Photon sphere

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

${\displaystyle ds^{2}=\left(1-{\frac {r_{\text{s}}}{r}}\right)c^{2}\,dt^{2}-\left(1-{\frac {r_{\text{s}}}{r}}\right)^{-1}\,dr^{2}-r^{2}(\sin ^{2}\theta \,d\phi ^{2}+d\theta ^{2}).}$ 

For a photon traveling at a constant radius r (i.e. in the φ-coordinate direction), ${\displaystyle dr=0}$ . Since it is a photon, ${\displaystyle ds=0}$  (a "light-like interval"). We can always rotate the coordinate system such that ${\displaystyle \theta }$  is constant, ${\displaystyle d\theta =0}$  (e.g., ${\displaystyle \theta =\pi /2}$ ).

Setting ds, dr and dθ to zero, we have

${\displaystyle \left(1-{\frac {r_{\text{s}}}{r}}\right)c^{2}\,dt^{2}=r^{2}\sin ^{2}\theta \,d\phi ^{2}.}$ 

Re-arranging gives

${\displaystyle {\frac {d\phi }{dt}}={\frac {c}{r\sin \theta }}{\sqrt {1-{\frac {r_{\text{s}}}{r}}}}.}$ 

To proceed, we need the relation ${\displaystyle {\frac {d\phi }{dt}}}$ . To find it, we use the radial geodesic equation

${\displaystyle {\frac {d^{2}r}{d\tau ^{2}}}+\Gamma _{\mu \nu }^{r}u^{\mu }u^{\nu }=0.}$ 

Non vanishing ${\displaystyle \Gamma }$ -connection coefficients are

${\displaystyle \Gamma _{tt}^{r}={\frac {c^{2}BB'}{2}},\quad \Gamma _{rr}^{r}=-{\frac {B^{-1}B'}{2}},\quad \Gamma _{\theta \theta }^{r}=-rB,\quad \Gamma _{\phi \phi }^{r}=-Br\sin ^{2}\theta ,}$ 

where ${\displaystyle B'={\frac {dB}{dr}},\ B=1-{\frac {r_{\text{s}}}{r}}}$ .

We treat photon radial geodesics with constant r and ${\displaystyle \theta }$ , therefore

${\displaystyle {\frac {dr}{d\tau }}={\frac {d^{2}r}{d\tau ^{2}}}={\frac {d\theta }{d\tau }}=0.}$ 

Substituting it all into the radial geodesic equation (the geodesic equation with the radial coordinate as the dependent variable), we obtain

${\displaystyle \left({\frac {d\phi }{dt}}\right)^{2}={\frac {c^{2}r_{\text{s}}}{2r^{3}\sin ^{2}\theta }}.}$ 

Comparing it with what was obtained previously, we have

${\displaystyle c{\sqrt {\frac {r_{\text{s}}}{2r}}}=c{\sqrt {1-{\frac {r_{\text{s}}}{r}}}},}$ 

where we have inserted ${\displaystyle \theta =\pi /2}$  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 ${\displaystyle \phi }$ -coordinate line, for the mass to be located directly in the centre of the photon's orbit, we must have ${\displaystyle \theta =\pi /2}$  radians).

Hence, rearranging this final expression gives

${\displaystyle r={\frac {3}{2}}r_{\text{s}},}$ 

which is the result we set out to prove.

 

In contrast to a Schwarzschild black hole, a Kerr (spinning) black hole does not have spherical symmetry, but only an axis of symmetry, which has profound consequences for the photon orbits, see e.g. Cramer^[2] for details and simulations of photon orbits and photon circles. There are two circular photon orbits in the equatorial plane (prograde and retrograde), with different Boyer–Lindquist radii:

${\displaystyle r_{\pm }^{\circ }=r_{\text{s}}\left[1+\cos \left({\frac {2}{3}}\arccos {\frac {\pm |a|}{M}}\right)\right],}$ 

where ${\displaystyle a=J/M}$  is the angular momentum per unit mass of the black hole.^[7]

There exist other constant-radius orbits, but they have more complicated paths which oscillate in latitude about the equator.^[7]

• Step by Step into a Black Hole
• Virtual Trips to Black Holes and Neutron Stars
• Guide to Black Holes
• Spherical Photon Orbits Around a Kerr Black Hole