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A Stable Model for the Positron

All positrons consist of electric charge q=1.602176565e-19 coul=e, mass m=9.10938356e-31 kg, and a magnetic dipole moment mq=9284.764e-27 m2coul/sec.

In this article we model a positron’s electric charge as a spinning disk with radius R, uniform charge density sq=e/pR2, and spinning with an angular speed of wq>0. The formula for the magnetic dipole moment of this disk is

mq=wqR2e/4.        (1)

Thus wqR2=4mq/e is a constant. We can solve for wq if we have a value for R. Let us estimate that R, the positron radius,  is R=1e-18. Then

wq=4mq/qR2=2.318037570664338e+32.                  (2)

We shall assume that a disk lies in the xy plane and that wq points 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 same direction as wq:

Bz=ewq/2peoc2R. (3)

Hence an increment of charge, d(e), 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:

dFmag x=d(e)(wqR)(ewq/2peoRc2)   (4)

=d(e)(wq2e/2peoc2).

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

Ex=wq2R-ewq2/2peoc2. (5)

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

dFelec x=d(e)Ex      (6)

= d(e)(wq2R-ewq2/2peoc2).

The total electromagnetic (or Lorentz) force on dq is

dFelecmag x=dFelec x+dFmag x,     (7)

=d(e)(wq2R-ewq2/2peoc2+ewq2/2peoc2).

=d(e)(wq2R).

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

The positron also has mass. Let us model the mass portion of the particle as a disk of mass m, radius R, uniform mass density sm=m/pR2, and also lying in the xy-plane and spinning with an angular speed of wm, where the value of wm at stability is to be determined. We shall assume that wm is parallel to wq. This mass theoretically engenders an imaginary gravitomagnetic field

Oz=2i|mgrav|wmG/Rc2.  ( 8)

Note that Oz is Imaginary and positive, since i|mgrav| is Imaginary and wm is positive. An increment of mass i(dm) on the disk’s periphery, therefore experiences a Real gravitomagnetic force:

dFgravmag x =i(dm)(vy)(Oz)     (9)

=i(dm)(wmR)(2i|mgrav|wmG/Rc2)

=-(dm)(2|mgrav|wm2G/c2).

Note in this equation that dFgravmag x is inward.

Of course the disk of mass also has a gravitational field. In the cited recent article it was determined that the rotating disk has a gravitational field, g, which on the periphery has a value of

gx=i(wm2R-2mwm2G/c2).                 (10)

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

dFgrav x = (dm)(gx)                 (11)

=-(dm)(wm2R-2Gmwm2/c2).

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

dFgrav-gravmag x=dFgrav x+dFgravmag x           (12)

=(dm)(-wm2R+2Gm/c2)+(dm)(-2mwm2G/c2)

=(dm)(-wm2R).

The positron will be stable if

dFelecmag x + dFgrav-gravmag x = 0,       (13)

or if

d(e)(wq2R)-(dm)(wm2R)=0,  (14)

Now

d(e)/dm=e/m.                (16)

Thus we have stability if

(e/m)(wq2R)-(wm2R)=0,        (17)

or if

wm2=-(e/m)wq2    (18)

This solves to

wm2=9.450665030982446e+75.   (19)

Thus

wm=9.721453096622153e+37.    (20)

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

wm/wq=419397.           (21)

In summary, a positron model of (a) a disk of charge, q=e, with radius R=1e-18, and spinning with an angular speed of wq, superposed on (b) a disk of mass, m, with radius R but spinning at a rate of wm>wq, is stable if

wm=419397 wq. (22(