The vacuum is teeming with real magnetic pairs; some are ‘invisible’ particle:antiparticles with no charge and no moment

Twitter comment on Physical Review Letters @PhysRevLett  Feb 13  Measurement of the Electron Magnetic Moment to 0.13 parts per trillion (replacing a limit from 14 years ago).  Letter: https://go.aps.org/3E3jjSX  Viewpoint: https://go.aps.org/3xfgeLT  Replying to @PhysRevLett  —  One of my favorite experiments. Thanks. I particularly like “according to quantum physics, the vacuum is teeming with virtual particles”. Except I now say “the vacuum is teeming with real magnetic pairs; some are ‘invisible’ particle:antiparticles with no charge and no moment”.

electron-electron magnetic pairs
proton-proton magnetic pairs
electron-positron magnetic pairs
proton-antiproton magnetic pairs

For identical charged magnetic particles bonded magnetically, the bond distance can be estimated from setting the magnetic and Coulomb energies equal.

(e^2)/(4*pi*e0*R) = (mu0*mu^2)/4*pi*R^3)

RmagneticElectronPair = mue/(e*c) = (9.2847647043E-24 Joules/Tesla)/(1.602176634E-19 Coulombs * 2.99792458E8 Meters/second) in femtoMeters = 193.303539 femtoMeters

RmagneticProtonPair = mup/(e*c) = (1.41060679736E-26 Joules/Tesla)/(1.602176634E-19 Coulombs * 2.99792458E8 Meters/second) in femtoMeters = 0.293680341 femtoMeters

e: Codata electron charge = 1.602176634E-19 Coulombs or (Joules/ElectronVolt)
c: Codata speed of light and gravity = 2.99792458E8 Meters/second
e0: Codata electric vacuum constant (permittivity) = 8.8541878128E-12 Farads/Meter or (Coulombs/Volt meter)  or (Coulomb/meter^2)/(Volt/meter)
mu0: Codata magnetic vacuum constant (permeability) = 1.25663706212E-6 Newtons/Ampere^2

mue: electron magnetic moment = 1.00115965218059*MuB = 1.00115965218059*9.2740100783E-24 Joules/Tesla = 9.2847647E-24 Joules/Tesla

MuB: Codata Bohr Magneton = 9.2740100783E-24 Joules/Tesla

mue: Codata electron magnetic moment = 9.2847647043E-24 Joules/Tesla
mup: Codata proton magnetic moment = 1.41060679736E-26 Joules/Tesla
mun: Codata neutron magnetic moment = 9.6623651E-27 Joules/Tesla


Notes and Sketches

The proton-antiproton works the same, except then you add rotational and vibrational energy because the rotation should be nonlinear quantized.  I do not know prccisely how vibration works yet.  The proton-antiproton has no charge, and no magnetic field showing.  It “has no hair”.  It is invisible to electromagnetic sensors.

The electron-positron pair is also invisible.  When the laser vacuum experiments (or routine electron capture) create electron positron pairs, they are likely hitting real particles and that happens in the strong magnetic fields of  other particles.

particle-antiparticle pairs are my favorite candidates for part of dark matter.  They have mass but are invisible to electromagnetic sensors.  But regular electron pairs and proton pairs and other charged particle pairs (muons, atoms with magnetic moments, etc etc) will also bind magnetically if they are charged and they can get close enough for magnetic binding. Alignment and timing are very precise. But hit the sweet spot and it should work almost every time.

Probably all the “quarks” are better handled with “parton” models where the magnetic energy and magnetic moment are handled more or less classically.  The neutrino is likely a nonlinear Schrodinger soliton.  But I can see how to create arbitrary synthetic matter in any shape from vacuum fields.  Long strings, complex lattices, spherical bubble. The 4D Fourier spectrum of the reactants is transformed into the 4D Fourier spectrum of the products.  That is completely lossless in theory.  Real experiments and processing units, channels and field generators will always have some error, however small.

All the scales are close.  Use this simple dipole to get near to the right answer and then refine.  This is a “dipole approximation” that needs full nonlinear Schrodinger or similar method for relativistic or gluon level calculations and models.

(9.2847647043E-24 Joules/Tesla)/(.602176634E-19 Coulombs * 2.99792458E8 Meters/second) in femtoMeters

RmagneticNeutronPair = (9.6623651E-27 Joules/Tesla)/(.602176634E-19 Coulombs * 2.99792458E8 Meters/second) in femtoMeters = 0.53522802 femtoMeters  [ this is not quite correct since it is two electron-protonpairs ]

RNeutron = sqrt(mue*mup)/(e*c) = (sqrt(9.2847647043E-24*1.41060679736E-26) Joules/Tesla)/(.602176634E-19 Coulombs * 2.99792458E8 Meters/second) in femtoMeters = 20.0467435 femtometers


Aluminum 27 pair. Aluminum with all electrons stripped is very small core.  The “nuclear radius” of Aluminum is approximately

NuclearRadius = R0*(AtomicMass in AtomicMassUnits)^(1/3) = 1.25 femtoMeters*(26.9815385)^(1/3) = 3.74914511 femtometers

I am a bit tired to do aluminum.  Let me see what two Nitrogen15s would do.

MuNitrogen15 = -0.2831892 MuNuclearMagnetons = 0.2831892 * 5.0507837461E-27 Joules/Tesla = 1.43032741E-27 Joules/Tesla

Codata NuclearMagneton = 5.0507837461E-27 Joules/Tesla

R_TwoNitrogen15’s = (mu/e*c) =( 1.43032741E-27 Joules/Tesla)/(1.602176634E-19 Coulombs * 2.99792458E8 Meters/second) in femtoMeters = 0.0297786061 femtometers so I made a mistake.

Forgot the charges

mu0*mux^2/(4*pi*R^3) = Z^2*e^2/(4*pi*e0*R)

mu0*mux^2/R^2 = Z^2*e^2/eo

R^2 = mu0*mux^2*eo/(Z^2*e^2)

R = mux/(Z*e*c)

As people have found, the proton-neutron, proton-proton, and deep proton-electron pairs matter.  The electron-electron pairs are large. And there can be currents in  fast rotating nuclei.

This works for partons but  I will have to do the full isotope table and all the mass energies and decays.  The kinds of things to look for is magnetic binding. But existing nuclei and isotopes were made in unknown circumstances, and measured by old methods that are mostly untraceable without weeks of work per measurement.  Sometimes years or decades. Easier to throw it all away and check it all anew.  Will see.  I will allocate two years.  But I have already gone through it several times.  And there are GitHub and shared datasets now.  Maybe, maybe, maybe.

Use SI units always
Use exact global reference values faithfully
Use full and clear units for all factors
Check every equation for dimensions and units
Any old formulas check the variations that have identical units because people are always throwing away the cross terms and “small” things and “noise”.

Richard K Collins

About: Richard K Collins

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