radius of curvature, but:
- why with that very curvature U would be considerd flat and keep isotropy?
- does it depend on the sensibility of your instruments or it is a principle?
- how do you derive that figure?
- do those properties apply also or a fortiori if the radius is twice that figure, say 30,000 Gly?

You explain that a curvature .000001 corresponds to a radius of 144,000 Gly, right, but my questions were:
- why with the current value you do not have isotropy and can't say U is flat?
- what so special about .000001? , if it is special, how did you determine it?
what did you mean by this?:

I don't believe it makes any different to real world computations (where there is always a limit on precision) whether one assumes Ω exactly = 1, or instead something like 1.000001.

do you mean that we can assume that value as real?, no problem?, is it really a matter of taste?
- can the radius be really 1000 greater?
- what concrete parameters are needed in order that 144,000 Gly be the real radius?
I must have missed something!

 

[timmdeeg]

Hi timmdeeg, you are right, and it is not a philosophical or abstract issue , but a concrete geometric issue: a line is straight if and only if it is always/ absolutely straight. It is an undeniable truth. You call it Ω, let's call it angle. The angle between any two adjacent segments must be 0°. Even if it is 57..°/10³¹ the resulting figure is finite and a circle/sphere (with a huge radius, of course).

Please tell me: Friedman equation is founded on the fact that matter/gravity curves spacetime (am I wrong?), doesn't that exclude that U can be flat, a priori?

Hi Bobie, in General Relativity a straight line isn't what you would naturally expect from your experiences. In GR a straight line means the worldline of an inertial (force free) object, called geodesic, see here. Without knowing this definition, you would hardly agree, that a satellite surrounding the Earth is moving along a "straight line".

I took reference to a philosophical issue perhaps in another sense, as you have interpreted that. Most probably we will never know whether the universe is spatially infinite or not. Even in case it has exactly euclidean geometry this question is still open. Now you can ask, what would nature prefer? A universe where you can move along a "straight line" and never will come back to the point where you started? Or a universe where you will come back, perhaps even on paths of different length, as in the case of the 3-torus. The first possibility is the most trivial and this might be the reason why it is preferred by cosmologists. One could assume a philosophical background such that nature prefers the most trivial shape of the universe. But this is a personal comment.

Friedman equation is founded on the fact that matter/gravity curves spacetime (am I wrong?), doesn't that exclude that U can be flat, a priori?

Yes, the Friedmann equations are special solutions of the Einstein equations (which describe the curvature of spacetime) based on the simplifying assumption that the energy density is homogeneous and isotropic. But no, GR und thus the Friedmann equations do not say anything about the shape (topology) of the universe. This question can only be solved empirically, at least in principle.

 

[bobie]

Now you can ask, what would nature prefer? A universe where you can move along a "straight line" ...? Or a universe where you will come back, ...as in the case of the 3-torus. The first possibility is the most trivial and this might be the reason why it is preferred by cosmologists. One could assume a philosophical background such that nature prefers the most trivial shape of the universe..

I am not aware of a single case where nature chooses something trivial, she always finds the best/most simple solution. All is simple for her, maths is a man-made artifice, probably she can only do +1 and -1.
As marcus says commenting on Anaximander, nature obeys the law of symmetry, timmdeeg.
I call it as the necessity of Being: order, equality, proportionality, balance;
Infinite is not trivial shape, but no shape/edge.
The sphere is the most simple yet the most complex and symmetric shape, the only locus where the contradiction between finite and infinite is solved, conciliated. It is not a coincidence that Hubble sphere and visible U are spheres.

...the Friedmann equations do not say anything about the shape (topology) of the universe. This question can only be solved empirically, at least in principle.

How do you solve it empirically? are you sure you can detect any curvature? why is Ω =1.000001 so special?

 

[PeterDonis]

Please tell me: Friedman equation is founded on the fact that matter/gravity curves spacetime (am I wrong?), doesn't that exclude that U can be flat, a priori?

No, because curvature of *spacetime* does not require curvature of *space*. The curvature can all be in the time dimension. In the case of the flat Friedmann model, that's exactly what happens: each spatial slice is flat, and the spacetime curvature is entirely contained in the fact that the universe is expanding, i.e., changing with time.

 

[bobie]

I don't believe it makes any different to real world computations (where there is always a limit on precision) whether one assumes Ω exactly = 1, or instead something like 1.000001.
That would correspond to a U which is spatially a 3D sphere. And the 3D spatial slice (a "hypersphere") would currently have a radius of curvature of 14400 billion light years.
In effect, no one could tell it from flat And with that one still has isotropy.

Can someone tell me why with such a radius one has isotropy, and not with the current one (144 Gly, 0.01)?

 

[Jorrie]

... why is Ω =1.000001 so special?

It is not special - it is just a value that falls within the present observational limits of roughly 0.95 ≤ Ω ≤ 1.05; Ω = 1 identically would be somewhat special. Some data-sets seem to indicate a very slight bias towards Ω > 1, but I do not think that is taken too seriously at present.

I think Marcus just chose any value within the limits to use as an example of how to calculate the radius of curvature for a very slight positive spatial curvature. The value itself has nothing to do with isotropy.

 

[bobie]

It is not special ... The value itself has nothing to do with isotropy.

. By making it finite, you lose isotropy. :

Thanks Jorrie, but marcus says:

"Good point about loss of isotropy!.14400Gly...with that one still has isotropy"

If radius is 14.4 or 14400 Gly U is still finite, the two statements are conflicting. Which is right?
marcus agrees with Bill, but adds that, by choosing that value, you rescue isotropy. Isn't he saying that ?
The model says that we do have isotropy anyway:

The Big Bang theory of the evolution of the observable universe assumes that space is isotropic

marcus adds:

... have a radius of curvature of 14400 billion light years...In effect, no one could tell it from flat

From what value you can't tell it from flat?
Lastly, marcus hints that that value would not imply great changes in the model

Oh, I guess it makes a difference to modeling the early universe.

could that be the actual, real value? could you remodel the early universe with no big problems?

 

[Jorrie]

Thanks Jorrie, but marcus says:
If radius is 14.4 or 14400 Gly U is still finite, the two statements are conflicting. Which is right?
The model says that we do have isotropy anyway:

Finite, like the surface of a sphere is finite, yet unbounded - you can go on and on around it without reaching an edge. Surface isotropy is then not a problem. Add one spatial dimension (to the surface) and you have something like a positively curved, isotropic, unbounded, 3-D universe. You run into trouble is if you should think "flat (or open), isotropic and finite". This cannot apply to the universe as a whole.

Besides that:
From what value you can't tell it from flat?
Lastly, marcus hints that that value would not imply great changes in the model

From the moment our observational evidence rules out Ω = 1, irrespective of how close to unity one of the limits is. The cosmic model we use allows virtually any value of Ω; it is just observations that can narrow it down. The best we have today is that it sits at or very close to 1, but if not exactly 1, we are not sure on which side it is.

 

[bobie]

Finite, like the surface of a sphere is finite, yet unbounded ..

You are describing the Earth situation.
But that does not fit the Hubble sphere nor the visible U, nor, in conclusion, the model. That would fit a true analogy with an inflating balloon, which is not accepted.
If we limit our speculations to visible U, as any rational man-of-science should ("whereof one cannot speak..."), the situation is the one I have descripted above, a sphere, flat inside and curved on the edge.
Is this relevant to my questions?

From the moment our observational evidence rules out Ω = 1,

That would rule out that U is flat. We know for sure then: any shape, but not flat.
And, if Bill is right:

By making it finite, you lose isotropy.

, then U must be infinite by postulate. Infinite (edit: infinite, not just finite-unbounded) and not flat, does that make sense?

 

[Jorrie]

You are describing the Earth situation.
But that does not fit the Hubble sphere nor the visible U, nor, in conclusion, the model. That would fit a true analogy with an inflating balloon, which is not accepted.
If we limit our speculations to visible U (as any rational man-of-science should "whereof one cannot speak...") the situation is the one I have descripted above, a sphere, flat inside and curved on the edge

Bobie, I think you are confusing yourself. The surface of an inflating balloon is a perfectly acceptable 2-D analogy for a positively curved universe. The observable universe is represented by a surface-circle around your vantage point (or for any balloon surface dweller). The radius (and 'curvature') of that circle is limited by the age of the present expansion (how far light could have traveled) and has nothing to do with the size or the spatial curvature of the cosmos.

The radius of the balloon itself represents the radius of spatial curvature. For a "flat" universe, let that radius tend to infinity... Think about it carefully and I'm sure you will eventually embrace the balloon analogy - correctly used, it is amazingly useful.

That would rule out that U is flat. We know for sure then: any shape, but not flat.
And if Bill is right: Then U must be infinite by postulate. Infinite and not flat, does that make sense?

We would also know whether the curvature is negative or positive (i.e. open or closed).
Open (negative curvature), infinite and isotropic makes sense, not so?

 

[bobie]

The radius (and 'curvature') of that circle is limited by the age of the present expansion (how far light could have traveled) and has nothing to do with the size or the spatial curvature of the cosmos.

Sorry, what is the cosmos, now ?
whatever you mean, how do you measure or conjecture its size/radius? Are you talking of what is outside the visible U? ..."...thereof one must be silent"

The radius of the balloon itself represents the radius of spatial curvature. For a "flat" universe, let that radius tend to infinity... the balloon analogy ...is amazingly useful.

A flat universe? you have just excluded the possibility that Ω=1
(Please, let's abandon any analogy, even if you think they're useful. That makes confusion. Surely they are not necessary to answer my simple question.)

Open (negative curvature), infinite and isotropic makes sense, not so?

Are we discussing Ω <1?, anyway, were you suggesting that Bill did not mean infinite, but finite-unbounded?
My question was:

"Good point about loss of isotropy!...14400 Gly..with that one still has isotropy"

Is it related, by any chance, to this?:

The latest research shows that even the most powerful future experiments (like SKA, Planck..) will not be able to distinguish between flat, open and closed universe if the true value of cosmological curvature parameter is smaller than 10⁻⁴

Probably I should wait for marcus to clarify what he meant. Thanks, anyway, Jorrie.

 

[Jorrie]

Bobie, to me you seem really confused about cosmology. I'm afraid the two of us do not communicate all that well and I'll rather leave it to others to help you out.

Sorry about that.