Derivation For A Schwarzschild Black Hole
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:
For a photon traveling at a constant radius r (i.e. in the Φ-coordinate direction), dr=0. Since it's a photon ds=0 (a "light-like interval"). We can always rotate the coordinate system such that θ is constant, dθ=0.
Setting ds, dr and dθ to zero, we have:
Re-arranging gives:
where Rs is the Schwarzschild radius.
To proceed we need the relation . To find it we use the radial geodesic equation
Non vanishing -connection coefficients are, where .
We treat photon radial geodesic with constant r and, therefore
.
Putting it all into r-geodesic equation we obtain
Comparing it with obtained previously, we have:
where we have inserted radians (imagine that the central mass, about which the photon is orbitting, is located at the centre of the coordinate axes. Then, as the photon is travelling along the -coordinate line, for the mass to be located directly in the centre of the photon's orbit, we must have radians).
Hence, rearranging this final expression gives:
which is the result we set out to prove.
Read more about this topic: Photon Sphere
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