# The curvature and size of the Universe

• I
Jaime Rudas
TL;DR Summary
Is it possible to determine the curvature and size of the universe from Ωκ?
According to the Planck 2018 results, the curvature component of the density parameter of the universe is Ωκ=0.001±0.002.

From this data, would it be possible to determine the greatest possible positive curvature of the universe and the radius of the corresponding 3-sphere?

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... radius of the corresponding 3-sphere
Why are you sure that the universe is a 3-sphere? Why not a hyper torroid, for example? Why not some other topology?

Jaime Rudas
Why are you sure that the universe is a 3-sphere? ...
I don't claim that the universe is a 3-sphere. I'm just asking what would be the minimum radius of the universe if it were a 3-sphere with that Ωκ.

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I don't claim that the universe is a 3-sphere. I'm just asking what would be the minimum radius of the universe if it were a 3-sphere with that Ωκ.
Ah. Well, you'll find that precise formulation of questions will fend off the nitpickers like me here on PF.

pinball1970 and Jaime Rudas
Mentor
Why not a hyper torroid, for example? Why not some other topology?
AFAIK the only topology consistent with positive curvature and homogeneity and isotropy is a 3-sphere.

ohwilleke, Jaime Rudas and phinds
Hornbein
TL;DR Summary: Is it possible to determine the curvature and size of the universe from Ωκ?

According to the Planck 2018 results, the curvature component of the density parameter of the universe is Ωκ=0.001±0.002.

From this data, would it be possible to determine the greatest possible positive curvature of the universe and the radius of the corresponding 3-sphere?
"The unobservable Universe, assuming there’s no topological weirdness, must be at least 23 trillion light years in diameter, and contain a volume of space that’s over 15 million times as large as the volume we can observe."

That's the minimum consistent with the data and some assumptions. There is no known maximum.

ohwilleke, Drakkith and Jaime Rudas
"The unobservable Universe, assuming there’s no topological weirdness, must be at least 23 trillion light years in diameter, and contain a volume of space that’s over 15 million times as large as the volume we can observe."
That's from Siegel's Forbes article. It'd have been nice if you had attributed the quote.

Why is it so large, though?
We can get the present radius of the curvature of the universe from this formula:
##R_c=\frac{c}{H_0} \frac{1}{\sqrt{\lvert{\Omega_k}\rvert}}##
(see e.g. Ryden, section 4.1)
If we use values from Planck 2018 for ##H_0=67.4 km/s/Mpc## and the maximal value of ##\Omega_k=0.003## from the error bars, we get ##14.5 Glyr/\sqrt{0.003}=265 Glyr##
Using this value as the radius of a hypersphere, the current unobservable universe should be at least ~3.3 1.7 trillion light years in diameter. Where's the extra order of magnitude coming from?

(edit: correction, see below)

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Vanadium 50, Jaime Rudas and Ibix
Cerenkov
Could either Hornbein or Bandersnatch please provide a link to the Ethan Siegel article on Forbes where his quote comes from?

Thank you,

Cerenkov.

ohwilleke
Cerenkov
Thank you!

Staff Emeritus
Ωκ=0.001±0.002.
Note that this is consistent with positive, negative, or zero. Makes it tough to draw conclusions.

Tertius, ohwilleke and phinds
Jaime Rudas
Ωκ=0.001±0.002.
Note that this is consistent with positive, negative, or zero. Makes it tough to draw conclusions.
Yes, that's right, but what we're trying to find is the minimum size consistent with that data.

Jaime Rudas
If we use values from Planck 2018 for ##H_0=67.4 km/s/Mpc## and the maximal value of ##\Omega_k=0.003## from the error bars, we get ##14.5 Glyr/\sqrt{0.003}=265 Glyr##
Using this value as the radius of a hypersphere, the current unobservable universe should be at least ~3,3 trillion light years in diameter. Where's the extra order of magnitude coming from?
Thanks for your answer. Could you explain to me why if the radius is 265 Gly, the diameter is 3300 Gly?

That's because I done goofed. It should be the circumference of a circle, not twice the circumference. So half as much as I wrote there.

ohwilleke, Jaime Rudas and pinball1970
Staff Emeritus
maximal value of 0.003 from the error bars
These are likely the 68% error bars. You might want to go to 2σ or 3σ.

If the point is to find the minimum size (not the maximum cuvature and size as originally posted, as the two go in opposite directions), I would use a bigger number here and derive a smaller radius - perhaps 150 Gly.

This assumes the error bars are more or less Gaussian. The actual limit may be even smaller.

ohwilleke
Jaime Rudas
Using this value as the radius of a hypersphere, the current unobservable universe should be at least ~3.3 1.7 trillion light years in diameter. Where's the extra order of magnitude coming from?

(edit: correction, see below)
1700 Gly is just 18.5 times the diameter of the observable universe. It seems to me that, with that size, a greater curvature would have to be detected in the observable universe.

Jaime Rudas
If the point is to find the minimum size (not the maximum cuvature and size as originally posted, as the two go in opposite directions)...
Obviously the confusion is caused by my poor knowledge of the English language: I wanted to ask about two aspects:
- the greatest possible positive curvature of the universe;
- the radius of the 3-sphere to which this curvature corresponds.

Jaime Rudas
Using this value as the radius of a hypersphere, the current unobservable universe should be at least ~3.3 1.7 trillion light years in diameter. Where's the extra order of magnitude coming from?

“Astronomers have studied the data and have determined that space is flat, or nearly so. However, this determination is a measurement, and measurements have uncertainty. It remains possible that the Universe has a very tiny curvature. But if it is curved, then the equivalent of the 'Universe’s equator' is at least 500 times bigger than the visible Universe. Or possibly bigger than that.”

I asked him if he could tell me how he got that result and he replied:

“The key is the 0.002 uncertainty in the curvature. 1/0.002 = 500. The central value is consistent with zero, however the uncertainty shows allowed deviation.
Accordingly, at a 68% confidence level, the radius of the universe is >500x bigger than the visible universe”

Hornbein
“Astronomers have studied the data and have determined that space is flat, or nearly so. However, this determination is a measurement, and measurements have uncertainty. It remains possible that the Universe has a very tiny curvature. But if it is curved, then the equivalent of the 'Universe’s equator' is at least 500 times bigger than the visible Universe. Or possibly bigger than that.”

I asked him if he could tell me how he got that result and he replied:

“The key is the 0.002 uncertainty in the curvature. 1/0.002 = 500. The central value is consistent with zero, however the uncertainty shows allowed deviation.
Accordingly, at a 68% confidence level, the radius of the universe is >500x bigger than the visible universe”
This is totally bogus. Each sentence in that last paragraph contains a fundamental error. If I were still teaching statistics I would teach my students to identify them. Shame on Don Lincoln.

"Dr. Don Lincoln is a Senior Scientist at Fermilab, America’s leading particle physics laboratory, who has coauthored over 1,500 scientific papers. He was a member of the teams that discovered the top quark in 1995 and the Higgs boson in 2012."

I wish I were shocked that such a distinguished physicist has such a superficial knowledge of statistics, but having taught statistics I can tell you this is not surprising. Understanding statistics is not on the curriculum. It's all about memorizing formulas and doing calculations, a skill that has been obsolete for fifty years.

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Jaime Rudas
Cerenkov
Hello Hornbein.

I've been following this thread and I was just wondering if you could explain the nature of your disagreement with Don Lincoln's quote please?

Please note that my level in this forum is Basic and so my grasp of statistics is also pegged at that level too.

Thank you for any help given.

Cerenkov.

Hornbein
Hello Hornbein.
Hello Hornbein.

I've been following this thread and I was just wondering if you could explain the nature of your disagreement with Don Lincoln's quote please?

Please note that my level in this forum is Basic and so my grasp of statistics is also pegged at that level too.

Thank you for any help given.

Cerenkov.
It's a bit too much to type with my thumb, but Bandersnatches post is statistically correct. Don's first and very basic error is that he ignores the mean of the measurement, 0.001. The other two errors are basic but not easy enough to explain.

Cerenkov
Thanks Hornbein.

Umm... so where does that leave us in respect of the initial question of this thread?

Given that Lincoln's comments suffer from statistical errors and the apparent 'order of magnitude' confusion over Siegel's calculations, where does that leave value of the minimum curvature of a 3-sphere, expressed in light years?

Please note that my level of understanding is Basic.

Thank you,

Cerenkov.

Cerenkov
OK, relax Hornbein... I see it now!

Bandersnatch's revised figure.

1.7 trillion light years in diameter.

Thanks,

Cerenkov.

Jaime Rudas
Bandersnatch's revised figure.

1.7 trillion light years in diameter.
Actually, the Bandersnatch's calculation contains an error in considering ##\Omega_k=0.003##, because for the curvature to be positive, ##\Omega_k## must be negative. Thus, the smallest possible value is ##\Omega_k=-0.001##, the radius of curvature is at least 459 Gly, and the maximum circumference of the universe is at least 2.9 trillion light-years.

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ohwilleke and Bandersnatch
Yes, that's correct. Well spotted.

Jaime Rudas
Hornbein
That's from Siegel's Forbes article. It'd have been nice if you had attributed the quote.

Why is it so large, though?
We can get the present radius of the curvature of the universe from this formula:
##R_c=\frac{c}{H_0} \frac{1}{\sqrt{\lvert{\Omega_k}\rvert}}##
(see e.g. Ryden, section 4.1)
If we use values from Planck 2018 for ##H_0=67.4 km/s/Mpc## and the maximal value of ##\Omega_k=0.003## from the error bars, we get ##14.5 Glyr/\sqrt{0.003}=265 Glyr##
Using this value as the radius of a hypersphere, the current unobservable universe should be at least ~3.3 1.7 trillion light years in diameter. Where's the extra order of magnitude coming from?

(edit: correction, see below)
Here is my view. The standard deviation of the measurement is 0.002. Let's use three standard deviations to get magnitudes of 0.001 + 3*0.002 = 0.007. I then can say with 99.87% confidence that the magnitude of omega is at least this high. Then the radius is at least 170 gigalightyears. We can bet safely that the radius is greater than this.

According to Stack Exchange the visible Universe has a radius of 46 Gly.

We can also say that we have 30% confidence that the Universe is infinite (omega is nonnegative), as a hyperbolic Universe is infinite. Indeed such a Universe is in a sense larger than a flat one.

Jaime Rudas
We can also say that we have 30% confidence that the Universe is infinite (omega is nonnegative), as a hyperbolic Universe is infinite. Indeed such a Universe is in a sense larger than a flat one.
In what sense is a hyperbolic universe larger than a flat one?

ohwilleke
Hornbein
In what sense is a hyperbolic universe larger than a flat one?
Take a sphere with a given radius. The more hyperbolic the Universe, the more volume inside that sphere.

Jaime Rudas
Take a sphere with a given radius. The more hyperbolic the Universe, the more volume inside that sphere.

Ok, but AFAIK the same can be said of the spherical universe: the volume inside a radius Ro sphere in a positive curvature space is greater than the volume inside a radius Ro sphere in a flat space, but that does not mean that a spherical universe is greater than a flat one.

ohwilleke
Hornbein
Ok, but AFAIK the same can be said of the spherical universe: the volume inside a radius Ro sphere in a positive curvature space is greater than the volume inside a radius Ro sphere in a flat space, but that does not mean that a spherical universe is greater than a flat one.
It could be said, but it would be wrong. The volume inside of a sphere in a positive curvature space is less than in a flat space.

Jaime Rudas
Jaime Rudas
It could be said, but it would be wrong. The volume inside of a sphere in a positive curvature space is less than in a flat space.
But of course! I don't know what I was thinking!

Jaime Rudas
It could be said, but it would be wrong. The volume inside of a sphere in a positive curvature space is less than in a flat space.
If the universe is flat, the volume of the observable universe (radius Ro = 46 Gly) is:

## \dfrac 43 \pi R_0^3 = 407,720 ~ Gly^3 ##

What would be its volume if the universe is spherical with curvature radius Rc = 459Gly?

Jaime Rudas
If the universe is flat, the volume of the observable universe (radius Ro = 46 Gly) is:

## \dfrac 43 \pi R_0^3 = 407,720 ~ Gly^3 ##

What would be its volume if the universe is spherical with curvature radius Rc = 459Gly?
I found that the volume of a sphere in a 3-spherical space is:

##V(r)=2\pi R^3 \left ( \dfrac r R-\dfrac{\sin \dfrac{2 r}R}2 \right )##

Where r is the radius of the sphere and R is the radius of curvature of the 3-spherical space.

For r=46 Gly and R=459 Gly, V=406,902 Gly³

The derivation of the formula can be seen in:

https://forum.lawebdefisica.com/for...-en-geometría-no-euclidea?p=361848#post361848

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