c-----------------------------------------------------------------------c
c Fortran program simulating the evolution of the universe. c
c Stig Sundman 2007-03-02. c
c-----------------------------------------------------------------------c
Revised 2008-08-06:
Phases 1 and 2: The variables r, ri, and rf were removed from
the calculation to demonstrate the fact that r is an arbitrary
parameter without precise physical meaning.
The basic fact is that the photon's wavelength lambda grows
as (t/ti)**(1/3), which follows from dV/dt = constant.
Therefore, ti alone defines the initial condition of each
phase. Thus, tf is the only output value from the previous
phase that is used as input value (ti) in the new phase.
Added 2009-02-14:
Calculations described in Appendix E.8.
See comments marked 'Pions'.
c-----------------------------------------------------------------------c
General:
The fundamental hydrodynamic equation, also known as the momentum equation,
is the mathematical expression for Newton's second law applied to a fluid.
The hydromechanical model for space directly follows from the simplest
imaginable meaningful solution to the momentum equation.
The hydromechanical model explains Dirac's large-number hypothesis, which
Paul Dirac himself deduced from observations.
Dirac noted that m_universe/m_electron is about 10**80 = 10**40
squared, and suggested that it is not by chance that:
* r_universe/r_electron is about 10**40,
* F_electric/F_gravity between two electrons is about 10**40,
* age_universe/(r_electron/c) is about 10**40.
To precisely describe the first three phases of the universe, path integrals
should be used. In the path-integral description, the number of particles at
the end of each phase is obtained as a weighted average over all possible
numbers.
In this program, I instead assume that the universe at every instant in time
contains a well-defined number N of particles (where N is an integer).
This means that no precise values can be obtained for the lepton mass ratios
and alpha. However, a quite precise value, 2.447 billion, is obtained for the
intitial photon-baryon ratio.
In addition, qualitative results are obtained that give new insight into
the evolution of the universe.
In particular, the simulation demonstrates a time paradox, which has the
consequence that G and H appear to decrease at an imperceptibly slow rate.
In other words, G and H appear to be constant in time.
Since a constant H implies acceleration, the universe is falsely believed to
be accelerating.
Details:
The hydromechanical model for space states that a volume V containing a fixed
number of particles (i.e., a given amount of energy), grows at a steady rate:
dV/dt = constant. Mathematical experiments, which are based on this equation,
lead to an unambiguous picture of the universe before structures begin to
form.
dV/dt = constant implies that r is proportional to t**(1/3).
In phase 1, energy conservation forces c to increase so that the spinless
tauon's rest energy (proportional to c**2) grows, thereby compensating for
the decrease in photon energy (proportional to 1/lambda) due to the expansion
of V.
Phase 2 is similar to phase 1, except that matter is represented by the
spinless muon.
In phase 3, matter is represented by the (spinning) electron. Planck's
constant h has appeared in the expression for photon energy: hc/lambda.
Since h is constant, the growing c causes the photon energy to decrease
at a much slower rate than before, and, consequently, the electron rest
energy to grow only slowly.
In phase 4, the formation of macroscopic structures, in particular black
holes, complicates the picture.
In all four phases, particle lifetimes (tau) grow with c.
This is the "true" picture of the universe in the sense that the energy
principle holds true and the universe is calculable. Since tau increases,
time intervals measured by the tick of an atomic clock increase in length.
In our practical system of measurement, we measure time by counting clock
ticks. In this system, the energy principle does not hold, the universe is
uncalculable, and it is impossible to reconcile Dirac's large-number
hypothesis with observations of dG/dt and dH/dt.
From the tauon-muon and muon-electron mass ratios it can be calculated how
much the rest energies of the massive particles must have grown in the first
two phases: about 16.919 times in phase 1, and 151.13631 times in phase 2.
These figures are target values used as input to the simulation. In a
path-integral approach, the exact values should result from the computation.
----------------------------------------------------------------------------
Data used as guidelines for the calculation
206.76826(3) = measured muon-electron (mu-e) mass ratio
205.75922 = uncorrected theoretical mu-e ratio = 1/(B*alpha)
151.13631 = mass increase of spin-0 muon in phase 2 (from mu-e ratio)
16.818(3) = measured tauon-muon mass ratio
16.919(4) = estimated ratio in the absence of radiative corrections
= estimated growth of spinless-tauon mass in phase 1
Variables used in the program:
tc time of creation (initial age of the universe)
rc initial radius of the universe at time of creation
cc initial value of the photon's velocity c (the speed of light)
Nf final number of particle pairs at the end of a phase
N number of massive-particle pairs
Nf-N number of radiative-particle (massless photon) pairs
Em total energy of matter (of the N massive-particle pairs)
Er total energy of radiation (total photon energy)
Em1 energy of a massive-particle pair, Em1 = Em/N
DelEr loss in radiation energy between annihilations
t time, or age of the universe
tN time when a pair annihilation takes place
ti initial time when phase begins
tf time of final annihilation (of last pair at end of phase)
r radius of the universe
ri initial radius of universe at beginning of phase
rf final radius of universe at end of phase
tau lifetime of massive particle pair (mean time or average time)
taui initial lifetime at beginning of phase
tauf final lifetime at end of phase
tauN tau at time of pair annihilation when N + 1 -> N
In phase 4:
tau any particle lifetime (atomic clock tick)
tau0 present particle lifetime (atomic clock tick)
t0 present age of the universe
Er0 present energy of photon radiation in volume V
Em0 present rest energy of matter in volume V
c0 present value of c
Emi Initial matter (proton + electron) rest energy in volume V,
1836 + 1 in units of initial electron rest energy
Eri Initial radiation energy in volume V, N3*2/x with N3 (about
2447000000) and x (about 10.89) calculated in phase 3
Boltz the Boltzmann constant, k = 8.617343(15)/10**5 eV/K
CBR present CBR temperature, 2.725(1) K
Without loss of generality one may choose:
tc = 1
rc = 1
cc = 1
Em1 = 1 for t = tc
Only the lifetime tau is subject to experimentation.
Its initial value and time dependence much be such that, at the
end of a phase, Nf = (tf/tc)**2 approximately, and the target
value for the final Em is met as closely as possible.
It turns out that (in units where tc = cc = 1):
tau = 1 initially in phase 1 and phase 2
tau = c for t >= ti
------------------------------------------------------------------------
BEGINNING OF PROGRAM
c.for options 32 6 12 3
c
c Note: I use my own homemade Fortran. In it,
c integers are integer*4, and real numbers are real*10
c (the math processor's internal tenbyte precision).
c
integer N, Nf, N1, N2, i, iter, nn1, nn2
integer imax
c
real Em, Er, DelEr, Em1
real t, tc, ti, tf, Delta
real tau, tauN, taui, taui1, taui2, taui3, tau1, tau2, tau3
real r, ri
real pi, alpha
real ratio, sum, help, x, x1, x2
real c, p, h, cN, cNf, cN3, cnn3
real E0, cntau, ck
real c0, tau0, Em0, Er0
real ci, Emi, Eri
real CBR, Boltz
real Pions, cme, cmpi, cmp, cmtau0, cmtau1, cmtau
real GF, y, ymu, z, y1
real targ1, targ2
c
pi = 3.14159265358979
alpha = 1/137.035999
Emi = 1837.15267261
Boltz = 8.617343/10**5
CBR = 2.725
c
tc = 1
c Initialize user-given values.
taui1 = 1
taui2 = 1
taui3 = 1/(2*pi**2*alpha**2)
N1 = 86
N2 = 1010
cN3 = 2447600000.0
c
open (9,file='temp0.tXt',status='new')
c
c V------------------------------------------ Pions ------------------V
Pions = 0
c
c To execute the modified code, remove statement 'go to 20':
c
go to 20
c
c Calculate in local picture the muon-electron mass ratio and
c the target ratios used in the simulation.
c
call muerat (ratio, targ1, targ2)
write (*,10) 'mass ratios: ', ratio, targ1, targ2
write (9,10) 'mass ratios: ', ratio, targ1, targ2
10 format (a, f15.9, f10.4, f15.9)
c
c Values used by modified code.
c Choose second cN3 value when speeding up the computation in phase3.
c
cN3 = 1393137450
cN3 = 1393096000
taui3 = 1/(8*pi*alpha**2)
x = 0.437
Pions = 1
go to 50
c A-------------------------------------------------------------------A
c
20 write (*,*) 'lifetimes =', taui1, taui2, taui3
write (*,*) 'Give new lifetimes, 0 to keep old, -1 0 0 = end: '
read (*,*) tau1, tau2, tau3
if (tau1.lt.0) go to 990
if (tau1.gt.0) taui1 = tau1
if (tau2.gt.0) taui2 = tau2
if (tau3.gt.0) taui3 = tau3
c
30 write (*,*) 'N1, N2, N3 =', N1, N2, cN3
write (*,*) 'Give N1, N2, and N3, 0 to keep old, -1 0 0 = back: '
read (*,*) nn1, nn2, cnn3
if (nn1.lt.0) go to 20
if (nn1.ne.0) N1 = nn1
if (nn2.ne.0) N2 = nn2
if (cnn3.ne.0) cN3 = cnn3
c
40 write (*,*) 'Give x in t = tN + x*Delta (x = 0.437, say): '
read (*,*) x
50 continue
c-----------------------------------------------------------------------c
c Phase 1. Follow the growth of the spinless tauon's self energy. c
c-----------------------------------------------------------------------c
c Em + Er = Nf. Between annihilations: N*Em1 + constant/r = Nf.
c
Nf = N1
taui = taui1
ti = tc
t = ti
N = Nf
Em = Nf
Er = 0
Em1 = 1
c Beginning of loop.
110 continue
c Calculate Delta = time to next pair
c annihilation. Note that Em1 = c**2.
tauN = taui*sqrt(Em1)
Delta = tauN/N
if (x.eq.0) go to 150
iter = 10
call itera (iter, Er, Em, N, t, tauN, taui, Delta, x)
150 t = t + Delta
c Wavelength prop to r prop to t**(1/3):
c Er proportional to t**(-1/3).
DelEr = Er*(1 - ((t - Delta)/t)**(1/3.0))
c Em + Er = Nf conserved:
Er = Er - DelEr
Em = Em + DelEr
c Rest energy of one massive pair:
Em1 = Em/N
c Annihilation. If final, jump.
if (N.eq.1) go to 160
N = N - 1
Em = Em - Em1
Er = Er + Em1
go to 110
c End of phase 1.
160 tf = t
write (*,*)
2' taui tauf Nf tf-sqrtNf tf Em x'
write (9,*)
2' taui tauf Nf tf-sqrtNf tf Em x'
write (*,180) taui, tauN, Nf, tf-sqrt(Nf*1.0), tf, Em, x
write (9,180) taui, tauN, Nf, tf-sqrt(Nf*1.0), tf, Em, x
180 format (f10.7, f10.3, i12, f12.3, f13.5, f9.3, f9.5)
c
if (N2.le.0) go to 30
c-----------------------------------------------------------------------c
c Phase 2. Follow the growth of the spinless muon's self energy. c
c-----------------------------------------------------------------------c
Nf = N2
taui = taui2
ti = tf
t = ti
N = Nf
Em = Nf
Er = 0
Em1 = 1
c Beginning of loop. See phase 1.
210 continue
tauN = taui*sqrt(Em1)
Delta = tauN/N
if (x.eq.0) go to 250
iter = 10
call itera (iter, Er, Em, N, t, tauN, taui, Delta, x)
250 t = t + Delta
c Wavelength prop to r prop to t**(1/3):
c Er proportional to t**(-1/3).
DelEr = Er*(1 - ((t - Delta)/t)**(1/3.0))
c Em + Er = Nf conserved:
Er = Er - DelEr
Em = Em + DelEr
Em1 = Em/N
c Annihilation. If final, jump.
if (N.eq.1) go to 260
N = N - 1
Em = Em - Em1
Er = Er + Em1
go to 210
c End of phase 2.
260 tf = t
write (*,280) taui, tauN, Nf, tf-sqrt(Nf*1.0), tf, Em, x
write (9,280) taui, tauN, Nf, tf-sqrt(Nf*1.0), tf, Em, x
280 format (f10.7, f10.3, i12, f12.3, f13.5, f9.3, f9.5)
c
if (cN3.le.0) go to 30
c-----------------------------------------------------------------------c
c Phase 3. Follow the growth of the electron's self energy. c
c-----------------------------------------------------------------------c
c Em + Er = Nf. Between annihilations: N*c**2 + constant*c/r = Nf.
c
cNf = cN3
taui = taui3
ti = tf
t = ti
ri = 1 Note that ri is arbitrary.
cN = cNf
Em = cNf
Er = 0
c
c = 1
p = 0
c Beginning of loop.
310 continue
c Calculate Delta = time to next pair
c annihilation.
tauN = taui*c
Delta = tauN/cN
c Speed up computation when testing program:
if (cN.gt.100000) Delta = 10*Delta
if (cN.gt.1000000) Delta = 10*Delta
if (cN.gt.10000000) Delta = 10*Delta
if (cN.gt.100000000) Delta = 10*Delta
if (cN.gt.1000000000) Delta = 10*Delta
t = t + Delta
r = ri*(t/ti)**(1/3.0)
c Solve Eq N*c**2 + p*c*ri/r = Nf.
h = ri/(2*cN*r)
c = - p*h + sqrt((p*h)**2 + cNf*1.0/cN)
if (cN.eq.1) go to 360
if (cN.gt.1000000000) cN = cN - 90000
if (cN.gt.100000000) cN = cN - 9000
if (cN.gt.10000000) cN = cN - 900
if (cN.gt.1000000) cN = cN - 90
if (cN.gt.100000) cN = cN - 9
cN = cN - 1
Em = cN*c**2
Er = cNf - Em
p = Er*r/(ri*c)
go to 310
c End of phase 3.
360 tf = t
c
write (*,380) taui, tauN, cNf, tf-sqrt(cNf), tf, Em
write (9,380) taui, tauN, cNf, tf-sqrt(cNf), tf, Em
380 format (f10.4, f10.3, f14.1, f10.3, f13.5, f9.3)
c
if (Pions.eq.0) go to 400
c---------------------------------------------- Pions ------------------c
c Phase pi. Calculate the growth in pion self energy. c
c-----------------------------------------------------------------------c
c
c Particle Data: PLB, Vol. 592, 15 July 2004 (Ref. [15]):
c
c cmtau = 1777.00(28) MeV/c^2 160 000 ppb
c cme = 0.510 998 918(44) MeV/c^2 86 ppb
c cmpi = 139.570 18(35) MeV/c^2 2 500 ppb
c cmp = 938.272 029(80) MeV/c^2 86 ppb
c GF = 1.16637(1) X 10^{-5} GeV^{-2} X (\hbar c)^3 8 500 ppb
c = 0.0000000000116637(1) MeV^{-2} X (\hbar c)^3
c
cmtau = 1777.00
cme = 0.510998918
cmpi = 139.57018
cmp = 938.272029
GF = 0.0000000000116637
c
c End of phase 3:
c
c Em is the self energy of the last pair of electron pairs in phase 3.
c Originally this energy was 1. Therefore, Em is a measure of the
c electron's growth in self energy in phase 3.
c
c Em total matter energy
c Er total radiation energy
c E0 total energy (matter plus radiation)
c X growth in electron self energy in phase 3
c
E0 = Em + Er
X = Em
c
c Initial values in the pion phase.
c
ti = t
ci = c
ri = 1
taui = ci*taui3
c
c The last pair of electron pairs transforms into a pair of
c dynamically interacting pion pairs.
c
Em = Em*cmpi/cme
Er = E0 - Em
c
c Assume that one pion pair immediately decays by strong interaction.
c
Em = Em/2
Er = E0 - Em
c+ write (9,*) 't = ', t
c+ 37325.757
c
c Balance equation for pion and proton creation. See Appendix E.8:
c
c 2*cN3*(2/X)*cme*ci**2*y*GF*cmtau**2 =
c 4*sqrt(2.0)*pi*(4*(cmpi - cme)*c**2 + 2*(cmp - cmpi)*c**2) (E.14)
c
c Note that (in the global picture) the particle's mass is conserved,
c while its rest energy grows with c**2.
c Also, note that real electrons do not exist in the pion phase,
c and that (2/X)*cme*ci**2 on the left side specifies the photon-pair
c energy at the end of phase 3.
c
c First, assume that also the remaining pion pair decays immediately,
c that is, c = ci:
c
z = 4*(cmpi - cme) + 2*(cmp - cmpi)
y = 4*sqrt(2.0)*pi*z/(2*cN3*(2/X)*cme*GF*cmtau**2)
c+ write (9,780) 'y = ', y
c+ 3.84454
c
c Compare with calculation in local picture.
c
y1 = 1 + 2*(cmp - cmpi)/(4*(cmpi - cme))
c+ write (9,780) 'y1 = ', y1
c+ 3.87181
c
c Next, assume that there is a time delay between the creation of
c the pion pairs and the proton pair.
c The assumption that the first pion pair annihilated immediately
c implies that the duration of the pion phase is defined by the
c lifetime tau of the last pion pair.
c
c Vary tau until y matches the value y1 of the local picture.
c
tau = 0
do 750 i = 1,50000
t = ti
tau = tau + 0.00001*taui
c
c Sum of matter energy and radiation energy = constant:
c
c Em*(c/ci)**2 + Er*(c/ci)*(ri/r) = E0, or
c
c c**2 + 2*h*c = (E0/Em)*ci**2, with h = 0.5*(Er/Em)*ci*ri/r
c
t = t + tau
r = ri*(t/ti)**(1/3.0)
h = 0.5*(Er/Em)*ci*ri/r
c = - h + sqrt(h**2 + (E0/Em)*ci**2)
c
c Modified balance equation:
c
c 2*cN3*(2/X)*cme*ci**2*y*GF*cmtau**2 =
c 4*sqrt(2.0)*pi*(4*(cmpi - cme)*ci**2 + 2*(cmp - cmpi)*c**2) (E.14)'
c
c The small decrease in photon energy in the pion phase is ignored.
c Also not considered is the effect (via the log of the tauon-muon
c mass ratio) of virtual muons.
c These errors are too small to be of practical interest.
c Totally negligible is the effect of virtual electrons.
c The largest uncertainty comes from the phase-3 ti value (the exact
c time when phase 2 ended and the electron was created).
c
z = 4*(cmpi - cme) + 2*(cmp - cmpi)*(c/ci)**2
y = 4*sqrt(2.0)*pi*z/(2*cN3*(2/X)*cme*GF*cmtau**2)
c
c Compare with value calculated in local picture:
c
if (y.ge.y1) go to 760
750 continue
go to 990
c Restore Em for phase 4.
760 Em = X
tf = t
c
c+ write (9,780) 'y = ', y
c+ 3.87181
c+ write (9,*) 't = ', t
c+ 37862.427
c
write (*,780) 'tau/taui = ', tau/taui
write (9,780) 'tau/taui = ', tau/taui
c+ 0.21639
780 format (a, f10.5)
c
go to 990
c-----------------------------------------------------------------------c
c Phase 4. Follow the growth in self-energy of a proton-electron c
c pair until present day when c has grown from 1 to c0. c
c-----------------------------------------------------------------------c
c Em + Er = E0. Between annihilations: N*c**2 + constant*c/r = Nf.
c See comments after end of source program.
c
c Typically at end of phase 3: tf = 49470
c Em = 10.889
c cN3 = 2447600000
400 ti = tf
x = Em
Eri = cN3*2/x
E0 = Emi + Eri
ci = 1
taui = 1
c Final values.
Er0 = cN3*Boltz*CBR
Em0 = E0 - Er0
c0 = sqrt(Em0/Emi)
c
c = 1
tau = 1
Em = 1
Er = Eri/Emi
E0 = Em + Er
t = ti
ri = 1
tau0 = c0
Em0 = c0**2
cntau = 0
ck = 10.0**12
imax = 2000000000
i = 0
p = Er
c Beginning of loop. Calculate
c age t = sum of time intervals tau,
c cntau = number of time intervals.
410 tau = c
Delta = ck*tau
t = t + Delta
cntau = cntau + ck
r = ri*(t/ti)**(1/3.0)
c Solve Eq c**2 + p*c*ri/r = E0.
h = 0.5*ri/r
c = - p*h + sqrt((p*h)**2 + E0)
if (c.ge.c0) go to 460
i = i + 1
if (i.ge.imax) go to 460
Em = c**2
Er = E0 - Em
p = Er*r/(ri*c)
go to 410
c End of phase 4.
460 ratio = cntau*tau/t
c
write (*,*) 'Phase 4:'
write (9,*) 'Phase 4:'
write (*,*)
2' ratio c0 Em t0/10**20 CBR T i'
write (9,*)
2' ratio c0 Em t0/10**20 CBR T i'
write (*,480) ratio, c, Em, t/10.0**20, CBR, i
write (9,480) ratio, c, Em, t/10.0**20, CBR, i
480 format (f9.5, f12.5, f13.3, f15.6, f8.3, i12)
write (*,*) ' '
write (9,*) ' '
go to 30
c
990 close (9)
stop
end
c-----------------------------------------------------------------------c
c Calculate and use value of tau midway to next annihilation. c
c-----------------------------------------------------------------------c
subroutine itera (iter, Er, Em, N, tN, tau, taui, Delta, x)
c
integer iter, N
real Er, Em, tN, tau, taui, Delta, x
c
integer i
real dEr, Em1, t
c
Delta = tau/N
do 110 i = 1,iter
t = tN + x*Delta
dEr = Er*(1 - (tN/t)**(1/3.0))
Em1 = (Em + dEr)/N
tau = taui*sqrt(Em1)
110 Delta = tau/N
return
end
c---------------------------------------------- Pions ------------------c
c Calculate in the local picture the muon-electron mass ratio. c
c-----------------------------------------------------------------------c
subroutine muerat (ratio, targ1, targ2)
c
real ratio, targ1, targ2
real pi, B, B0, alpha, GF
real cme, cmmu, cmmu0, cmmu1, cmtau, cmtau0, cmtau1, cmpi, cmp
real rmue, rmue0, rmue1, E0, E1, E2, y, ymu, z
integer i
c
c Standalone calculation performed in the local picture.
c
c Calculate the muon-electron mass ratio and the target mass ratios
c for phase 1 and phase 2.
c
c Numerical constants:
c
pi = 3.14159265358979
B = 0.666001731498
B0 = 0.9783964019
c
c Review of Particle Physics, PLB, Vol. 592, 15 July 2004 (Ref. [15]):
c
c cme = 0.510 998 918(44) MeV/c^2 86 ppb
c cmmu = 105.658 369(9) MeV/c^2 85 ppb
c rmue = 206.768 2837(56) - not used 27 ppb
c cmtau = 1777.00(28) MeV/c^2 160 000 ppb
c GF = 1.16637(1) X 10^{-5} GeV^{-2} X (\hbar c)^3 8 500 ppb
c = 0.0000000000116637(1) MeV^{-2} X (\hbar c)^3
c cmp = 938.272 029(80) MeV/c^2 86 ppb
c cmpi = 139.570 18(35) MeV/c^2 2 500 ppb
c
c alpha = 1/137.035 999 084(51) PRD 78, 053005 (2008-09) 0.37 ppb
c 1/B/alpha = 205.759 223 442(77) = zeroth-order rmue 0.37 ppb
c
c cmmu0, cmtau0: Initial lepton masses in phase 4.
c cmmu1, cmtau1: Lepton masses after pion creation.
c rmue0: Initial muon-electron mass ratio in phase 4.
c rmue1: Muon-electron mass ratio after pion creation.
c
alpha = 1/137.035999084
cme = 0.510998918
cmmu = 105.658369
cmtau = 1777.00
GF = 0.0000000000116637
cmp = 938.272029
cmpi = 139.57018
c Initialize variables.
cmmu0 = cmmu
cmmu1 = cmmu
cmtau0 = cmtau
cmtau1 = cmtau
c Energy for e-pi transformation.
E1 = 4*(cmpi - cme)
c Energy for pi-p transformation.
E2 = 2*(cmp - cmpi)
c Iterate cmmu0, etc.
do 110 i = 1,3
c
c E0 Tauon contribution to E1 and
c 3*E0 - z*E0 E2.
c
c E0*(cmmu0/cmtau0)**2 Muon contribution to E1 and
c 3*E0*(cmmu1/cmtau1)**2 - z*E0*(cmmu1/cmtau1)**2*log(cmtau1/cmmu1)) E2.
c
c E1 = E0*(1 + (cmmu0/cmtau0)**2)
c E2 = 3*E0*(1 + (cmmu1/cmtau1)**2)
c - z*E0*(1 + (cmmu1/cmtau1)**2*log(cmtau1/cmmu1))
c Solve for E0 and z.
E0 = E1/(1 + (cmmu0/cmtau0)**2)
z = (3*(1 + (cmmu1/cmtau1)**2) - E2/E0)
2 / (1 + (cmmu1/cmtau1)**2*log(cmtau1/cmmu1))
c
y = 4 - z
ymu = 4 - z*log(cmtau1/cmmu1)
c
c Lepton masses before proton creation.
c
cmtau1 = cmtau/(1 - (y - 1)*GF*cmtau1**2/(4*sqrt(2.0)*pi*alpha))
cmmu1 = cmmu/(1 - (ymu - 1)*GF*cmmu1**2/(4*sqrt(2.0)*pi*alpha))
c
c Lepton masses before pion creation.
c
cmtau0 = cmtau1/(1 - GF*cmtau0**2/(4*sqrt(2.0)*pi*alpha))
cmmu0 = cmmu1/(1 - GF*cmmu0**2/(4*sqrt(2.0)*pi*alpha))
c
c Theoretical mu-e mass ratio initially.
rmue0 = 1/(B*alpha) + 1/(1 - 2*B*alpha)
c Mass ratio after pion creation.
rmue1 = rmue0*(1 - GF*(rmue0*cme)**2/(4*sqrt(2.0)*pi*alpha))
c Mass ratio after proton creation.
rmue = rmue1*(1 - (ymu-1)*GF*(rmue1*cme)**2/(4*sqrt(2.0)*pi*alpha))
c
c+ write (9,120) cmtau0/cmmu0, cmtau1/cmmu1, cmtau/cmmu
c+ result: 16.8368 16.8321 16.8184
c+ write (9,120) rmue0, rmue1, rmue
c+ 206.769038949 206.768831341 206.768283185
c+ write (9,120) y
c+ 3.8726
110 continue
120 format (3f15.9)
c
c Vary input: rmue
c 206.768283185 = ref
c 1/alpha 137.035999084 + 51 206.768283261 + 76
c 137.035999084 - 51 206.768283108 - 77
c GF 0.0000000000116637 + 1 206.768283178 - 7
c 0.0000000000116637 - 1 206.768283191 + 6
c cmpi 139.57018 + 35 206.768283190 + 5
c 139.57018 - 35 206.768283180 - 5
c cmtau 1777.00 + 28 206.768283189 + 4
c 1777.00 - 28 206.768283180 - 5
c
c Conclusion: 206.768283185(77)(7)(5)(5)
c = 206.768283185(78)
c = 206.76828318(8)
c = 206.7682832(1)
c Ref. [15]: 206.7682837(56)
c 2006 CODATA: 206.7682823(52)
c
c Target values for use in simulation:
c
c Phase 1: (cmtau0/cmmu0)*rmue0/(1/(B*alpha)) = 16.9195(27)
c
c Phase 2: c2 = (1/alpha)/(2*B**2/B0) = 151.136306673(56) Eq. (7.11)
c
ratio = rmue
targ1 = (cmtau0/cmmu0)*rmue0/(1/(B*alpha))
targ2 = (1/alpha)/(2*B**2/B0)
return
end
c.asm skipc
c include /home/stig1/fairI.fasm
Output from program when default input values are used:
taui tauf Nf tf-sqrtNf tf Em x
1.0000000 3.442 86 1.000 10.27351 16.839 0.43700
1.0000000 9.770 1010 1.000 32.78015 149.771 0.43700
951.3484 3139.439 2447600000.0 1.042 49474.26731 10.890
Phase 4:
ratio c0 Em t0/10**20 CBR T i
1.00032 494.33718 244369.247 28.592049 2.725 5785768
COMMENTS
Because of the sensitivity of the results to variations in input, it can be
seen that no consistent picture is possible unless tau is proportional to c
in the first three phases, and unless taui = 1 in both phase 1 and phase 2.
The phase-4 calculation is, of course, highly unrealistic, since all kinds
of particle interaction are ignored (from neutron production to the
formation of black holes).
In a more realistic simulation, one would expect the ratio 1.00032 (which
measures the effect of the time paradox) to be an order of magnitude
greater than 1, and the age t0 several orders of magnitude greater than
10**21.