3pi/Lambda = area of cosmological horizon

[marcus]

It just turns out that the dark energy density Λ or cosmological constant is of such a size that

3pi/Λ = the surface area of the observable universe

Smolin calls the surface of the observable universe "the cosmological horizon" and treats it in the same paragraphs with black hole event horizons. Both are kinds of horizons. According to LQG both sorts of areas must be integers when expressed in natural units ("quantized in steps of the Planck area")

Also according to LQG the reciprocal of Λ is quantized.
Smolin says that 6pi/Λ must be an integer. It is the dimension of some finite dimensional Hilbert space important in quantum cosmology.

The usual figure for the (current) radius of the observable is about 40 billion lightyears. So the cosmological horizon is this expanding spherical surface with (current) area 7.3E123
And meanwhile 3pi/Λ turns out to be currently the same 7.3E123

The cosmological horizon is receding at about 3c (Ned Wright's tutorial has a good explanation, its also in the FAQ). No connection is assumed between Lambda and the area, at least by me, but the coincidence of the two numbers is striking and if they are connected then, since the area is expanding this means Lambda is diminishing----asymptotic to zero----a declining rate of acceleration leading to continued expansion (but no longer noticeably accelerating expansion)

Smolin has a good survey of QG

arXiv:hep-th/0303185

quite recent, John Baez reviewed it in his latest column

 

[marcus]

I wrote an expert in quantum gravity about this.
It turns out there is an underlying connection between the two numbers, so this is probably not so interesting. It could be,
but probably isnt.

The ratio of 1/Λ the reciprocal cosmo. constant and the area of the horizon has to be

2/3 divided by the dark energy fraction of rho crit and divided by the square of the radius factor (R_observable/R_Hubble)

This radius factor (R_observable/R_Hubble) is commonly taken to be 3, which squared is 9. And the value of the dark energy fraction of rho crit derived from current observation(WMAP) is 0.73.
So the relations between 1/Λ and the area of the horizon has to be 2/3 divided by 9 and divided by 0.73.
So it looks like it might be just a temporary coincidence that these numbers are of the same magnitude. Or so it seems for now.

Originally posted by marcus
It just turns out that the dark energy density Λ or cosmological constant is of such a size that

3pi/Λ = the surface area of the observable universe

Smolin calls the surface of the observable universe "the cosmological horizon" and treats it in the same paragraphs with black hole event horizons. Both are kinds of horizons. According to LQG both sorts of areas must be integers when expressed in natural units ("quantized in steps of the Planck area")

Also according to LQG the reciprocal of Λ is quantized.
Smolin says that 6pi/Λ must be an integer. It is the dimension of some finite dimensional Hilbert space important in quantum cosmology.

The usual figure for the (current) radius of the observable is about 40 billion lightyears. So the cosmological horizon is this expanding spherical surface with (current) area 7.3E123
And meanwhile 3pi/Λ turns out to be currently the same 7.3E123

The cosmological horizon is receding at about 3c (Ned Wright's tutorial has a good explanation, its also in the FAQ). No connection is assumed between Lambda and the area, at least by me, but the coincidence of the two numbers is striking and if they are connected then, since the area is expanding this means Lambda is diminishing----asymptotic to zero----a declining rate of acceleration leading to continued expansion (but no longer noticeably accelerating expansion)

Smolin has a good survey of QG

arXiv:hep-th/0303185

quite recent, John Baez reviewed it in his latest column