The curvature and size of the Universe

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SUMMARY

The forum discussion centers on the curvature and size of the universe, specifically referencing the Planck 2018 results which indicate a curvature density parameter of Ωκ=0.001±0.002. Participants debate the implications of this data for determining the universe's curvature and the minimum radius of a corresponding 3-sphere. Calculations suggest that the unobservable universe must be at least 23 trillion light years in diameter, with a minimum radius of curvature estimated at 459 Gly. The conversation also highlights statistical misunderstandings regarding curvature measurements and their implications for cosmological models.

PREREQUISITES
  • Understanding of cosmological parameters, specifically Ωκ (curvature density parameter).
  • Familiarity with the Hubble constant (H₀) and its significance in cosmology.
  • Basic knowledge of spherical geometry and hyperspheres.
  • Statistical analysis principles, particularly regarding measurement uncertainty.
NEXT STEPS
  • Research the implications of the Hubble constant (H₀=67.4 km/s/Mpc) on cosmic expansion.
  • Explore the mathematical derivation of the radius of curvature using the formula R_c=c/H₀ * 1/√|Ω_k|.
  • Study the different topologies of the universe, including 3-spheres and hyperbolic models.
  • Investigate the statistical methods used in cosmological measurements and their uncertainties.
USEFUL FOR

Astronomers, cosmologists, physicists, and anyone interested in the geometric and statistical properties of the universe will benefit from this discussion.

  • #31
Hornbein said:
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!
 
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  • #32
Hornbein said:
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?
 
  • #34
Jaime Rudas said:
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
 
  • #35
Temporarily closed for moderation.
 
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  • #36
Thread has been cleaned up and is now reopened.
 
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  • #37
And now the thread is closed again for moderation.
 
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