Spinning Spiral Galaxies and the Gravitomagnetic Force

The Milky Way

As the Sun orbits the Milky Way’s center, Newton 2 indicates that the total force acting on the Sun must point toward the galaxy’s center and have the magnitude

F_total =|m|ω²r,          (1)

where m is the Sun’s (imaginary) gravitational mass, |m| is its real inertial mass, ω is the angular rate of galactic spin, and r is the Sun’s displacement from the galaxy’s center.

In this article we shall assume that

F_total=F_grav+F_gravmag,         (2)

where F_grav is the gravitational force exerted upon the Sun in the galaxy’s gravitational field, and F_gravmag is the gravitomagnetic force experienced by the Sun in the galaxy’s theoretical gravitomagnetic field. 

Let us begin by deriving  F_gravmag. According to   gravitomagnetic theory the rotating Milky Way engenders a gravitomagnetic field which, at internal points, has the same direction as the galaxy’s spin vector. We can model the galaxy as a spinning disk of matter of uniform mass density σ=M/πR² where M is the mass of the Milky Way and R is its radius. (N.B. The galaxy density is not uniform, and for this and other reasons our model is only an approximation.) Analogous to a rotating disk of positive electric charge’s B field, at internal points there theoretically is a gravitomagnetic field that points in the same direction as the galaxy’s spin vector. We can derive the gravitomagnetic field for the Milky Way by applying the gravitomagnetic theory transformation rules to the well-known B field of a rotating disk of positive charge. At the center of the disk of charge the B field has the magnitude

B = σωR/2ε_oc² (3)

          = qω/2πε_oRc².

It is noteworthy that B is single-valued at all points on the disk. B is normal to the disk surface for all r<=R.

The gravitomagnetic equivalent of Eq. 3 would then be

O = 2GMω /Rc².     (4)

Viewed from the inertial frame in which the center of the Milky Way is at rest, the Sun should experience an inward-pointing gravitomagnetic force of magnitude

F_gravmag =mvO   (5)

                =mω²r(2GMω/Rc²).

(F_gravmag points inward because m and M are both imaginary.) And, since F_grav and F_gravmag both point toward the galaxy’s center, we should haveF_grav= F_total -F_gravmag.     (6)