An Electron Stable Model, Spinning Disks

J. J. Thomson, discoverer of the electron

All electrons consist of electric charge q=-e=-1.602176565e-19 coul, mass m=9.10938356e-31 kg, and a magnetic dipole moment μ_q=-9284.764e-27 meter²coul/sec.

In this article we shall model an electron’s electric charge as a spinning, infinitesimally thin disk with radius R, uniform charge density σ_q=-e/πR², and spinning with an angular velocity of ω_q>0. The formula for the magnetic dipole moment of such a disk is

μ_q=ω_qR²(-e)/4.    (1)

Thus ω_q R²=-4μ_q/e is a constant.

We can solve for ω_q if we have a constant value for R. Let us estimate that R, the disk’s radius,  is R=1.40898e-15. Then

ω_q=-4μ_q/eR²=1.167645e26.             (2)

We shall assume that our disk lies in the xy plane with ω_q pointing in the positive z-direction. It is readily shown that, given such a disk, the magnetic field in the disk is uniform and points in the negative z-direction:

B=-eω_q/2πε_oc²R. (3)

Hence an increment of charge, dq=d(-e)<0, momentarily coincident with the positive x-axis and at a distance R from the disk’s center, is subject to an outward-pointing  magnetic force:

dF_mag=d(-e)(ω_qR)( -eω_q/2πε_oc²R)          (4)

        =d(-e)(-eω_q²/2πε_oc²).

In the article “On the Fields of a Spinning Disk of Charge” it is found that the electric field, E, points radially inward. On the disk’s periphery:

E=-ω_q²R+eω_q²/2πε_oc².    (5)            

Thus in addition to the magnetic force, d(-e) also experiences an outward-pointing electric force:

dF_elec=d(-e)(-ω_q²R+eω_q²/2πε_oc²).   (6)

The total electromagnetic (or Lorentz) force on dq is

dF_elecmag=dF_elec+dF_mag      (7)

        = d(-e)(-ω_q²R+eω_q²/2πε_oc²)+d(-e)(-eω_q²/2πε_oc²)

        = d(-e)(-ω_q²R).

dF_elecmag points outward (in the positive-x direction). If the electron is to be stable, some other force must be equal and oppositely directed. An answer is suggested by gravitomagnetic theory.

The electron also has mass. Let us model the mass portion of the particle as a disk of mass m, radius R, uniform mass density σ_m=m/πR², and also lying in the xy-plane, but spinning with an angular speed of ω_m, where the value of ω_m at stability is to be determined. We shall assume that ω_m is parallel to ω_q. 

Such a mass theoretically engenders an imaginary gravitomagnetic field

O=2mω_mG/Rc².  ( 8)

Note that O points in the positive z-direction and is imaginary (since m is imaginary). An increment of mass dm on the disk’s periphery therefore experiences a real gravitomagnetic force:

dF_gravmag =(dm)(v_y)(O_z)      (9)

        =(dm)(ω_mR)(2mω_mG/Rc²)

        =(dm)(2mω_m²G/c²).

dF_gravmag points inward since dm and m are both imaginary.

Of course the disk of mass also has an imaginary gravitational field. In the cited  article it was determined that the rotating disk has such a field that (on the disk’s periphery) has a value of

g=(iω_m²R-2ω_m²Gm/c²).            (10)

Hence dm also experiences a Real gravitational force inward which has the value:

dF_grav = (dm)(g_x)         (11)

        =(dm)(iω_m²R-2Gmω_m²/c²).

The total grav-gravitomagnetic force acting on dm points inward:

dF_grav-gravmag=dF_grav+dF_gravmag                (12)

        =(dm)(iω_m²R-2Gmω_m²/c²+2Gmω_m²/c²)

        =(dm)(iω_m²R).

The electron will be stable if

dF_elecmag + dF_grav-gravmag = 0,    (13)

or if

d(-e)(-ω_q²R)+(dm)(iω_m²R)=0,   (14)

or if

d(-e)/(idm)(-ω_q²)=ω_m².        (15)

Now

(d(-e)/(idm)=e/|m|.        (16)

Thus we have stability if

e/|m|(ω_q²)=ω_mz²      (17)

This solves to

ω_m²=2.397966e63.     (18)

Thus

ω_m=4.896903e31.                (19)

At stability the disk of mass spins faster than the disk of charge:

ω_m/ω_q=419383.           (20)

 In summary, an electron model of (a) a disk of charge, q=-e, with radius R=1.40898e-15, and spinning with an angular speed of ω_q, superposed on (b) a disk of mass, m, with radius R but spinning at a rate of ω_m>ω_q, is stable if ω_m=419383 ω_q.