- #1

Spinnor

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Does this get me close? First let us estimate the total length of the line segments, Ʃ. Use an average separation distance D between each pair of points. The distance D is of order one half the length of the volume V, V = L^3, D = L/2.

The total number of segments is N(N-1)/2 so an estimate for the length of line-segments,

Ʃ = D*N(N-1)/2 for large N this is about D*N^2/2

Ʃ ≈ D*N^2/2

Assume this total length is evenly divided into each small volume v. The length in volume v is the fraction [d^3/D^3] times Ʃ,

[d^3/D^3]*Ʃ = [d^3/D^3]*D*N^2/2 = σ

Assume the average length of the line-segments that intersect the little volume v is one-half the length the little volume v, d/2.

Then the average number of line-segments in v, ω, is,

ω = σ/[d/2] = {[d^3/D^3]*D*N^2/2}/[d/2] = d^2*N^2/D^2

using Wolfram calculator,

http://www.wolframalpha.com/

Using our Universe as an example, let d = 1m, N = 10^80, D = [3.5*10^80m^3]^.3333 ≈ 7*10^26

ω = 2*10^106 segments intersecting a volume of 1m^3.

We can ask what must the size of the volume v above be so that on average there will be only one line-segment intersecting it.

Set ω = 1 = d^2*N^2/D^2 now d is unknown and we use N and D above,

d = D/N = 7*10^26/10^80 = 7*10^-54m. If we are too near a point this estimate is bad. If we enclose a point ω jumps by about N

We can also ask how many points must a volume v have so that ω above changes significantly because of the additional line-segments from the enclosed points.

I made many bad estimates but I think I'm within a factor of a billion above?

Thanks for any help!