 Schwarzschild circular-ring energy shift

For the receding edge of a circular emitter ring, the direct-projection model gives $1+z=(1-3r_g/r{)}^{-1/2}[1+\sqrt{r_g/(r-2r_g)}\sin i]$, with $r_g=GM/c^2$. Gravitational and transverse shifts combine with the line-of-sight Doppler factor. Mapping the distant inclination directly to a local photon direction neglects gravitational ray bending.