What the \({F}_{n}\), \(F\)?

Let's say a photon is propel forward by a force \({F}\), at terminal velocity a drag force develops that just opposes this force such that its velocity is \(c\), light speed.  However,  there are two other perpendicular degrees of freedom.  A drag force also develop along one of the two axis such that the velocity is also a constant in that direction.  At the same time a third force, acting as centripetal force \({F}_{c}\), brings the system into circular motion.  The resulting circular motion has no movement along \({F}_{c}\) and so the drag force in that direction is zero.  That means the centripetal force is the same force \(F\) that drives the photon forward.

A photon being setup

This can be true if the medium surrounding the photon is providing the propelling force, that such a force has no directional dependence.  On earth, photons bounce off region of higher gravity, that means photons are always in motion upwards.

The relationship,

\(F={m}_{p}\cfrac{c^2}{r}\)

is a precarious one.  It suggest that given a medium that provides \({F}_{n}\), and given photon mass \({m}_{p}\),

\(F_n={m}_{p}\cfrac{c^2}{n^2 .r}\)

where n is the refractive index of the medium.  Then \(r\) the radius of the circular path is fixed, at least initially.  But the post "Wave Front and Wave Back" shows that \(r\) can vary continuously with incident angle. One possibility is that \(r\) changes temporarily,  under the force \({F}_{n}\) acting as the centripetal force, \(r\) returns to the value given by the equation above.  If polarization is the result of changing \(r\), this would mean polarization is also temporary.  Along the path of the polarized beam, the beam looses its polarization.

The force in circular motion does not do work but the photons has rotation kinetic energy in addition to \(\cfrac{1}{2}{m}_{p}{c}^{2}_n\).  The photon is constantly being acted upon by a force from the medium it is in, such that it eventually reaches its terminal speed \({c}_{n}\) in the medium and resume a helix path, where \(r\) is given by the last equation.

The path of the photon is straight down the helix.  The circular plane containing the circular motion at any instance is always perpendicular to the direction of travel.  Both velocities are at light speed, \(c\) or in some medium of refractive index \(n\), \(\cfrac{c}{n}\).  So if light speed is \(c\) in free space, the photons may actually have speed greater than light speed instantaneously.