Schwarzschild circular-ring energy shift
= Schwarzschild circular-ring energy shift
{c}
{title2=$1+z=(1-3r_g/r)^{-1/2}(1+\beta\sin i)$}
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.