Schwarzschild circular-ring energy shift ($1+z=(1-3r_g/r{)}^{-1/2}(1+\beta \sin i)$)

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.

At $r=6r_g$, the local circular speed is $c/2$, so the receding direct-projection shift is $\sqrt{2}(1+\sin i/2)$. The upper envelope over inclinations is $3/\sqrt{2}$. It is not the shift for a face-on or arbitrary azimuthal emitter, and it excludes plunging-region emission and ray-tracing corrections.