On the Fields of a Spinning Disk of Charge

Let us begin by modeling a spiral galaxy as a disk of single-valued mass density s=mgalaxy/pR2, lying in the xy-plane and rotating with an angular velocity w that points in the +z direction. In the article Galaxies and the Gravitomagnetic Force it was found that such a galaxy theoretically has an Imaginary uniform gravitomagnetic field O that points in the z direction:

Oz = 2pGswR/c2.         (1)

          =2G i|mgalaxy| w / Rc2.

Given a star of mass mstar, momentarily on the x-axis and at a fixed distance r from the galaxy’s center, the gravitomagnetic force on the star is then (according to the GTM version of the Lorentz force law)

Fgravmag x = i|mstar| (vyOz) (2)

          = i|mstar| wrx 2G i|mgalaxy| w/Rc2

          = - (2G|mstar| |mgalaxy| w2rx) / Rc2.

And the total force on the star is theoretically the sum of the gravitomagnetic force plus the gravitational force (which also points toward the galactic center):

Ftotal x = Fgravmag x + Fgrav x.               (3)

Now Newton 2 indicates that Ftotal x is:

Ftotal x = |mstar| ax    (4)

            = - |mstar| w2rx.  

Assuming that

Fgrav x = i|mstar| gx    (5)

where gx is the gravitational field of the galaxy at the star’s position, we find that

gx = Fgrav x / i|mstar|    (6)

    = -i (Ftotal x - Fgravmag x) / |mstar|

    =-i (- |mstar| w2rx - (2G |mstar| |mgalaxy| w2rx /Rc2)) / |mstar|

    = -i (- w2rx)(- 2G |mgalaxy| w2rx / Rc2))

    = (i w2rx)( -2G |mgalaxy| w2rx / Rc2).

Using the GMT rules that (1) g transforms to E, (2) G transforms to 1/4peo, and (3) O transforms to B, we find that, given a spinning disk of positive charge with uniform charge density s, the electric field in the disk’s plane points away from the center and has the magnitude

Ex = w2rx(-1/2peo)(qw2rx / Rc2).     (7)   

And, applying the same transformations to Eq. 1, we find that

Bz=moswR/2.      (8)

Eq. 8 agrees with the value for B specified on the Internet.