On Wednesday, March 25, 2020 at 1:28:21 AM UTC-5, Alan Grayson wrote:
> On Tuesday, March 24, 2020 at 6:21:50 PM UTC-6, Lawrence Crowell wrote:
>> On Tuesday, March 24, 2020 at 6:19:41 PM UTC-5, Alan Grayson wrote:
>>> On Monday, March 23, 2020 at 9:35:48 AM UTC-6, Lawrence Crowell wrote:
>>>> Inflation was initiate 10^{-35}sec after the quantum fluctuation 
>>>> appearance of the observable cosmos, and this had a duration of 
>>>> 10^{-30}sec. The cosmological constant averaged around Λ = 10^{48}m^{-2}. 
>>>> If I divide by the speed of light squared this comes to 10^{32}s^{-2} and 
>>>> we get √(Λ)T = 10^{2}. This means any spatial region expanded by a factor 
>>>> of 10^{√(Λ)T} which is large. The natural log of this is 230 and this 
>>>> is not too far off from the more precise calculation of 60 e-folds. The 60 
>>>> e-folds is a phenomenological fit that matches inflation with the observed 
>>>> universe.
>>>> How much of the universe is unavailable depends upon whether k = -1, 0 
>>>> or 1. The furthest out some quantum might emerge and have an influence is 
>>>> for a Planck scale quantum to now be inflated to the CMB scale. I know I 
>>>> have gone through this here before, but the result is the furthest we can 
>>>> detect anything is around 1800 billion light years, which would be a 
>>>> graviton or quantum black hole that leaves an imprint or signature on the 
>>>> CMB. It is not possible from theory to know what percentage this is of the 
>>>> entire shebang, and for k = -1 or 0 it is an infinitesimal part.
>>>> LC
>>> For k=0, a flat universe, we know the answer since, as you've 
>>> acknowledged, it's infinite in spatial extent.  Consequently, since the 
>>> observable universe is finite in spatial extent, the unobserved universe 
>>> must be infinite in extent (for a flat universe). Can you estimate the size 
>>> of the unobservable universe for a positively curved universe? AG
>> The cosmological constant is a Ricci curvature with Λ = R_{tt} for the 
>> flat k = 0 case. for k = 1 there is a spatial Ricci curvature R_{rr}. This 
>> contributes to the occurrence of the cosmological constant, but it is 
>> tiny. So R_{rr} = δR_{tt} for δ a rather small number. The spatial sphere 
>> has a radius R = 1/√(R_{rr}} ≈ 1/√(δΛ). This is then for Λ = 
>> 10^{-52}m^{-2} R ≈ δ^{-1/2} 10^{26}m, or about the distance to the 
>> cosmological horizon multiplied by the reciprocal of a small number. 
>> The problem is that we really do not what that small number is. For 
>> various reasons I think it is δ < 5×10^{-5}This gives a radius where a 
>> Planck frequency is redshifted to a CMB scale. If it is smaller then there 
>> are regions of the universe completely inaccessible to us even as Planck 
>> modes redshifted to the cosmic horizon scale.
>> LC
> FWIW, another reason I think our universe has a positive curvature is that 
> if it were flat, with zero curvature, and we made many measurements, we'd 
> get a distribution of measured values above and below zero due to 
> unavoidable measurement errors. But I think we invariably get a small 
> positive number. Is this what we actually get; values always positive but 
> close to zero, but no negative values? TIA,AG 

As yet attempt to find optical results due to spatial curvature have not 
found anything. The curvature of spacetime is mostly due to how space is 
embedded in spacetime.


>>> On Sunday, March 22, 2020 at 11:55:19 PM UTC-5, Alan Grayson wrote:
>>>>> According to some cosmologists, Krauss?, the duration of inflation is 
>>>>> about 10^-35 seconds, which is presumably the duration necessary to 
>>>>> create 
>>>>> isotropy and homogeneity in a universe of age 13.8 BY. If these 
>>>>> assumptions 
>>>>> are correct, can we use them to compute the fraction of the universe 
>>>>> which 
>>>>> is now UN-observable? TIA, AG

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