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At the photon sphere (where photon exhibit circular orbit around a black hole), the radius is given by $r={\frac {3}{2}}r_{\rm {s}}$, and the angular velocity is $({\frac {d\phi }{dt})^{2}=\frac {c^{2}r_{\rm {s}}}{2r^{3}\sin ^{2}\theta}}$, so $\frac {d\phi }{dt}=\frac {c\sqrt{r_{\rm {s}}}}{\sqrt{2r^{3}}}$ for $\theta=2\pi$. The velocity would then be given by $v=r\frac {d\phi }{dt}=\frac {c\sqrt{r_{\rm {s}}}}{\sqrt{2r}}$. Substitute the expression for $r$ gives $v=\frac{c}{\sqrt{3}}$.

My question is: if the photon sphere describes the circular orbit of a photon around a black hole, why is the speed of the orbit given by $v=\frac{c}{\sqrt{3}}$ instead of the speed of light $c$?

ACB
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Jim RF
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    Does this answer your question? Speed of light in a gravitational field? – John Rennie Sep 14 '21 at 04:56
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    Hi Jim. What we mean by the speed of light is a surprisingly subtle business. See Does light really travel more slowly near a massive body? for the gory details. In this case from the perspective of the observer far from the black hole one way of interpreting this is that the gravitational field is causing time dilation, and the light moves more slowly because time runs more slowly at the photon sphere. – John Rennie Sep 14 '21 at 05:01
  • You may enjoy the beautiful animations of photon trajectories near the photon sphere in this recent Astronomy answer: https://astronomy.stackexchange.com/a/45114/16685 – PM 2Ring Sep 14 '21 at 05:37
  • this question and answer in Quora may help understand https://www.quora.com/Why-does-the-speed-of-light-have-a-ds-0-null-geodesic – anna v Sep 14 '21 at 08:54
  • Imagine orbiting in your spaceship laboratory, just above the photon sphere. Since its in orbit, the lab provides a free-falling inertial reference frame. In that lab, you would measure the speed of light as being completely normal, same as it ever was. You could try orbiting closer and closer to the photon sphere, and get the same results.There are no stable orbits below the photon sphere; you would just fall in. As you fell in, you'd still be in a free-falling reference frame, and get the normal usual speed of light. – Linas Nov 08 '23 at 23:41

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