WOLFRAM NOTEBOOK

The Exponential Hypothesis

Incorrect:

Given an expansion parameter per step:
κ^(
t
H
/T)
L = (c tH)/
κ^(
t
H
/T)
Lp/Sqrt[Ξ]=LH/κ^(Sqrt[Ξ] tH/tP)
tH/tP Sqrt[Ξ] =κ^(Sqrt[Ξ] tH/tP)
New version:
10^60Sqrt[Ξ]=κ^(10^60Sqrt[Ξ])
(LP/(c tH)/
κ^(
t
H
/T)
)^2 = Ξ
(tP/( tH)/
κ^
Ξ
t
H
tP
)^2 = Ξ
κ^(-10^60Sqrt[Ξ])=10^60Sqrt[Ξ]
In[]:=
Solve[k^-xx,x]
Solve
:Inverse functions are being used by Solve​, so some solutions may not be found; use Reduce for complete solution information.
Out[]=
x
ProductLog[Log[k]]
Log[k]
In[]:=
Plot
ProductLog[Log[k]]
Log[k]
,{k,1,4}
Out[]=
In[]:=
Plot[{x,2^-x},{x,0,4}]
Out[]=

Correct version:
L
must change with time

L = (c tH)/
(κ^(
t
H
/T)
^(1/d))
L = (LH)/
(κ^(
t
H
/T)
^(1/d))
L = c T
tH / T = (κ^(tH/ T )^(1/d))
In[]:=
Solve[xk^x,x]
Solve
:Inverse functions are being used by Solve​, so some solutions may not be found; use Reduce for complete solution information.
Out[]=
x-
ProductLog[-Log[k]]
Log[k]
In[]:=
ReImPlot-
ProductLog[-Log[k]]
Log[k]
,{k,1,4},PlotRangeAll
Out[]=
In[]:=
Solve[xk^(x/d),x]
Solve
:Inverse functions are being used by Solve​, so some solutions may not be found; use Reduce for complete solution information.
Out[]=
x-
dProductLog-
Log[k]
d
Log[k]
In[]:=
ReImPlotEvaluate-
dProductLog-
Log[k]
d
Log[k]
/.d3,{k,1,4}
Out[]=

Try κ = 1 + ϵ

Jonathan’s version

T = tH/
(κ^(
t
H
/T)
^(1/d))
T/tHκ^-(tH/dT)
In[]:=
xk^(-1/x)
Out[]=
x
-1/x
k
In[]:=
Solve[%,x]
Solve
:Inverse functions are being used by Solve​, so some solutions may not be found; use Reduce for complete solution information.
Out[]=
x-
Log[k]
ProductLog[-Log[k]]
In[]:=
ReImPlot-
Log[k]
ProductLog[-Log[k]]
,{k,1,4}
Out[]=
xk^(-1/(3x))

If there is refinement, then T (as measured in seconds) must vary with time

T(now) = T(0) κ^-(tH/T(now))
T[t]T[0]κ^-Integrate[1/T[τ],{τ,0,t}]
L(now) = c T(now)
If T(t) decreases rapidly, it takes more time steps for any fixed thing to happen.
κ tells you for each event how many edges are produced
Also κ gives you the number of branchings per event
Assume that merging is not so important.....
M(now) : volume of the multiway system
M(now) = κ ^ (tH/T(now))
M(now) = M(0) Integrate[κ’^(τ / T[τ]), {τ,0, tH}]
Ξ = M(now)
L(now) = Lp/Sqrt[M(now)]
First approx:
Ξ = κ^(tH/T(now))
T(now) = tP / κ^(tH/T(now))
T(now)/tP = k^-(tH/tP (tP/T(now)))
tr = k^-(tH/tP * 1/tr)

Try 2

T(0)/T(now) is of order κ^steps ; Ξ ~ κ^κ^steps
T(now)/T(0) ~ κ^-steps
T(now)/TP ~ 1/Sqrt[Ξ]
T(now)/T(0) ~ 1/Ξ
T(now)/TP ~ 1/Sqrt[Ξ]
T(0)/TP ~ Sqrt[Ξ]
T(0) ~ Sqrt[Ξ] TP
T(now) ~ TP/Sqrt[Ξ]
Ξ ~ κ^κ ^ (TH / T(now) )
Ξ ~ κ^κ^ ( (TH/TP) Sqrt[Ξ] )

Compare spatial vs multiway

At a given step, size S spatial graph goes to size κ S ;
multiway graph goes to κ^S

Multiway evolution

What is the growth rate of a multiway graph that has resolutions?

What is a generation?
​
It is the individual single updating event time....
​
“Generational time” = Ξ
branchlike slice to the next does Ξ Tbar event times
Observer perceives these in parallel
“Observer time” i.e. generational time
“Event time” : individual update event times
event energy = ℏ / generational time
Ξ Ebar = hbar / tevent
tgeneration = Ξ tevent
Let’s define Tbar to be the elementary event time
​
To get to the next branchial hypersurface, every node in the multiway graph has to be updated
Multiway graph has a separate node for every configuration
Ξ is size of branchial graph....
In terms of generational time. the branchial graph grows ~ κ ^( t / tgeneration )
Ξ ~ κ^( tH / tgeneration)
tgeneration ~ Ξ tevent
Ξ ~ κ ^ ( tH / ( Ξ tevent ) )
tevent = tP/Sqrt[Ξ]
Ξ ~ κ ^ ( (tH/tP) / Sqrt[Ξ] )
Value of x
For κ close to 1, Ξ ~ tH/tP
Ξ ~ (tH/tP)^κ
Ξ ~ κ ^ ( (tH/tP) / Sqrt[Ξ] )

Substitute values

Main result:

Ξ ~ κ ^ ( (tH/tP) / Sqrt[Ξ] )

Interpretation

Ξ ~ κ^( tH / tgeneration)
Over time, tgeneration gets longer ( time to measure everything in the universe gets bigger ... )
Meanwhile, tH goes up

Implications

Number of nodes across the universe:
Maximum speed of entanglement:
Ξ E0 / ℏ
In nodes in multiway graph.....
black hole at center of galaxy
4.31 million solar masses
Gulp the galaxy in half an hour
[ Traversing multiway space is faster than traversing spacetime; because more densely connected ]
Probably lots of particles lighter than electron
Particles 10^-30 times electron mass, what happens???
Axions go down to 10^-5
Dilaton ??
Graviton mass: XXX https://arxiv.org/pdf/1706.01812.pdf
Photon mass: XXXX
E0 is 10^-30 eV

Oligon physics

Depending on when it decouples, its temperature might be 1 K
10^-5 eV / 10^-30 eV

CMB, CNB

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