Since a Schwarzschild black hole has spherical symmetry, all possible axes for a circular photon orbit are equivalent, and all circular orbits have the same radius. This derivation involves using the
Schwarzschild metric, given by : ds^2 = \left(1 - \frac{r_\text{s}}{r}\right) c^2 \,dt^2 - \left(1 - \frac{r_\text{s}}{r}\right)^{-1} \,dr^2 - r^2 (\sin^2\theta \,d\phi^2 + d\theta^2). For a photon traveling at a constant radius
r (i.e. in the
φ-coordinate direction), dr = 0. Since it is a photon, ds = 0 (a "light-like interval"). We can always rotate the coordinate system such that \theta is constant, d\theta = 0 (e.g., \theta = \pi/2). Setting
ds,
dr and
dθ to zero, we have : \left(1 - \frac{r_\text{s}}{r}\right) c^2 \,dt^2 = r^2 \sin^2\theta \,d\phi^2. Re-arranging gives : \frac{d\phi}{dt} = \frac{c}{r \sin\theta} \sqrt{1 - \frac{r_\text{s}}{r}}. To proceed, we need the relation \frac{d\phi}{dt}. To find it, we use the radial
geodesic equation : \frac{d^2r}{d\tau^2} + \Gamma^r_{\mu\nu} u^\mu u^\nu = 0. Non vanishing \Gamma-connection coefficients are : \Gamma^r_{tt} = \frac{c^2 BB'}{2}, \quad \Gamma^r_{rr} = -\frac{B^{-1} B'}{2}, \quad \Gamma^r_{\theta\theta} = -rB, \quad \Gamma^r_{\phi\phi} = -Br\sin^2\theta, where B' = \frac{dB}{dr},\ B = 1 - \frac{r_\text{s}}{r}. We treat photon radial geodesics with constant
r and \theta, therefore : \frac{dr}{d\tau} = \frac{d^2r}{d\tau^2} = \frac{d\theta}{d\tau} = 0. Substituting it all into the radial geodesic equation (the geodesic equation with the radial coordinate as the dependent variable), we obtain : \left(\frac{d\phi}{dt}\right)^2 = \frac{c^2 r_\text{s}}{2r^3\sin^2\theta}. Comparing it with what was obtained previously, we have : c \sqrt{\frac{r_\text{s}}{2r}} = c \sqrt{1 - \frac{r_\text{s}}{r}}, where we have inserted \theta = \pi/2 radians (imagine that the central mass, about which the photon is orbiting, is located at the centre of the coordinate axes. Then, as the photon is travelling along the \phi-coordinate line, for the mass to be located directly in the centre of the photon's orbit, we must have \theta = \pi/2 radians). Hence, rearranging this final expression gives : r = \frac{3}{2} r_\text{s}, which is the result we set out to prove. ==Photon orbits around a Kerr black hole==