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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 i|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 GMT version of the Lorentz force law)

Fgravmag x = i|mstar| (vyOz)     (2)

          = i|mstar| (wr) Oz.

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| a    (4)

            = - |mstar| w2r. 

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)

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

    = - |mstar| w2r + (2Gi|mstar|)(i|mgalaxy| w2r) / Rc2) /i |mstar|.

    = i(w2r - 2G|mgalaxy| w2r / Rc2).

Using the GMT rules that (1) g/i 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

E = w2r+(qw2r /2peo 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.