Do we have evidence the universe is finite in size?

[rjbeery]

I've seen various, wildly different, estimates of the size of the Universe. Do we have evidence demanding that the Universe is finite in size? If so, what are the clues that lead us to estimate that size beyond absolute speculation?

 

[jerromyjon]

estimate that size beyond absolute speculation

https://www.physicsforums.com/insights/radius-observable-universe-light-years-greater-age/

 

[marcus]

I've seen various, wildly different, estimates of the size of the Universe. Do we have evidence demanding that the Universe is finite in size?

No evidence as far as I know.
It's common to use a version of the standard cosmic model ΛCDM which is spatially flat and take for granted that it is spatially infinite---because that is comparatively easy to compute with and is consistent with observations.

...what are the clues that lead us to estimate that size beyond absolute speculation?

Estimates of size, if you read the fine print, usually involve making additional assumptions---and often involve a measurement of spatial curvature.

You can see how assuming a value for the spatial curvature (conventionally denoted Ω_k), if you add to that the uniformity assumption called the "Cosmological Principle" so you assume that the rest of the U has the SAME spatial curvature as the region where we can measure, could lead to a figure for the size.

 

[rjbeery]

No evidence as far as I know.
It's common to use a version of the standard cosmic model ΛCDM which is spatially flat and take for granted that it is spatially infinite---because that is comparatively easy to compute with and is consistent with observations.

==quote==
...what are the clues that lead us to estimate that size beyond absolute speculation?
==endquote==
Estimates of size, if you read the fine print, usually involve making additional assumptions---and often involve a measurement of spatial curvature.

You can see how assuming a value for the spatial curvature (conventionally denoted Ω_k), if you add to that the uniformity assumption called the "Cosmological Principle" so you assume that the rest of the U has the SAME spatial curvature as the region where we can measure, could lead to a figure for the size.

Curvature, of course. Makes sense thanks.

 

[marcus]

Curvature, of course. Makes sense thanks.

Yes! It's neat! There is a distance called the "radius of curvature" and (in the positive curvature case where large triangles add up to more than 180º) the formula for it is the Hubble radius divided by the square root of |Ω_k|

Because of some cockeyed historical accident which has never been rectified, Ω_k was defined with a rogue minus sign so that positive spatial curvature is expressed by -Ω_k, so you need the absolute value to take the square root. Anyway if you see a confidence interval for Ω_k it will be around zero (the flat case) and it will say something like -Ω_k < 0.01. That is the LARGEST the curvature could be (an upper bound) so it tells you the SMALLEST a spatial 3-sphere universe could be (a lower bound on the radius of curvature). So you can multiply that by 2π and get a kind of circumference. If you could pause expansion to make circumnavigating possible, how long would it take to go around...

So then if -Ω_k < 0.01 the square root is 0.1 and you know the Hubble radius is 14.4 billion LY, so you divide by 0.1 and give 144 billion LY, the RoC. And multiply by 2π to get the circumf.
Half that would be the farthest away anything could be at this moment.

 

[rjbeery]

Yes! It's neat! There is a distance called the "radius of curvature" and (in the positive curvature case where large triangles add up to more than 180º) the formula for it is the Hubble radius divided by the square root of |Ω_k|

Because of some cockeyed historical accident which has never been rectified, Ω_k was defined with a rogue minus sign so that positive spatial curvature is expressed by -Ω_k, so you need the absolute value to take the square root. Anyway if you see a confidence interval for Ω_k it will be around zero (the flat case) and it will say something like -Ω_k < 0.01. That is the LARGEST the curvature could be (an upper bound) so it tells you the SMALLEST a spatial 3-sphere universe could be (a lower bound on the radius of curvature). So you can multiply that by 2π and get a kind of circumference. If you could pause expansion to make circumnavigating possible, how long would it take to go around...

So then if -Ω_k < 0.01 the square root is 0.1 and you know the Hubble radius is 14.4 billion LY, so you divide by 0.1 and give 144 billion LY, the RoC. And multiply by 2π to get the circumf.
Half that would be the farthest away anything could be at this moment.

Well that explains the various estimates of the size of the universe; differing estimates for global curvature possibly based on different methods.

 

[Chalnoth]

Well that explains the various estimates of the size of the universe; differing estimates for global curvature possibly based on different methods.

Not really different methods. Just different data.

 

[rjbeery]

Not really different methods. Just different data.

Right, that's what I meant, calculating curvature based off of data collected through different methods...

 

[Chronos]

This is also a good way to humor check intrinsic uncertainties between data sets. When a single data set stands out from the others as an outlier, it is usually a good sign that data has unresolved errors and is less reliable than the others in some respect.