Is this result published in the literature?

On Jan 15, 2012, at 4:06 PM, JACK SARFATTI wrote:


(1 + phi/c^2) + A.c'/c + c'^2/c^2 = 0

A.c' = Ac'cos(A,c')

x = c'/c

x^2 + Acos(A,c')x + (1 + phi/c^2)  = 0

x(+,-) = [-Acos(A,c') +,- [A^2cos^2(A,c') - 4(1 + phi/c^2)]^1/2/2

At a horizon

(1 + phi/c^2) = 0

x(+,-) = [-Acos(A,c') +,- [A^2cos^2(A,c')]^1/2/2

therefore one root is c'(+) = 0

the other root is c'(-) = -cAcos(A,c')

that also vanishes when the gravimagnetic field is perpendicular to the light ray.