B What is the probability that the Universe is absolutely flat?

[Buzz Bloom]

A flat 3-torus universe with appropriate parameters is finite, flat, and isotropic. There can also be flat 3-torus geometries that are not isotropic, as discussed earlier, but that does not mean it's impossible for a flat 3-torus to be isotropic.

Hi Peter:

Do you know of any particular specified parameters that mathematically defines a theoretical 3-torus topologically shaped universe model which is finite, flat, isotropic, and homogeneous? For the purpose of defining such a model the homogeneous requirement can just be assumed.

I found some suggestions on the internet that some people believe it is possible, but I could not find any actual example. Nor could I find any argument supporting this concept that seemed rational to me.

In the absence of any even partially defined plausible model, I feel it necessary (for the present) to disbelieve it is possible. But, perhaps it is just a difficult problem, like the Riemann hypothesis, and someday I might be persuaded to change my mind.

Regards,
Buzz

 

[George Jones]

I understand that the inflation model assumes a finite universe with a positive radius of curvature.

No.

So I would rather state it as: the null hypothesis that comes from inflationary models is that the spatial curvature is positive.

Again, no.

See, for example Baumann's lectures on inflation,

https://arxiv.org/abs/0907.5424
Equations (43), (44), (45), as well as the discussion preceding these equation, all use absolute values. Other examples that don't restrict inflation to universe that have positive spatial curvature: "Cosmology" by Weinberg; "The Primordial Density Perturbation: Cosmology, Inflation, and the Origin of Structure" by Lyth and Liddle.

That implies a total value of $\Omega$ (i.e., including all forms of energy plus curvature) that is greater than 1.

As @Buzz Bloom notes, the sum of all the $\Omega$ s, including curvature is always unity. This means that $\Omega_k > 0$ for open (non-flat) universes, and $\Omega_k < 0$ for closed universes. Always.

 

[PeterDonis]

Do you know of any particular specified parameters that mathematically defines a theoretical 3-torus topologically shaped universe model which is finite, flat, isotropic, and homogeneous?

Sure, the specified parameters that the space is a 3-torus which is finite, flat, isotropic, and homogeneous. Unless you have a proof that these parameters are not all consistent with each other, then such a model must exist.

[Edit: Erroneous portion deleted.]

 

[PeterDonis]

If you then assume "homogeneous", then the space must also be isotropic

Oops, sorry, I got this backwards. Isotropy (strictly speaking, isotropy about every point) implies homogeneity, but not the other way around. I have edited the previous post to remove this portion.

 

[PeterDonis]

Equations (43), (44), (45), as well as the discussion preceding these equation, all use absolute values.

Yes, but those are general constraints from known data, not parameters specific to any inflation model.

The actual models discussed in this reference don't seem to make any assumption about spatial geometry, though, so they would indeed be consistent with either a small positive or a small negative spatial curvature today, as far as I can tell.

the sum of all the $\Omega$ s, including curvature is always unity.

Ah, yes, I misstated it. The sum of all the $\Omega$ s except curvature is greater than 1 if the spatial curvature is positive.

 

[George Jones]

Generally I understand that Ω is defined as the sum of a combination of model parameters: Ω_r, Ω_m, Ω_k, and Ω_Λ. When all four are summed, the result is Ω=1.

I don't think that I have ever seen this convention for an unsubscripted $\Omega$.

If you exclude Ω_k then Ω−1 = -Ω_k. Is this your intended meaning?

This is the convention with which I am familiar.

Yes, but those are general constraints from known data, not parameters specific to any inflation model.

And thus to be consistent with known data, models of inflation must work in universes that have negative spatial curvature. The Lyth and Liddle reference that I gave above explicitly notes that inflation smooths inhomogeneities in both $\Omega > 1$ and $\Omega < 1$ universes.

 

[Buzz Bloom]

Sure, the specified parameters that the space is a 3-torus which is finite, flat, isotropic, and homogeneous. Unless you have a proof that these parameters are not all consistent with each other, then such a model must exist.

Hi Peter:

I think you are using the term "parameters" in a difference sense than I use it. The terms flat, finite and isotropic are resulting behaviors/properties of a model. I use "parameters" to mean the variables (and possibly inital values) which are inputs to a model and from which the resulting behavior can be determined. For example, the models based on the Friedmann equation has five parameters: 4 Ωs and H₀. From values for these parameters ons can calculate radius of curvature as a function of time, the scale factor "a" as a function of time, age as the length of time from a=0 to a=1, etc. I also believe isotropy can also be deduced, but I am not up to explaining how in general, but I can for special cases, such as rhe one i describe in my post #25.

Regards,
Buzz

 

[PeterDonis]

And thus to be consistent with known data, models of inflation must work in universes that have negative spatial curvature.

I don't think a model that didn't work in a universe with negative spatial curvature would be inconsistent with the known data (since the known data does not rule out zero or positive spatial curvature). But it might be considered more unlikely given the known data. However, that appears to be a moot point for inflation models, since the reference you give notes that they do in fact work with either positive or negative spatial curvature.

 

[timmdeeg]

In this context, the assumption that the sum of the four Ω parameters equals 1 implies that Ω_k is negative for a closed finite universe.

This follows directly from $\Omega_k=-\frac{kc^2}{H_0^2}$

 

[timmdeeg]

The model parameter that corresponds to your Ω is Ω_m.

No, $\Omega$ in $(\Omega^{-1}-1)\rho{a}^2=-\frac{3c^2}{8\pi{G}}k$ (post #13) is the ratio of the actual density to the critical density. From this $k=0$ requires $\Omega =1$ which is definitively not a prediction of inflation.

 

[metastable]

Is it necessary or only convenient to reference a 3-space?

I define an arc with n=arbitrarily high # turns. I make a straight line from point t=0 to point t=1. Next I bisect line t=0, t=1 creating a point A. The vector is created first by a line from point A, through the referenced spiral point B to point C, a point in space. Points in space are referenced as Vector: t=0.14563... Distance: 686,739,974.97969...km

If we're talking about the universe, then yes, since spatial slices of the universe are 3-spaces.

I don't know what you're trying to get at with the arcs.

wolframalpha.com:

ParametricPlot3D [ { [//math:cos(2*pi*10^27*t)*sin(pi*t)//] , [//math:sin(2*pi*10^27*t)*sin(pi*t)//] , [//math:cos(pi*t)//] } , { t , [//number:0//] , [//number:1//] } ]

if on this scale 1 = Earth radius, loops = 10^27

unless I've made a mistake in my calculation we have:

~0.000020037... femtometers position specification accuracy & separation distance between loops at Earth radius

 

[Ibix]

...so you are trying to use distance along your spiral to specify angular position and radius as in regular spherical polars? How do you specify a point that does not lie on a line through your spiral?

 

[metastable]

How do you specify a point that does not lie on a line through your spiral?

I make a straight line from point t=0 to point t=1. Next I bisect line t=0, t=1 creating a point A. The vector is created first by a line from point A, through the referenced spiral point B to point C, a point in space.

^With a point A at the bisection of a straight line from point t=0 to t=1

 

[Buzz Bloom]

the ratio of the actual density to the critical density

Hi timmdeeg:

The above quote is the definition of Ω_M.
Please see

https://en.wikipedia.org/wiki/Friedmann_equations​

in the section Density parameter.

Regards,
Buzz

 

[Buzz Bloom]

And thus to be consistent with known data, models of inflation must work in universes that have negative spatial curvature.

However, that appears to be a moot point for inflation models, since the reference you give notes that they do in fact work with either positive or negative spatial curvature.

Hi George and Peter:

I apologize for my error. My bad. My previous careless exposure to discussions of inflation have involved explanatory pictures of finite shapes getting bigger. I failed to realize that the metaphor in the pictures intended the inclusion of both kinds of non-flat universes.

https://en.wikipedia.org/wiki/Inflation_(cosmology)#/media/File:History_of_the_Universe.svg​

http://www.ctc.cam.ac.uk/outreach/origins/inflation_zero.php
Regards,
Buzz

 

[Ibix]

^With a point A at the bisection of a straight line from point t=0 to t=1

And the position of A along that line is your third number.

 

[timmdeeg]

The above quote is the definition of Ω_M.
Please see

https://en.wikipedia.org/wiki/Friedmann_equations​

in the section Density parameter.

I'm not sure how you come to this conclusion. This article states "The density parameter (useful for comparing different cosmological models) is then defined as:" ... and describes thereafter $\Omega$ as the ratio of the actual density to the critical density.

If $\Omega$ in the expression in brackets (post #60) were $\Omega_m$ then how could this expression be identical zero in case of exact flatness?

 

[metastable]

And the position of A along that line is your third number.

Are you saying if I want to specify 3 chosen subsequent particle positions it would take me more than 6 numbers? (3 spiral points and 3 distances from point A)

 

[Ibix]

Are you saying if I want to specify 3 chosen subsequent particle positions it would take me more than 6 numbers? (3 spiral points and 3 distances from point A)

Yes. Because you cannot specify a position that does not lie on a line through your spiral.

 

[metastable]

Because you cannot specify a position that does not lie on a line through your spiral.

In a 3 space can I specify any irrational position?

 

[Buzz Bloom]

k=0 requires Ω=1

If Ω in the expression in brackets (post #60) were Ω_m then how could this expression be identical zero in case of exact flatness?

Hi timmdeeg:

It is Ω_k = k = 0 then

Ω_m = 1, and the universe is flat.​

When Ω_m > 1,

k=+1 and Ω_k < 0.​

The universe is then finite and hyperspherical.​

When Ω_m < 1,

k=-1 and Ω_k > 0.​

The universe is then infinite and hyperbolic.​

Regards,
Buzz

 

[George Jones]

No, $\Omega$ in $(\Omega^{-1}-1)\rho{a}^2=-\frac{3c^2}{8\pi{G}}k$ (post #13) is the ratio of the actual density to the critical density.

The above quote is the definition of Ω_M.
Please see

https://en.wikipedia.org/wiki/Friedmann_equations​

in the section Density parameter.

I'm not sure how you come to this conclusion. This article states "The density parameter (useful for comparing different cosmological models) is then defined as:" ... and describes thereafter $\Omega$ as the ratio of the actual density to the critical density.

I agree with @timmdeeg

In this section, density is given by
$$\rho = \rho_r + \rho_m + \rho_\Lambda$$

and

$$\begin{align} \Omega &= \frac{\rho}{\rho_c} \nonumber \\
&= \frac{\rho_r + \rho_m + \rho_\Lambda}{\rho_c} \nonumber \\
&= \frac{\rho_r}{\rho_c} + \frac{\rho_m}{\rho_c} + \frac{\rho_\Lambda}{\rho_c} \nonumber \\
\Omega &= \Omega_r + \Omega_m + \Omega_\Lambda . \nonumber
\end{align}$$

 

[Ibix]

In a 3 space can I specify any irrational position?

If you can specify the irrational number you mean, then sure. $(x,y,z)=(\pi,e,\sqrt{2})$, for example.

 

[metastable]

Can I write a complete list of the x,y,z positions an electron had as it traveled a meter, even in principle?

 

[Buzz Bloom]

In this section, density is given by

ρ=ρ_r+ρ_m+ρ_Λ​

and

Ω=ρ_r+ρ_m+ρ_Λ/ρ_c​

=ρ_r/ρ_c+ρ_m/ρ_c+ρ_Λ/ρ_c​

Ω=Ω_r+Ω_m+Ω_Λ.​

Hi George:

I edited the format of quote so that I could read it more clearly.

I confess I was not previously familiar with this usage of Ω, but I now see at the bottom of the Wiki section:

Ω_0,k - 1-Ω₀.​

So I yield, and I am now convinced I was previously mistaken. Thank you for pointing out my error.

Regards,
Buzz

 

[Buzz Bloom]

Hi @metastable:

I am puzzled by the figure with a spiral on a sphere in your post #35. Please explain in words what this is intended to represent.

Regards,
Buzz

 

[metastable]

Hi @metastable:

I am puzzled by the figure with a spiral on a sphere in your post #35. Please explain in words what this is intended to represent.

Regards,
Buzz

In rough terms: suppose you take a standard desk globe of the earth, and you attach a motorized gantry to the arch which supports it at the poles, and you spin the globe at a constant velocity, and you move the gantry from the north pole to the south pole at constant angular speed, with a pen attached to it, the pen will trace out this spiral on the globe depending on the relative speeds of the gantry and the globe.

 

[PeterDonis]

suppose you take a standard desk globe of the earth, and you attach a motorized gantry to the arch which supports it at the poles, and you spin the globe at a constant velocity, and you move the gantry from the north pole to the south pole at constant angular speed, with a pen attached to it, the pen will trace out this spiral on the globe depending on the relative speeds of the gantry and the globe.

What does this have to do with the thread topic?

 

[metastable]

What does this have to do with the thread topic?

I was asking if 3 spaces are necessary or only convenient, because I thought, via the illustration, I could specify an infinite subset points in space, not all along the same plane, with only 2 dimensions: 1) distance along spiral from pole and distance along straight line from point A through 1)

 

[PeterDonis]

I thought, via the illustration, I could specify an infinite subset points in space, not all along the same plane, with only 2 dimensions

Since the motion in your scenario is restricted to the surface of a 2-sphere, obviously it is only describing points in a space with 2 dimensions, since a 2-sphere is a 2-dimensional manifold. The fact that you are describing the 2-sphere using its embedding in 3-dimensional space does not change that.

 

[metastable]

Is the spiral not 1 dimensional since no area no volume?

 

[Ibix]

Can I write a complete list of the x,y,z positions an electron had as it traveled a meter, even in principle?

$\vec r(t)=\int_0^t\vec v(t')dt'$

And a critical problem with your approach is that you cannot use calculus to describe motion across the grain of your spiral because you don't have a smooth map from space to your coordinate system. So you've gone a long way towards disabling the only tool you can use to describe things moving in rigorous mathematical terms. And you haven't achieved your goal of specifying points in 3-space using two numbers because it's impossible to do so.

 

[metastable]

And a critical problem with your approach is that you cannot use calculus to describe motion across the grain of your spiral because you don't have a smooth map from space to your coordinate system.

After specifying a position, I was toying with the idea in my head of describing its heading with a second spiral at the specified point and its instantaneous motion as a curve which would be represented by the circumference of a circle with a specified radius and orientation, but I am far from claiming any of this represents reality, but I mention it because you mentioned not being able to use calculus with the approach and I wondered if specifying instantaneous motions as such curves could work with calculus?

 

[Buzz Bloom]

In rough terms: suppose you take a standard desk globe of the earth, and you attach a motorized gantry to the arch which supports it at the poles, and you spin the globe at a constant velocity, and you move the gantry from the north pole to the south pole at constant angular speed, with a pen attached to it, the pen will trace out this spiral on the globe depending on the relative speeds of the gantry and the globe.

Hi metastable:

I am wondering why you associate this concept with the topic of this thread?

Regards,
Buzz

 

[Ibix]

After specifying a position, I was toying with the idea in my head of describing its heading with a second spiral at the specified point and its instantaneous motion as a curve which would be represented by the circumference of a circle with a specified radius and orientation, but I am far from claiming any of this represents reality, but I mention it because you mentioned not being able to use calculus with the approach and I wondered if specifying instantaneous motions as such curves could work with calculus?

Then you have three numbers to specify positions in 3d space. Inevitably. I suspect there's more than one way to identify a location using this system, so this has nasty mathematical properties. But knock yourself out.

 

[Ibix]

Hi metastable:

I am wondering why you associate this concept with the topic of this thread?

Regards,
Buzz

That's a good question. I'll stop contributing to this hijack now.

 

[metastable]

Hi metastable:

I am wondering why you associate this concept with the topic of this thread?

Regards,
Buzz

asking if 3 spaces are necessary or only convenient

^I was trying to ascertain whether "3 spaces" are a "requirement" & "necessary" to describe the universe or merely "mathematically convenient."

 

[George Jones]

^I was trying to ascertain whether "3 spaces" are a "requirement" & "necessary" to describe the universe or merely "mathematically convenient."

These "3 spaces" can be shown mathematically to be the only cosmological spaces that are both spatially isotropic and spatially homogeneous, but the mathematics (of Killing vectors) is appropriate for a thread that has an "A" label.

More spaces result if one/both of these conditions is/are relaxed.

 

[PeterDonis]

I was trying to ascertain whether "3 spaces" are a "requirement" & "necessary" to describe the universe or merely "mathematically convenient."

And I already answered that way back in post #36. Please do not discuss this question, or your questions about whether it takes 3 numbers to specify a point in 3-space (it does), any further in this thread.

 

[timmdeeg]

And thus to be consistent with known data, models of inflation must work in universes that have negative spatial curvature. The Lyth and Liddle reference that I gave above explicitly notes that inflation smooths inhomogeneities in both $\Omega > 1$ and $\Omega < 1$ universes.

So regardless the global curvature locally the universe can have small areas like our observable universe which are positively or negatively curved depending on local inhomogeneities?

If this is correct so far would some kind of averaging over local curvatures in principle yield the global curvature? It would also mean that even exact knowledge about local curvature provides no clue to global curvature.

 

[Buzz Bloom]

So regardless the global curvature locally the universe can have small areas like our observable universe which are positively or negatively curved depending on local inhomogeneities?

Hi timmdeeg:

The assumption that the quote question is answered "yes" implies the corresponding universe is not (for large scales) isotropic and homogeneous. The means it is not compatible with a GR universe.

Regards,
Buzz

 

[PeterDonis]

The assumption that the quote question is answered "yes" implies the corresponding universe is not (for large scales) isotropic and homogeneous.

Not quite. It means that if the universe on large enough scales is isotropic and homogeneous, "large enough" must mean "larger than our observable universe". If the universe were spatially infinite, or spatially finite but much, much, much larger than our observable universe, this would not be a problem from a modeling standpoint; but it would mean that we would not be able to test the assumption of global isotropy and homogeneity directly by observations, since those are by definition restricted to our observable universe.

The means it is not compatible with a GR universe.

GR does not require that the universe is homogeneous and isotropic on any scale. The assumptions of global homogeneity and isotropy are made because they are simple (i.e., we know exact solutions for this case) and, at least so far, consistent with the data. But if for some reason global homogeneity and isotropy were ruled out, that would not in any way mean GR could no longer model the universe. It would just be a lot harder since we would not be able to use any exact solutions and would have to do the modeling numerically.

 

[metastable]

Are there any conceivable objects that could be large enough outside the observable universe to have any measurable effects within our observable universe?

 

[Jaime Rudas]

Do you know of any particular specified parameters that mathematically defines a theoretical 3-torus topologically shaped universe model which is finite, flat, isotropic, and homogeneous?

I think Friedman was referring to this when, in his 1924 paper On the Possibility of a World with Constant Negative Curvature of Space, he mentioned:

"In the present Notice it will be shown that it actually is possible to derive from the Einstein world equations a world with constant negative curvature of space.
[...]
At the end of this Notice we will touch upon the question of whether on the grounds of the curvature of space one is allowed at all to judge on its finiteness or infinitude.
[...]
We have convinced ourselves that the Einstein world equations possesses solutions that correspond to a world with constant negative curvature of space. This fact points out that the world equations taken alone are not suficient to decide the question of the finiteness of our world. Knowledge of the curvature of space gives us still no immediate hint on its finiteness or infinitude. To arrive at a definite conclusion on the finiteness of space, one needs some supplementary agreements."

 

[Buzz Bloom]

To arrive at a definite conclusion on the finiteness of space, one needs some supplementary agreements.

Hi Jaime:

Thank you for the Friedmann quotes.

I interpret this as saying that something that scientists currently do not have (after almost 100 years) is a reason to believe a flat finite universe is possible, except for the fact that such a model has not (yet) been proved to be impossible.

Regards
Buzz

 

[Jaime Rudas]

I interpret this as saying that something that scientists currently do not have (after almost 100 years) is a reason to believe a flat finite universe is possible, except for the fact that such a model has not (yet) been proved to be impossible.

Hi Buzz:

I interpret this as that general relativity plus the cosmological principle aren't a suitable instrument to determine if the universe is finite or infinite. That is, the fact that the universe is flat or negatively curved does not imply that is infinite, as it is frequently seen in popular science texts.

On the other hand, it seems logical to consider possible everything that we can't prove impossible.

Regards.

 

[Buzz Bloom]

On the other hand, it seems logical to consider possible everything that we can't prove impossible.

Hi Jaime:

I do not disagree with the quote. However, I do not think it is practical to give serious consideration to a possibility for which there is no evidence that the possibility is true except that there is no proof that it cannot be true.

In another thread there was a point that many scientists want a 4 sigma (99.99% confidence level) before taking a conclusion seriously. I find this to be a reasonable attitude. To take seriously a possibility just because it has not been proved impossible is like deciding to take a possibility seriously if it has 0.01% probability of being true.

Regards,
Buzz